Integral Calculus 3D Geometry and Vector Booster with Problems and Solution
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f(x)
A
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b a
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yz-plane
y
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f(x) dx
x xy-plane
first octant
a
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Integral Calculus, 3D Geometry & Vector Booster 11 March 2017 11:32:07 AM
f(x)
A
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f(x) dx
Integral Calculus, 3D Geometry & Vector Booster with Problems & Solutions
JEE
Main and Advanced
About the Author REJAUL MAKSHUD (RM) Post Graduated from Calcutta University in PURE MATHEMATICS. Presently, he trains IIT Aspirants at RACE IIT Academy, Jamshedpur.
Integral Calculus, 3D Geometry & Vector Booster with Problems & Solutions
JEE
Main and Advanced
Rejaul Makshud M. Sc. (Calcutta University, Kolkata)
McGraw Hill Education (India) Private Limited chennai
McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto
McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai - 600 116, Tamil Nadu, India Integral Calculus, 3D Geometry & Vector Booster Copyright © 2017, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-5260-576-7 ISBN (10): 93-5260-576-4 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Bharati Composers, D-6/159, Sector-VI, Rohini, Delhi 110 085, and text and cover printed at Cover Designer: Creative Designer Visit us at: www.mheducation.co.in
Dedicated to My Beloved Mom and Dad
Preface
INTEGRAL CALCULUS, 3D GEOMETRY & VECTOR BOOSTER with Problems & Solutions for JEE Main and Advanced is meant for aspirants preparing for the entrance examinations of different technical institutions, especially NIT/IIT/BITSAT/ IISc. In writing this book, I have drawn heavily from my long teaching experience at National Level Institutes. After many years of teaching I have realised the need of designing a book that will help the readers to build their base, improve their level of mathematical concepts and enjoy the subject. This book is designed keeping in view the new pattern of questions asked in JEE Main and Advanced Exams. It has six chapters. Each chapter has the concept booster followed by a large number of exercises with the exact solutions to the problems as given below: Level - I
: Problems based on Fundamentals
Level - II
: Mixed Problems (Objective Type Questions)
Level - III
: Problems for JEE Advanced
Level - IV
: Tougher problems for JEE Advanced
(0.......9)
: Integer type Questions
Passages
: Comprehensive Link Passages
Matching
: Matrix Match
Reasoning
: Assertion and Reason
Previous years’ papers : Questions asked in Previous Years’ IIT-JEE Exams Remember friends, no problem in mathematics is difficult. Once you understand the concept, they will become easy. So please don’t jump to exercise problems before you go through the Concept Booster and the objectives. Once you are confident in the theory part, attempt the exercises. The exercise problems are arranged in a manner that they gradually require advanced thinking. I hope this book will help you to build your base, enjoy the subject and improve your confidence to tackle any type of problem easily and skillfully. My special thanks goes to Mr. M.P. Singh (IISc. Bangalore), Mr. Yogesh Sindhwani (Head of School, Lancers International School, Gurugram), Mr. Manoj Kumar (IIT, Delhi), Mr. Nazre Hussain (B. Tech.), Dr. Syed Kashan Ali (MBBS) and Mr. Shahid Iyqbal, who have helped, inspired and motivated me to accomplish this task. As a matter of fact, teaching being the best learning process, I must thank all my students who inspired me most for writing this book. I would like to convey my affectionate thanks to my wife, who helped me immensely and my children who bore with patience my neglect during the period I remained devoted to this book. I also convey my sincere thanks to Mr Biswajit Das of McGraw Hill Education for publishing this book in such a beautiful format.
viii Preface I owe a special debt of gratitude to my father and elder brother, who taught me the first lesson of Mathematics and to all my learned teachers— Mr. Swapan Halder, Mr. Jadunandan Mishra, Mr. Mahadev Roy and Mr. Dilip Bhattacharya, who instilled the value of quality teaching in me. I have tried my best to keep this book error-free. I shall be grateful to the readers for their constructive suggestions toward the improvement of the book. Rejaul Makshud M. Sc. (Calcutta University, Kolkata)
Contents
Preface
1. Indefinite Integrals
1.1–1.161
Definition 1.1 Geometrical interpretation of Integration 1.1 Basic formulae on integration 1.1 Standard Methods of Integration 1.2 Integration by Parts 1.5 Choice of the first function and the second function 1.5 Partial Fractions 1.6 Integration of Irrational Functions 1.8 Euler’s Substitution 1.9 Integration by Reduction Formula 1.9 Inexpressible integrals 1.12 Exercises 1.12 Answers 1.39 Hints and Solutions 1.41
2. Definite Integrals
vii
What is definite Integral? 2.1 Evaluation of Definite Integrals 2.1 Evaluation of definite integrals by substitution 2.1 Geometrical interpretation of definite integral 2.1 Definite integral as the limit of sum 2.2 Evaluation of the limit of the sum using Newton-Leibnitz Formula 2.2 Properties of Definite Integrals 2.2 Mean Value of a Function over an Interval 2.7 Improper Integrals 2.7 Gamma Function 2.7 Beta Function 2.7 Walli’s Formula 2.7 Exercises 2.8 Answers 2.37 Hints and Solutions 2.38
2.1–2.131
x Contents
3. Area Bounded by the Curves
4. Differential Equation
4.1–4.103
Introduction 4.1 Definition 4.1 Ordinary Differential Equation 4.1 Partial differential equation 4.2 Order of a differential equation 4.2 Degree of a differential equation 4.2 Linear and non-linear differential equation 4.3 Formation of a differential equation 4.3 Differential equation of first order and first degree 4.3 Orthogonal Tragectories 4.6 First Order Higher Degree Differential Equation 4.6 Higher Order Differential Equation 4.7 Applications of Differential Equation 4.8 Exercises 4.9 Answers 4.26 Hints and Solutions 4.28
5. Vectors
3.1–3.62
Rules to Draw Different Types of Curves 3.1 Area of the Cartesian Curve 3.2 area of the region bounded by a single curve and the co-ordinate axes 3.2 Area between two curves 3.3 Area of the region bounded by the several curves 3.4 Least value of a variable area 3.4 Exercises 3.4 Answers 3.17 Hints and Solutions 3.18
Introduction 5.1 Physical quantities 5.1 Representation of vectors 5.1 types of vectors 5.2 Algebra of vectors 5.3 Left and right handed orientation 5.4 Position vector of a point in a space 5.4 Linear Combination 5.4 Linearly dependent vectors 5.4 Linearly independent vectors 5.4 Section formulae 5.4 Bisector of angle between vectors A and b 5.6 straight line 5.6 Plane 5.6 Definition 5.7 Geometrical interpretation of a ◊ b 5.7 Properties of dot product of two vectors 5.7 __› __› Component of a vector B along and perpendicular to vector A 5.8
5.1–5.80
Contents
Physical significance of the dot prodcut of two vectors 5.8 Introduction 5.9 _› _ › Geometrical Interpretation of a × b 5.9 Properties of vector product of two vectors 5.9 Introduction 5.10 Properties of scalar triple product of vectors 5.10 Introduction 5.11 Geometrical significance of a × (b × c) 5.11 Explanation of Vector Triple Product 5.11 Properties of vector triple product 5.11 Scaler product of four vectors 5.12 Vector product of four vectors 5.13 Geometrical interpretation of (a × b) × (c × d) 5.13 Reciprocal system of vectors 5.13 Exercises 5.15 Answers 5.35 Hints and Solutions 5.36
6. 3D-Co-ordinate Geometry
xi
Introduction 6.1 Rectangular co-ordinate system 6.1 Position vector of a point in a space 6.1 Distance between two points in space 6.2 Section formulae 6.2 Direction cosines and direction ratios of a vector or a line 6.3 Definition 6.4 Equation of a line passing thorugh a point and parallel to a vector 6.4 Equation of a Line Passing Through Two Points A (r1) and B (r2) is r = r1 + l(r2 – r1) 6.5 Angle between two straight lines 6.5 Skew lines 6.5 Definition 6.5 General Form 6.5 Equation of a plane passing through a point (x, y, z,) 6.6 Equation of a plane passing through three non-collinear points 6.6 Coplanarity of four points 6.6 Intercept form of a plane 6.6 Normal to a plane 6.6 Vector form 6.6 Equation of a plane in normal form 6.6 Normal form of a Plane 6.7 Theorem 6.7 Angle between two planes 6.7 Angle between a line and a plane 6.7 Equation of a plane parallel to a given plane 6.7 Equation of a plane parallel to the axes 6.7 Equation of a plane passing thorugh (x1, y1, z1), (x2, y2, z2) and parallel to the line having direction ratios a, b, c 6.8
6.1–6.69
xii Contents
Equation of the plane passing through a point (x 1, y1, z1) and parallel to two lines having direction ratios (a1, b1, c1) and (a2, b2, c2) is 6.8 Equation of a plane passing through the line of intersection of planes 6.8 Distance of a point from a plane 6.9 Distance between two parallel planes 6.9 Sides of a plane 6.9 Intersection of a line and a plane 6.9 Condition for a line to lie in a plane 6.9 Condition of coplanarity of two lines 6.9 Equation of the planes bisecting the angle between the planes 6.9 Bisector of the acute and obtuse angles between two planes 6.9 Foot of perpendicular of a point w.r.t. a plane 6.10 Image of a point w.r.t. a plane 6.10 Equation of the plane containing a given line and parallel to a given line 6.10 Equation of the plane containing two given lines 6.10 Exercises 6.10 Answers 6.24 Hints and Solutions 6.25
Chapter
1
Indefinite Integrals
Concept Booster
1. Definition The inverse process of differentiation is called integration. d Let g (x) be a differentiable function of x such that ___ dx (g(x) + c) = f (x). Then (x) dx = g(x) + c. Thus g(x) is called a primitive or anti-derivative or an indefinite integral or simply integral of f (x) with respect to x, where f (x) is called the integrand, c is called the constant of integration.
2. Geometrical interpretation
of Integration
Let f (x) be a given continuous function and g(x) one of its anti-derivatives such that y = If y =
(x) dx = g(x) + c, then y = g(x) + c represents
3. Basic formulae
on integration
Formula 1
xn + 1 1. Ú xndx = _____ + c n+1 2. Ú 0 . dx = c
3. Ú k . dx = kx + c 1 4. Ú __ x . dx = log |x| + c 5. Ú ex = ex + c ax 6. Ú ax dx = ____ + c log a Formula 2 1 1 1. Ú __2 dx = – __ x + c x __ 2 3/2 __ 2. Ú ÷x dx = x + c 3
(x) dx = g(x) + c.
__ 1__ 3. Ú ___ dx = 2 ÷x + c x ÷ Formula 3
1. Ú sin x dx = – cos x + c 2. Ú cos x dx = sin x + c 3. Ú sec2 x dx = tan x + c 4. Ú cosec2 x dx = – cot x + c 5. Ú sec x ◊ tan x dx = sec x + c 6. Ú cosec x ◊ cot dx = – cosec x + c Formula 4
dx 1. Ú ______ _____ = sin– 1 x + c 2 ÷ 1 – x
dx 2. Ú _____ = tan– 1 x + c 1 + x2
dx 3. Ú ________ _____ = sec– 1 x + c 2 |x|÷x – 1
Type 1: Integrals of the form dx Ú _______ 1 ± sin x sin x Ú _______ dx, 1 ± sin x dx Ú ________ , 1 ± cos x
(
)
(
)
cos x Ú ________ dx, 1 ± cos x sec x dx Ú ___________ , sec x ± tan x cosec x Ú _____________ dx cosec x ± cot x
1.2 Integral Calculus, 3D Geometry & Vector Booster Rule: Simply rationalize the denominator. f (x) Type 2: Integrals of the form Ú ____ dx, g (x)
( )
where f (x) and g(x) be two polynomials such that deg of f(x) ≥ deg of g(x). Rule: Simply divide the numerator by the denominator. Type 3: An integral is related to inverse trigonometric functions. Rule: Simply write the simplest form of the expression under the inverse term.
4. Standard Methods
cos x 2. Ú cot x dx = Ú ____ dx sin x = log |sin x| + c sec x(sec x + tan x) 3. Ú sec x dx = Ú ________________ dx (sec x + tan x) = log |sec x + tan x| + c
of Integration
There is no general method to find the integral of a function. If the integral is not a derivative of some known functions, the corresponding integrals cannot be determined. In general, we use the following three types of integration: (i) Integration by substitution (ii) Integration by parts (iii) Integration by partial fractions. Now we shall discuss about the integration by the substitution method. Type 4: Integrals of the form Ú f (ax + b) dx Rule: Simply put a x + b = t. Type 5: Integrals of the form dx ______ ______ . Ú _________________ ax + b ± ÷ ax – d ÷ Rule: Simply rationalize the denominator. Type 6: Integrals of the form ____
f (x)
____ dx, Ú _____ dx. Ú f (x) ÷g(x) g(x)
Rule: Put
g(x) = t2
÷
f ¢(x) Type 7: Integrals of the form Ú ____ dx f(x) Rule: Put f (x) = t Formula 5
|
| | |
|
1 + sin x = log ________ + c cos x p 1 + cos __ – x 2 = log ______________ + c p __ sin – x 2 p x 2cos2 __ – __ 4 2 = log _____________________ + c p x p x 2 sin __ – __ cos __ – __ 4 2 4 2
(
(
|
| |
Proof sin x 1. Ú tan x dx = Ú ____ dx cos x = – log |cos x| + c = log |sec x| + c
|
( ) ( ) ( ) ( ) |
|
Type 8: Integrals of the form
Ú (f (x))n ◊ f ¢(x) dx
( ) |
|
1 – cos x = log _______ + c sin x x 2sin2 __ 2 = log ______________ x x + c 2 sin __ cos __ 2 2 x = log tan __ + c 2
2. Ú cot x dx = log |sec x| + c
)|
)
= log |cosec x – cot x| + c
(
) (
)
cosec x(cosec x – cot x) 4. Ú cosec x dx = Ú ____________________ dx (cosec x – cot x)
Rule: Put f (x) = t Type 9: Integrals of the form
| |
|
( ) | = log tan ( __ | p 2 – ( __ p 4 – __ 2x ) ) | + c = log tan ( __ | p 4 + __ 2x ) | + c
1. Ú tan x dx = log |sec x| + c
3. Ú sec x dx = log |sec x + tan x| + c p x = log tan __ + __ + c 4 2 4. Ú cosec x dx = log |cosec x – cot x| + c x = log tan __ + c. 2
(
)
)
p x = log cot __ – __ + c 4 2
(
Ú sinm x cos nx dx, where m and n are real numbers.
Case I: When m is odd and n is even Rule: Put cos x = t Case II: When m is even and n is odd Rule: put sin x = t. Case III: When m and n both are odd Rule: Put either sin x = t or cos x = t Case IV: When m and n both are even Rule: In this case, we shall use the following formulae:
Indefinite Integrals
1 + cos 2x = 2 cos2x 2
or 1 – cos 2x = 2 sin x. Case V: When m is odd and n is zero Rule: put cos x = t Case VI: When m is zero and n is odd Rule: Put sin x = t Case VII: When m is even and n is zero. Rule: In this case we shall use the following formulae: 1 + cos 2x = 2 cos2x
1 – cos 2x = 2sin2x.
Case VIII: When m is zero and n is even Rule: In this case, we shall use the following formulae:
1 + cos 2x = 2 cos2x
= Р2k (say), where k ΠN
Rules 1. Divide the numerator and denominator by cos2k x 2. Put tan x = t. Formula 6 dx x 1 1. Ú ______ = __ tan– 1 __ a + c x2 + a2 a
|
x–a 1 ___ log _____ + c x+a 2a
dx 2. Ú ______ = 2 x – a2
dx ______ 3. Ú _______ = log x + ÷ x 2 + a2 + c 2 2 ÷ x + a
dx ______ 4. Ú ________ = log x + ÷x 2 – a2 + c. 2 2 ÷ x – a
|
_______
|
______
|
|
( )
dx x ______ 5. Ú ________ = sin– 1 __ a + C. 2 2 ÷ a – x Note: You should keep in mind that, the coefficient of x2 must be unity every time. If not, first we make the coefficient of x2 be unity.
dx Type 10: Integrals of the form Ú ___________ ax2 + bx + c Rules 1. Make the co-efficient of 2 unity 2. Express the denominator as a sum or difference of two perfect squares. Type 11: Integrals is of the form
Type 12: Integrals of the form
dx ___________ . Ú ____________ 2 ÷ax + bx + c
Rules 1. Make the coefficient of x2 unity. 2. Express the term under the square root as a sum or difference of two perfect squares. dx dx ______ 3. Use Ú ______ 2 or Ú _______ 2 2 x ±a ÷a – x2
|
_______
( )
|
x = log x + ÷ x 2 ± a2 + c or sin–1 __ a + c
Type 13: Integrals of the form
f ¢(x)dx
___________________ , Ú ____________________ 2
÷a{f (x)} ± b{f (x)} ± c
dx ___________ i.e. reducible to Ú ____________ 2 + bx + c ÷ ax
( )
|
d (Denominator) = Numerator + k 2. Then separate the Numerator into two terms and use dx . Ú x______ 2 ± a2
1 – cos 2x = 2 sin2x. Case IX: When m + n = – ve even integer
1.3
px + q dx. Ú ___________ 2 ax + bx + c
Rules 1. Reduce the denominator in such a way that
Rules 1. Put f (x) = t 2. Apply (Type 12) Type 14: Integrals of the form
px + q
___________ dx Ú _____________ 2
÷ax + bx + c
Rules 1. Reduce the term under the square root in such a way that
d (ax2 + bx + c) = numerator + k
2. Separate the numerator into two terms and use
|
_______
|
dx ______ = log x + ÷x 2 ± a2 + c Ú ________ 2 2 ÷ x ± a
Type 15: Integrals of the form
dx , Ú ______________ 2 a sin x + b cos2x
dx dx , Ú _________ , Ú a_________ ± b sin2x a ± b cos2x
dx 2 , Ú __________________ 2 a sin x ± b cos x ± c
1.4 Integral Calculus, 3D Geometry & Vector Booster
dx _______________ , Ú (a sin x ± b cos x)2
dx , Ú ____________________________ (a sin x ± b cos x)(c sin x ± d cos x)
dx . Ú __________ a + b sin2x
Rules
÷ a2 + b2
sin x cos x , ________ Ú _______ sin 3x dx Ú cos 3x dx
Rules 1. First we cancel the common factors from the numerator and the denominator. 2. Divide the numerator and the denominator by cos2x. 3. Put tan x = t. Type 17: Integrals of the form
dx dx , _________ , Ú _____________ a cos x + b sin x Ú a + b sin x
dx dx , ________________ . Ú a_________ + b cos x Ú a cos x + b sin x + c
Rules
|
Type 19: Integrals of the form
|
(
( ) |
)|
a sin x + b cos x dx. Ú _____________ c sin x + d cos x
Rules 1. Write the numerator = l (denominator) + m (denominator). 2. Compare the coefficients of sin x and cos x 3. Solve for l and m. Type 20: Integrals of the form
a sin x + b cos x + c dx. Ú ________________ p sin x + q cos x + r
Rules 1. Write numerator = l (denominator) + m (denominator) + n 2. Compare the coefficient of sin x, cos x and the contant term. 3. Find the values of l, m and n. Type 21: Integrals of the form
( ) ( )
x 2 tan __ 2 1. Replace sin x by ___________ , x 2 __ 1 + tan 2
Rules
( ) ( )
x 1 – tan __ 2 cos x by ___________ x . 1 + tan2 __ 2 2
2. Reduce the denominator as sin (A ± B) or cos (A ± B) 3. use p x + __ + c Ú sec x dx = log tan __ 4 2 or Ú cosec x dx = log tan __ 2x + c
Rules 1. Divide the numerator and denominator by the highest power of cos x. 2. Put tan x = t. Type 16: Integrals of the form
1. Divide the numerator and the denominator by ______
(cos x ± sin x)
dx Ú ___________ f (sin2x)
1. Put sin x ± cos x = t
2. Use (sin x ± cos x)2 = 1 ± sin2 x
Type 22: Integrals of the form 2
x x 2. Replace 1 + tan __ in the numerator by sec2 __ 2 2 x 3. Put tan __ = t 2
x +1 dx, Ú ___________ x4 ± l x2 + 1
dx x –1 dx, Ú ___________ , Ú ___________ 4 2 4 x ± lx + 1 x ± lx2 + 1
dx dx 4. Use Ú _______ or Ú ______ . x2 + a2 x2 – a2
where
l ΠR
2
( )
( )
Type 18: Integrals of the form
dx , Ú _____________ a cos x + b sin x _______
where
0 < ÷a 2 + b2 £2
( )
2
Rules 1. Divide the numerator and the denominator by x2. 2. Reduce the denominator in such way that 1 d x ± __ x = numerator ± k dx 3. Use Ú _______ . 2 x ± a2
(
)
1.5
Indefinite Integrals
Type 23: Integrals of the form
Type 24: Integrals of the form
Ú tan x dx, Ú cot x dx
Ú (tan x + cot x) dx
Type 25: Integrals of the form
Type 26: Integrals of the form
Rules 1. Put tan x or cot x = t2 2. Apply the form 24. by
Ú ex {( f + g) + ( f ¢ + g¢ )} dx = ex ( f + g) + c
Ú (tan x – cot x) dx.
5. Integration
Ú ex {f (x) + f ¢(x)} dx = ex f (x) + c
Ú ekx {k f (x) + f ¢(x)} dx = ekx f (x) + c
Type 27: Integrals of the form
Parts
If u and v be two functions of x, then
{
}
du Ú (u v) dx = u Ú v dx – Ú ___ Ú v. dx dx dx
Ú eax sin bx dx, Ú eax cos bx dx.
Rule 1
Ú eax sin bx dx eax = _______ 2 2 (a sin bx – b cos bx) + c a + b
Proof: We know that, d ___ ( f (x).g(x)) = f (x) . g¢(x) + g(x) . f ¢(x) dx fi
Ú [f (x) . g¢(x) + g(x) . f ¢(x)] dx = f (x) . g(x)
fi
Ú ( f (x) . g¢(x)dx = f (x) g(x) – Ú g(x) . f ¢(x)dx
Put
d f (x) = u, and ___ [g(x)] = v dx
fi Thus,
of the
eax = _______ (a cos bx – b sin bx) + c a2 + b2
Proof: Let
u = Ú eax cos bx dx
and
v = Ú eax sin bx dx
Then
u + i v
= Ú eax cos bx dx + i Ú eax sin bx dx
Ú v dx = g(x)
= Ú eax (cos bx + i sin bx) dx
du Ú v . dv } dx. Ú (u v) dx = u Ú v dx – Ú { ___ dv
= Ú eax◊ eibxdx
= Ú eax + ibxdx
= Ú e(a + ib)xdx
e(a + ib)x = ______ + c a + ib
(a – ib)e(a + ib)x = _____________ + c (a2 + b2)
i.e. the integral of the product of two function = (First function) × (Integral of the second function) – Integral of [(derivative of the first function) × (Integral of the second function)]
6. Choice
Ú eax cos bx dx.
Rule 2
first function
and the
second function
1. We can choose the first fnction as the function which comes first in the word ILATE, where I stands for inverse trigonometric functions L stands for logarithmic functions A stands for algebraic functions T stands for trigonometric functions E stands for exponential functions. 2. If the integrand be logarithmic functions or inverse trigonometric functions alone, take the second function as unity. 3. If both the functions are trigonometric, consider the second function, whose integral being simpler and the other as the first function.
Comparing the real and imaginary part, we get the required result. Formula 7 ______
______
x a2 x 1. Ú ÷a 2 – x2 dx = __ ÷a 2 – x2 + __ sin– 1 __ a + c 2 2
2. Ú ÷x 2 + a2 dx
( )
_______
_______
______
x a2 = __ ÷x 2 + a2 + __ log x + ÷x 2 – a2 + c 2 2 ______ 3. Ú ÷x 2 – a2 dx _______ ______ x a2 = __ ÷x 2 – a2 – __ log x + ÷ x 2 – a2 + c 2 2
| |
|
|
1.6 Integral Calculus, 3D Geometry & Vector Booster Proof:
1. Let
fi fi fi fi fi fi fi
2. Let
______
I = Ú ÷a 2 – x2 dx ______
(
)
1 ______ × – 2x I = ÷a 2 – x2 ◊ x dx Ú dx – Ú ________ 2 ÷a 2 – x2 ______
(
)
2 _______ I = x ÷a 2 – x2 + Ú x ________ dx 2 ÷ a – x2
______
(
______
)
2
2
2
fi 3. Let
______
______
|
______
|
______
|
______
|
|
fi 2I = x ÷x 2 – a2 – a2 log x + ÷ x 2 – a2 + c
|
___________
( ) ( ) ( )
I = Ú ÷x 2 + a2 dx
___________
_______
(
_______
(
)
)
(
2 + bx + c dx Ú ( px + q) ÷ ax
Rules 1. Reduce (px + q) as a derivative of (ax2 + bx + c) 2. Use ______
)
_______ (x2 + a2) – a2 ______ = x ÷x 2 + a2 – Ú ____________ dx ÷ x 2 + a2
_______
dx Ú ÷x 2 ± a2 dx. Ú ÷a 2 – x2
7. Partial Fractions
A special type of Aational (proper) function is known as the partial fraction, where the degree of the numerator < degree a dx ______ = dx + Ú _______ of denominator. Ú ÷x 2 + a2 ÷ x 2 + a2 ______ ______ f (x) h (x) = ____ . = x ÷x 2 + a2 – I + a2 log x + ÷ x 2 + a2 + c Let g(x) _______ x ÷x 2 + a2 –
_______
2
|
2I =
_______ x ÷x 2 + a2 + a2 log x
|
|
______
|
+ ÷x 2 + a2 + c
_______
_______
x a2 I = __ ÷x 2 + a2 + __ log |x + ÷ x 2 + a2 |+c 2 2 ______
= x ÷x 2 – a2 – I – a2 log x + ÷ x 2 – a2 + c
2
x2 = x ÷x 2 + a2 – Ú _______ ______ dx ÷ x 2 – a2
fi
dx 2 + bx + c dx Ú ÷ax ______ I = x ÷a – x – Ú ÷a – x dx + a Ú _______ 2 2 ÷ a – x Rules ______ x 1. First we make the coefficient of x2 unity I = x ÷a 2 – x2 – I + a2 sin–1 __ a + C 2. Express the term under the square root as a sum or ______ x difference of two perfect squares. 2 2 2 – 1 __ 2I = x ÷a – x + a sin a + c _______ ______ 2 2 ______ 2 3. Use x ± a d x or a 2 – x2 dx ÷ ÷ Ú Ú x a x 2 2 –1 I = __ ÷ a – x + __ sin __ + c a 2 2 Type 29: Integrals of the form ______ 2
Type 28: Integrals of the form
1_______ × 2x = ÷x 2 + a2 ◊ x dx Ú dx – Ú _________ 2 ÷x 2 + a2
a2 ______ = x ÷x 2 – a2 – Ú ÷x 2 – a2 dx – Ú ________ dx ÷ x 2 – a2
______
______
x a2 fi I = __ ÷ x2 – a2 – __ log x + ÷ x 2 – a2 + c 2 2
(a2 – x2) – a2 ______ I = x ÷a 2 – x2 – Ú ____________ dx ÷ a 2 – x2 ______
______
I = Ú ÷x – a dx 2
2
______
(
)
1______ × 2x = ÷x 2 – a2 ◊ x dx Ú dx – Ú ________ 2÷x 2 – a2 ______
(
)
x2 = x ÷x – a – Ú _______ ______ dx ÷ x 2 – a2
______ (x2 – a2) + a2 ______ = x ÷x 2 – a2 – Ú ____________ dx ÷ x 2 – a2
2
2
(
)
Type I: When the denominator is expressible as the product of non-repeating linear factors. Let g(x) = (x – a1) (x – a2) (x – a3) ... (x – an) f (x) Then ____ g(x) A3 An A1 A2 = _______ + _______ + _______ + ... + _______ (x – a1) (x – a2) (x – a3) (x – an) where A1, A2, ..., An are constant and can be determined by equating the numerator on RHS to the numerator on LHS and then substituting x = a1, a2, a3, ..., an Type II: When the denominator is expressible as the product of linear factors such that some of them are repeating. Let
g(x) = (x – a1)k(x – a2)(x – a3)...(x – an)
Indefinite Integrals
A1 A1 A1 f (x) _______ Then ____ = + _______ + ... + _______ g(x) (x – a1) (x – a1)2 (x – a1)k Bn B1 B2 + _______ + _______ + ... + _______ (x – a2) (x – a3) (x – an) Type III: When the denominator is expressible as the product of linear and quadratic factors but non-repeating. Let
g(x) = (x – a1)(x – a2)(x – a3) ... (ax2 – bx + c)
A3 A1 A2 f (x) _______ Then ____ = + _______ + _______ g(x) (x – a1) (x – a2) (x – a3) Type IV: Integrals of the form
Ú cosec2r + 1x dx = Ú cosec2r – 1x . cosec2x dx and then integrate it by parts, where consider cosec2x as the first function. consider sec2x as the second function.
2
Rules 1. Put x2 = t 2. Donot find its derivative 3. Use the concept of partial fractions. Type V: Integrals of the form
Rules 1. If m is even or odd integer and n is even positive integer, put cot x = t. 2. If m is odd positive integer and n Πeven positive integer, put cosec x = t. 3. If m = 0 and n = 2 r + 1, " r ΠN, write
Bx + C + ... + ___________ Type 3: Integrals of the form ax2 + bx + c dx dx , Ú ____________ , Ú ________ (n – 1) ______ x(xn + 1) 2 n n x (x + 1)
x dx Ú ______________ 2 (x + a)(x2 + b)
(x2 + a)(x2 + b) dx Ú ______________ (x2 + c)(x2 + d)
Rules 1. Put x2 = t 2. Do not find its derivative 3. Reduce it into a partial fraction (degree of the numerator < Degree of denominator) 4. Use the concept of partial fractions.
dx , where n Œ N Ú ___________ n x (1 + xn)1/n
Rules 1. Take common xn from the denominator.
1 + x– n = t.
2. Put
Type 4: Integrals of the form m
x dx, where m, n ŒN Ú ________ (ax + b)n
Rule put
ax + b = t.
Type 5: Integrals of the form dx Ú __________ , where m, n ŒN xm(ax + b)n
(
)
ax + b Rule put ______ x = t
Advanced Level
Type 6: Integrals of the form
Type 1: Integrals of the form
Ú tanmx . secnx dx
Rules 1. If is even or odd integer and n is even positive integer, put tan x = t 2. If m is odd positive integer and n Πeven positive integer, put sec x = t. 3. If m = 0 and n = 2 r + 1, " r ΠN, write
Ú sec2r + 1x dx = Ú sec2r – 1x . sec2x dx
and then integrate it by parts, where Type 2: Integrals of the form
Ú cotmx . cosecnx dx
1.7
Rules
( (
dx , where m, n ŒN Ú (x_____________ – a)n(x – b)n
) )
x–a 1. _____ = t, when m < n x–b x – b 2. _____ x – a = t, when m > n
Type 7: Integrals of the form dx , where n ŒN Ú _________ x(a + bx n) 1 xn = __ t Type 8: Integrals of the form Rule: put
2m + 1
x ________ dx Ú (ax 2 + b)n
1.8 Integral Calculus, 3D Geometry & Vector Booster Rule Put
(a x2 + b) = t
Rule Put (ax + b) = t p, where p is the LCM of m and n.
Type 9: Integrals of the form
Type 3: Integrals of the form
(a sin x + b)
dx Ú ___________ (a + b sin x)2
1
Rules 1. Divide the numerator and the denominator by cos2x 2. Put a sec x + b tan x = t. Type 10: Integrals of the form
Rule Put
dx ___________ Ú (a + b sin x)2
Rule Put
L(x) = t2
dx_____ Ú __________ L (x) Q (x)
Rule Put
1 L (x) = __ t
(
)
)
______ n
) } dx Ú { f ( x ± ÷x 2 + a2
Type 8: Integrals of the form
Functions
( (
dx , where m + p = N Ú __________ xm(a + bx)p
and
m+p>1
Rule Put
a + bx = t x.
Type 9: Integrals of the form
Type 1: Integrals of the form
))
a n ax + b __ ______ f x , dx, Ú cx + d
where
a, b, c, d, a ,
ΠR
)
ax + b Rule Put ______ = t n cx + d Type 2: Integrals of the form Ú f ( x, (ax + b)
Q2(x) 1 x = __ or t2 = _____ t Q1(x)
______
Rules 1. Express the numerator = l (Denominator) + m × derivative of (Denominator) 2. Compare the coefficients of ex and e– x 3. Find l and m.
(
2
Rule Put ( x ± ÷x 2 + a2 ) = t
a e + b e x dx Ú __________ p e + q e– x
of Irrational
÷
Type 7: Integrals of the form
Type 13: Integrals of the form
8. Integration
dx_____ Ú ___________ Q (x) Q (x) 1
a cos x + b t = __________ a + b cos x – x
÷
Type 6: Integrals of the form
Rule Put
x
÷
Type 5: Integrals of the form
Type 12: Integrals of the form dx ___________ Ú (a + b cos x)2
(
L2(x) = t 2 dx_____ Ú _________ Q(x) L (x)
2
a sin x + b Rule Put _________ =t a + b sin x
Rule put
÷
Type 4: Integrals of the form
(a cos x + b)
dx Ú ____________ (a + b cos x)2
Rules 1. Divide the numerator and the denominator by sin2x 2. Put a cosec x + b cot x = t. Type 11: Integrals of the form
dx_____ Ú __________ L (x) L (x)
a /n
, (ax + b)
where m, n are positive integers.
) dx,
b /m
dx , where m, n ŒR Ú _____________ L (x))m(L (x))n 1
2
Rules
L1(x) 1. If n > m, Put _____ = t L2(x)
L 2 (x) 2. If m > n, Put _____ =t L 1(x)
dx ______ Type 10: Integrals of the form Ú ________ x ÷ax n + b Rule Put
axn + b = t2
Indefinite Integrals
Type 11: Integrals of the form dx where a, b Œ R – {0} Ú (a__________ + bx2)3/2 Rule Put
1 x = __ t
dx ___________ , Ú __________________ r (x – k) ÷ax 2 + bx + c
where
r ΠN and k ΠR Р{0}
1 Rule Put x – k = __ t Type 13: Integrals of the form
___________
2 + bx + c ) dx Ú R ( x, ÷ax
Rules
___________
__
1. Put ÷ax 2 + bx + c = t ± x ÷a , if a > 0
2. Put ÷ax 2 + bx + c = t x ± ÷c , if c > 0,
3. Put (x – a) t, or (x – b) t, where a and b are the real roots of ax 2 + bx + c.
___________
Type 12: Integrals of the form
(ax + b)
___________ dx Ú ___________________ 2
(cx + d) ÷px + qx + r
Rules 1. Put (ax + b) = A(cx + d) + B 2. Find the values of A and B 3. Reduce the given integral into two separate integrals. Type 14: Integrals of the form (ax2 + bx + c) ___________ dx Ú ___________________ (dx + e) ÷px 2 + qx + r
1.9
__
___________ ÷a x 2 + bx + c =
10. Integration
by
Reduction Formula
A reduction formula is defined as a formula or a connection by means of which the power of the integrand is reduced, therefore, making the integration easier. The basic technique of obtaining a reduction formula is the integration by parts. In some cases the method of differentiation or other special devices is adopted. Type I: Reduction formula for Ú sinnx dx Soln. Let In = Ú sinnx dx
= Ú sinn – 1x ◊ sin x dx = sinn – 1x Ú sin x dx – Ú (n – 1)sinn – 2x (– cos2x) dx = sinn – 1x(– cos x) + (n – 1) Ú sinn – 2x (1 – sin2x) dx = sinn – 1x(– cos x) + (n – 1) I n – 2 – In
Rules 1. Put (ax2 + bx + c) = L (dx + e) (2px + q) + M(dx + e) + N 2. Compare the co-efficents of the like terms of both the sides and find L, M and N. 3. Integrate the given integral.
Thus, (1 + n – 1) In = – cos x ◊ sinn – 1x + (n – 1)In – 2
Type 15: Integrals of the form
Type 2: Reduction formula for Ú cosnx dx
Ú xb (a + bx g )a dx
Rules 1. If a Œ I +, expand the integral by the concept of binomial expansion. 2. If a Œ I –, we put x = t p, where p is the LCM of the denominator of b and g.
b+1 3. If _____ ΠI and a is a fraction, put (a + bxg ) = t p, where p is the denominator of a . b + 1 4. If _____ + a ΠI, put (a + bxg ) = t pxg, where p is the denominator of a .
9. Euler’s Substitution Type 16: Integrals of the form
fi
nIn = – cos x ◊ sinn – 1x + (n – 1) In – 2
cos x ◊ sinn – 1x n–1 In = ___________ + _____ n In – 2 n
( )
which is the required reduction formula.
Soln. Let In = Ú cosnx dx = Ú cosn – 1x ◊ cos x dx = cosn – 1x Ú cos x dx + (n + 1) Ú cosn – 2x ◊ sinnx dx = cosn – 1x ◊ sin x + (n + 1) Ú cosn – 2x ◊ (1 – cos2x) dx = cosn – 1x ◊ sin x + (n + 1) I n – 2 – In Thus, (1 + n + 1)In = cosn – 1x ◊ sin x + (n + 1) I n – 2 + C
(
)
cosn – 1x ◊ sin x n+1 fi In = ___________ + _____ I + C (n + 2) n + 2 n – 2 which is the required reduction formula. Type 3: Reduction formula for Ú tann x dx Soln. Let
In = Ú tann x dx
1.10 Integral Calculus, 3D Geometry & Vector Booster 2 sin(n – 1) x = ___________ + In – 2 (n – 1)
= Ú tann – 2x ◊ tan2 x dx
= Ú tann – 2x ◊ (sec2 x – 1)dx
which is the required reduction formula.
= Ú tann – 2x ◊ sec2 x dx – Ú tann – 2x dx
dx Type 7: Reduction formula for Ú ________ 2 (x + k)n
tann – 1x = ______ – In – 2 + C n–1
dx Soln. Let In = Ú ________ (x2 + k)n
(
)
which is the required reduction formula.
dx Then In – 2 = Ú __________ 2 (x + k)n – 1
Type 4: Reduction formula for Ú cot x dx n
Soln. Let
(
In = Ú cotnx dx
)
1 = Ú ___________ 2 dx (x + k)n – 1 ◊1
= Ú cotn – 2x ◊ cot2 x dx
= Ú cotn – 2x ◊ (cosec2 x – 1) dx
– (n – 1) 1 = _________ 2 n – 1 Ú 1 ◊ dx – Ú ________ ◊ 2x ◊ x dx (x + k) (x2 + k)n
= Ú cotn – 2x ◊ cosec2 x dx – Ú cotn – 2x dx
x x2 = __________ 2 + 2(n – 1) Ú _______ 2 n dx n – 1 (x + k) (x + k)
(
(
)
(
n – 1
cot x = – ______ – I n – 2 + C n–1
)
)
(x2 + k) – k x ___________ = _________ + 2 (n – 1) dx Ú (x2 + k)n – 1 (x2 + k)n
which is the required reduction formula.
(
)
x k 1 = _________ 2 n – 1 + 2 (n – 1) Ú __________ 2 n – 1 – ________ dx 2 (x + k) (x + k) (x + k)n
Type 5: Reduction formula for Ú secnx dx Soln. Let In = Ú sec x dx n
= Ú secn – 2x ◊ sec2 x dx
= secn – 2x Ú sec2 x dx – Ú (n – 2)secn – 2x ◊ tan2 x dx
= secn – 2x tan x – (n – 2) Ú secn – 2x ◊ (sec2 x – 1)dx
= secn – 2x tan x – (n – 2) Ú (secnx – secn – 2 x)dx
= secn – 2x tan x – (n – 2) In + (n – 2) In – 2
x = _________ + 2 (n – 1) (In – 1 – k In) (x2 + k)n – 1 x = _________ + 2(n – 1)In – 1 – 2(n – 1)k In (x2 + k)n – 1 fi 2 (n – 1) k In
x = _________ + 2(n – 1) In – 1 – In – 1 (x2 + k)n – 1
secn – 2x tan x n–2 fi In = __________ + _____ I +C n–1 n – 1 n – 2
x = _________ 2 n – 1 + (2n – 3) In – 1 (x + k)
which is the required reduction formula.
x fi In = __________________ 2 + 2(n – 1) k (x + k)n – 1
fi (n – 2) In = secn – 2x tan x – (n – 2) In – 2
(
)
sin nx Type 6: Reduction formula for Ú _____ dx sin x sin nx Soln. Let In = Ú _____ dx sin x
(
]
dx x = _________ 2 n – 1 + 2 (n – 1) In – 1 – In – 1 – k Ú _______ 2 (x + k) (x + k)n
(
)
1 2n – 3 __ ______ I k 2n – 2 n – 1
which is the required reduction formula. Type 8: Reduction formula for Ú x m(log x)n dx
)
sin(n – 2)x = Ú 2cos(n – 1)x + __________ dx sin x
Soln. Let Im, n = Ú x m(log x)n dx
[ sin nx – sin (n – 2)x = 2 cos (n – 1) x sin x
]
(
( )
)
1 x m + 1 = (log x)n Ú x m dx – Ú n(log x)n – 1 __ x _____ dx m+1
sin(n – 2)x sin n x fi _____ = 2 cos (n – 1)x + __________ = (log x)n, sin x sin x
x m + 1 n _____ – _____ ( x m(log x)n – 1 ) dx m+1 m+1 Ú
Indefinite Integrals
1.11
x m + 1 n = (log x)n, _____ – _____ I m + 1 m + 1 m, n – 1
m m – 1 – __ n Ú – cos x(sin nx cos x – sin(n – 1)x) dx
x m + 1 n fi Im, n – 1 = (log x)n, _____ – _____ I m + 1 m + 1 m, n – 1 which is the required reduction formula.
cosmx ◊ cos nx __ m = – ___________ – n Ú cos mx sin nx dx n m m – 1 + __ n Ú cos x sin (n – 1) x dx
Type 9: Reduction formula for
Ú x m(1 – x)n
cosmx ◊ cos nx __ m m = – ___________ – n Im, n + __ n Im – 1, n – 1 n
Im, n = Ú x m(1 – x)n dx
Soln. Let = (1 – x)
n
(
Ú x dx + Ú n(1 – x) m
m + 1
(
m fi 1 + __ n Im, n
)
x ◊ _____ dx m+1
n – 1
x m + 1 n = (1 – x)n _____ + _____ [x m + 1 ◊ (1 – x)n – 1] dx m + 1 m + 1Ú x m + 1 n = (1 – x)n _____ + _____ x m + 1 . (1 – x)n – 1dx m + 1 m + 1Ú
cosmx ◊ cos nx __ m = – ___________ + n Im – 1, n – 1 n
fi
cosmx cos nx _______ m Im, n = – ___________ + I (m + n) (m + n) m – 1, n – 1
Note: Similarly, we can easily formulate the reduction formula for m x sin nx m __________ + _______ I Ú cosmx cos nx dx = cos m+n (m + n) m – 1, n – 1
x m + 1 _____ n = (1 – x)n _____ + m+1 m+1
Type 11: Reduction formula for
Ú [x m ◊ (1 – x)n – 1 ◊ {– 1(1 – x)}] dx
x m + 1 n = (1 – x) _____ + _____ [I – Im, n] m + 1 m + 1 m, n – 1 n
m + 1
x n n = (1 – x)n _____ + _____ I – _____ – Im, n m + 1 m + 1 m, n – 1 m + 1 n fi 1 + _____ I m + 1 m, n
(
which is the required reduction formula.
x m + 1 n = (1 – x)n _____ + _____ [x m . (1 – x)n – 1 ◊ x] dx m + 1 m + 1Ú
)
)
m + 1
x n _____
n = (1 – x) + _____ I m + 1 m + 1 m, n – 1
Ú sinmx sin nx dx
Soln. Let Im, n = Ú sinmx sin nx dx
= sinmx Ú sin nx dx
(
)
cos n x – Ú m sinm – 1x ◊ cos x ◊ – ______ dx n
sinmx ◊ cos nx = – ___________ n
x m + 1 (1 – x)n n fi Im, n = ____________ + _________ I (m + n + 1) m + n + 1 m, n – 1
m m – 1 + __ n Ú sin x cos x cos nx dx
hich is the required reduction formula. w Type 10: Reduction formula for
Ú cosmx sin nx dx Soln. Let Im, n = Ú cosmx sin nx dx
m m – 1 + __ n Ú sin x (cos(n – 1)x – sin nx sin x)dx
= cosmx Ú sin nx dx
(
)
cos nx – Ú – _____ ◊ m ◊ cosm – 1x◊ – sinx dx n cos nx = cosmx – _____ n
(
)
m m – 1 – __ n Ú – cos x (cos nx sin x) dx
cosmx ◊ cos nx = – ___________ n
sinmx ◊ cos nx = – ___________ n
sinmx ◊ cos nx = – ___________ n
m m – 1 + __ n Ú sin x (cos(n – 1) x dx m m – __ n Ú sin x sin nx dx sinmx ◊ cos nx m fi 1 + __ n Im, n = – ___________ n
(
)
m + __ n Im – 1, n – 1
1.12 Integral Calculus, 3D Geometry & Vector Booster fi
sinmx ◊ cos nx _______ m Im, n = – ___________ + I (m + n) (m + n) m – 1, n – 1
Note: Similarly, we can easily formulate the reduction formula for sin nx m + _______ I Ú sinmx cos nx dx = sinmx _______ (m + n) (m + n) m – 1, n – 1 Type 12: Reduction formula for
which is the required reduction formula.
11. Inexpressible integrals functions, the integral is known as computable. But if an
n + 1
Soln. Let P = sin x cos x dp fi ___ = (m – 1) sinm – 2 x cos n + 2x dx – (n + 1) sinm x cos nx
sinm – 1x cosn + 1x ______ m–1 = – Ú _____________ + I m+n m + n m – 2, n
If an integral Ú f (x) dx as expressible in terms of elementary
Ú sinmx cos nx dx m – 1
Ú sinmx cosnx dx
fi
integral Ú f (x) dx is not expressible in terms of elementary functions, the integral is known as inexpressible or ‘cannot be found’. Some inexpressible integrals are
= (m – 1) sinm – 2x cosnx ◊ cos 2x
(i) Ú ex dx 2
dx – (n + 1) sinmx cos nx (ii) Ú e– x 2
____
(iii) Ú ÷sin x dx
= (m – 1) sinm – 2 x cosnx ◊ (1 – sin 2x )
– (n + 1) sinmx cos nx (iv) ÷_____ dx Ú cos x m – 2 n 2 = (m – 1) sin x cos x ◊ (1 – sin x ) (v) Ú x tan x dx
– (n + 1) sinmx cos nx
sin x (vi) Ú ____ dx x = (m – 1) sin x cos x cos x (vii) Ú ____ dx x – (m – 1) sinmx cosnx – (n + 1) sinmx cos nx dx (viii) Ú ____ = (m – 1) sinm – 2 x cosnx log x _____ – (m – 1 + n + 1) sinmx cosnx (ix) Ú ÷1 + x3 m – 2 n = (m – 1) sin x cos x ______ 3 1 + x3 (x) Ú ÷ dx – (m + n) sinmx cosnx (xi) Ú sin (x2) dx On integration, we get m – 2
(xii) Ú cos (x2) dx
P = (m – 1) Ú sinm – 2x cosnx dx – (m – n) Ú sinmx cosnx dx
fi
n
(m – n) Ú sinmx cosnx dx
__________
= – P + (m – 1) Ú sinm – 2x cosnx dx
x2 (xiii) Ú _____ dx 1 + x5
2 (xiv) Ú ÷1 – k2sin x dx
__
__
(xv) Ú ÷x cos ÷x dx.
Exercises
(Problems based on Fundamentals) ABC of Integration
1. Evaluate: Ú logx x dx
2. Evaluate: Ú ( 3log x2 – 2 log x3 )
(
m 3. Evaluate: Ú xm + mx + mm + __ x dx
4. Evaluate: Ú 2 ◊ 3 ◊ dx
5. Evaluate: Ú tan2x dx
6. Evaluate: Ú cot2x dx
dx 7. Evaluate: Ú _________ 2 sin x cos2x
x
)
x
Indefinite Integrals
(
(
) ) (
(
))
3p p 8. Evaluate: Ú 1+ tan x + ___ 1 + tan __ – x dx 8 8 9. Evaluate: Ú (tan x + cot x)2dx
dx 10. Evaluate: Ú _________ 1 + cos2x
dx 31. Evaluate: Ú ________________________ (tan x + cot x + sec x + cos x)
11. Evaluate: Ú ( 3log 5x – 2 log 5x ) dx
(
) ) (
(
(
Type 2
))
p p 12. Evaluate: Ú 1+ tan __ – x 1+ tan __ + x dx 8 8
(
)
81 + x + 41 + x 13. Evaluate: Ú __________ dx 22x x m 14. Evaluate: Ú __ m + __ x + xm + mx dx
(
)
(ax + b x)2 15. Evaluate: Ú ________ dx axbx (2x + 3 x)2 16. Evaluate: Ú ________ 2x ◊ 3xdx p 1 1 17. If f ¢(x) = __ x + ______ ______ and f (1) = __ , find f (x). 2 2 ÷ 1 – x 18. If f ¢(x) = a cos x + b sin x and
( )
p f ¢(0) = 4, f ¢(0) = 3, f __ = 5, find f (x). 2 dx 19. Evaluate: Ú _______ 1 – sin x
) )
sin4x + cos4x 20. Evaluate: Ú ____________ dx sin2x cos2x 6
6
sin x + cos x 21. Evaluate: ____________ 2 dx sin x cos2x
(
)
cos 2x – cosa 22. Evaluate: Ú ____________ dx cosx – cosa
( Ú (
)
cos4x – sin4x 23. Evaluate: Ú ___________ _________ dx 1 + cos4x ÷
)
1 + tan2x 24. Evaluate: ________ dx 1 + cot2x
(
(1 + x)2 33. Evaluate: Ú ________ dx x (1 + x2) x2 – 2 34. Evaluate: Ú ______ dx x2 + 1 x–1 35. Evaluate: Ú _____________ dx (x2/3 + x1/3 + 1)
( Ú ( Ú ( Ú ( Ú ( Ú ( Ú (
) ) )
x4 + 2 36. Evaluate: Ú ______ 2 dx x +2 x4 – 3 37. Evaluate: ______ 2 dx x +1 x6 – 1 38. Evaluate: ______ 2 dx x +1
)
___________
)
x4 40. Evaluate: _____ 2 dx x +1
)
x4 + x2 + 1 41. Evaluate: __________ dx x2 + x + 1
)
x6 + 1 42. Evaluate: ______ 2 dx x + 1 Type 3 43. Evaluate: Ú sin– 1(sin x) dx 44. Evaluate: Ú sin– 1(cos x) dx
( ÷
_________
)
cos x – cos 2x 25. Evaluate: Ú ___________ dx 1 – cos x
(
x 32. Evaluate: Ú _____ dx x+1
x8 + x4 + 1 39. Evaluate: __________ 4 dx x + x2 + 1
Type 1
( Ú (
( ) cos x – cos2x 30. Evaluate: Ú ( ____________ dx 1 – cosx )
cos 5x + cos 4x 29. Evaluate: Ú _____________ dx 1 + 2 cos3x
)
)
1 – cos 2x 45. Evaluate: Ú tan _________ dx 1 + cos 2x – 1
(
)
+ 2 ÷ x 4 + x– 4 26. Evaluate: Ú ____________ dx 3 x
sin 2x 46. Evaluate: Ú tan– 1 _________ dx 1 + cos 2x
5cos3x + 3sin3x 27. Evaluate: Ú ______________ dx cos2x sin2x
1 – cos 2x 47. Evaluate: Ú tan– 1 _________ dx 1 + cos 2x
(
(
)
)
cos x – sin x 28. Evaluate: Ú ___________ (1 + sin2x) dx cos x + sin x
( ÷
_________
(
)
)
sin x 48. Evaluate: Ú tan– 1 ________ dx 1 – cos x
1.13
1.14 Integral Calculus, 3D Geometry & Vector Booster
( ÷
________
)
dx _________ _________ 71. Evaluate: Ú ______________________ 2x + 2014 + ÷2x + 3013 ÷ Type 6
1 – sin x 49. Evaluate: Ú tan ________ dx 1 + sin x –1
(
)
sin x 50. Evaluate: Ú tan–1 ________ dx 1 + cos x
x _____ 72. Evaluate: Ú ______ dx x – 1 ÷
( ) 1 – sin x ( ________ cos x ) dx
cos x 51. Evaluate: Ú tan–1 ________ dx 1 – sin x 52. Evaluate: Ú tan– 1 53. Evaluate: Ú
(
__
x ÷ 73. Evaluate: Ú _____ dx x+1
_________ _________ ÷(1 + sin x) + ÷(1 – sin x) – 1 ______________________ _________ dx tan _________ ÷(1 + sin x) – ÷(1 + sin x)
)
54. Evaluate: Ú tan– 1 (sec x + tan x) dx
(
)
sin 2 x 55. Evaluate: Ú tan– 1 __________ dx 1 + cos 2 x Type 4 56. Evaluate: Ú (3x + 2) dx dx 57. Evaluate: Ú ______ 2x – 3 dx 58. Evaluate: Ú ______ 5 – 2x 59. Evaluate: Ú eax + b dx 60. Evaluate: Ú 34x + 5 dx 61. Evaluate: Ú cos (5x + 3) dx 62. Evaluate: Ú sin 2x dx ______
63. Evaluate: Ú ÷3x + 2 dx dx ______ 64. Evaluate: Ú _______ 3x + 4 ÷
x ______ 74. Evaluate: Ú _______ dx 3x + 1 ÷ x______ +1 75. Evaluate: Ú _______ dx 2x – 1 ÷ x_____ –1 76. Evaluate: Ú ______ dx x + 4 ÷ x 77. Evaluate: Ú _____ dx x2 + 1 Type 7 cos x – sin x 78. Evaluate: Ú ___________ dx sin x + cos x 3 cos x 79. Evaluate: Ú _________ dx 2 sin x + 5 cos x – sin x 80. Evaluate: Ú __________ dx 2 + sin 2x xe x + ex 81. Evaluate: Ú ________ dx cos2(xex) dx 82. Evaluate: Ú _________ x(1 + ln x)2 cos x – sin x + 1 – x 83. Evaluate: Ú _________________ dx ex + sin x + x dx 84. Evaluate: Ú _____ 1 + ex
Type 5
dx 85. Evaluate: Ú ________ 3 x(x + 1)
dx _____ _____ 65. Evaluate: Ú ______________ x + 2 – ÷x + 1 ÷
dx 86. Evaluate: Ú ________ 4 x(x + 1)
dx ______ 66. Evaluate: Ú _________________ ______ (÷2x + 5 – ÷2x + 3 ) dx ______ ______ 67. Evaluate: Ú ________________ 3x + 4 – ÷3x + 1 ÷ dx ______ ______ 68. Evaluate: Ú ________________ 2x + 3 + ÷2x – 3 ÷ dx _____ 69. Evaluate: Ú ___________ __ x + 1 + ÷x ÷ dx _____ 70. Evaluate: Ú _______________ _____ x + a + ÷x + b ÷
dx 87. Evaluate: Ú _______ 5 x(x – 1) sin 2x 88. Evaluate: Ú __________ dx sin 5 x sin 3x dx 89. Evaluate: Ú _________________ sin(x – a) sin (x – b) e x – e– x 90. Evaluate: Ú ________ dx e x + e– x
Indefinite Integrals
dx 91. Evaluate: Ú ______ 1 + e– x
dx 112. Evaluate: Ú ____________ __ __ 2 x (4 + 3÷x ) ÷
dx 92. Evaluate: Ú _____ 1 + e x
113. Evaluate: Ú 33 ◊ 33 dx
sin 2 x 93. Evaluate: Ú ______________ dx 2 a sin x + b cos2x
114. Evaluate: Ú tan3 x ◊ sec2xdx
sin(x – a) 94. Evaluate: Ú _________ dx sin x sin x 95. Evaluate: Ú ________ dx sin (x – a) sin(x + a) 96. Evaluate: Ú _________ dx sin(x + b)
3x
x
115. Evaluate: Ú sin3 x ◊ cos xdx (log x)3 116. Evaluate: Ú ______ dx x sin x _________ 117. Evaluate: Ú ___________ dx 3 + 2 cos x ÷ ________
dx 97. Evaluate: Ú _________ __ __ x (÷ x + 1) ÷
÷2 + log x 118. Evaluate: Ú _________ dx x dx 119. Evaluate: Ú ______ __ 1 + ÷x
1 + tan x 98. Evaluate: Ú ___________ dx x + log sec x
120. Evaluate: Ú x3sin x4 dx
sin 2x 99. Evaluate: Ú ___________ dx sin 5x . sin 3x
121. Evaluate: Ú 55 ◊ 55 ◊ 5x dx
cos x – sin x 100. Evaluate: Ú ___________ dx 1 + sin 2x dx 101. Evaluate: Ú _________ sin x . cos2x dx 102. Evaluate: Ú __________ . sin2x . cos2x dx 103. Evaluate: Ú _________________ sin (x – a) sin (x – b) dx 104. Evaluate: Ú __________________ cos (x – a) cos (x – b) dx 105. Evaluate: Ú _________________ sin (x – a) sin (x – b) x x(1 + ln x) 106. Evaluate: Ú __________ dx xx + 1 cos x – sin x + 1 – x 107. Evaluate: Ú _________________ dx ex + sin x + x 108. Evaluate: sin3x Ú _________________________________ dx 4 2 (cos x + 3cos x + 1)tan– 1(sec x + cos x) Type 8 109. Evaluate: Ú 3x2 sin (x3) dx (1 + ln x)3 110. Evaluate: Ú _________ dx x dx 111. Evaluate: Ú __________ 2 x (1 + x4)3/4
5x
x
sin x – cos x 122. Evaluate: Ú __________ dx ex + sin x dx 123. Evaluate: Ú _________ x (1 + x 3) Type 9 case I: 124. Evaluate: Ú sin3 x ◊ cos4x dx 125. Evaluate: Ú sin x ◊ cos6x dx 126. Evaluate: Ú sin5x ◊ cos9x dx case II: 127. Evaluate: Ú sin2x ◊ cos3x dx 128. Evaluate: Ú sin4x ◊ cos3x dx 129. Evaluate: Ú sin6x ◊ cos5x dx case III: 130. Evaluate: Ú sin3x ◊ cos3x dx 131. Evaluate: Ú sin5x ◊ cos5x dx 132. Evaluate: Ú sin5x ◊ cos7x dx case IV: 133. Evaluate: Ú sin2x ◊ cos2x dx 134. Evaluate: Ú sin2x ◊ cos4x dx
1.15
1.16 Integral Calculus, 3D Geometry & Vector Booster 135. Evaluate: Ú sin4x ◊ cos2x dx case V: 136. Evaluate: Ú sin3x dx 137. Evaluate: Ú sin5x dx 138. Evaluate: Ú sin7x dx case VI: 139. Evaluate: Ú cos5x dx 140. Evaluate: Ú cos3x dx 141. Evaluate: Ú cos7x dx case VII: 142. Evaluate: Ú sin4x dx 143. Evaluate: Ú sin2x dx 144. Evaluate: Ú sin6x dx case VIII: 145. Evaluate: Ú cos x dx 6
146. Evaluate: Ú cos2x dx 147. Evaluate: Ú cos4x dx case IX: dx 148. Evaluate: Ú ____________ 1/2 sin x cos3/2x dx 149. Evaluate: Ú ____________ 3/2 sin x cos5/2x ____ ÷tan x ________
150. Evaluate: Ú dx sin x cos x sin x 151. Evaluate: Ú _____ dx. cos5x dx 152. Evaluate: Ú _________ 3 sin x cos5x
dx 156. Evaluate: Ú ___________ 7 1 __ __ 2 2 sin x ◊ cos x dx __________ 157. Evaluate: Ú ___________ 3 5 x ◊ cos x ÷ sin ABC of Formula 6 dx 158. Evaluate: Ú ______ x 2 + 4 dx 159. Evaluate: Ú _______ 9x 2 + 1 dx 160. Evaluate: Ú ______ 2 x – 4 x 4 – 1 161. Evaluate: Ú _______ x2 + 5dx dx _____ 162. Evaluate: Ú _______ 4 ÷ x + 4 dx ______ 163. Evaluate: Ú ________ 2 + 1 ÷ 4x
(
)
x+4 164. Evaluate: Ú _______ dx x3 + 4x
( )
x4 + 1 165. Evaluate: Ú ______ 2 dx x +1 dx __________ 166. Evaluate: Ú ___________ – x)2 + 1 ÷ (2 x+9 167. Evaluate: Ú _______ dx x3 + 9x 1 + x 168. Evaluate: Ú _____2 dx 1+x 1+x 169. Evaluate: Ú _____ dx x3 + x dx 170. Evaluate: Ú _____ x4 + 1 dx 171. Evaluate: Ú _____ 3 x +x Type 10
sin2x dx 153. Evaluate: Ú _______ cos6x
dx 172. Evaluate: Ú __________ 3 x + 4x + 4
dx 154. Evaluate: Ú _________ sin x cos3x
dx 173. Evaluate: Ú ___________ 2 x + 6x + 10
dx 155. Evaluate: Ú _________ 2 sin x cos4x
dx 174. Evaluate: Ú ___________ 2 2x + 5x + 6
Indefinite Integrals
dx 175. Evaluate: Ú _________ x2 + x + 1 dx 176. Evaluate: Ú _________ 1 + x + x2 dx 177. Evaluate: Ú __________ 2 x + 4x + 3
3x + 2 195. Evaluate: Ú __________ dx x2 – 3x + 4 x 195. Evaluate: Ú _________ dx 2 x + x + 1 4x + 1 196. Evaluate: Ú __________ dx x2 + 3x + 2
dx 178. Evaluate: Ú ____________ 2 4x + 7x + 10
dx 197. Evaluate: Ú ____________ 2e2x + 3ex + 1
dx 179. Evaluate: Ú _______ x2 – 2ax
(3 sin x – 2) cos x dx 198. Evaluate: Ú _________________ (5 – cos2x – 4sin x)
dx 180. Evaluate: Ú _______ 2 x + 2ax
ax3 + bx 199. Evaluate: Ú ________ dx. x4 + c2
dx 181. Evaluate: Ú ________ 2 a + 2a x
cos x – sin x 200. Evaluate: Ú ___________ × (2 + 2 sin 2 x) dx cos x + sin x
dx 182. Evaluate: Ú _______ 2a x – x2
sin x + cos x 201. Evaluate: Ú ___________ dx 5 + 3sin2x
dx 183. Evaluate: Ú ______________ 2 (x + 1) (x2 + 4)
sin x – cos x 202. Evaluate: Ú ___________ dx 3 + 5sin 2x
x2 184. Evaluate: Ú ________ 6 x + 1dx
Type 12
cos x dx 185. Evaluate: Ú _______________ 2 sin x + 3sin x + 2 x
x (1 + log x) 186. Evaluate: Ú ___________ dx x2x + x x + 1 dx 187. Evaluate: Ú ________ 4 x(x + 1) x dx 188. Evaluate: Ú __________ 4 x + x2 + 1 ex dx 189. Evaluate: Ú ___________ e2x + 6ex + 5 5
x 190. Evaluate: Ú _______ dx 1 + x12 dx 191. Evaluate: Ú ________ 4 x(x + 1) dx 192. Evaluate: Ú ________ x(x3 + 1)
(
dx _________ 203. Evaluate: Ú ___________ 2 ÷ x + x + 1 dx ________ 204. Evaluate: Ú _________ 2 ÷ x – 2ax dx _______ 205. Evaluate: Ú ________ – x2 ÷ 4x dx _________ 206. Evaluate: Ú __________ ÷ 6 – x – x2 dx _________ 207. Evaluate: Ú __________ ÷ 1 + x + x2 dx _________ 208. Evaluate: Ú __________ ÷ 1 + x – x2 dx ________ 209. Evaluate: Ú _________ 2 ÷ x + 2ax dx ________ 210. Evaluate: Ú _________ – x2 ÷ 2ax Type 13
dx 193. Evaluate: Ú ________ x(xn + 1)
x dx __________ 211. Evaluate: Ú ___________ 4 + 1 ÷ x – x2
Type 11
212. Evaluate: Ú ÷sec x – 1 dx
2x + 3 194. Evaluate: Ú __________ dx 2 x + 4x + 5
)
________
dx _______ 213. Evaluate: Ú ___________ 3/4 1/2 x ÷ x – 1
1.17
1.18 Integral Calculus, 3D Geometry & Vector Booster
÷
_________
sin (x – a) 214. Evaluate: Ú _________ dx sin (x + a)
dx 233. Evaluate: Ú ______________ . 3 sin2x + 4 cos2x
ex ______ 215. Evaluate: Ú _______ dx ÷ 4 – e2x
dx 234. Evaluate: Ú ______________ (2sin x + 3 cos x)2
sec2x _________ 216. Evaluate: Ú __________ dx ÷16 + tan x ________
217. Evaluate: Ú ÷sec x + 1 dx _________
218. Evaluate: Ú ÷cosec x – 1 dx dx ______ 219. Evaluate: Ú ________ ÷ 1 – e2x
÷
_________
sin (x – a) 220. Evaluate: Ú _________ dx sin(x + a) dx ______ 221. Evaluate: Ú ___________ 2/3 x ÷x 2/3 – 4
÷
______
x 222. Evaluate: Ú ______ dx x 3 – x3
sin x cos x 235. Evaluate: Ú ____________ dx sin4x + cos4x dx 236. Evaluate: Ú _____________ (sin x + 2 cos x)2 dx 237. Evaluate: Ú _____________ (sin x + 2sec x)2 sin 2 x dx 238. Evaluate: Ú ____________ sin4x + cos4x dx 239. Evaluate: Ú _______________ (2 sin x + 3 cos x)2 dx 240. Evaluate: Ú _________ 2 + cos2 x Type 16
sinq – cosq _________ 224. Evaluate: Ú ___________ dq ÷2 – sin 2q
sin x 241. Evaluate: Ú _____ dx sin 3 x cosec 3x 242. Evaluate: Ú _______ dx cosec x sec 3x 243. Evaluate: Ú _____ dx sec x
Type 14
Type 17
x–1 __________ 225. Evaluate: Ú ___________ dx 2 + 2 ÷ x – 3x
dx 244. Evaluate: Ú _________ 1 + 2 sin x
(
)
cosq + sinq 223. Evaluate: Ú ___________ _________ dq ÷5 + sin 2q
3x + 4 __________ 226. Evaluate: Ú ___________ dx 2 + 2 ÷ x + 5x x+2 __________ 227. Evaluate: Ú ___________ dx 2 + 6 ÷ x + 5x 6x – 5 ___________ 228. Evaluate: Ú ____________ dx 2 – 5x + 1 ÷ 3x _____ a – x 229. Evaluate: Ú _____ dx a+x
÷
dx 245. Evaluate: Ú _________ 3 cos x + 4 dx 246. Evaluate: Ú ______________ 1 + sin x + cos x 1 + sin x 247. Evaluate: Ú _____________ dx sin x (1 + cos x) dx 248. Evaluate: Ú _________ 1 + 2 sin x
a2 – x2 230. Evaluate: Ú x2 _______ 2 dx a + x2
dx 249. Evaluate: Ú ________________ 3 sin x + 4 cos x + 5 dx 250. Evaluate: Ú ____________ cos x + cos a
4 – x3 231. Evaluate: Ú x2 ______ dx 4 + x3
dx 251. Evaluate: Ú _______________ tan x + 4 cot x + 4
Type 15
Type 18
dx 232. Evaluate: Ú _________ 3 + 4 sin2x
dx 252. Evaluate: Ú __________ sin x + cos x
÷ ÷
_______
______
Indefinite Integrals
dx __ 253. Evaluate: Ú _____________ 3 sin x + cos x ÷ dx__ 254. Evaluate: Ú _____________ sin x + ÷3 cos x dx __ 255. Evaluate: Ú _____________ 3 sin x + cos x ÷ dx __ 256. Evaluate: Ú _____________ . 3 sin x – cos x ÷ Type 19 2 sin x + cos x 257. Evaluate: Ú _____________ dx 3 sin x + 2 cos x sin x 258. Evaluate: Ú ___________ dx sin x + cos x 2 sin x + 3 cos x 259. Evaluate: Ú _____________ dx 3 sin x + 4 cos x
(
)
sin x 260. Evaluate: Ú __________ dx sin x + cos x
( ) 2 cos x – sin x 276. Evaluate: Ú ( ____________ dx 9 + 16 sin 2x ) cos x – sin x 275. Evaluate: Ú ___________ dx 5 – 7sin 2x
Type 22 x2 + 1 277. Evaluate: Ú ______ dx x4 + 1 x2 – 1 278. Evaluate: Ú ______ dx x4 + 1 x4 + 1 279. Evaluate: Ú ______ dx x6 + 1 x4 + 3x + 1 280. Evaluate: Ú __________ dx x4 + x2 + 1
( )
dx 1 – x2 ____________ 281. Evaluate: Ú ______ _________ 2 1+x ÷ 1 + x2 + x4
( )
cos x 261. Evaluate: Ú __________ dx sin x + cos x 1 262. Evaluate: Ú _______ dx 1 + tan x
dx x–1 282. Evaluate: Ú _____ × ___________ __________ 3 x+1 + x ÷ x + x2 2 dx x – 1 _______ 283. Evaluate: Ú ______ × _____ x2 + 1 ÷x 4 + 1
1 263. Evaluate: Ú _______ dx 1 – tan x
x2 – 1 ___________ 284. Evaluate: Ú ______________ dx x3 ÷ x 4 – 2x2 + 1
1 264. Evaluate: Ú _______ dx 1 + tan x
x4 – 1 __________ 285. Evaluate: Ú _____________ dx x2 ÷ x 4 + x2 + 1
(
Type 20 3 sin x + 2 cos x + 4 265. Evaluate: Ú _________________ dx 3 cos x + 4 sin x + 5 3 cos x + 2 266. Evaluate: Ú _______________ dx sin x + 2 cos x + 3 2 cos x + 3 sin x 267. Evaluate: Ú ________________ dx 2sin x + 3 cos x + 5 Type 21
(
)
sin x + 2 cos x 268. Evaluate: Ú ____________ dx 9 + 16 sin 2x 269. Evaluate: Ú (cos x – sin x) (2 + 3 sin 2x) dx sin x + cos x 270. Evaluate: Ú ___________ dx 9 + 16 sin 2x dx 271. Evaluate: Ú ____________ cos x + cosec x 272. Evaluate: Ú (sin x + cos x) (2 + 3 sin 2x) dx 273. Evaluate: Ú (sin x – cos x) (3 – 4 sin 2x) dx
(
)
cos x – sin x 274. Evaluate: Ú ___________ dx 3 + 2 sin 2x
)
dx______ 286. Evaluate: Ú _____________ 2 x (x + ÷ 1 + x2 ) x2 – 3x – 1 287. Evaluate: Ú __________ dx x4 + x2 + 1 dx 288. Evaluate: Ú ___________ 4 sin x + cos4x dx 289. Evaluate: Ú _____ x4 + 1 x2dx 290. Evaluate: Ú _____ x4 + 1 2dx 291. Evaluate: Ú _____ . 4 x +1
( )
x4 + 1 292. Evaluate: Ú ______ 6 dx x +1 dx 293. Evaluate: Ú __________ x2 (1 + x4)3/4
(
)
x4 – 1 __________ 294. Evaluate: Ú ______________ dx x2 ÷x 4 + x2 + 1
1.19
1.20 Integral Calculus, 3D Geometry & Vector Booster dx 295. Evaluate: Ú __________ dx x2(1 + x5)4/5
(
315. Evaluate: Ú x2 sin x dx 316. Evaluate: Ú log x dx
)
x2_____ –1 296. Evaluate: Ú ________ dx x ÷1 + x4
317. Evaluate: Ú (log x)2dx
dx ___________ 297. Evaluate: Ú _____________ 4 x ÷x + 3x2 + 1
318. Evaluate: Ú log(x2 + 1) dx
( )
dx x2 + 1 298. Evaluate: Ú ______ × ___________ __________ 4 1 – x2 + 1 ÷ x + x2
(
)
dx x2 + 1 299. Evaluate: Ú ______ x × ____________ ___________ 4 + 1 ÷ x + 3x2
319. Evaluate: Ú tan– 1x dx
( )
1 – x2 320. Evaluate: Ú cos– 1 ______ dx 1 + x2 __
321. Evaluate: Ú e÷ x dx
(
)
xx(x2x + 1) (ln x + 1) 300. Evaluate: Ú _________________ dx (x4x + 1)
x + sin x 322. Evaluate: Ú ________ dx 1 + cos x
x2 – 1 ___________ 301. Evaluate: Ú _____________ dx x3 ÷x 4 – 2x2 + 1
1 – x 323. Evaluate: Ú tan– 1 _____ dx 1+x
(
)
x2009 302. Evaluate: Ú __________ dx (1 + x2)1006
x tan– 1x 326. Evaluate: Ú _________ dx (1 + x2)3/2
Type 23
(
____
)
÷cot x – ÷tan x 304. Evaluate: Ú _____________ dx 1 + 3 sin 2x ____
305. Evaluate: Ú ÷tan x dx ____ ____ (÷ tan x – ÷cot x ) dx
306. Evaluate: Ú
______
307. Evaluate: Ú ÷cot x dx
( ) 1 309. Evaluate: Ú ( cot x + _____ dx cot x ) 1 310. Evaluate: Ú ( tan x + _____ dx tan x ) ____ 1 308. Evaluate: Ú ÷cot x – _____ ____ dx cot x ÷ ____ ÷
____ ÷
____ ÷
____ ÷ 4 ____ 4 ____ 2 311. Evaluate: ( ÷tan x + ÷cot x ) dx
Ú
dx 312. Evaluate: Ú _______________ 4 dx ____ _____ (÷sin x + ÷ cos x )
Integration by Parts ABC of integration by parts 313. Evaluate: Ú x ex dx 314. Evaluate: Ú x sin x dx
__
__
– 1 sin– 1 ÷x – cos ÷x 324. Evaluate: Ú ________________ dx __ – 1 – 1 __ sin ÷x + cos ÷x
x2 325. Evaluate: Ú _____________ dx (x sin x + cos x)2
(2x + 1) dx 303. Evaluate: Ú _____________ 2 (x + 4x + 1)3/2 ____
( ÷ ) _____
_____
( ÷
)
x 327. Evaluate: Ú sin– 1 _____ dx a+x 328. Evaluate: Ú log(1 + x) dx x – sin x 329. Evaluate: Ú ________ dx 1 – cos x ____
330. Evaluate: Ú sin ÷x dx 331. Evaluate: Ú x log (1 + x) dx 332. Evaluate: Ú (sin– 1x)2 dx x – sin x 333. Evaluate: Ú ________ dx 1 – cos x 334. Evaluate: Ú x sin3x dx
(
)
sec 2 x – 1 335. Evaluate: Ú x _________ dx sec 2 x + 1
( ) ( )
2x 336. Evaluate: Ú sin–1 _____ 2 dx 1+x 2x 337. Evaluate: Ú tan–1 _____ 2 dx 1–x x3sin– 1(x2) _____ dx 338. Evaluate: Ú _________ ÷ 1 – x4
Indefinite Integrals
sec x (2 + sec x) 339. Evaluate: Ú ______________ dx (1 + 2sec x)2
359. Evaluate: Ú ex(2sec2x – 1) tan x dx 360. Evaluate: Ú ex (log (sec x + tan x) + sec x) dx
1–x 340. Evaluate: Ú ______ dx. ex + x
(
Type 24
Type 25
341. Evaluate: Ú ex(sin x + cos x) dx
(
{
)
( (
)
)
(
} )
{ } 2 + sin 2x 349. Evaluate: Ú e { _________ dx 1 + cos 2x }
367. Evaluate: Ú e3x(3sin x + cos x) dx 368. Evaluate: Ú e2x(sec2x + 2 tan x) dx
x
(
)
370. Evaluate: Ú e2x (– sin x + 2 cos x) dx
(
log x 351. Evaluate: Ú __________ 2 dx (1 + log x)
)
)
373. Evaluate: Ú e2x(2 × log (sec x + tan x) + sec x) dx Type 27
}
( )
374. Evaluate: Ú ex sin 3x dx 375. Evaluate: Ú e4x cos 3x dx
1 – x 2 355. Evaluate: Ú e _____ dx 1 + x2
(
376. Evaluate: Ú e2x sin 3x dx
)
377. Evaluate: Ú e–x cos x dx
1 – x 2 356. Evaluate: Ú ex _____ dx 1+x
(
)
378. Evaluate: Ú e2xcos (3x + 4) dx
x2 + 1 357. Evaluate: Ú e _______ dx (x + 1)2 x
(
)
1 – sin x 372. Evaluate: Ú e– x/2 ________ dx 1 + cosx
1 1 354. Evaluate: Ú ____ – ______ dx log x (log x)2 x
( ÷
________
x–1 353. Evaluate: Ú ex _______ dx (x + 1)3
{
)
1 + sin 2x 371. Evaluate: Ú e2x _________ dx 1 + cos 2x
1 352. Evaluate: Ú ex log x + __ x dx
(
)
2 sin 4x – 4 369. Evaluate: Ú e2x __________ dx 1 – cos 4x
x2 + 1 350. Evaluate: Ú e _______ dx (x + 1)2
(
)
Type 26
1 – sin x 348. Evaluate: Ú ex ________ dx 1 – cos x
(
)
x3 – x + 2 366. Evaluate: Ú ex _________ 2 dx (x – 1)2
1 1 346. Evaluate: Ú ex __ x – __ 2 dx x x 347. Evaluate: Ú ex _______ dx (x + 1)2
x
(
esin x(x cos3x – sin x) 365. Evaluate: Ú _________________ dx cos2x
x2 + 1 345. Evaluate: Ú e _______ dx (x + 1)2
{ (
}
x4 + 2 364. Evaluate: Ú ex _________ dx (1 + x2)5/2
1 + x + x dx 344. Evaluate: Ú e _________ (1 + x2)3/2 x
)
1 363. Evaluate: Ú log (log x) + ______ 2 dx (log x)
2 + sin 2x 343. Evaluate: Ú ex _________ dx 1 + cos 2x 3
(
1 362. Evaluate: Ú ex log x + __2 dx x
x ex 342. Evaluate: Ú _______ dx (x + 1)2
x
)
x cos3x – sin x 361. Evaluate: Ú esin x ____________ dx cos2x
_____
)
(x + 1) + ÷1 – x2 _____ 358. Evaluate: Ú ex _______________ dx 2 2 (x + 1) ÷1 – x
379. Evaluate: Ú excos2x dx 1 380. Evaluate: Ú __3 sin(log x) dx. x
1.21
1.22 Integral Calculus, 3D Geometry & Vector Booster ______
2x 404. Evaluate: Ú _____________ dx (x2 + 1)(x2 + 2)
_______
cosq 405. Evaluate: Ú __________________ dq (2 + cosq )(3 + cosq )
Type 28 381. Evaluate: Ú ÷4 – x2 dx 382. Evaluate: Ú ÷1 – 9x2 dx _____
383. Evaluate: Ú ÷x 2 + 1 dx ______
384. Evaluate: Ú ÷3x 2 + 1 dx ______
dx 407. Evaluate: Ú ___________ sin x – sin 2 x
_______
Type 2
385. Evaluate: Ú ÷x 2 – 9 dx 386. Evaluate: Ú ÷4x 2 – 1 dx
__________ ÷x 2 + 2x + 3 dx __________ ÷3 – 4x – x2 dx ________ ÷2ax – x2 dx ________ ÷2ax + x2 dx
387. Evaluate: Ú 388. Evaluate: Ú 389. Evaluate: Ú 390. Evaluate: Ú
(1 – cos x) 406. Evaluate: Ú _____________ dx cos x (1 + cos x)
_______
391. Evaluate: Ú ÷x – 4x2 dx Type 29 392. Evaluate: Ú (2x +
__________ 1) ÷x 2 + 3x + 4 dx ______
393. Evaluate: Ú (x – 5) ÷x 2 + x dx
_________
394. Evaluate: Ú (3x – 2) ÷x 2 + x + 1 dx _________ (4x + 1) ÷x 2 – x – 2 dx _________ x ÷1 + x – x2 dx.
395. Evaluate: Ú 396. Evaluate: Ú
Partial Fractions Type 1 (2x + 1) dx 397. Evaluate: Ú ____________ (x + 2)(x + 3) dx 398. Evaluate: Ú ____________ (x – 1)(x – 2) dx 399. Evaluate: Ú __________________ (x + 1)(x + 2)(x + 3) dx 400. Evaluate: Ú ________________________ (x + 1) (x + 2) (x + 3) (x + 4) x–1 401. Evaluate: Ú ____________ dx (x + 1)(x – 2)
dx 408. Evaluate: Ú _____________ (x + 1)(x + 1)2 2x + 1 409. Evaluate: Ú ____________ dx (x + 2)(x – 3)2 3x + 1 410. Evaluate: Ú ____________ dx (x – 2)2(x + 2) x2 + 1 411. Evaluate: Ú _____________ dx (x – 1)2 (x + 3) x2 412. Evaluate: Ú ____________ dx (x – 1)3 (x + 1) (x – 1) 413. Evaluate: Ú ________ dx x2(x + 4) (2x – 1) 414. Evaluate: Ú _______ dx x3(x – 2) Type 3 (2x + 3) dx 415. Evaluate: Ú _____________ (x + 1)(x2 + 4) (3x – 2) dx 416. Evaluate: Ú ____________ (x – 1)(x2 + 9) 2x – 1 417. Evaluate: Ú _____________ dx (x + 1)(x2 + 2) x 418. Evaluate: Ú _____________ dx (x + 1) (x2 + 4) 8 419. Evaluate: Ú _____________ dx (x + 2) (x2 + 9) x 420. Evaluate: Ú ____________ dx (x + 1)(x2 + 1) Type 4
2x – 1 402. Evaluate: Ú ___________________ dx (x + 1) (x + 2) (x + 3)
x2 421. Evaluate: Ú _____________ dx (x2 – 1) (x2 + 1)
x3 403. Evaluate: Ú ____________ dx (x – 1)(x – 2)
x2 422. Evaluate: Ú _____________ dx (x2 – 3) (x2 + 4)
Indefinite Integrals
(x2 + 4) 423. Evaluate: Ú _____________ dx (x2 + 5)(x2 + 7) x2 424. Evaluate: Ú _____________ dx (x2 – 1)(x2 – 2) x2 425. Evaluate: Ú ___________________ dx (x2 – 1)(x2 – 2)(x2 – 3) Type 5
(
)
(x2 + 3)(x2 + 1) 426. Evaluate: Ú _____________ dx (x2 – 1)(x2 + 2) 2
2
(x + 1)(x + 2) 427. Evaluate: Ú _____________ dx (x2 + 3)(x2 + 4) (x2 – 1)(x2 + 3) 428. Evaluate: Ú _____________ dx. (x2 + 2)(x2 + 1)
Type 3 dx 17. Evaluate: Ú ________ 7 x(x + 1) dx 18. Evaluate: Ú __________ 2 4 x (x + 1)3/4 dx 19. Evaluate: Ú ________ 5 x (x + 1) dx 20. Evaluate: Ú ________ 4 x (x + 1) dx 21. Evaluate: Ú __________ 2 7 x (x + 1)6/7 dx 22. Evaluate: Ú __________ 3 x (1 + x3)1/3 Type 4 x2 23. Evaluate: Ú _______ dx (x + 3)2 x3 dx 24. Evaluate: Ú ________ (2x + 3)2
Type 1
x2 25. Evaluate: Ú _______ dx (x + 2)3
1. Evaluate: Ú tan x ◊ sec x dx
2. Evaluate: Ú sec3x dx
3. Evaluate: Ú sec5x dx
4. Evaluate: Ú tan2x ◊ sec4x dx
dx 27. Evaluate: Ú _________ x (1 + x3)2
5. Evaluate: Ú tan3x ◊ sec6x dx
6. Evaluate: Ú tan3x ◊ sec5x dx
x4 28. Evaluate: Ú ________ dx (3x – 2)3
7. Evaluate: Ú sec7x dx
4
2
8. Evaluate: Ú sec9 x dx
Type 2
9. Evaluate: Ú cosec2 x ◊ cot2 x dx
10. Evaluate: Ú cot3 x ◊ cosec3 x dx 11. Evaluate: Ú cosec3 x dx 12. Evaluate: Ú cot2 x ◊ cosec4 x dx 13. Evaluate: Ú tan–5x ◊ sec6 x dx 14. Evaluate: Ú cot3 x ◊ cosec–8 x dx
x2 26. Evaluate: Ú ________ dx (ax + b)2
Type 5 dx 29. Evaluate: Ú _________ 2 x (3x + 2)3 dx 30. Evaluate: Ú _________ 3 x (b + ax)2 dx 31. Evaluate: Ú ________ 2 3 x (x + 2) dx 32. Evaluate: Ú _________ 3 x (a + bx)2 dx 33. Evaluate: Ú _________ 2 x (1 + x2)3
15. Evaluate: Ú cosec x dx
dx 34. Evaluate: Ú _________ 2 x (a – bx)2
16. Evaluate: Ú cosec7 x dx
dx 35. Evaluate: Ú _________ 4 x (2x + 1)3
5
1.23
1.24 Integral Calculus, 3D Geometry & Vector Booster Type 6 dx 36. Evaluate: Ú _____________ (x – 1)3 (x – 2)4 dx 37. Evaluate: Ú _______________ 4 _____________ – 1)3 (x + 2)5 ÷ (x dx 38. Evaluate: Ú _____________ (x – 1)3 (x – 2)2 dx 39. Evaluate: Ú ______________ (x – 3)4 (x – 2)5 dx 40. Evaluate: Ú ________________ 3/2 (x – 3) (x – 2)7/2 dx ______________ 41. Evaluate: Ú _______________ 5 + 1)4 (x + 3)6 ÷ (x dx ______________ 42. Evaluate: Ú _______________ 5 ÷(x + 1)4 (x + 3)7 Type 7 dx 43. Evaluate: Ú _________ x (2 + 3x3) dx 44. Evaluate: Ú _________ x (3 + 5x5) dx 45. Evaluate: Ú _________ x (2 + 3x2) dx 46. Evaluate: Ú _________ x (3 + 4x3) dx 47. Evaluate: Ú _________ x (2 – 5x3) dx 48. Evaluate: Ú _________ x (1 – 4x4) dx 49. Evaluate: Ú _________ x (3x4 + 1) Type 8 x5 50. Evaluate: Ú _______ 2 4 dx (x – 1) x9 51. Evaluate: Ú _________ 5 dx 2 (2x + 3) x3 52. Evaluate: Ú _______ 2 4 dx (x + 1) 5
8x9 55. Evaluate: Ú ________ 2 5 dx (3x – 2) 10x11 56. Evaluate: Ú ________ 2 dx (3x + 5)4 Type 9 2 sin x + 3 57. Evaluate: Ú __________ dx (3sin x + 2)2 (2 sin x + 5) 58. Evaluate: Ú ___________ dx (2 + 5 sin x)2 (3 sin x – 2) 59. Evaluate: Ú ___________ dx (2 – 3 sin x)2 Type 10
(
)
4 cos x + 3 60. Evaluate: Ú ___________ dx (3 cos x + 4)2 (cos x + 2) 61. Evaluate: Ú ___________ dx (1 + 2 cos x)2 3 cos x + 4 62. Evaluate: Ú ___________ dx (3 + 4 cos x)2 Type 11 dx 63. Evaluate: Ú ___________ (3 + 4 sin x)2 dx 64. Evaluate: Ú ___________ (5 + 4 sin x)2 dx 65. Evaluate: Ú __________ (1 – 2 sin x)2 Type 12 dx 66. Evaluate: Ú ___________ (2 + 3 cos x)2 dx 67. Evaluate: Ú _____________ (12 + 13 cos x)2 dx 68. Evaluate: Ú ___________ (3 – 4 cos x)2 Type 13
(
)
4ex + 6e–x 69. Evaluate: Ú _________ x dx. 9e – 4e–x
x 53. Evaluate: Ú _______ 2 4 dx (x – 3)
3ex – 2e–x 70. Evaluate: Ú _________ dx 2ex + 5e–x
x7 54. Evaluate: Ú ________ 2 4 dx (3x – 2)
4ex + 3e–x 71. Evaluate: Ú _________ dx 3ex + 7e–x
Indefinite Integrals
Integration of Irrational Functions Type 1 dx ______________ 72. Evaluate: Ú _______________ + 2)5 (x + 1)3 ÷ (x
(
)
dx x + 2 1/2 ___ ◊ x 73. Evaluate: Ú ______ 2x + 3 (x + 1) 74. Evaluate: Ú ______________ dx (x + 2) (x + 3)3/2
÷
_____
32 – x 2 75. Evaluate: Ú _______ 2 _____ dx (2 – x) 2 + x
Type 2 dx 76. Evaluate: Ú ______________ _____ _____ 3 x + 1 x + 1 + ÷ ÷ dx 77. Evaluate: Ú _________________ _____ 1/4 x + 1 – (x + 1) ÷ 1 + x1/2 – x1/3 78. Evaluate: Ú ____________ dx 1 + x1/3
dx 90. Evaluate: Ú _________ 2 __ (x + 1) ÷x x dx 91. Evaluate: Ú _________________ _____ 2 (x + 2x + 2) ÷x + 1 dx 92. Evaluate: Ú _________ 2 __ (x – 1) ÷x x dx 93. Evaluate: Ú _________________ _____ (x2 – 2x + 2) ÷x – 1 Type 5 dx _____ 94. Evaluate: Ú ____________ (x + 1) ÷x 2 + 1 dx __________ 95. Evaluate: Ú _________________ (x + 1) ÷x 2 + 2x + 2 dx 96. Evaluate: Ú ________ _____ x ÷x 2 + 4 dx _____ 97. Evaluate: Ú ____________ (x – 1) ÷x 2 + 4
dx 79. Evaluate: Ú ________ 1/2 1/3 x +x
dx _____ 98. Evaluate: Ú _____________ (2x – 1)÷x 2 + 1
x ÷ 80. Evaluate: Ú 4_______ _____ dx 3 ÷x + 1
dx _____ 99. Evaluate: Ú _____________ (3x + 2) ÷x 2 – 4
x ÷ 81. Evaluate: Ú _______ __ dx __ x + ÷ 3 x ÷
Type 6
__
__
Type 3
dx 100. Evaluate: Ú ________ _____ 2 2 x ÷ x – 1
dx _____ 82. Evaluate: Ú ___________ (x – 1) ÷x + 3
dx _____ 101. Evaluate: Ú ______________ 2 (x + 1) ÷x 2 + 2
dx _____ 83. Evaluate: Ú ___________ (x + 3) ÷x + 2
dx _____ 102. Evaluate: Ú _____________ 2 (x – 1) ÷x 2 + 2
__
x ÷ 84. Evaluate: Ú _____ dx x+1 dx 85 .Evaluate: Ú _______ _____ x÷x – 2 dx 86. Evaluate: Ú ________ __ (x + 3) ÷x
dx ______ 87. Evaluate: Ú ____________ (x + 3) ÷2x + 1 Type 4 dx 88. Evaluate: Ú ________ 2 _____ x ÷ x – 1 dx _____ 89. Evaluate: Ú ____________ 2 (x – 4) ÷x + 1
dx _____ 103. Evaluate: Ú _____________ 2 (1 + x ) ÷1 – x2 x dx_____ 104. Evaluate: Ú _____________ 4 (x – 1) ÷x 4 + 3 dx __________ 105. Evaluate: Ú __________________ 2 (x – 1) ÷x 2 + 4x + 5 Type 7 _____
106. Evaluate: Ú (x + ÷ x 2 + 1 )10 dx _____
107. Evaluate: Ú (x – ÷x 2 + 4 )5 dx _____
108. Evaluate: Ú (x + ÷ 1 + x2 )n dx
1.25
1.26 Integral Calculus, 3D Geometry & Vector Booster dx _____ 109. Evaluate: Ú _____________ (x + ÷x 2 – 4 )5/3
dx 127. Evaluate: Ú ________ ______ x ÷3x – 2
dx _____ 110. Evaluate: Ú _____________ (x – ÷x 2 + 9 )10
dx 128. Evaluate: Ú _________ ______ x ÷3x 9 – 2
dx_____ 111. Evaluate: Ú ____________ 2 x (x – ÷ x 2 + 9 )
dx 129. Evaluate: Ú __________ _______ x ÷2 x 10 – 3
Type 8
Type 11
dx 112. Evaluate: Ú ____________ x1/2 (2 + 3x)3/2
dx 130. Evaluate: Ú __________ (2 + 3x2)3/2
dx 113. Evaluate: Ú ____________ x2/3 (2 + 3x)4/3
dx 131. Evaluate: Ú _________ (c + dx2)3/2
dx 114. Evaluate: Ú ____________ 3/4 x (3x – 1)5/4
dx 132. Evaluate: Ú __________ (3 + 5x2)3/2
dx 115. Evaluate: Ú ____________ 1/3 x (2x + 1)5/3
dx 133. Evaluate: Ú _________ (3 – 4x2)3/2
dx 116. Evaluate: Ú ____________ 1/2 x (2 + 3x)5/2
x dx 134. Evaluate: Ú _________ (2 – 5x4)3/2
dx 117. Evaluate: Ú ___________ x2(2 + 3x2)5/2
x2 dx 135. Evaluate: Ú _________ (1 – 4x6)3/2
Type 9
Type 12
dx 118. Evaluate: Ú _____________ (x – 1)3 (x + 2)4
dx _________ 136. Evaluate: Ú _________________ 2 2 (x – 2) ÷ x – 4x + 7
dx 119. Evaluate: Ú _____________ (x – 1)3 (x – 2)2
dx _________ 137. Evaluate: Ú _________________ 3 2 (x + 1) ÷ x + 2x + 4
dx 120. Evaluate: Ú _____________ (x – 1)2 (x – 2)3
dx ____________ 138. Evaluate: Ú ____________________ 3 (x – 2) ÷4x 2 – 16x + 20
dx ______________ 121. Evaluate: Ú 4_______________ ÷ (x – 1)3 (x + 2)5
dx _____________ 139. Evaluate: Ú _________________________ 2 (x – 6x + 9) ÷4x 2 – 24 x + 20
dx 122. Evaluate: Ú ________ 2 x (x + 5)4
dx ___________ 140. Evaluate: Ú ________________________ 2 (4x + 4x + 1) ÷4x 2 + 4x + 7
dx 123. Evaluate: Ú ________________ 3/2 (x – 1) (x + 1)5/2
dx __________ 141. Evaluate: Ú __________________ 3 (x + 1) ÷x 2 + 2x – 4
Type 10
Type 13
dx 124. Evaluate: Ú _________ ______ x ÷3x 3 + 4
(2x + 3) __________ 142. Evaluate: Ú __________________ dx (3x + 4)÷x 2 + 2x + 4
dx 125. Evaluate: Ú _________ ______ x ÷5x 4 + 3
(2x + 3) __________ 143. Evaluate: Ú _________________ dx (x + 1) ÷x 2 + 2x + 9
dx 126. Evaluate: Ú _________ ______ x ÷2 – 5x6
(4x + 7) __________ 144. Evaluate: Ú _________________ dx (x + 2) ÷x 2 + 4x + 7
Indefinite Integrals
Type 14 2
x + __________ 4x + 2 145. Evaluate: Ú _________________ dx 2 (x + 1) ÷x + 2x + 3 x2 + __________ 5x + 6 146. Evaluate: Ú _________________ dx 2 (x + 2) ÷x + 5x + 4 x2 + 10x +6 __________ 147. Evaluate: Ú _________________ dx 2 (x + 2) ÷x + 4x + 9
dx __________ 164. Evaluate: Ú ____________ 2 x ÷x – 3x + 2 dx ______ 165. Evaluate: Ú __________ x+÷ x 2 – 1 x dx _____ 166. Evaluate: Ú _________ x+÷ x 2 – 1
Reduction Formulae Type 1
Type 15 __
167. Evaluate: Ú sin5 x dx
__
3 3 148. Evaluate: Ú ÷ x (1 + ÷ x ) dx
__
149. Evaluate: Ú ÷x 2 (3 + x–2/3)–2 dx 3
(
______ __
)
x ÷ 1 +__÷4 150. Evaluate: Ú _______ dx 3 4 ÷ x dx 151. Evaluate: Ú ________ 2 _____ 7 x ÷ 1 + x4
170. Evaluate: Ú cos8 x dx Type 3 171. Evaluate: Ú tan5 x dx
153. Evaluate: Ú x–1/2 (2 + 3x1/3)–2 dx 154. Evaluate: Ú
________ __ 3 1 + ÷x 4
÷( 7
)
155. Evaluate: Ú x–6 (1 + 2x3)2/3 dx dx 156. Evaluate: Ú ________ 3 _____ x ÷ 1 + x5 dx 157. Evaluate: Ú _________ 2 _____ 11 x ÷ 1 + x4
(
)
________ __ _________ (1 + 4÷x ) 4 __ dx ÷x 3
158. Evaluate: Ú
÷
3
Type 2 169. Evaluate: Ú cos7 x dx
dx 152. Evaluate: Ú ___________ __ __ 4 10 x ( ÷ x + 1 ) ÷
__ 3÷x ×
168. Evaluate: Ú sin6 x dx
Type 16 dx _________ 159. Evaluate: Ú _______________ 2 (1 + ÷ x + x + 1 ) dx _________ 160. Evaluate: Ú _____________ x+÷ x 2 – x + 1 x dx ___________ 161. Evaluate: Ú ____________ – 10 – x2 ÷ 7x dx _________ 162. Evaluate: Ú _____________ 2 x – ÷x – x + 2 dx __________ 163. Evaluate: Ú ______________ 2 x–÷ x – 2x + 4
172. Evaluate: Ú tan6 x dx Type 4 173. Evaluate: Ú cot7 x dx 174. Evaluate: Ú cot6 x dx Type 5 175. Evaluate: Ú sec3 x dx 176. Evaluate: Ú sec4 x dx 177. Evaluate: Ú sec5 x dx 178. Evaluate: Ú sec7 x dx Type 6 sin 3 x 179. Evaluate: Ú _____ dx sin x sin 5 x 180. Evaluate: Ú _____ dx sin x sin 6 x 181. Evaluate: Ú _____ dx sin x sin 8 x 182. Evaluate: Ú _____ dx sin x Type 7 dx 183. Evaluate: Ú _______ 2 2 (x + 2)
1.27
1.28 Integral Calculus, 3D Geometry & Vector Booster dx 184. Evaluate: Ú _______ (x2 + 3)3
x + 1 185. Evaluate: Ú ____________ dx 2 2 (x + 3x + 2)
Type 8 186. Evaluate: Ú x2 log x dx 187. Evaluate: Ú x2 (log x)2 dx
188. Evaluate: Ú x3 (log x)2 dx 189. Evaluate: Ú x2 (1– x)3 dx
Mixed Problems
190. Evaluate: Ú sin8 x dx
sin x 3. Ú _____ dx = sin 3 x
| |
|
__
3 + tan x ÷ 1__ __ (a) ___ log _________ + c 3 ÷3 – tan x ÷ __
|
3 + tan x ÷ 1__ __ (b) ____ log _________ + c 2÷3 ÷3 – tan x
|
__
|
3 + tan x ÷ __ (c) log _________ + c 3 – tan x ÷
|
__
|
3 + tan x ÷ 1__ __ (d) ___ log _________ + c 3 ÷3 – tan x ÷ dx 4. Ú __________________ = sin (x – a) cos (x – b)
191. Evaluate: Ú cos10 x dx
1 (a) _________ (log |sin (x – a)| + log |sec (x – b)|) + c cos (a – b)
192. Evaluate: Ú tan8 x dx
(b) ( log | sin (x – a) | + log | sec (x – b) | ) + c
(c) (log |sin (x – a)| + c)
(d) (log |sec (x – b )| + c)
193. Evaluate: Ú cot8 x dx 194. Evaluate: Ú cot x dx
195. Evaluate: Ú sec9 x dx
196. Evaluate: Ú xn ex dx
9
dx 197. Evaluate: Ú _______ (x2 + 2)2
1.
Ú x2 cos x dx =
5. Ú tan3 2 x ◊ sec 2x dx = 1 1 (a) __ sec3 2 x – __ sec 2 x + c 3 2 1 1 3 (b) – __ sec 2 x – __ sec 2 x + c 2 6 1 1 (c) __ sec3 2 x – __ sec 2 x + c 2 6 1 1 3 (d) __ sec 2 x + __ sec 2 x + c 3 2 1 6. If Ú f (x) sin x ◊ cos x dx = _______ 2 2 log ( f (x)) + k, then (b – a ) f (x) is
1 (a) ________________ + c 2 2 a sin x + b2 cos2 x
1 (b) _______________ + c a2 sin2 x – b2 cos2 x
Ú ex cos2 x dx =
1 (a) ex + ___ ex (cos 2 x + 2 sin 2 x) + c 10
1 (c) _______________ + c a2 cos2 x + b2 sin2 x
1 (b) __ ex + ex (cos 2 x + 2 sin 2 x) + c 2
1 (d) _______________ + c a2 cos2 x – b2 sin2 x
(a) x2 sin x (b) x2 sin x (c) x2 sin x (d) x2 sin x
2.
+ + + +
2 x cos x – 2 sin x + c 2 x cos x + 2 sin x + c 2 x cos x + sin x + c 2 cos x + sin x + c.
1 (c) __ ex + (cos 2 x + 2 sin 2 x) + c 2
1 1 (d) __ ex + ___ ex (cos 2 x + 2 sin 2 x) + c 2 10
7. If
cos 4 x + 1 __________ dx = k cos 4 x + c, then Ú ( cot x – tan x )
1 (a) k = – __ 2 1 (c) k = – __ 4
1 (b) k = – __ 8 1 (d) k = __ 6
dx 8. Ú ______ = 6 x + x4
1 1 (a) – ___ 2 + __ x + cosec–1x + c 3x 1 1 (b) – ____ 2 + __ x + cot–1 x + c 3 x 1 1 (c) – ___ 3 + __ + tan–1 x + c x 3x 1 1 (d) ____ 2 – __ x + sin–1 x + c 3 x
|
|
sin x – (x + 1) cos x – 1 1 (a) __ log _____________________ + c 2 sin x – (x + 1)cos x + 1 1 (b) __ tan–1 {sin x – (x + 1) cos x} + c 2 1 (c) __ sin–1 {sin x – (x + 1) cos x} + c 2 1 (d) __ sin–1 (sin x + cos x) + c 2
{
cos x (a) _________ + c 2 + 5 sin x
– cos x (b) _________ + c 2 + 5 sin x
1 (c) _________ + c 2 + 5 sin x
sin x (d) _________ + c 2 + 5 sin x
}
(log x – 1) 10. Ú __________ dx = 1 + (log x)2
x ex (a) _____ + c 1 + x2
x ex (b) __________ + c 1 + (log x)2
log x (c) __________ + c 1 + (log x)2
x (d) _____ + c 1 + x2
}
( ( ( (
) ) )
x 1 (a) ex _______ _____ + _________ _______ + c 2 ÷(1 + x2)3 ÷ 1 + x
x 1 (b) ex _______ _____ – _________ _______ + c 2 ÷(1 + x2)3 ÷ 1 + x
x 1 (c) ex _______ _____ + _________ _______ + c 2 ÷ ÷ 1 + x (1 + x2)5
x 1 (d) ex ______ _____ – ________ _______ + c 2 ÷(1 + x2) ÷ 1 – x
)
x cos x + 1 ___________ 15. Ú ____________ dx = 3 e sin x + x2 ÷ 2x
( ( ( (
_________
) ) ) )
2x e sin x + 1 – 1 ÷ __________ (a) log ______________ + c sin x + 1 + 1 ÷ 2 xe
2
cosec x – 2005 11. Ú _____________ dx = cos2005 x
_________
cot x (a) _______ + c cos2005 x
tan x (b) _______ + c cos2005 x
sin x – 1 + 1 ÷ 2x e (b) log ______________ __________ + c sin x + 1 + 1 ÷ 2 xe
– tan x (c) _______ + c cos2005 x
– cot x (d) _______ + c cos2005 x
2x e sin x + 1 +1 ÷ _________ (c) log ______________ + c sin x – 1 + 1 ÷ 2 xe
sin x _________ + 1 + 1 ÷ 2x e (d) log ______________ + c sin x – 1 – 1 ÷ 2 xe
sin 2 x + 2 tan x 12. Ú __________________ dx = ( cos6x + 6 cos2 x + 4 )
÷
1.29
2x2 1 14. Ú ex _______ _____ + 1 – _________ ________ dx = ÷ 1 + x2 ÷ (1 + x2)5
2 sin x + 5 9. Ú ___________ dx = (2 + 5 sin x)2
{
Indefinite Integrals
________
_________
_________
1 + cos2 x (a) 2 ________ +c cos7x
16. Let f (x) be a function such that f (0) = f ¢(0) = 0, f ¢¢(x)
1 + cos2 x 1 (b) tan ___ __ × ________ +c 2 ÷ cos7x
1 + cos2 x 1 (c) ___ log ________ + c 12 cos7x
6 1 4 (d) ___ log 1 + _____ 4 + _____ 6 + c 12 cos x cos x
–1
( ) ÷
= sec4 x + 4, the function is
________
(
(
)
)
(1 + x) sin x 13. Ú __________________________ dx = 2 (x + 2x) cos2 x – (1 + x) sin 2 x
1 (a) log (sin x) + __ tan3 x + cx 3 2 1 (b) __ log (sec) + __ tan2 x + 2x2 3 6 x2 1 (c) log (cos x) + __ cos2 x + __ 6 5 (d) none.
17. The value of the integral I = where p x Π0, __ 2
( )
____
____
+ ÷cot x ) dx, Ú (÷tan x
1.30 Integral Calculus, 3D Geometry & Vector Booster
__
(a) ÷2 sin–1 (cos x – sin x) + c
(b) ÷ 2 sin (sin x – cos x) + c
ln |x| _______ 23. Ú ____________ dx = x × ÷1 + ln |x|
(c) ÷2 sin–1 (sin x + cos x) + c
(d) – ÷2 sin–1 (sin x + cos x) + c
__
–1
__
__
____
____
18. The value of the integral I = Ú ( ÷tan x + ÷cot x ) dx,
3p where x Πp, ___ 2
( )
(a) ÷2 sin–1 (cos x – sin x) + c
(b) ÷2 sin–1 (sin x – cos x) + c
__ __ __
(c) ÷2 sin (sin x + cos x) + c (d) – ÷2 sin–1 (sin x + cos x) + c ____
____
)
p 3p where x Œ 0, __ » p, ___ 2 2
____ ____ __ tan x – ÷cot x –1 ÷ ____________ __ (a) ÷2 tan + c ÷2 ____ ____ __ ÷tan x + ÷cot x –1 _____________ __ (b) ÷2 tan + c ÷2 ____ ____ __ ÷tan x – ÷cot x –1 _____________ __ (c) – ÷2 tan + c ÷2 ____ ____ __ ÷tan x + ÷cot x __ (d) – ÷2 tan–1 _____________ + c ÷2
) )
( (
+ 2) + c
(a) – 1/9 (b) – 1/3 (c) 1/3 (d) Non existent. 25. The number of values of x satisfying the equation 3 __ x x + 1 28 2 2 ___ _______________ ______ Ú 8t + t + 4 dt = , 3 log(x + 1) ( ÷(x + 1) ) –1
(
)
is (a) 0 (c) 2
(b) 1 (d) 3.
________________________
26. Ú ÷(1 + 2 cot x (cot x + cosec x)) dx =
) )
(
– 2) + c
0
__
( (
+ 2) + c
24. If Ú t2 dt = x cos (p x), the value of f ¢(9) is
19. The value of the integral I = Ú (÷tan x + ÷ cot x )dx,
– 2) + c
f (x)
–1
( ) (
2 _______ (a) __ ÷1 + ln |x| × (ln |x| 3 2 _______ (b) __ ÷ 1 + ln |x| × (ln |x| 3 1 _______ (c) __ ÷ 1 + ln |x| × (ln |x| 3 1 _______ (d) __ ÷ 1 + ln |x| × (ln |x| 3
)
x4 cos3 x – x sin x + cos x 20. Ú e(x sin x + cos x) ____________________ dx x2 cos2 x
( ) 1 ( ______ x cos x ) + c 1 ( ______ x cos x – x ) + c 1 ( ______ x cos x + x ) + c
( ( (
( ) ) ( ) ) ( ) )
x (a) 2 ln cos __ + c 2 x (b) 2 ln sin __ + c 2 x 1 (c) __ ln sin __ + c 2 2 (d) None.
1 (a) e(x sin x + cos x) x – ______ x cos x + c
27. A differentiable function satisfies 3f 2(x) f ¢(x) = 2x. Given f (2) = 1, the value of f (3) is
(b) e(x sin x + cos x)
(a) ÷24
(b) ÷6
(c) e(x sin x + cos x)
(c) 6
(d) 2.
(d) e(x sin x + cos x)
1 x 28. Ú __ x log __ x dx = e
x
dt _____ 21. Let f (x) = Ú _______ and g be the inverse of f. Then 2 ÷ 1 + t4
the value of g¢(0) (a) 1 ___ (c) ÷17
(b) 17 (d) None.
3 ___
3
( ( ) )
1 (a) __ ex – ln x + c 2 1 (c) __ ln 2 x – ex + c 2
29.
Ú x × 2 ln(x + 1) dx =
t dt 22. If f (x) = eg(x) and g (x) = Ú _____ 4 , then f ¢(2) is 21 + t
2 ln (x + !) (a) ________ + c 2 2(x + 1)
(x2 + 1) 2 ln (x + !) (b) _____________ + c 2 (x2 + 1)
x
(a) 2/17 (c) 1
(b) 0 (d) cannot be determined.
2
2
2
__
1 (b) __ ln x – ex + c 2 (d) None.
Indefinite Integrals
(
)
(x2 + 1) 2 ln2 + 1 (c) ____________ + c 2 (ln 2 + 1)
(d) None.
(
)
(
)
x (a) __________ _________ + c 2 ÷ x + rx + 1 x (b) __________ _________ + c ÷ x 2 + rx + 1
(
)
(
)
1 (d) cos–1 __ (sin x + cos x) + c. 3
}
(b) B = cos a
(c) A = cos a
(d) B = sin a
36.
Ú x log (x2 + 1) dx = f (x) log (x2 + 1) + g (x) + c, then
(a) ex tan x + c
(b) ex cot x + c
(c) ex cosec2 x + c
(d) ex sec2 x + c.
{
(
_____
)}
2 etan x ( –1 1 – x2 32. Ú _______2 sec ÷1 + x2 ) + cos–1 _____2 dx (1 + x ) 1+x
(a) etan–1 x ◊ tan–1 x + c
etan–1 x ◊ (tan–1 x)2 (b) ______________ +c 2
( ( ) x ◊ ( cosec ( ÷1 + x ) ) + c 2
(c) etan–1 x ◊ sec–1 ÷1 + x2 ) + c
(d) etan–1
_____
2
( ( (
__
)
(1 + ÷x ) (b) ________ + c (1 – x)2
__
__
) )
(1 – ÷x ) (c) ________2 + c (1 – x)
2 (1 – ÷ x ) _____ (d) ________ + c – ÷1 – x
(
(
__
)
(a) A = 1/2 (c) f (x) = tan x
(b) A = 1/3 (d) f (x) = tan 2 x.
dx 38. If Ú __________________ cos (x – a) cos (x – b)
2
(a) A = sin (a – b) (b) f (x) = cos (x – a) (c) g (x) = cos (x – b) (d) A = sin (b – a).
Ú tan4 x dx = k tan3 x + L tan x + f (x), then
(a) k = 1/3
(b) L = –1
(c) f (x) = x + c
(d) k = 2/3.
(More than one options are correct)
(
)
x + sin x 40. If Ú ________ dx = f (x) tan (g (x)) + c, then 1 + cos x
)
(a) f (x) = x2
(b) f (x) = x
x2 (c) g (x) = __ 2
x (d) g (x) = __ 2
(
)
x–1 ex 41. If Ú ex _______ dx = ______ + c, then 3 (g (x))m (x + 1)
)
cos x – sin x ________ 34. Ú __________ dx = 8 – sin 2 x ÷
(a) sin–1 (sin x + cos x) + c
1 (b) sin–1 __ (sin x + cos x) + c 3
(
(
39. If
dx _____ 33. Ú _____________ = __ (1 + ÷ x )÷x – x2 2 (1 + ÷x ) (a) _________ + c 2 (1 – x)
1+ x2 (a) f (x) = _____ 2 1 + x2 (b) g (x) = _____ 2 1 + x2 (c) g (x) = – _____ 2 x2 – 1 (d) f (x) = _____ . 2 1 + sin 2 x 37. If Ú e2x _________ dx = Ae2x ◊ f (x) + c, then 1 – sin 2 x
1 = __ (log |f (x)| + log |g (x)| + c, then A
_____
–1
)
(a) A = sin a
cos x (2 sin x + 1) 2 sin2 x – 1 ______________ 31. Ú ex _________ + dx = cos x 1 + sin x
–1
(
)
{
x2 (c) __________ _________ + c ÷ x 2 + rx + 1 1 _________ (d) __________ + c 2 ÷ x + rx + 1
(
(c) cos–1 ((sin x + cos x)) + c
sin x 35. If Ú _________ dx = Ax + B log |sin (x – a)| + c, then sin (x – a)
2x + 1 30. Ú _____________ dx = (x2 + 4x + 1)3/2 3
1.31
)
(a) g (x) = x (c) m = 1
(b) g (x) = x + 1 (d) m = 2.
1.32 Integral Calculus, 3D Geometry & Vector Booster
(
)
dx 1 B __ 42. If Ú __________ = A 1 + + c, then x2 (x4 + 1)3/4 x4
cos4 x 50. If Ú ___________________ 3 dx sin x {sin5 x + cos5 x}3/5 m __ 1 = – __ (B + cot5 x) n + c, then A (a) A + B = 3 (b) m + n = 7
(a) A = –1 (c) A = 1/2
(b) B = 1/4 (d) B = 1/2.
(
)
logx e + logex e + log 2 e ex 43. If Ú _____________________ dx x = A (log (loge x)) + B loge (1 + loge x) + C log (2 + loge x) + K, then
(a) A + B = 2 (c) A – B = 0
(
____
(
(b) L – M = 6 (d) L + M – N = 14.
)
(c) m + n = 8
(d) A + B = 4.
(b) A – C = 0 (d) A + B + C = 3
)
tan x ÷ 44. If Ú ________ dx = A (f (x))1/m + c, then sin x cos x
(a) A = 2
(b) m = 2
(c) f (x) = tan x
(d) A + m = 5
(
(a) L + M = 18 (c) L + M + N = 22
cos 2 x – cos x 1. Evaluate: Ú ____________ dx 1 – cos x cos 5 x + cos 4 x 2. Evaluate: Ú _____________ dx 1 – 2 cos 3 x
____ cot x ÷ 45. If Ú ________ dx = A ( – ÷f (x) ) + c, then sin x cos x
sin6 x + cos6 x 3. Evaluate: Ú ___________ dx sin2 x ◊ cos2 x
(a) A = 2
(b) f (x) = cot x
(c) f (x) = tan x
(d) A = 4.
cos 2 x 4. Evaluate: Ú __________ dx cos2 x ◊ sin2 x
sin 2 x 5. Evaluate: Ú ___________ dx sin 5 x ◊ sin 3 x
sin x 6. Evaluate: Ú __________ dx sin x + cos x
dx 7. Evaluate: Ú ____________ sec x + cosec x
8. Evaluate: Ú tan 3 x ◊ tan 2 x ◊ tan x dx
b 9. Evaluate: Ú ______ dx a + c ex
____
(
)
)
( )
______
x9/2 L 46. If Ú _______ ______ dx = __ log xN/P + ÷1 + x11 + c, 11 M ÷ 1 + x then
(a) L (b) L (c) L (d) L
(
+ – + +
M M M M
= = + +
)
÷
_________
L P __ N + __ 4 + c , then M x
dx 10. Evaluate: Ú _____ 1 + ex
(a) L = 1 (b) M = 2 (c) L + M + N + P = 5 (d) L + M + N – P = 3.
( )
tan–1 x 48. If Ú ______ 4 dx = x
(a) A = 3 (c) C = 6
(
x+9 11. Evaluate: Ú _______ dx x3 + 9 x sin x – cos x 12. Evaluate: Ú __________ dx ex + sin x
| |
tan–1 x __ x2 + 1 1 1 – ______ + log _____ + ____ 2 + k, 3 B A x x2 C x
then
|
13 9 N + P = 26 N = 15.
dx 47. If Ú ________ _____ = 3 x ÷ 1 + x4
|
(b) B = 6 (d) A + B + C = 15.
)
2 (cos x + sec x) sin x 49. If Ú _________________ 6 dx cos x + 6 cos2 x + 4 N 1 M = __ log 1 + _____ 4 + _____ 6 + c, L cos x cos x then
( (
))
cos x – sin x + 1 – x 13. Evaluate: Ú ________________ dx ex + sin x + x sin (x + a) 14. Evaluate: Ú _________ dx sin (x + b) dx 15. Evaluate: Ú _________________ sin (x – a) sin (x – b) dx 16. Evaluate: Ú _________________ sin (x – a) cos (x – b) ____
tan x ÷ 17. Evaluate: Ú ___________ sin x ◊ cos x dx
Indefinite Integrals 3 ____
dx _____ 18. Evaluate: Ú ________ x ÷x 4 – 1
40. Evaluate: Ú ÷tan x dx dx 41. Evaluate: Ú _______ (ex – 1)2
dx 19. Evaluate: Ú _______ (x2 + 1)2
tan–1 x 42. Evaluate: Ú ______ dx x4 dx ____ 43. Evaluate: Ú _______________ 4 _____ ( ÷ cos x + ÷ sin x )
÷
_________
sin (x – a) 20. Evaluate: Ú _________ dx sin (x + a) x 21. Evaluate: Ú _________ dx x4 + x2 + 1
1 + x– 2/3 44. Evaluate: Ú _______ dx 1+x
22. Evaluate: Ú sec3 x dx 23. Evaluate: Ú cosec3 x dx
dx 45. Evaluate: Ú ___________ 2 sin x + sec x
x2 – 1 __________ 24. Evaluate: Ú ____________ dx x ÷x 4 + 3x2 + 1
x4 + 1 46. Evaluate: Ú ______ dx x6 + 1
dx 25. Evaluate: Ú ____________ x4 + 18x2 + 81
dx 47. Evaluate: Ú _______ (ex – 1)2
(
)
x + sin x 26. Evaluate: Ú ________ dx 1 + cos x
dx _____ 48. Evaluate: Ú ____________ (x – 1) ÷x + 2
27. Evaluate: Ú sin 4 x etan2 x dx
dx _____ 49. Evaluate: Ú ________ x2 ÷ x + 1
(
)
2x + 2 ____________ 28. Evaluate: Ú sin–1 _____________ dx 2 + 8x + 13 ÷ 4x
x4 – 1 _________ 50. Evaluate: Ú ____________ dx x2÷ x 4 + x2 + 1
x2 29. Evaluate: Ú _____________ dx (x sin x + cos x)2
x2 – 1 ___________ 51. Evaluate: Ú ______________ dx x3 ÷2x 4 – 2x2 + 1
x2 (x sec2 x + tan x 30. Evaluate: Ú _______________ dx (x tan x + 1)2
______
cos 2 x ÷ 52. Evaluate: Ú _______ dx sin x
secx (2 + sec x) 31. Evaluate: Ú _____________ dx (1 + 2 sec x)2
(
)
cos q + sin q 32. Evaluate: Ú cos (2 q) × log ___________ dq cos q – sin q
(
)
x4 + 2 33. Evaluate: Ú ex _________ dx (1 + 2)5/2
(
sec2 x 54. Evaluate: Ú _____________ dx (sec x + tan x)9/2 tan 2 q ____________ 55. Evaluate: Ú ______________ dq 6 6 q + cos q ÷ sin
)
x cos3 x – sin x 34. Evaluate: Ú esinx _____________ dx cos2 x dx 35. Evaluate: Ú ____________ cos x + cosec x
Q. Evaluate the following integrals.
36. Evaluate: Ú cot (1 + x + x ) dx –1
cos q + sin q ___________ 53. Evaluate: Ú ____________ dq ÷ 5 + sin q (2q)
2
37. Evaluate: Ú tan–1 (1 + x + x2) dx sin x 38. Evaluate: Ú ________ sin 4 x dx
x2 + 6 1. Ú _______________ dx (x sin x + 3 cos x)6
log (1 + sin2 x) 2. Ú _____________ dx cos2 x
3. Ú x 3 1 + x3 dx
39. For any natural number m, evaluate Ú (x
3m
2m
+x
m
2m
+ x ) (2x
m
1/m
+ 3x + 6)
dx.
2 – __
(
)
2 –1 __
1.33
1.34 Integral Calculus, 3D Geometry & Vector Booster
(
(
) ) ( (
(
)))
5p p 4. Ú 1 + tan ___ – x 1 + tan – ___ + x dx 16 16
5. Ú tan (x – a) ◊ tan (x + a) ◊ tan2 x dx
( ( ÷ ) ) _____
1–x 6. Ú cos 2 cot _____ dx 1+x
(x – 1) ÷x 4 + 2x3 – x2 + 2x + 1 7. Ú __________________________ dx x2 (x + 1)
p tan __ – 4 4 __________________ 8. Ú ________________________ dx 2 3 cos x ÷tan x + tan2 x + tan x
tan–1 x 9. Ú ______ 4 dx x
–1
(
)
x + 2 2 25. Ú ex _____ dx x+4 ______ ÷1 + x8 _______
___________________
(
dx 24. Ú ____________________ __________________ 3 + a) cos (x – b ) ÷ sin (x
)
26. Ú
dx x13 dx 27. Ú ___________ 3 sin x + cos3 x _____
(x + ÷1 + x2 )3 _____ 28. Ú ____________ dx ÷ 1 + x2 (x + 1) 29. Ú __________ dx x (1 + x ex)2
÷
_____________
(1 – x sin x) 10. Ú ____________ dx x (1 – x3 e3cosx)
sin x 30. Ú _____________ dx 2 sin x + 3 cos x
(1 + x cos x) 11. Ú ____________ dx x (1 – x2 e2sinx)
cos 4 x + 1 31. Ú __________ dx cot x – tan x
dx 12. Ú ____________ 2 sin x + sec x
cos3 x + cos5 x 32. Ú ____________ dx sin2 x + sin4 x
13.
dx
dx 33. Ú ____________ 6 sin x + cos6 x
_____________________ ________________ Ú cos x sin (2x + a) + sin a
÷
(
( )
( ) )
f (x) 34. Evaluate Ú _____ 3 dx, where f (x) is a polynomial of x – 1 degree 2 in x such that f (0) = f (1) = – 3 = 3 f (2).
1 1 14. Ú 3 2 tan __ x – x sec2 __ x dx __________
3 cos 2 x – 1 ÷ 15. Ú ____________ dx cos x
(
) ( (
)
( ) )
Integer Type Questions
1 2x 2x 16. Ú _____ 8 cos–1 _____ 2 + tan–1 _____ 2 dx 1–x 1+x 1–x x2 – x3 17. Ú ___________________ dx (x + 1) (x3 + x2 + x)3/2
cos x + cos 2 x 1. If of Ú ____________ dx = A sin x + Bx + c, find the 2 cos x – 1 value of A + B + 3.
4
cos x 18. Ú ___________________ sin3x (sin5 x + cos5 x)3/5
sin3 x dx 19. Ú _________________________________ 4 (cos x + 3 cos2 x + 1) tan–1 (sec x + cos x)
2 ÷ p x = ___ log tan __ + __ L M N
____ ____ ÷cot x – ÷tan x _____________
20. Ú
dx 4 + 3 sin2 x
x2 (x sec2 x + tan x 21. Ú _______________ dx (x tan x + 1)2 ___________
3 – sin x ÷ sin x 22. Ú ____________ dx 3 1 – sin x x
__
|
(
)|
P + __ tan–1 (sin x – cos x) + c, 3 then find the value of L + M – N – P.
dx 3. If Ú ____________ sin6 x + cos6 x
= tan–1 (L tan x + M cot x) + c, find the value of L + M + N + 4.
2
e (2 – x ) _____ 23. Ú ____________ dx (1 – x) ÷1 – x2
dx 2. If Ú ___________ 3 sin x + cos3 x
x+1 4. Let Ú ______ dx x3 + x
( )
| |
xM 1 = tan–1 x + __ log ______ N + c. L x +1
)
L + M + N Find the value of __________ 3 sin x 5. Let Ú __________ dx sin x + cos x = Ax + B log |sin x + cos x| + c Find the value of A + B + 1
(
(
)
x2 6. Let Ú ________ dx (2x + 3)2
(3 + 2x) __ 9 1 = _______ + log |3 + 2x| – ________ + C. L M 8(2x + 3) Find the value of L + M + 4
(
2
)
(
)
(b) x (ln x) + c
x (c) ___ + c ln x
(d) x + ln x + c
x ex dx 2. Ú _______ = (1 + x)2
(a) x ex + c
ex (b) __________ (1 + x)2 + c
1 (c) ex – _____ + c 1+x
ex (d) _____ + c 1+x
find the value of L + M + N + P + Q.
(
)
cos4 x 8. If Ú ___________________ 3 dx sin x (sin5 x + cos5 x)3/5
=
(
)
M __ 1 1 + tan x N – __ ________ + 5 L 5
tan x
c,
9. If Ú x13/2 (1 + x5/2)1/2 dx
8M L = ___ (1 + x5/2)7/2 – ___ (1 + x5/2)5/2 35 25 N + ___ (1 + x5/2)3/2 + c, 15 find the value of L + M + N. 10. If Ú cos4 x dx = L x + M sin 2 x + N sin 4 x + c, find the value of 8L + 4M + 32 N.
Comprehensive Link Passages Passage I In some of the cases we can split the integrand into the sum of the two functions such that the integration for which one of them integrate by parts and other one be fixed.
( (
) )
) ( ) (
) )
x4 – 2 _________ 1. Ú ____________ dx = x2 ÷x 4 + x2 + 2
÷ 2 (b) x + __ + 1 + c ÷ x 1 (c) x + __ + c ÷ x 2 (d) x + __ ÷ x + c. 1 (a) x2 + __ 2 + 1 + c x __________
2
2
_______
= Ú e f (x) dx + Ú e f ¢(x) dx
= ex f (x) – Ú ex f ¢(x) dx + Ú ex f ¢(x) dx + c = e f (x) + c
)
_________
)
) ( ) (
( (
Ú ex {f (x) + f ¢(x)} dx
x
( (
) )
a a a Ú f x – __ x 1 + __ 2 dx, put x – __ x = t, x a a a __ __ Ú f x + x 1 – 2 dx, put x + __ x = t, x a a a Ú f x2 – __ 2 x + __ 3 dx, put x2 – __ 2 = t x x x a a a Ú f x2 + __ 2 x – __ 3 dx, put x2 + __ 2 = t x x x The following integrands can be brought into above forms by suitable transformations.
Consider
x
)
( (
For integrals
x
(
1 3. Ú ex log x – __ 2 dx x 1 x (a) e log x – __ x + c 1 (b) ex log x + __ x + c 1 (c) ex log x + __ 2 + c x 1 (d) ex log x + ____ + c log x
Passage II
find the value of L + M + N.
)
(a) ln (ln x) + c
xM + xN + P 1 = __ tan–1 ___________ + k, L Q
(
1.35
1 1 1. Ú ___ – _____ dx = ln x (ln x)2
3x + 2x 7. If Ú _________________________ dx 6 5 x + 2x + x4 + 2x3 + 2x2 + 5
Indefinite Integrals
2
2
______
2
2
(x – 1) dx __________ 2. Ú _________________ = (x + 1) ÷x 3 + x2 + x
1.36 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 (a) tan–1 x + __ x + 1 + c __________
÷(
)
1 (b) tan–1 x + __ x + 1 + c
1 (c) 2 tan x + __ x + 1 + c
(d) tan–1
__________
) ÷( 1 ÷( x – __ x + 1 ) + c –1
__________
(
4
5
)
5x + 4x 3. Ú __________ dx = (x5 + x + 1)2
(a) x5 + x + 1 + c
x5 (b) ________ 5 + c x +x+1
( ( (
__
__
)
Matrix Match (For JEE-Advanced Examination only)
) ) )
4
(
3
x + ÷x 2 + ÷4 x __ 3. Ú ___________ dx 3 4 x+÷ x 3 6 __ (a) __ x2/3 + 6 tan–1 (÷ x ) + C 2 __ 3 (b) __ x2/3 – 6 tan–1 (÷ 6 x ) + C 2 __ 3 (c) – __ x2/3 – 6 tan–1 (÷ 6 x ) + C 2 __ 3 6 (d) – __ x2/3 + 6 tan–1 ( ÷6 ) + C 2
Q1. Match the following columns: Column I
Column II x 1 (P) ___ __ tan–1 ___ __ + c 2 ÷2 ÷
dx
x (c) ________ 5 + c x +x+1
(A)
__ Ú ________ (x + 2)÷x
x4 (d) ________ 5 + c x +x–1
(B)
dx __ __ Ú _________ (÷x + 2)÷x
(C)
Ú _______ x (x + 2)
(D)
2 Ú _______ (x + 2)
Passage III Let us consider the integrals of the form
(
Ú R x,
b __ xq ,
a __ xp ,
)
__g x r dx
In this case, we shall put x = t m, where m is the LCM of (p, q, r). __
÷x __ 1. Ú _______ dx = 4 3 ÷ x + 1 __
___
(
__
__
)
__
2 2 12 (b) ___ 4Rx9 – ___ ÷x 13 + C 27 13
13 2 12 (c) ___ 4Rx9 – ___ ÷x 13 + C 27 13
___
(
___ 2 ___ ÷x +
2 (d) – ___ 4Rx9 – 27 13
12 13
)
x 1 (S) __ + c log _____ 2 x+2
(
)
(B)
Ú sin4 x dx
3 3 (Q) – __ (tanx)– 8/3 + __ (tan x)4/3 8 4 3 – __ (tan x)–2/3 + c 2
(C)
Ú sin7 x cos3 x dx
sin5 x _____ sin7 x (R) _____ – + c 7 5
(D)
Ú cosec11/3 x ◊
sin8 x _____ sin10x (S) _____ – + c 8 10
Q.3 Match the following columns: (Write the suitable substitutions)
___
(R) 2 log (2 + ÷ x ) + c
dx
sec7/3 x dx
2 4 2 12 (a) ___ ÷x 9 + ___ ÷x 13 + C 27 13
Column II sin 4 x 1 _____ (A) sin4 x ◊ cos3 x dx (P) __ 3 x + – 2 sin 2 x Ú 8 4 + c,
___
__
Q.2 Match the following columns:
÷ x 3 – ÷3 x 2. Ú ________ dx = __ 6 – 4÷x
__
(Q) ÷2 tan–1
__
dx
Column I
4 3 4 4 (a) __ ( ÷x 3 – ln ( ÷ x + 1 ) ) + C 3 __ ___ 2 3 4 2 3 (b) __ ( ÷ x – ln ( ÷ x + 1 ) ) + C 3 __ ___ 2 3 5 2 3 (c) __ ( ÷ x – ln ( ÷ x + 1 ) ) + C 3 __ ___ 4 3 8 4 (d) __ ( ÷x 3 – ln ( ÷ x + 1 ) ) + C 3
( ) x ( __ ÷ 2 ) + c
C
Column I (A)
sin x cos x
dx Ú ____________ sin2 x + cos2 x
Column II (P) put sin x = t
Indefinite Integrals
(B)
sin x◊cos x dx Ú ___________________ 2 2
(Q) put sin x = t
dx
(C)
Ú 3___________ + sin x cos x
(D)
cos x – sin x dx Ú _____________ 2 2
2
(R) put cos x = t
2
(S) put sin x cos x=t
1 – sin x cos x
Q.4 Match the following columns: Column I
(
Column II p (P) __ 2
)
x + sin x (A) If f (x) = ________ dx and Ú 1 + cos x p f (0) = 0, then f __ 2
( )
(B) Let
(
Q.6 Match the following columns: Column I
sin x + 2 cos x + cos x
Column II x 1 __ _____ dx (A) _____ Ú x2 + x (P) – x – log x + 1 + c x _____ dx (B) ______ + c (Q) log x + 1 Ú x3 + x2 x 1 1 __ 1 ___ ___ _____ dx (C) ______ + c (R) – 3 + 2 – x – log x + 1 Ú x4 + x3 3x 2x (D)
)
x _____ dx f (x) = Ú esin–1 x 1 – ______ ÷ 1 – x__2 k ÷3 ep/6 1 and f (0) = 1, if f __ = _______ , p 2
( )
p dx (C) Let f (x) = Ú _____________ (R) __ and 4 (x2 + 1) (x2 + 9) __ 5 f (0) = 0, if f (÷3 ) = ___ k, then k 56 is
x 1 1 – ____ 2 + __ x + log _____ + c x +1 2 x
dx
(S) Ú ______ x5 + x4
| | | | | | | |
(S) p. ) dx and ( sin x cos x
Let f (x) = Ú
( )
p 2k f (0) = 0, if f __ = ___ p , then k is 4
Column II 1 dx _____ (P) – ______ _____ (A) ___________ Ú (x – 1)÷ x – 1 ÷ _____ x + 4 –1 1__ –1 x_____ ___ – tan + c 2 2 ÷
| (
( ÷ ) )
1 1 dx ___ __ (B) ____________ _____ (Q) – __ log t – 3 Ú (x2 – 1)÷ 3 ÷ x – 1 __________ 2 1 __ t2 – __ t + + + c, 3 3 1 _____ t = x+2
÷
_____
(
_____
)
dx
x3 1 __ _____ + c (Q) log 3 3 x +1
Ú ________ x (x5 + 1)
dx
x2013 1 log _______ 2013 + c (R) _____ 2013 x +1
Ú dx/x (x2013 + 1)
x2 1 log _____ 2 + c (S) __ 2 x +1
Ú ________ x (x3 + 1)
(C) (D)
Q.8 Match the following columns: Column I 3
Column II
2
x – 3x
(A)
dx Ú _______ ex + x3
(B)
dx Ú _____ ex + x
(C)
dx Ú __________ ex + sin x
(D)
dx Ú ___________ ex + tan x
(P) x – log |x + ex| + c
x–1
x (Q) log |sin x + e | – x + c
cos x – sin x
tan x – sec2 x
x (R) x – log |tan x + e | + c
3 x (S) x – log |x + e | + c
Q.9 Match the following columns:
(A)
dx _____ (R) __ ÷x + 4 – 3 1 (C) ____________ _____ log _________ + c Ú 3 ÷x + 4 + 3 (x + 2) ÷x 2 – 1 dx ______ (S) ___ (D) _____________ ÷x 2 + 4 1 __ Ú 2 – __ tan–1 _______ + c 2 ÷3 x ÷3 (x + 1) ÷ x + 4
x5 1 log _____ 5 + c (P) __ 5 x +1
(B)
|
|
dx
Ú ________ x (x2 + 1)
Q.5 Match the following columns: Column I
Column II
(A)
____ ÷ tan x ________
|
| | | |
Q.7 Match the following columns:
then k is
(D)
| | | |
Column I p (Q) __ 3
1.37
(B)
(
Column I
5p – x ) Ú 1 – tan ( ___ 4
(
(
(
(
)
))
(P)
2x + c
))
(Q)
3x + c
3p 1 + tan ___ – x dx 2
Ú ( 1 + tan ( __ p8 – x ) )
Column II
p 1 + tan __ + x dx 8
1.38 Integral Calculus, 3D Geometry & Vector Booster
(C)
(
5p – x ) Ú 1 + tan ( ___ 4
16. The value of the integral
)
(R)
(
(
))
(
(
(S)
))
5x + c
Questions asked in Previous Years’ JEE-Advanced Examinations
dx 1. Ú _______ 1 – cot x x dx 2. Ú _____ 1 + x4
[IIT-JEE, 1978] [IIT-JEE, 1979]
÷
( )
[IIT-JEE, 1980]
6. Ú (elogx + sin x) cos x dx
[IIT-JEE, 1981]
x–1 dx 7. Ú ex _______ (x + 1)3
[IIT-JEE, 1983]
dx 8. Ú __________ 2 4 x (x + 1)3/4
[IIT-JEE, 1984]
(
÷
1 – ÷x dx 9. Ú ______ __ 1+÷ x
(
__
dx ____________ 18. Ú ______________ is equal to – p)3 (x – q) ÷ (x
__
)
–1 sin–1 ÷x – cos ÷x 10. Ú _______________ dx __ –1 –1 __ sin ÷x + cos ÷x
______
____
12. Ú (÷tan x + ÷ cot x ) dx
(
1 ___________ (c) _____________ + c – p) (x – q) ÷ (x
(d) none. [IIT-JEE, 1996]
dx 19. If Ú _________________ (sin x + 4) (sin x – 1)
( ( ) )
4 tan x + 1 1 2___ (a) A = __ , B = – _____ , f (x) = _________ 5 5 5÷15
[IIT-JEE, 1986]
4 tan (x/2) + 1 1 2___ ___ (b) A = – __ , B = – _____ , f (x) = ____________ 5 5÷15 ÷15
[IIT-JEE, 1987]
4 tan x + 1 2 2___ (c) A = __ , B = – _____ , f (x) = _________ 5 5 5÷15
[IIT-JEE, 1988]
4 tan (x/2) + 1 2 2___ ___ (d) A = __ , B = – _____ , f (x) = ____________ 5 5÷15 ÷15
)
then A = ..., B = ... and C = ...
[IIT-JEE, 1990]
log (1 + x1/6) 1 14. Ú ________ 1/3 1/4 + ___________ dx __ 1/3 x +x x + x ÷
[IIT-JEE, 1992]
(
)
)
cosq + sinq 15. Ú cos 2q log __________ dq cosq – sinq
A = ___________ + B tan–1 (f (x)) + C, x __ tan – 1 2
[IIT-JEE, 1985]
4 ex + 6 e–x 13. If I = __________ x dx = Ax + log |(9 e2x – 4)| + C 9e – 4 e–x
(
______
cos 2 x ÷ 11. Ú _______ sin x dx ____
÷( ) x–p 2 (b) – ______ ( _____ + c (p – q) ÷ x – q ) ______
x–p 2 (a) ______ _____ + c (p – q) x – q
then
______ __
[IIT-JEE, 1996]
[IIT-JEE, 1980]
)
6 tan–1 (sin x) + c 2 (sin x)–1 + c 2 (sin x)–1 – 6 tan–1 (sin x) + c 2 (sin x)–1 – 5 tan–1 (sin x) + c
x+1 17. Ú __________ dx x 2 x (1 + x e )
x dx 4. Ú 1 + sin __ 2 x2 _____ 5. Ú ______ dx 1 – x ÷
– – – –
[IIT-JEE, 1995]
[IIT-JEE, 1979]
___________
(a) sin x (b) sin x (c) sin x (d) sin x
x2 dx 3. Ú ________ (a + bx)2
)
cos3 x + cos5 x Ú ____________ dx is sin2 x + sin4 x
(1 + tan x) dx
7p – x Ú 1 + tan ___ 8 (D) 3p 1 + tan ___ + x dx 8
(
4x + c
[IIT-JEE, 1994]
[IIT-JEE, 1997] cos x – sin x 20. Ú ___________ (2 + 2 sin 2 x) is equal to cos x + sin x
(
(a) sin 2 x + c (c) tan 2 x + c
÷
21. Ú
)
_______ __ 1 – ÷x dx ______ × ___ x __ 1+÷ x
(b) cos 2 x + c (d) None [IIT-JEE, 1997]
dx ___________ 22. Ú ___________________ is equal to (2x – 7) ÷x 2 – 7x + 12
[IIT-JEE, 1997]
Indefinite Integrals
(a) 2 sec–1 (2x – 7) + c (b) sec–1 (2x – 7) + c (c) 1/2 sec–1 (2x – 7) + c (d) None of these.
[IIT-JEE, 1997]
2
x + 3x + 2 23. Ú _____________ dx (x2 + 1)2 (x + 1)
(
[IIT-JEE, 1999]
)
2x + 2 ____________ 24. Ú in–1 _____________ dx 2 + 8x + 12 ÷ 4x 3m
25. (x
2m
+x
2m
m
+ x ) (2 x
[IIT-JEE, 2001] m
+ 3x + 6)
2
____________
2 4 – 2x + 1 ÷ 2x (a) _____________ +c x2
(b)
4 – 2x 2 + 1 ÷ 2x (c) _____________ +c x
( ( ( (
e4x – e2x + 1 1 (a) __ log ___________ + c 2 e4x + ex + 1
e2x + ex + 1 1 (b) __ log __________ + c 2 e2x – ex + 1
e2x – ex + 1 1 (c) __ log __________ + c 2 e2x + ex + 1
e4x + e2x + 1 1 (d) __ log ___________ + c 2 e4x – e2x + 1
+c
x3
[IIT, 2008]
sec2 x 29. The integral Ú _____________ dx equals (sec x + tan x)9/2
____________
____________
2 4 – 2x + 1 ÷ 2x (d) _____________ + c [IIT, 2006] 2 2x x 27. Let f (x) = _________ for n ≥ 2 and g (x) = ( f O fO (1 + xn)1/n
{ { { {
}
1 1 (a) – ______________ ___ 1 – __ (sec x + tan x)2 + K (sec x + tan x)11/2 11 7
1 1 (b) ______________ ___ 1 – __ (sec x + tan x)2 + K (sec x + tan x)11/2 11 7
1 1 (c) – ______________ ___ 1 + __ (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x) 1 1 (d) ______________ ___ 1 + __ (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x)
n–2
... O f ) (x), then x
) ) ) )
____________ ÷ 2x 4 – 2x2 + 1 _____________
For an arbitrary constant C, the value of J – I is
dx where m > 0
x – 1 dx ____________ 26. Ú _______________ 3 4 x÷ 2x – 2x2 + 1
[IIT, 2007]
e–x J = ____________ –4x dx. e + e–2x + 1
1/m
[IIT-JEE, 2002]
1 __ 1 (d) ______ (1 + n xn)1 + n + c (n + 1) ex 28. Let I = Ú __________ dx, e4x + e2x + 1
1.39
g (x) dx is
1 __ 1 (a) _______ (1 + n xn )1 – x + c n (n – 1)
1 __ 1 (b) ______ (1 + n xn)1 – x + c (n – 1)
1 __ 1 (c) _______ (1 + n xn)1 + n + c n (n – 1)
} } }
[IIT, 2012]
Answers Level IIB 1. (a) 6. (b) 11. (a) 16. (d) 21. (c)
2. (b) 7. (a) 12. (b) 17 (a) 22. (d)
3. (c) 8. (b) 13. (a) 18. (b) 23. (a)
4. (d) 9. (c) 14. (b) 19. (a) 24. (b)
5. (a) 10. (d) 15. (c) 20. (b) 25. (a)
26. (b) 27. (c) 28. (d) 29. (a) 31. (a) 32. (b) 33. (c) 34. (d) 36. (b) 37. (a) 38. (b) 39. (c) 41. (b, d) 42. (a, b) 43. (a, b, c, d) 45. (a, b) 46. (a, b, c) 47. (a, b, c, d) 49. (a, b, c, d) 50. (a, c)
30. (b) 35. (a) 40. (b, d) 44. (a, b, c) 48. (a, b)
1.40 Integral Calculus, 3D Geometry & Vector Booster 1 18. – __ (1 + cot5 x)2/5 + c 2 19. log |tan–1 (cos x + sec x)| + c
Level IV
– x sec x 1. ____________ + tan x + c x sin x + 3 cos x __
2
2. tan x log (1 + sin x) – 2 ÷2 tan (t÷2 ) + c, where t = tan x –1 1/3
3. 3 tan (x ) + c 1 4. – __ log |cos (2x)| + log |cos (x – a)| 2
( ÷ ) _____
_____
6. – ÷1 – x + c
(t + 1) – ÷2 1 1 7. __ log |t2 + 2t – 3| – ____ __ log ___________ __ 2 2÷2 (t + 1) + ÷ 2
|
( ) 1 ( t + __ ÷ t + 1 ) + c
__
(
(
|
)
+1 1__ –1 3t + – ___ sin _____ + c 2 3 ÷
(
( )
| | | |
(x ex)3 – 1 1 10. __ log ________ + c 3 (x ex)3
_________
where t = ÷ 2 + 3 cot x
(x esin x)2 1 11. __ log __________ sin x 2 + c 2 (x e ) – 1
( ) ( )
1 31. – __ cos (4x) + c 8
)|
x p 1__ 1 12. ____ log tan __ + __ – ___________ + c 2 8 (sin x + cos x) 2÷2
32. sin x – 2 cosec x – 6 tan–1 (sin x) + c 33. tan–1 (tan x – cot x) + c
|
tan x cos a + sin a __ 13. _______________ + c 2 cos a ÷
( )
1 14. x3 tan __ x + c
( ÷
__________
)
]
)
_________
(
)
Integer type Questions
2 (x2 + 1) (x + 1) p 1 1 _______ 16. __ Ú _____ 2 + _____ 2 + Ú _______ – Ú dx 4 4 8 1–x x +1 x +1 1+x
( ÷
|
(x2 + x + 1) 2x +__ 1 2 34. log __________ + ___ __ tan–1 ______ + c (x – 1) 3 ÷3 ÷
__ __ __ cosec2 x – 3 15. ÷ 6 sin–1 (÷3 s in x) + 2÷2 tan–1 __________ + c 2
( ÷
| |
__ 1 t + __ – ÷2 t 1__ –1 1 1 30. – ____ tan t – __ + ____ __ log ___________ __ + c t 1 3÷2 6÷2 t + __ + ÷ 2 t
where tanq = x
and then you do it.
)|
| |
)
[ (
(
(x + ÷1 + x2 )3 28. ____________ +c 3 t 1 29. log ____ + ______ + c, where t = x ex t+1 (t + 1)
q 1 1 9. – ______ – __ ______ + log |sin q| + c 3tan3q 3 2 sin2 q
|
|
_____
(
( )
1 1 + x8 3/2 26. – ___ _____ + c 12 x8 x p 2__ 2 27. ____ log tan __ + __ – __ tan–1 (sin x – cos x) + c 2 8 3 3÷2
1 where t = x + __ x
)
x 25. ex ______ + c (x + 4)
________
8. – 2 tan–1
____________________________
24. 2 sec (a + b) ÷ cos (a + b) tan (x – b) + sin (a + b) + c
1 where t = x + __ x
2 22. __ sin–1 (sin3/2 x) + c 3 1+x 23. ex _____ + c 1–x
+ log |cos (x + a) + c
|
x 21. – __________ + 2 log |x sin x + cos x| + c (x tan x + 1)
1 5. – __ log |cos (2x)| + log| cos (x – a)| 2
2
|
__
(tan x + cot x) ÷2 – 1 1__ __ 20. ____ log __________________ + c 2÷2 (tan x + cot x) ÷2 + 1 2
+ log |cos (x + a)| + c
__
–1
)
)
1 1 17. 2 _________ ________ + tan–1 x + __ x + 1 + c 1 __ x + x + 1
1. 5 6. 4
2. 7 7. 9
3. 5 8. 9
4. 2 9. 7
Comprehensive Link Passages Passage I: Passage II: Passage III:
1. (b) 1. (b) 1. (b)
2. (d) 2. (c) 2. (b)
3. (a) 3. (b) 3. (a)
5. 1 10. 5
Indefinite Integrals
Matrix Match 1. (A)Æ(Q), (B)Æ(R), 2. (A)Æ(R), (B)Æ(P), 3. (A)Æ(P, Q, R), (C)Æ(R),
(C)Æ(S), (D)Æ(P) (C)Æ(S), (D)Æ(Q) (B)Æ(Q), (D)Æ(R, S)
Hints
Ú logx x dx = Ú 1 ◊ dx = x + c 2 – 2log 3) dx = Ú (2log 3 – 2log 3)dx Ú (3log x
x
x
3. we have, m x ) dx Ú ( xm + mx + mm + __
x m + 1 = _____ + m+1
mx _____ + mmx + m log |x| + c log m
4. we have,
Ú 2 ◊ 3 dx = Ú (2 ◊ 3) dx = Ú 6 dx
6x = ____ + c log 6
x
x
dx
dx
1 = __ tan x + c 2 11. The given integral is
x – xlog 3) dx = Ú (3log x – 3log x) dx Ú (3log 5
5
dx sin2x + cos2x 2 2 = Ú ____________ dx Ú _________ sin x cos x sin2x cos2x
(
)
Ú ( 1 + tan ( __ p8 – x ) ( 1 + tan ( __ p8 + x ) ) ) dx
= Ú (2) ◊ dx
13. The given integral is
= 2x + C
= – cot x – x + c
(
1 + x
1 + x
= 8 Ú 2xdx + Ú 4 dx 8 ◊ 2x = ____ + 4x + c log 2
= Ú (sec2x + cosec2x) dx
14. The given integral is
= tan x – cot x + c
) ) (
(
))
2 + 2x
= Ú (23 + x + 22) dx
(
3 + 3x
3p p 8. Ú 1 + tan x + ___ 1 + tan __ – x dx 8 8
)
(
)
+4 2 +2 dx = Ú ___________ dx Ú 8__________ 2x 2 22x
1 1 = Ú _____ + _____ dx cos2x sin2x
(
6
= Ú (0) ◊ dx = c
6
12. The given integral is
= tan x – x + c
1 = ______ = __ sec2x dx Ú 1_________ + cos 2x Ú 2cos2x Ú 2
Ú cot2x dx = Ú (cosec2x – 1) dx
= tan x – cot x + c
6. we have,
7. we have,
10. We have,
Ú tan2x dx = Ú (sec2x – 1) dx
= Ú (sec2x + cosec2x) dx
x
5. we have,
x
Ú (tan x + cot x)2dx = Ú (tan2x + cot2x + 2) dx
x
= Ú 0 ◊ dx = c
(D)Æ(S) (D)Æ(Q) (D)Æ(R) (D)Æ(R) (D)Æ(R) (D)Æ(P)
9. we have,
(C)Æ(R), (C)Æ(S), (C)Æ(S), (C)Æ(P), (C)Æ(Q), (C)Æ(P),
solutions
2. we have,
(B)Æ(Q), (B)Æ(P), (B)Æ(P), (B)Æ(Q), (B)Æ(P), (B)Æ(P),
p = Ú 2dx [ if A + B = __ , 4 then (1 + tan A) (1 + tan B) = 2] = 2x + c.
1. We have,
and
4. (A)Æ(P), 5. (A)Æ(R), 6. (A)Æ(Q), 7. (A)Æ(S), 8. (A)Æ(S), 9. (A)Æ(P),
1.41
m m + __ x + xm + mx ) dx Ú ( __ x
x2 xm + 1 = ___ + m log x + _____ + 2m m+1
mx _____ + c log m
1.42 Integral Calculus, 3D Geometry & Vector Booster
)
((sin2x + cos2x)2 – 2sin2x cos2x) = Ú ___________________________ dx sin2x cos2x
a x b x = Ú __ + __ a + 2 dx b
(1 – 2 sin2x cos2x) = Ú _______________ dx sin2x cos2x
(a/b)x (b/a)x = _______ + _______ + 2x + c log(a/b) log(b/a)
(sin2x + cos2x – 2sin2x cos2x) = Ú ________________________ dx sin2x cos2x
= Ú (sec2x + cosec2x – 2) dx
= tan x – cot x – 2x + c
15. The given integral is
(
(ax + bx)2
2x
2x
x x
a + b + 2a b dx = Ú _______________ dx Ú ________ x x ab a xbx
( ( ) ( ) )
16. The Given integral is (2x + 3x)2 dx = Ú _________ 2x ◊ 3x
(
Ú ( __ 23 ) + ( __ 32 ) + 2 dx x
x
)
(2/3)x (3/2)x = _______ + _______ + 2x + c log (2/3) log (3/2)
21. We have,
17. It is given that 1 1 _____ f ¢(x) = __ x + ______ ÷ 1 – x2 On integration, we get
f (x) = log |x| + sin–1x + c
p x = 1 and f (1) = __ , then c = 0 2 Hence, the function is f (x) = log |x| + sin–1x 18. It is given that f ¢(x) = a cos x + b sinx On integration, we get f (x) = a sin x – b cos x + c When x = 0 and f ¢(0) = 4, then a = 4 p When x = 0, f (0) = 3 and f __ = 5 2 then, we get b = – 2 and c = 1 Hence, the function is
(
sin6x + cos6x
f (x) 4 cos x + 2sin x + 1
19. We have, (1 + sin x) dx = Ú _________________ dx Ú 1_______ – sin x (1 – sin x) (1 + sin x)
(1 + sin x) 1 + sin x = Ú _________ dx = Ú _______ dx (1 – sin2x) cos2 x
= Ú (sec2x + sec x ◊ tan x) dx
= tan x + sec x + c
sin4x + cos4x
dx Ú ____________ sin2x cos2x
)
(sin2x)3 + (cos2x)3 = Ú _______________ dx sin2x ◊ cos2x
(sin4x + cos4x – sin2x ◊ cos2x) = ________________________ dx sin2x cos2x
(1 – 3sin2x ◊ cos2x) = _______________ dx sin2x cos2x
(sin2x + cos2x – 3sin2x cos2x) = ________________________ dx sin2x cos2x
= Ú (sec2 x + cosec2 x – 3)dx
= tan x – cot x – 3x + c
)
)
)
22. We have,
– cosa ____________ dx Ú ( cos 2x cos x – cosa )
(
)
2cos2x – 1 – 2cos2a + 1 = Ú ____________________ cos x – cosa dx
cos2x – cos2a = Ú 2 ____________ dx cos x – cosa
= Ú 2 (cos x + cosa) dx
= 2 (sin x + x cosa) + c
)
(
23. We have,
20. We have,
( Ú ( Ú ( Ú (
When
( )
)
dx Ú ___________ sin2x ◊ cos2x
(
cos4x – sin4x
)
________ dx Ú ____________ ÷1 + cos4x
(
)
(cos2x – sin2x) (cos2x + sin2x) _________ = Ú _________________________ dx 1 + cos 4x ÷
Indefinite Integrals
( (
cos 2x = Ú ________ __ dx 2 cos 2x ÷
x = ___ __ + c 2 ÷
( )
24. We have,
(
27. We have,
)
)
(cos2x – sin2x) __ = Ú ____________ dx 2 cos 2x ÷
)
+ tan2x 1 + tan2x dx = Ú ________ dx Ú 1________ 2 1 1 + cot x 1 + _____ 2 tan x
(
(
5cos3x + 3sin3x
)
dx Ú ______________ cos2x sin2x
(
)
5cos3x 3sin3x _________ = Ú _________ 2 + dx cos x sin2x cos2x sin2x
= 5 Ú cosec x cot x dx + 3 Ú sec x tan x dx
= – 5 cosec x + 3 sec x + c
28. We have,
)
cosx – sin x __________ (1 + sin 2x) dx Ú ( cos x + sin x )
tan2x (1 + tan2x) = Ú ______________ dx (1 + tan2x)
= Ú (tan x) dx
cos x – sin x = Ú __________ (cos x + sin x)2dx cos x + sin x
= Ú (sec x – 1) dx
= Ú (cos x – sin x) (cos x + sin x) dx
= Ú (cos2x – sin2x) dx
= tan x – x + c
2
2
– cos 2x ___________ dx Ú ( cos x 1 – cos x )
( Ú ( Ú ( Ú (
)
cos x – 2cos2x + 1 = Ú ________________ dx 1 – cos x
2cos2x – cos x – 1 = – _______________ dx 1 – cos x
)
(cos x – 1) (2cos x + 1) = – ___________________ dx 1 – cos x
(1 – cos x) (2cos x + 1) = ___________________ dx 1 – cos x
=
= 2 sin x + x + c.
)
)
Ú ((2 cos x + 1))dx
26. We have,
(
___________
(
)
= Ú cos2x dx sin 2x = _____ + c 2 29. The given integral is
25. We have,
) ( ÷ Ú ( ÷
__________
) )
+ 2 ÷ x 4 + x– 4 x4 + x– 4 + 2 dx = Ú __________ dx Ú ____________ 3 x x6 ___________
dx Ú ( _____________ 1 – 2cos 3x ) cos 5x + cos 4x
sin 3x (cos 5x + cos 4x) = Ú ___________________ dx sin 3x – sin 6x
( ) ( ) ( ) 3x x = – Ú 2cos ( ___ ) cos ( __ ) dx 2 2 =
Ú
( ) ( ) ( )
9x 3x 3x x 2sin ___ cos ___ 2cos ___ cos __ 2 2 2 2 _____________________________ dx 9x 3x ___ ___ – 2cos sin 2 2
= – Ú (cos 2x + cos x) dx
sin 2x = – _____ + sin x + c 2
(
)
30. The given integral is
– cos 2x ___________ dx Ú ( cos x 1 – cos x )
x8 + 2x4 + 1 = __________ dx x10
x4 + 1 = Ú _____ dx x5
2cos2x – cosx – 1 = – Ú _______________ dx 1 – cos x
1 = Ú x– 5 + __ x dx
(cos x – 1) (2cos x + 1) = – Ú ___________________ dx 1 – cos x
1 = log |x| – ___ 4 + c 4x
= Ú (2cos x + 1) dx
= (2sin x + x) + C
( ) (
1.43
)
1.44 Integral Calculus, 3D Geometry & Vector Booster 31. The given integral is
dx
Ú __________________________ (tan x + cot x + sec x + cosec x) sin x cos x dx = Ú ______________ 1 + sin x + cos x
=
_____________ Ú sec x + tan x + 1
=
Ú _____________________ (1 + tan x)2 – sec2x
=
= Ú (x1/3 – 1) dx
3 = – __ x– 2/3 – x + c 2
36. We have,
sin x(1 + tan x – sec x)dx
(
(
)
4
– 1 + 3 3 x –1 dx = Ú _____ + _____ 2 dx Ú x_________ 2 2 x +1 x +1 x +1
Ú ____________________ 2tan x
1 = __ Ú cos x (1 + tan x – sec x) dx 2
(x2 – 1) (x 2 + 1) _____ 3 = Ú _____________ + 2 dx x2 + 1 x +1
3 = Ú (x2 – 1) + _____ 2 dx x +1
x3 = __ – x + 3 tan–1x + c 3
(
sin x(1 + tan x – sec x)dx
32. We have,
(x + 1 – 1) x dx = Ú __________ dx Ú x_____ +1 (x + 1)
1 = Ú 1 – _____ dx x+1
= x – log |x + 1| + C
(
)
(1 + x)2 ________ dx = Ú x (1 + x2)
Ú
(1 + x2 + 2x) ___________ dx x (1 + x2)
)
37. We have,
)
(
(
x4 – 1 – 2
[ ( ) ] ) ]
)
4
x –1 2 dx = Ú _____ – _____ dx Ú _________ x2 + 1 x2 + 1 x2 + 1
[ (
(x2 + 1) (x2 – 1) 2 = Ú ______________ – _____ dx 2 2 x +1 x +1
2 = Ú (x2 – x) – _____ 2 dx x +1
x3 = __ – x – 2 tan–1x + c 3
(
33. We have,
)
4
1 = __ Ú (cos x + sin x – 1) dx 2 1 = __ (sin x + cos x – x) + C 2
=
sin x dx
(x1/3 – 1) (x2/3 + x1/3 + 1)
dx Ú _____________________ (x2/3 + x1/3 + 1)
38. we have,
(
)
) [ ( ) Ú [ ( )
]
(1 + x2) + 2x = ____________ dx x (1 + x2)
dx dx = Ú ___ x + 2 Ú _____ 1 + x2
= log |x| + 2 tan–1x + c
(x2 + 1) (x4 – x2 + 1) 2 = _________________ – _____ dx 2 2 x +1 x +1
=
x5 = __ – 5
34. We have,
Ú
)
(
)
3 = Ú 1 – _____ dx x2 + 1
–1
= x – 3 tan x + c
35. We have,
(
x2 – 2 x2 + 1 – 3 ______ 2 dx = Ú ________ dx x +1 x2 + 1
(x1/3)3 – 1 x–1 _____________ d x = dx Ú _____________ Ú (x 2/3 + x1/3 + 1) (x2/3 + x1/3 + 1)
x6 + 1 – 2
(
(
]
)
2 2 dx Ú (x4 – x2 + 1) – _____ x +1 x3 __ + x – 2 tan–1x + c 3
39. We have,
6
x +1 2 dx = Ú _____ – _____ dx Ú _________ 2 2 2 x +1 x +1 x +1
x8 + x4 + 1
)
dx Ú __________ x4 + x2 + 1
(
)
(x4 – x2 + 1) (x4 + x2 + 1) = Ú ______________________ dx (x4 + x2 + 1)
= Ú ((x4 + x2 + 1)) dx
Indefinite Integrals
x5 x3 = __ + __ + x + c 3 5
( )
(x4 – 1) + 1
4
x 2 dx = Ú __________ dx Ú _____ x2 + 1 x +1
(
)
1 = Ú x2 – 1 + _____ 2 dx x +1
x3 = __ – x + tan–1x + c 3
(
)
41. The given integral is
(
4
)
2
(x2 – x + 1) (x2 + x + 1) _____________________ dx (x2 + x + 1)
=
= Ú (x2 – x + 1)dx
x3 x2 = __ – __ + x + C 3 2
(
)
( )
(x2 + 1) (x4 – x2 + 1)
x6 + 1
dx = Ú __________________ dx Ú ______ x2 + 1 (x2 + 1) = Ú (x4 – x2 + 1) dx
(
5
3
)
43. We have,
x2 + c Ú sin–1(sin x) dx = Ú x dx = __ 2
44. We have,
Ú sin–1(cos x) dx = Ú sin–1 [ sin ( __ p2 – x ) dx ]
(
p = Ú __ – x dx 2 p x2 = __ x – __ + c 2 2
)
45. We have,
(
= Ú tan–1(tan x) dx
= Ú x dx
x2 = __ + c 2
( ÷
– cos 2x dx Ú tan–1 ÷ 11_________ + cos 2x _________
( )
= Ú tan–1(tan x) dx
= Ú x dx
x2 = __ + c 2
)
( ÷ )
sin x dx Ú tan–1 ( _______ 1 – cos x )
(
)
2sin (x/2) cos (x/2) = Ú tan–1 _______________ dx 2sin2(x/2)
[ ( ) ]
x = Ú tan–1 tan __ dx 2 x = Ú __ dx 2
( )
x2 = __ + c 4
49. The given integral is
Ú tan
( ÷
________
)
1 – sin x ________ dx 1 + sin x
–1
( (
)
cos (x/2) – sin (x/2) = Ú tan–1 _________________ dx cos (x/2) + sin (x/2)
1 – tan(x/2) = Ú tan–1 __________ dx 1 + tan(x/2)
_______
[ (
)
)]
p x = Ú tan–1 tan __ – __ dx 4 2 p x = Ú __ – __ dx 4 2
(
2 sin2x = Ú tan–1 ______ dx 2 cos2x
= Ú tan (tan x) dx = Ú x dx
x2 = __ + c 2
50. The given integral is
–1
)
– cos 2x sin x dx = Ú tan–1 ____ cos x dx Ú tan–1 11_________ + cos 2x
x x = __ – __ + x + c 3 5
)
2sin x cos x = Ú tan–1 _________ dx 2 cos2x
48. The given integral is
42. The given integral is
(
_________
sin 2x dx Ú tan–1 ( _________ 1 + cos 2x )
47. The given integral is
x + x + 1 dx Ú __________ x2 + + 1
Ú
46 We have,
40. The given integral is
1.45
(
)
)
p x x2 = ___ – __ + c 4 4
1.46 Integral Calculus, 3D Geometry & Vector Booster
sin x dx Ú tan–1 ( 1________ + cos x )
53. The given integral is
(
)
2 sin (x/2) cos (x/2) = Ú tan–1 ________________ dx 2cos2(x/2)
= Ú tan–1 [tan (x/2)] dx
x = Ú __ dx 2
( )
x2 = __ + c 4 51. The given integral is
( (
(
(
)
)
((( (( )) (( )))) (( (( )) (( ))))) ( (( )) )
)
= Ú tan–1
)
p x p x 2 sin __ – __ cos __ – __ 4 2 4 2 _____________________ dx p x 2sin2 __ – __ 4 2 p x cot __ – __ dx 4 2
–1
x cos __ 2 _______ dx x sin __ 2
[ ( ) ] p x [ tan ( __ – __ dx 2 2 )]
(
p x = Ú __ – __ dx 2 2
)
p x x2 = ___ – __ + c 2 4 54. The given integral is
Ú tan–1 (sec x + tan x)dx
(
) p 1 + cos ( __ – x ) 2 _____________ dx p __ sin ( – x ) 2 p x 2cos ( __ – __ 4 2) ____________________ dx p x p x 2sin ( __ – __ ) cos ( __ – __ ) 4 2 4 2 p x [ cot ( __ – __ dx 4 2 )] p p x [ tan { __ – ( __ – __ dx 2 4 2 )} ]
1 + sin x = Ú tan–1 _______ dx cos x
= Ú tan–1
–1
(
( (
)
= Ú tan–1
– sin x ) dx Ú tan–1 ( 1_______ cos x
= Ú tan–1
= Ú tan–1
p = Ú __ + 4
p x x2 = ___ + __ + c 4 4
p x x2 = ___ + __ + c 4 4
52. The given integral is
= Ú tan–1
= Ú tan–1
( (
(
)
p 1 – cos __ – x 2 _____________ dx p sin __ – x 2 p x 2sin2 __ – __ 4 2 _____________________ dx p x p x 2 sin __ – __ cos __ – __ 4 2 4 2
( ( ( )
(
(
(
))
)
)
) ) (
p x = Ú tan–1 tan __ – __ dx 4 2 p x = Ú __ – __ dx 4 2 p x x2 = ___ – __ + c 4 4
)
x = Ú tan–1 cot __ dx 2
) ( ) = Ú tan ( ) = Ú tan ( ( )) p p x = Ú tan ( tan ( __ – ( __ – __ dx 2 4 2 ))) p x = Ú tan ( tan ( __ + __ dx 4 2 )) p x = Ú ( __ + __ dx 4 2) –1
(
_________
x x x x cos __ + sin __ + cos __ – sin __ 2 2 2 2 __________________________________ dx x x x x cos __ + sin __ – cos __ – sin __ 2 2 2 2
= Ú tan
p sin __ – x 2 _____________ = Ú tan dx p 1 – cos __ – x 2 –1
= Ú tan
–1
(
_________
+ sin x) + ÷(1 – sin x) ÷ (1 _________ dx ______________________ _________ + sin x) –÷ (1 + sin x) ÷ (1
–1
cos x dx Ú tan–1 ( _______ 1 – sin x ) –1
Ú tan
–1
)
)
(
2
)
x __ dx 2
55. The given integral is
sin 2x dx Ú tan–1 ( _________ 1 + cos 2x )
(
)
2sin x cos x = Ú tan–1 _________ dx 2cos2x
)
)
1.47
Indefinite Integrals
= Ú tan–1 (tan x) dx
= Ú x dx 2
x = __ + c 2
56. We have,
+c Ú (3x + 2) dx = Ú ________ 2.3
(3x + 2)2 = Ú ________ + c 6 1 __ log |2x – 3| + c 2
= Ú (÷x + 2 + ÷ x + 1 ) dx
2 2 = __ (x + 2)3/2 + __ (x + 1)3/2 + c 3 3
ax + b
e a + c Ú eax + bdx = _____ 4x + 5
3 +c Ú 34x + 5dx = _____ 4log 3
sin (5x + 3) +c Ú cos (5x + 3) dx = __________ 5 62. We have, cos 2x + c Ú sin 2x dx = – _____ 2
(÷2x + 5 + ÷2x + 3 ) dx = Ú ____________________ (2x + 5) – (2x + 3)
______ ______ 1 = __ Ú (÷2x + 5 +÷ 2x + 3 ) dx 2
1 2 __ 1 2 1 = __ __ × (2x + 5)3/2 + __ × __ (2x + 3)3/2 + c 2 3 2 3 2
1 = __ ((2x + 5)3/2 + (2x + 3)3/2) + c 6
(3x + 2)3/2
+ 2 dx = _________ +c Ú ÷3x 3 __
◊ 3 2
64. We have,
Ú (3x +
1 – __ 4) 2 dx
(3x + = ________ +c 1 __ ◊ 3 2 1 __ 4)2 +
______
______ ______ 1 = __ Ú (÷3x + 4 +÷ 3x + 1 ) dx 3
+ 4)3/2 (3x + 1)3/2 1 (3x = __ _________ + _________ + c 3 9/2 9/2
(
)
dx
______ ______ Ú _________________ ÷2x + 3 + ÷2x – 3 ______
(÷2x + 3 – ÷2x – 3 ) = Ú _________________ dx (2x + 3 – 2x + 3)
______ ______ 1 = __ Ú (÷2x + 3 –÷ 2x – 3 ) dx 6
+ 3)3/2 (2x – 3)3/2 1 (2x = __ _________ – _________ + c 3 3 6
c
(
)
69. The given integral is
dx
_____ _____ Ú ______________ ÷x + 2 – ÷x + 1 _____
dx
______ ______ Ú ÷ _______________ 3x + 4 – ÷3x + 1
______
65. We have,
)
______
1 __ 4)2
2 = __ (3x + 3
(
68. The given integral is
2 = __ (3x + 2)3/2 + c 9
______
(÷3x + 2 + ÷3x + 1 ) = Ú __________________ dx (3x + 4 – 3x– 1)
63. We have, ______
______
(÷2x + 5 +÷ 2x + 3 )dx _____ ______ ______ ______ = Ú __________________________________ (÷x + 5 – ÷ 2x + 3 ) (÷2x + 5 +÷ 2x + 3 )
67. The given integral is
61. We have,
dx
______ ______ Ú _________________ (÷2x + 5 – ÷2x + 3 )
______
60. We have,
_____
______
dx 1 = – __ log |5 – 2x| + c Ú ______ 5 – 2x 2 59. We have,
_____
57. We have, dx = Ú ______ 2x – 3 58. We have,
_____
66. We have,
(3x + 2)2
_____
(÷x + 2 + ÷x + 1 ) dx = Ú _________________ (x + 2) – (x + 1)
_____
(÷x + 2 + ÷ x + 1 ) dx _____ _____ _____ _____ = Ú _______________________________ (÷x + 2 – ÷x + 1 ) (÷x + 2 + ÷ x + 1 )
dx
_____
__
_____ __ = Ú (÷x + 1 – ÷ x )dx Ú ÷__________ x + 1 + ÷ x 2 = __ ((x + 1)3/2 – (x)3/2 ) + c 3
1.48 Integral Calculus, 3D Geometry & Vector Booster 70. The given integral is
74. We have,
dx
_____ _____ Ú ÷______________ x + a + ÷x + b
(
_____
_____
)
x + a – ÷x + b ÷ = Ú ______________ dx x+a–x–b
_____ _____ 1 = ______ Ú (÷x + a – ÷ x + b ) dx (a – b)
2 = _______ ((x + a)3/2 – (x + b)3/2) + c 3(a – b)
_________
_________
_________
_________
÷2x + 2014 – ÷2x + 2013 = Ú _______________________ dx (2x + 2014 – 2x – 2013)
2 t3 = __ __ – t + c 3 3
3/2 ______ 2 (3x + 1) = __ _________ –÷ 3x + 1 + c 3 3
( )
(
)
1 = __ [(2x + 2014)3/2 – (2x + 2013)3/2] + c 3
fi 2 dx = 2t dt fi dx = t dt
= 2 Ú ( + 1) dt
( )
t3 = 2 __ + t + c 3
(
2/3
_____
)
(x – 1) = 2 ________ + ÷ x – 1 + c 3
__
x ÷ t dx = Ú _____ 2 ◊ 2t dt, Ú _____ x+1 t + 1
Let x = t2
fi dx = 2t dt
t2 = 2 Ú _____ dt t2 + 1 2
(t + 1 – 1) = 2 Ú __________ dt t2 + 1
(
)
1 = 2 Ú 1 – _____ 2 dt t +1
= 2 (t – tan–1t) + c
–1 = 2 (÷x – tan (÷x ) ) + c
__
__
(t2 + 1)
t dt Ú ______ 2t
\
=
1 = __ Ú (t2 + 1) dt 2
1 t3 = __ __ + t + c 2 3
3/2 1 (2x – 1) = __ _________ + (2x – 1)1/2 + c 2 3
( )
2
Let 2x – 1 = t2
2 = __ Ú (t2 – 1) dt 3
= Ú (÷2x + 2014 – ÷2x + 2013 ) dx
2 (t______ + 1) x ______ _____ d x = × 2t dt. Ú ÷ x – 1 Ú t Let x – 1 = t2 fi dx = 2t dt
73. We have,
Let 3x + 1 = t2 2t dt fi dx = ____ 3
75. We have, x______ +1 dx Ú _______ 2x + 1 ÷
72. We have,
(Ú )
t2 – 1 _____ 3 _______ × 2t dt, t
dx
_________ _________ Ú ÷______________________ 2x + 2014 + ÷2x + 3013
=
71. The given integral is
x
______ dx Ú _______ ÷3x + 1
(
)
76. We have,
x–1
_____ dx Ú ÷______ x + 4
Let x + 4 = t 2
fi dx = 2t dt
( ) 2
–5 × t dt Ú t_____ t
=
= Ú (t2 – 5) dt
t3 = __ – 5t + c 3
(x – 1)3/2 = ________ – 5(x – 1)1/2 + c 3
(
)
77. We have,
x
2 dx Ú _____ x +1
Let x2 + 1 = t
Indefinite Integrals
1.49
fi 2x dx = dt
Let xex = t
1 fi x dx = __ dt 2
fi (xe x + ex) dx = dt
fi (x + 1) ex dx = dt
1 dt = __ Ú __ t 2 1 = __ log |t| + c 2 1 = __ log |x2 + 1| + c 2
78. We have,
cos x – sin x __________ dx Ú sin x + cos x
Let sin x + cos x = t
fi (cos x – sin x)dx = dt
dt = Ú _____ 2 cos t
= Ú sec2t dt
= tan t + c
= tan(xex) + c
82. We have,
dx
Ú _________ x (l + ln x)2
dt = Ú __ t = log |t| + c
= log |sin x + cos x| + c
dt = __ 2 t
1 = – __ + c t
1 = – __ + c t
1 = – _______ +c 1 + ln x
79. We have,
Let 1 + ln x = t 1 fi __ x dx = dt
3cos x
________ dx Ú 2sin x +5
Let 2 sin x + 5 = t fi 2 cos x dx = dt 1 fi cos x dx = __ dt 2
83. We have, cos x – sin x + 1 – x
3 dt = __ Ú __ t 2
dx Ú _________________ ex + sin x + x
3 = __ log |t| + c 2
(e x + cos x + 1) – (ex + sin x + x) = Ú ____________________________ dx ex + sin x + x
3 = __ log |2sin x + 5| + c 2
(e x + cos x + 1) (ex + sin x + x) = Ú _____________ – _____________ dx x e + sin x + x (e x + cos x + 1)
(ex + cos x + 1) (ex + sin x + x) = _____________ – _____________ dx x (e + sin x + x) (ex + sinx + x)
80. We have,
cos x – sin x
cos x – sin x
dx = Ú _____________ dx Ú __________ 2 + sin 2x 1 + (1 + sin 2x)
( Ú ( Ú (
)
cosx – sin x = Ú ________________ dx 1 + (sin x + cos x)2
(ex + cos x + 1) = _____________ – 1 dx (e x + sin x + x)
Let sin x + cos x = t
= log |ex + sin x + x| – x + c
fi (cos x – sin x) dx = dt
84. We have,
dt = Ú _____ 1 + t2
= tan–1(t) + c –1
= tan (t) + c
81. We have,
xe x + ex
(x + 1)ex
dx = Ú _________ dx Ú ________ cos2(xe x) cos2 (x e x)
dx
) )
e– x
x = Ú ______ – x dx Ú _____ 1+e e +1
Let e– x + 1 = t
fi – e– xdx = dt
dt = – Ú ____ t+1
= – log |t + 1| + c
= – log |e– x + 1| + c
1.50 Integral Calculus, 3D Geometry & Vector Booster 85. We have,
dx x2dx 3 = Ú ________ 3 3 Ú ________ x (x + 1) x (x + 1) dt 1 = __ Ú ______ 3 t(t + 1)
(
Put x3 = t
1 1 ____ 1 = __ Ú __ – dt t t+1 3
t 1 = __ log ____ + c 3 t+1
x3 1 = __ log _____ 3 + c 3 x +1
| |
dt 1 = __ Ú _______ Let x4 = t 4 t (t + 1) dt fi x3dx = __ 4
)
1 1 1 = __ Ú __ – ____ dt t t+1 4
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
)
| |
sin (5x – 3x)
dx = Ú ___________ dx Ú __________ sin 5x sin 3x sin 5x sin 3x
Let ex + e– x = t
fi (ex – e– x) dx = dt
dt = Ú __ t = log |t| + c
= log |e x + e–x| + c
91. The given integral is
ex
dx
= Ú ______ x dx = log |ex+ 1| + c Ú 1______ + e– x e + 1 e– xdx
dx
x = Ú ______ Ú 1_____ +e e– x + 1
= – log |e– x + 1| + c
ex = log _____ x + c e +1
| |
93. The given integral is
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 5x sin 3x sin 5x cos 3x __________ cos 5x sin 3x = Ú __________ – dx sin 5x sin 3x sin 5x sin 3x
(
ex – e– x
dx Ú _______ ex + e–x
x –1 1 = __ log _____ + c 5 x5 sin 2x
|)
92. The given integral is
5
88. We have,
1 = _________ [log |sin (x – b)| – log |sin (x – a)|] + c sin (b – a)
t–1 1 = __ log ____ + c t 5
1 = _________ [cot (x – b) – cot (x – a)] dx sin (b – a) Ú
dt 1 = __ Ú ______ Let x5 = t 5 t (t – 1) dx __ dt ___ x = 5t 1 1 1 = __ Ú ____ – __ dt 5 t–1 t
| |
)
90. The given integral is
sin (x – a) cos (x – b) – cos (x – a)sin (x – b)
( |
dx x4dx ________ 5 = Ú ________ 5 5 Ú x (x – 1) x (x – 1)
(
(
dx Ú ______________________________________ sin (x – a) sin (x – b)
sin (x – b) 1 = _________ log _________ + c sin (b – a) sin (x – a)
| |
87. We have,
dx
Ú __________________ sin (x – a) sin (x – b)
1 = _________ sin (b – a)
| |
89. We have,
4 = Ú ________ 4 4 Ú ________ x (x + 1) x (x + 1)
(
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
sin ((x – a) – (x – b)) dx 1 = _________ Ú ____________________ sin (x – a) sin (x – b) sin (b – a)
x3dx
dx
sin (b – a)dx 1 = _________ Ú __________________ sin (b – a) sin (x – a) sin (x – b)
| |
86. We have,
= Ú (cot 3x – cot 5x) dx
)
)
sin 2x
_____________ 2 dx Ú a sin x + b cos2x
Let a sin2x + b cos2x = t
Indefinite Integrals
(a – b) sin 2x dx = dt
dt sin 2x dx = ______ (a – b) dt 1 = ______ __ (a – b) Ú t 1 = ______ log |t| + c (a – b)
1 = ______ log |a cos2x + b sin2x| + c (a – b)
94. The given integral is
sin (x – a)
sin x cos a – cos x sin a
dx = Ú __________________ dx Ú _________ sin x sin x = Ú (cos a – sin a cot x) dx
1.51
98. The given integral is 1 + tan x dx Ú __________ x + log sec x
Let x + log sec x = t
fi (1 + tan x) dx = dt
dt = Ú __ t = log |t| + c
= log |x + log (sec x)| + c 99. The given integral is
sin(5x – 3x)
sin 2x
dx = Ú ___________ dx Ú __________ sin 3x sin 5x sin 3x sin 5x
= x cos a = sin a log |sin x| + c 95. The given integral is sin (t + a) sin x ________ dx = Ú _________ dt, Ú sin (x – a) sin t
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 3x sin 5x
= Ú (cot 3x – cot 5x) dx
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
100. The given integral is
Let t = x – a fi dt = dx
cos x – sin x
(cos x – sin x)
dx = Ú ________________ dx Ú ___________ 1 + sin 2x 1 + (sin x + cos x)2
= t cos a + sina log |sin t| + c = (x – a) cos a + sin a log |sin(x – a)| + c 96. The given integral is
Put (sin x + cos x) = t
sin (x + a) Ú _________ dx sin (x + b)
fi (cos x – sin x) dx = dt
sin [t + (a – b)] = Ú _____________ dt, sin t
dt = Ú _____ 1 + t2
Let t = (x + b)
= tan–1(t) + c
fi dt = dx
= tan–1(sin x + cos x) + c
sin (t) cos (a – b) + cos t sin (a – b) = Ú _____________________________ dt sin t
101. The given integral is
= Ú (cos (a – b) + sin (a – b) cot t) dt
= t cos (a – b) + sin (a – b) log |sin t| + c
= Ú (sec x tan x + cosec x)dx
x = sec x + log tan __ + c 2
= (x + b) cos (a – b) + sin (a – b) log |sin (x + b)| + c
|
97. The given integral is
dx
__ __ Ú ÷__________ x (÷x + 1)
dt
Ú __ t
=2
= 2 log |t| + c
= 2 log |÷x + 1| + c
__
(sin2x + cos2x) dx
dx
= Ú ______________ Ú _________ sin x cos2x sin x cos2x
( ) |
102. The given integral is __
Let (÷x + 1) = t dx fi ____ = dt __ 2÷x dx __ fi ___ = 2dt x ÷
(sin2x + cos2x)
dx
2 2 = Ú ____________ dx Ú _________ sin x cos x sin2x ◊ cos2x = Ú (sec2x + cosec2x) dx
= tan x – cot x + c. 103. The given integral is
dx
__________________ Ú sin x (x – a) sin (x – b)
1.52 Integral Calculus, 3D Geometry & Vector Booster sin (a – b) 1 = _________ Ú _________________ dx sin (x – a) sin (x – b) sin (a – b) sin [(x – b) – (x – a)] 1 = _________ ___________________ dx sin (a – b) Ú sin (x – a) sin (x – b)
= log |t| + c
107. The given integral is
= log |xx + 1| + c
1 = _________ sin (a – b)
cos x – sin x + 1 – x
dx Ú _________________ ex + sin x + x (cos x + 1) – (sin x + x) = Ú ____________________ dx (e x + sin + x)
sin (x – b) cos (x – a) – cos (x – b)sin (x – a) Ú ______________________________________ sin (x – a) sin (x – b) dx 1 = _________ [cot (x – a) – cot (x – b)] dx sin (a – b) Ú
=
1 = _________ [log |sin (x – a)| – log |sin (x – b)|] + c sin (a – b)
( |
|)
sin (x – a) 1 = _________ log ________ + c sin (a – b) sin (x – b)
(ex + cos x + 1) – (ex + sin x + x)
dx Ú _____________________________ (ex + sin x + x)
(
)
(ex + cos x + 1) = Ú ________________ dx (ex + sin x + x) – 1
= log |ex + sin x + x| – x + c
108. The given integral is
104. The given integral is
sin3x
4 dx Ú __________________________________ (cos x + 3cos2x + 1) tan–1(sec x + cos x)
dx Ú _________________ cos (x – a) cos (x – b)
sin (a – b) 1 = _________ Ú _________________ dx sin (a – b) cos (x – a) cos (x – b)
Let tan–1(sec x + cos x) = t
sec x tan x – sin x fi ________________ dx = dt 1 + (sec x + cos x)2
sin3x fi ______________________ dx = dt cos2x (1 + (sec x + cos x)2)
sin3x fi ____________________ dx = dt 2 2 cos x (sec x + cos2x + 3)
sin3x fi _________________ dx = dt 4 (cos x + 3cos2x + 1)
sin [(x – b) – (x – a)] 1 = _________ ___________________ dx sin (a – b) Ú cos (x – a) cos (x – b) 1 = _________ [tan(x – b) – tan (x – a)] dx sin (a – b) Ú 1 = _________ [– log |cos (x – b)| + log |cos (x – a)|] + c sin (a – b) cos (x – a) 1 = _________ log _________ + c sin (a – b) cos (x – b)
( |
|)
dt = Ú __ t
= log |t| + c
= log |tan–1(sec x + cos x)| + c
cos (a – b) 1 = _________ _________________ dx cos (a – b) Ú sin (x – a)cos (x – b)
cos [(x – b) – (x – a)] 1 = _________ Ú ___________________ dx cos (a – b) sin (x – a) cos (x – b)
= – cos (t) + c
1 = _________ [tan (x – b) tan (x – b)]dx cos (a – b) Ú
= – cos (x3) + c
cos (x – a) 1 = _________ log _________ + c cos (a – b) cos (x – b)
105. The given integral is dx Ú _________________ sin (x – a) cos (x – b)
( |
|)
106. The given integral is x
x (1 + ln x)
dt
x dx = Ú __ , Let t = xx Ú __________ t x +1
109. We have,
Ú 3x3sin (x3) dx = Ú sin(t) dt
where x3 = t
110. We have,
(1 + ln x)3
dx = Ú t3dt Ú _________ x
Let (1 + ln x) = t dx fi ___ x = dt
t4 = __ + c 4
(1 + ln x)4 = _________ +c 4
Indefinite Integrals
111. We have,
114. The given integral is dx
dx
x x 4 + 1 x
Let tan x = t
fi sec2x dx = dt
2 = Ú _______________ Ú __________ 3/4 1 2 4 __ x (1 + x4)3/4
=
( ( ) ) Ú ( ) Ú ( ) dx ___________ 3/4 1 5 __ x 4 + 1 x
4t3 1 = – ___ 3 dt, Let __ 4 + 1 = t4 t x – 4 dx fi _____ 5 = t3dt x
= – 4 Ú dt
= – 4t + c
1/4 1 = – 4 __ 4 + 1 + c x
(
)
__
Let 4 + 3÷x = t
fi
3 ____ __ dx = dt 2÷x
fi
dx ___ __ = x ÷
2 __ dt 3
dt __ 2 t 2 1 = – __ × __ + c t 3 2 = __ Ú 3
Ú 33 ◊ 33 ◊ 3xdx 3x
x
Let 3 x = t
1 fi 3xdx = _____ dt loge 3
t4 = __ + c 4
tan4x = _____ + c 4 115. The given integral is
Ú sin3x ◊ cos x dx
= Ú t3dt,
(log x)3 ______ dx Ú x
= Ú t3dt,
1 = ______ 2 Ú 33 ◊ 3tdt, (log3) t
3v = ______ +c (log3)3
33 = ______ +c (log3)3
3x
Let t = log x dx fi dt = ___ x
sin x
_________ dx Ú __________ ÷3 + 2cos x
Let 3 + 2 cos x = t2
fi – 2 sin x dx = 2t dt
fi sin x dx = – t dt
t dt = – Ú ___ t
= – Ú dt
Let 3t = v
= – t + c
fi logs 3dt = dv
= – ÷3 + 2 cos x + c
1 = ____ Ú 3 3 ◊ 3tdt log3
t4 = __ + c 4
t
Let t = sin x fi dt = cos x dx
sin4x = _____ + c 4 116. The given integral is
113. We have,
t4 = __ + c 4 (log x)4 = ______ + c 4 117. The given integral is
2 1 = __ × ________ __ + c 3 (4 + 3÷x )
= Ú t3dt
dx Ú ÷ __________ __ __ 2 x (4 + 3 ÷x )
Ú tan3x ◊ sec2x dx
112. We have,
1.53
_________
118. The given integral is ________
÷ 2 + log x dx Ú _________ x
Let 2 + log x = t2 dx fi ___ x = 2t dt
1.54 Integral Calculus, 3D Geometry & Vector Booster
= Ú (t ◊ 2t) dt
122. The given integral is
= 2 Ú t2 dt
t3 = 2 __ + c 3
( )
2 = __ [(2 + log x)3/2] + c 3 119. The given integral is
sin x – cos x
dx Ú __________ ex + sin x
(ex + sin x) – (ex + cos x) = Ú _____________________ dx (ex + sin x)
(ex + cos x) = Ú 1 – _________ dx (ex + sin x)
= (x – log |ex + sin x| + c)
(
dx
__ Ú 1______ +÷ x Let x = t2 fi dx = 2t dt
2t = Ú ____ dt 1+t
1 = 2 Ú 1 – ____ dt 1+t
= 2 (t – log |1 + t|) + c
(
__
Let x4 = t
dt fi x dx = __ 4
1 = __ Ú sin (t)dt 4 1 = – __ cos (t) + c 4 1 = – __ cos (x4) + c 4
Ú 5
x
5x
x
◊ 5 ◊ 5 dx
Put 5x = t
dt fi 5xdx = ____ log 5
t 1 = ____ 55 ◊ 5tdt log5 Ú
1 = ______ 2 Ú 5z ◊ dz, z = 5t (log5)
1 = ______ 2 5z ◊ dz Ú (log5)
5z = ______ +c (log5)3
= Ú (1 – cos2x)cos4x sin x dx
= – Ú (1 – t2) t4dt
55 = ______ +c (log5)3
Let cos x = t fi sin xdx = dt
= Ú (t6 – t4) dt
t7 t5 = __ – __ dt 7 5
cos7x _____ cos5x = _____ – + c 7 5
( (
)
)
127. We have,
Ú sin2x cos3x dx = Ú sin2x(1 – sin2x) cos x dx = Ú t2(1 – t2) dt, Let sin x = t
fi cos xdx = dt
= Ú (t2 – t4) dt
t3 t5 = __ – __ + c 3 5
sin3x _____ sin5x = _____ – + c 3 5
5x
Ú sin3 x cos4 x dx
121. The given integral is 5 5
| |
| |
124. We have,
3
)
x3 1 = __ log _____ 3 + C 3 x +1
3
fi 4x dx = dt
(
1 1 1 = __ Ú __ – ____ dt t t+1 3 t 1 = __ log ____ + C 3 t+1
x2dx
d
________ = Ú ________ 3 3 Ú x (1 + x3) x (x + 1)
Ú x3sin x4 dx
dt 1 = __ Ú _______ , Let t = x3 3 t (t + 1) fi dt = 3x2
= 2 (÷x – log |1 + ÷ x | ) + c 120. The given integral is
123. The given integral is
)
__
)
( (
)
)
1.55
Indefinite Integrals
130. We have,
Ú sin3x cos3x dx
= Ú sin x (1 – sin )cos x dx
= Ú t (1 – t ) dt
3
3
2
Let sin x = t fi cos xdx = dt
= Ú (t3 – t5) dt
sin3x _____ sin5x = _____ – + c 3 5
(
)
134. we have,
Ú sin2x cos2x dx = __ 14 Ú (2 sin2x) (2 cos2x)dx
1 = __ Ú (1 – cos 2x) (1 – cos 2x)dx 4
1 = __ Ú (1 – cos22x)dx 4
1 = __ Ú sin22x dx 4
1 = __ Ú 2 sin22x dx 8 1 = __ Ú (1 – cos 4x) dx 8 sin 4x 1 = __ x – _____ + c 8 4
(
)
135. We have,
Ú sin3x dx = Ú sin2x ◊ sin x dx = Ú (1 – cos2x) ◊ sin x dx = – Ú (1 – t2)dt, Let cos x = t
139. We have,
fi –sin xdx = dt
= Ú (t2 – 1)dt
( ) (
t3 = __ – t + c 3 3 cos x = _____ – cos x + c 3
)
Ú cos5x dx = Ú cos4x ◊ cos dx
= Ú (1 – sin2x)2 ◊ cos dx
= Ú (1 – t ) dt,
t5 = __ – 5
sin5x __ 2 = _____ – sin3x + sin x + c 3 5
( (
2 2
Let sin x = t fi cos xdx = dt
)
2 __ t3 + t + c 3
)
142. We have,
= Ú (t4 – 2t2 + 1)dt
2
Ú sin4 x dx = __ 14 Ú (2 sin2x)2dx
1 = __ Ú (1 – cos 2x)2dx 4 1 = __ Ú (1 – 2 cos 2x + cos22x) dx 4 1 1 = __ Ú (1 – 2 cos 2x)dx + __ Ú (2cos22x) dx 4 8 1 1 = __ Ú (1 – 2 cos 2x) dx + __ Ú (1 + cos 4x) dx 4 8 2 sin 2x sin 4x 1 1 = __ Ú x – ______ + __ x + _____ + c 4 2 8 4 3 1 1 = __ x – __ sin 2x + ___ sin 4 x + c 8 4 32
(
) (
)
145. We have,
Ú cos6x dx = __ 18 Ú (2 cos2x)3dx
1 = __ Ú (1 + cos 2x)3dx 8 1 = __ Ú (1 + 3 cos 2x + 3 cos22x + cos32x) dx 8 3 1 = __ Ú (1 + 3 cos 2x) dx + ___ Ú (2 cos22x) dx 8 16 1 + ___ Ú (4 cos32x) dx 32
3 1 = __ Ú (1 + 3 cos 2x)dx + ___ Ú (1 + cos 4x) dx 8 16 1 + ___ Ú (cos 6x + 3 cos 2x) dx 32
(
(
)
)
3 3 sin 4x 1 = __ x + __ sin 2x + ___ x + _____ 8 2 4 16
(
)
3 1 sin 6x __ + ___ _____ + sin 2x + c 32 2 6
5 15 3 1 = ___ x + ___ sin 2x + ___ sin 4x + ____ sin 6x + c 192 16 64 64 148. We have,
dx
1/2 Ú ___________ sin x cos3/2x
Divide the numerator and the denominator by cos2x, we get
1.56 Integral Calculus, 3D Geometry & Vector Booster 2t dt sec2x ____ dx = Ú ____ , Let tan x = t2 Ú _____ t tan x ÷ fi sec2 xdx = dt
= 2 Ú dt
= 2t + c
=
t2 t4 = __ + __ + C 2 4
tan2x _____ tan4x = _____ + + C 2 4
152. The given integral is
____ 2÷tan x +
c
sec8x
dx
3 5 = Ú _____ 3 dx Ú _________ tan x sin x cos x
149. We have, dx 3/2 Ú ___________ sin x cos5/2x Divide the numerator and the denominator by cos4x, we get
Ú
(1 + tan2x) sec2x sec4x ______ 3/2 dx = Ú ______________ dx tan x tan3/2x (1 + t2)dt = Ú ________ Let tan x = t2 t3/2 fi sec2 xdx = dt
=
Ú (t– 3/2 + t1/2)dt
_ 2 = 2 ÷t + __ t3/2 + c 3 ____ 2 = 2 ÷tan x + __ (tan x)3/2 + c 3
150. The given integral is
____ ÷tan x ________
dx = Ú dx Ú sin x cos x sin x cos x ____ sec x ÷tan x __________ 2
= Ú
=
2t dt = Ú ____ , Let tan x = t2 t
dx tan x
sec2x
____ dx Ú _____ ÷tan x
fi sec2 xdx = 2t dt
= Ú 2 dt
= 2t + c
=
_____ 2÷ tan x +
c
sin x
5 dx = Ú sec4x tan x dx Ú _____ cos x
= Ú (1 + tan2x) tan x sec2x dx
= Ú (1 + t ) t dt Let tan x = t fi sec2 xdx = dt 2
= Ú (t + t3) dt
(
)
where t = tan x 153. The given integral is
sin2x dx
= Ú tan2x sec4x dx Ú _______ 6 cos x =
=
Ú tan2x (1 + tan2x) sec2x dx Ú t2(1 + t2) dt Let tan 2 x = t
fi sec x dx = dt
= Ú (t4 + t2) dt
t 5 t3 = __ + __ + c 3 5
tan5x ____ tan3 = _____ + + c 3 5
154. The given integral is sec4x
dx
__________ = Ú _____ dx Ú sin x cos x 3 tan x x
Ú
(1 + tan2x)sec2 x _____________ dx tan x
Ú
(1 + t2) ______ dt Let tan x = t t
=
=
=
t2 = __ + log |t| + c 2
tan2x = _____ + log |tan x| + c 2
Let t = tan x fi dt = sec2 xdx
(1 + 3t2 + 3t4 + t6) = Ú _________________ dt t3 1 3 = Ú __ 3 + __ + 3t + t3 dt t t 3t2 t4 1 = – ___2 + 3log |t| + ___ + __ + c 2 4 2t
151. The given integral is
(1 + t2)3 = Ú _______ dt, t3
____ ÷tan x ________
(1 + tan2x)3sec2 = Ú _____________ dx tan3x
Ú ( 1 + __ 1t ) dt
Indefinite Integrals
155. The given integral is
dx sec x 2 4 = Ú _____ 2 dx Ú _________ sin x cos x tan x (1 + tan2x)2sec2x = Ú ______________ dx tan2x
(1 + t2)2 = Ú _______ dt Let tan x = t t2 fi sec2 xdx = dt
(
)
158. We have, x 1 2 = Ú ______ = __ tan–1 ( __ ) + c Ú _____ 2 x +4 x2 + 22 2 dx
dx
159. We have, dx
dx
1 1 2 = __ Ú ________ 2 = __ × 3tan–1 (3x) + c Ú _______ 9 1 __ 9 x + 1 9 2
( )
x + 3
(t4 + 2t2 + 1) = Ú ___________ dt t2
(
tan2x = 2 _____ + log |tan x| + C 2
6
160. We have,
)
1 = Ú t2 + __ 2 + 2 dt t
t3 1 = __ – __ + 2t + c t 3
x–2 1 = ___ log _____ + c 2.2 x+2
x–2 1 = __ log _____ + c 4 x+2
tan3x = _____ + 2tan x – cot x + c 3 156. The given integral is
dx 1 = Ú __________ 7 __ __ sin2 x cos2 x
sec4x ____ dx Ú ÷_____ tan x 2
=
2
(1 + tan x)sec x
____ dx Ú _____________ ÷tan x
(1 + t2)2t = Ú ________ dt t
= 2 Ú (1 + t ) dt
t3 = 2 t + __ + c 3
Let tan x = t fi sec2 xdx = dt
dx
=
dx Ú ____________ x2 + 4
=
dx Ú ____________________ x2 + 4
5 = Ú x2 – 4 + ______ dx 2 x + 22
5 x3 x = __ – 4x + __ tan–1 __ + c 3 2 2
sec x
dx
(1 + tan2x) sec2x (1 + t )2t
dt, Ú ________ 3 t
(1 + t2) = 2 Ú ______ dt t
1 = 2 Ú t + __ dt t
t2 = 2 __ + log |t| + C 2
(
dx
÷x + 4 ÷ x +2
)
= log (x + ÷x 2 + 4 )+c
163. We have, dx
Let tan x = t 2
÷ ( ) 1 = __ log x + x + __ 2 | ÷ 14 | + c
÷4x + 1
x2 + __ 2 ______ 2
2
fi sec x dx = 2t dt
dx
1 ______ = __ Ú _________ ________ Ú ________ 2 2 1 2
dx Ú ______________ tan3/2x
( )
( )
÷sin x cos x
)
_____
dx
=
((x2 + 4) (x2 – 4)) + 15
_____ = Ú _______ ______ Ú _______ 2 2 2
4
2
(x4 – 16) + 15
(
162. We have,
_________ = Ú ______ 3/2 dx Ú __________ 3 5 tan x =
x4 – 1
dx Ú _____ x2 + 4
157. The given integral is
| | | |
161. We have
3
tan x = 2 tan x + _____ + c 3
dx
2 = Ú _______ 2 2 Ú _____ x –4 x – (2)
2
( ) ( )
1.57
164. We have,
(
)
(
)
x+4 x 4 Ú ______ dx = Ú ______ 3 + ______ dx x3 + 4x x + 4x x3 + 4x
(
)
1 4 = Ú _____ 2 + ________ dx x + 4 x (x2 + 4)
1.58 Integral Calculus, 3D Geometry & Vector Booster
169. The given integral is
dx 4x = Ú _____ 2 + Ú ________ 2 2 dx x +4 x (x + 4)
( )
x 1 = __ tan–1 __ + 2 2 2
dt
, Let t = x2 Ú _______ t (t + 4) fi dt = 2x2dx
(1 + x) dx
1+x
dx = Ú _________ 2 Ú ______ x3 + x x (x + 1) (1 + x2) + x – x2
=
dx Ú _______________ x (x2 + 1)
=
x + Ú _____ 2 – Ú _____ dx Ú ___ x +1 x2 + 1
x 1 2 1 ____ 1 = __ tan–1 __ + __ Ú __ – dt t t+4 2 2 4
1 = __ tan–1 2
) ( ) ( x 1 ( __ ) + __ log ____ 2 2 | t +t 4 | + c
1 = __ tan–1 2
x __ + 2
1 = log |x| + tan–1x – __ log |x2 + 1| + c 2 170. The given integral is
( )
| |
x2 1 __ log _____ 2 + c 2 x +4
165. The given integral is
4 +1 dx = Ú xx_____ 2 +1
(x4 – 1) + 2 ___________ dx Ú x2 + 1
( (
) )
2 = Ú x2 – 1 + _____ dx 2 x +1
x3 = __ – x + 2 tan–1x + c 3
=
dt
_____ , Ú ______ 2
Let t = (x – 2)
÷t + 1
fi dt = dx
_____ ÷t 2 + 1 |+
= log |t +
= log |(x – 2) + ÷ (x – 2)2 + 1 | + c
c
dx
(x – 1) (x + 1)
(
=
Ú
x + (9 + x2) – x2 _______________ dx x (x2 + 9)
dx = Ú _______ 2 + (x + 9)
1 = __ tan–1 3
dx
x dx
x – Ú _______ 2 Ú ___ (x + 9)
( )
x 1 __ + log |x| – __ log| x2 + 9| + c 3 2
1 1 1 = __ Ú _____ – _____ dx 2 x2 – 1 x2 + 1
x–1 1 __ 1 = __ log _____ – tan–1x + c 2 2 x+1
( |
1+x
dx
x
dx = Ú _____ 2 + Ú _____ dx Ú ______ 2 1 + x 1+x 1 + x2
1 = tan–1(x) + __ log |1 + x2| + c 2
|
)
171. The given integral is
dx
dx
3 = Ú ________ 2 Ú _____ x +x x(x + 1) x dx = Ú ________ 2 2 x (x + 1)
dt 1 = __ Ú _______ , Let x2 = t2 2 t (t + 1) 2x dx = 2t dt
(
)
1 1 ____ 1 = __ Ú __ – dt t t+1 2
t 1 = __ log ____ + c 2 t+1
x2 1 = __ log _____ 2 + c 2 x +1
| |
| |
172. We have,
dx
dx = Ú _______ 2 2 Ú __________ 2 (x + 2)
1 = – ______ + c (x + 2)
x + 4x + 4
173. We have,
168. The given integral is
)
167. The given integral is
x
dx 4 = Ú ______________ Ú x_____ 2 2 –1
__________
x+9 x+9 dx = Ú ________ dx Ú _______ 3 x + 9x x (x2 + 9)
dx
166. The given integral is dx dx __________ = Ú ____________ __________ Ú ____________ 2 ÷(2 – x) + 1 ÷(x – 2)2 + 1
dx
dx dx = Ú ________________ Ú ___________ 2 2 x + 6x + 10
(x + 3) + (10 – 9)
dx = Ú __________ (x + 3)2 + 1
= tan–1(x + 3) + c
1.59
Indefinite Integrals
174. We have,
178. The given integral is
1 dx dx = __ Ú ___________ Ú ___________ 2 2 2 __ 5 2x + 5x + 6
x + x + 3 2 1 dx = __ Ú _________________ 2 2 5 25 x + __ + 3 – ___ 4 16
( ) (
)
dx 1 ___ = __ Ú _______________ 2 ÷23 2 5 2 __ ____ x + + 4 4
( ) ( ) 5 ( x + __ ) 4 1 ____ 4 __ _______ = × tan + c
23 ÷ ____ 4
)
( ) ( ) 1 ( x + __ ) 2 2 = ___ tan _______ + c –1
__
÷3
( ) __
(
3 ÷ ___ 2
)
dx
dx
( ) ( ) ÷5 1 ( x – __ ) – ___ 2 2 1 = – ___ log ___________ + c ÷5 ÷5 1 __ ( x – ) + ___ 2 2
|
| |
dx Ú (x__________ + 2)2 – 1
| | | |
|
(x – a) – a 1 = ___ log __________ + c 2a (x – a) + a
(x – 2a) 1 = ___ log _______ + c x 2a
|
|
180. The given integral is
dx
dx
2 = Ú __________ Ú x_______ + 2ax (x + a)2 – a2
|
|
x 1 = ___ log ______ + c 2a x + 2a
dx
1 2 = ___ log |a2 + 2ax| + c Ú a_______ 2a + 2ax
182. The given integral is
dx
dx = – Ú _______ Ú ________ 2 x2 – 2ax 2ax – x
dx = – Ú ___________ (x – a)2 – a2
x – a – a 1 = – ___ log ________ + c x–a+a 2a
x 1 = ___ log ______ + c 2a x – 2a
__
dx 2 = Ú _________ x + 4x + 3
(x – a) – a
dx __ = – Ú ______________ ÷5 2 1 2 __ ___ x – – 2 2
177. The given integral is
dx
dx 2 = Ú ____________ Ú x_______ 2 2 2 – 2ax
181. The given integral is
= – Ú ________ Ú 1________ + x – x2 x2 – x – 1
__
)
179. The given integral is
2x +__ 1 2 = ___ __ tan–1 ______ + c 3 3 ÷ ÷
)
(4x + 5) 2 ___ = ____ ___ tan–1 _______ + c 23 ÷23 ÷
(
___
__
(
( )
–1
___ ÷32
176. The given integral is
( ) ( )
7 2 = _____ ____ tan–1 8x + _____ ____ + c 111 111 ÷ ÷
dx __ dx = Ú ______________ Ú _________ 2 2 ÷3 2 1 x +x+1 x + __ + ___ 2 2
dx 1 ____ = __ Ú ________________ 4 2 111 2 ÷ 7 __ _____ x + + 8 8
(
)
x + x + 4 2
8 7 1 = __ × _____ ____ tan–1 8x + _____ ____ + c 4 ÷ 111 111 ÷
175. The given integral is
(
4x + 7x + 10
2
dx
1 dx = __ Ú ____________ Ú ____________ 2 4 5 7 2 __ __
|
183. The given integral is
(x + 2) – 1 1 = __ log __________ + c 2 (x + 2) + 1
(x + 1) 1 = __ log ______ + c 2 (x + 3)
|
|
|
(
)
1 1 1 2 = ___ Ú _____ 2 – _____ 2 dx Ú (x_____________ 2 3a + 1) (x + 4) x +1 x +4 dx
(
1 1 = __ tan–1x – __ tan–1 3 2
( ) )
x __ + c 2
1.60 Integral Calculus, 3D Geometry & Vector Booster 184. We have, x2 1 3x2 6 dx = __ _____ dx Ú _____ 3 x6 + 1 x +x 1 dt = __ _____ , Let t = x3 3 t2 + 1 fi dt = 3x3 dx
1 = __ tan–1 (t) + c 3
1 = __ tan–1(x3) + c 3
cos x dx _______________ 2 Ú sin x + 3 sin x + 2
dt = Ú _________ 2 t + 3t + 2
Let t = sin x fi dt = cos x dx
)
1 1 ____ 1 = __ Ú __ – t t+1 4
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
| |
| |
194. We have,
185. We have,
(
(2x + 4) – 1
2x + 3 dx = Ú ___________ dx Ú __________ 2 2
(2x + 4) = Ú __________ dx – 2 x + 4x + 5
(2x + 4) dx = Ú __________ dx – Ú __________ 2 (x + 2)2 + 1 x + 4x + 5
= log |x2 + 4x + 5| – tan–1(x + 2) + c
x + 4x + 5
x + 4x + 5
dx
2 Ú __________ x + 4x + 5
dt = Ú ___________ (t + 1) (t + 2)
1 1 = Ú ____ – ____ dt t+1 t+2
t+1 = log ____ + c t+2
sin x + 1 = log _______ + c sin x + 2
9 3 (2x – 3) + 2 + __ __ 2 2 = Ú __________________ dx 2 x – 3x + 4
(2x – 3) 3 13 dx = __ Ú __________ dx + ___ Ú __________ 2 x2 – 3x + 4 2 x2 – 3x + 4
(
195. We have,
)
| | | |
186. We have,
xx(1 + log x)
dx = Ú ________ dt Ú ___________ 2x x 2 x
+x +1
t + t + 1
dt __ = Ú ______________ 3 2 ÷ 1 2 __ ___ t + + 2 2
( ) ( ) 1 ( t + __ ) 2 2 = ___ tan ______ + c
__
–1
÷3
( ) (
)
2x x __ +1 2 = ___ __ tan–1 ______ + c 3 3 ÷ ÷
187. We have,
dx
x3dx
________ 4 = Ú ________ 4 4 Ú x (x + 1) x (x + 1) 3
4x dx 1 = __ Ú ________ 4 x4(x4 + 1)
dt 1 = __ Ú _______ , Let t = x4 4 t (t + 1) fi dt = 4 x3 dx
x – 3x + 4
(
)
(2x – 3) dx 3 13 __ = __ Ú __________ dx + ___ Ú ______________ 2 2 x – 3x + 4 2 ÷7 2 3 2 – __ + ___ 2 2
( ) ( ) 3 ( x – __ ) 2 13 ______ ___ – 3x + 4| + tan + c
( )
3 = __ log |x2 2
3 13 2x –__ 3 = __ log |x2 – 3x + 4| + ___ __ tan–1 ______ + c 2 7 ÷7 ÷
__
3 ÷ ___ 2
3x + 2 dx Ú __________ 2
–1
__
÷7
__
7 ÷ ___ 2
(
)
195. The given integral is 1 (2x + 1) – 1 x Ú _________ dx = __ Ú ___________ dx 2 2 (x2 + x + 1) x + x + 1 (2x + 1) dx 1 1 __ = __ Ú __________ dx – __ Ú ______________ 2 (x2 + x + 1) 2 ÷3 2 1 2 __ ___ x + + 2 2
( ) ( )
(
)
2x +__ 1 1 1 = __ log |x2 + x + 1| – ___ __ tan–1 ______ + C 2 3 3 ÷ ÷
Indefinite Integrals
196. The given integral is
200. The given integral is.
1 2x + __ 4x + 1 2 _________ ___________ dx = 2 Ú dx Ú x2 + 3x +2 (x2 + 3x + 2)
(2x + 3) – (5/2) = 2 Ú ______________ dx (x2 + 3x + 2)
(2x + 3) dx = 2 Ú ___________ dx – 5 Ú ___________ 2 (x + 1) (x+ 2) (x + 3x + 2)
| |
____________ 2x = Ú ______________ x Ú 2e (2e + 1) (ex + 1) + 3ex + 1
x+1 = 2 log |x2 + 3x + 2| – 5 log _____ + c x+ 2
dx
dx
( Ú (
) )
× (2 + 2 sin2 x)dx Ú ( __________ cos x + sin x ) cos x – sin x
( ) 1+t = Ú ( _____ dt, Let cos x + sin x = t t )
cos x – sin x = Ú __________ (1 + (sin x + cos x)2) dx cos x + sin x 2
197. The given integral is
fi (– sin x + cos x) dx = dt
( )
1 = Ú t + __ dt t t2 = __ + log |t| + c 2 (sinx + cos x)2 = ____________ + log |sin x + cos x| + c 2
201. The given integral is (sin x + cos x) dx
2 1 = Ú ________ x – _______ dx (2e + 1) (ex + 1)
2e–x e– x = _______ – ________ – x dx –x (2 + e ) (e + 1)
(sin x + cos x)dx = Ú ______________ 8 – 3 (1 – sin 2x)
= log |e– x + 1| – 2 log |e– x + 2| + c
(sin x + cos x)dx = Ú _________________ 8 – 3 (sin x – cos x)2
198. The given integral is
(3sin x – 2) cos x = Ú _______________ dx sin2x – 4sin x + 4
(3t – 2) = Ú _________ dt, 2 t – 4t + 4
sin x + cos x dx = Ú __________________ Ú ___________ 5 + 3 sin 2x 5 + 3 (1 – 1 + sin 2x)
dt = Ú ______ 2 , Let t = (sin x – cos x) 8 – 3t dt = (cos x + sin x)dx
(3sin x – 2) cos x dx Ú _________________ (5 – cos2x – 4sin x)
Let t = sin x fi dt = cos x dx
dt = Ú ______ 2 8 – 3t dt 1 __ = – __ Ú _________ 3 2 8 2 __ t – 3
( ÷ )
|
(3t – 2) = Ú _______ dt (t – 2)2
(3t – 6 + 4) = Ú __________ dt (t – 2)2
dt dt = 3 Ú ______ + 4 Ú ______ (t – 2) (t – 2)2
4 = 3 log |t – 2| – ______ + c (t – 2) 4 = 3 log |sin x – 2| – ________ + c (sin x – 2)
sin x – cos x = Ú ________________ dx 5 (sin x + cos x)2 – 2
cos x – sin x = Ú ________________ dx 2 – 5 (sin x + cos x)2
=
ax3 + bx
x3
x dx
dx = a Ú ______ 4 2 dx + b Ú ______ 4 2 Ú _______ x4 + c2 x +c x +c
( )
a b x2 = __ log |x4 + c2| + ___ tan–1 __ c + c 4 2c
___
|
t – (÷8/3 ) 1 ___ = – _____ ___ log ________ + c 2÷24 t + (÷8/3 )
where t = sin x + cos x 202. The given integral is
199. The given integral is
1.61
sin x – cos x
sin x – cos x
dx = Ú _______________ dx Ú ___________ 3 + 5 sin 2x 5 (1 + sin 2x) – 2
dt
2 , Let t = (sin x + cos x) Ú 2______ – 5t
fi dt = (cos x – sin x)dx
1.62 Integral Calculus, 3D Geometry & Vector Booster dt
2 Ú ______ 2 – 5t
=
dt 1 ___ = – __ Ú __________ 2 5 t2 – (÷2/5 )
t – (÷2/5 ) 1 ___ = – ____ ___ log _________ + c 10 t + (÷2/5 ) ÷
=–
| |
Ú
|
| (
)
1 = log x + __ + 2 dx
dx = Ú _________________ ________________ 25 1 2 ___ – x + __ – 2 4
÷ { ( )
÷( ) ( ) 1 ( x + __ ) 2 _______ = sin + c –1
(
|
= log (x – a) +
dx
dx
dx
dx
______ = Ú _______________ ______________ Ú _______ 2 2 ÷4x – x
c
÷( ) ( )
÷1 + x + x
3 ÷ 1 ___ x + __ + 2 2
| (
dx = Ú ____________ ___________ 2 ÷2 – (x – 2)2
x – 2 = sin–1 _____ + c 2
( )
dx = Ú ________________ _______________ 5 1 2 __ – x – __ – 2 4
÷ { ( ) }
dx = Ú ______________ _____________ – 2)2 – 4} ÷ – {(x
=
dx ____________ __________ ÷4 – (x – 2)2
x – 2 = sin _____ + c 2
Ú
206. The given integral is
–1
( )
dx
_______________ __ Ú ________________ 2 2
÷( ) ( )
=
1 x – __ 2 __ = sin–1 ______ + c ÷5 ___ 2
2x –__ 1 = sin–1 ______ + c ÷5
÷5 1 __ ___ – x – 2 2
(( )) (
205. The given integral is dx dx ______ = Ú __________ _________ Ú _______ 2 2 – x ÷– (x – 4x) ÷ 4x
|
208. The given integral is dx dx _________ = Ú ______________ ____________ Ú __________ 2 2 + x – x ÷– {x – x – 1} ÷ 1
÷4 – 4 + 4x – x
dx = Ú _________________ ________________ 2 2 ÷2 – (x – 4x + 4)
)
_________
1 = log x + __ + ÷x 2 + x + 1 + c 2
|
)
_________ = Ú ________________ _______________ __ Ú ___________ 2 2 2
÷x – 2ax ÷x – 2ax + a – a
_______ + ÷ x 2 – 2ax
5 __ 2
207. The given integral is
dx
dx = Ú _____________ ___________ ÷(x – a)2 – a2
( )
2x + 1 = sin–1 ______ + c 5
|
}
dx = Ú _______________ ______________ 2 5 1 2 __ __ – x + 2 2
_________ ÷x 2 + x + 1 + c
205. We have,
÷6 – x – x ÷– (x + x – 6)
c
_______ = Ú _________________ ________________ Ú ________ 2 2 2 2
dx
|
÷( ) ( )
204. We have,
___
dx dx __________ ________ = Ú ________________ _______________ __ 2 2 3 2 ÷ ÷ x + x + 1 1 x + __ + ___ 2 2
dx
_________ = Ú _____________ ___________ Ú __________ 2 2
___ (sin x + cos x) – (÷2/5 ) 1 ____ ___________________ ___ + ___ log ÷10 (sin x + cos x) + (÷2/5 )
203. We have,
)
209. The given integral is
dx
dx
2 = Ú ____________ __________ Ú x_______ 2 2 + 2ax ÷(x + a) – a
________
|
210. The given integral is
|
= log (x + a) + ÷ x 2 + 2ax + c
dx
dx
________ = Ú ____________ ___________ Ú _________ 2 2 ÷2ax – x ÷ – {x – 2ax}
Indefinite Integrals
dx = Ú _______________ ______________ – a)2 – a2} ÷ – {(x
dx = Ú ____________ ___________ 2 a)2 ÷ a – (x –
x–a = sin–1 _____ + c a
(
dt 1 _______ = __ Ú ________ , 2 2 ÷ t + t +1
dt fi x dx = __ 2
(sin xcos a – cos x sin a) ___________ = Ú ___________________ dx 2 2x – sin a ÷ sin
sin x = cos a Ú ____________ ___________ dx 2 2 x – sin a ÷ sin
|
1 1 = __ log t + __ + ÷ t 2 + t + 1 + c 2 2
1 1 = __ log x2 + __ + ÷x 4 + x2 + 1 + c 2 2
_________
|
_______
– 1 dx Ú ÷sec x
cosx – sin a Ú ____________ ___________ dx 2 2 x – sin a ÷ sin
212. We have,
÷
____________________
sin (x – a) ___________ = Ú ____________ dx 2 2x – sin a ÷ sin
)
sin x = cos a Ú ____________ ___________ dx 2 2 a – cos x ÷ cos
– sin a
÷ Ú ÷
___________________
(1 + cos x) 1 – cos x __________ = Ú _______ × dx cos x (1 + cos x) ______________
1 – cos2x = _____________ dx cos x (1 + cos x)
sin x = Ú ______________ _____________ dx + cos x) ÷ cos x (1
dt = – Ú ________ _______ Let cos x = t + 1) ÷ t (t fi – sin x dx = dt
_________
sin (x – a) sin (x – a) _________ dx = Ú _________ dx Ú sin (x + a) sin (x + a)
_________
)
÷
_________
Let x2 = t
÷( ) ( )
| ( | (
÷
c
sin (x – a) _________ sin (x – a) = Ú _________ × dx sin (x + a) sin (x – a)
dt 1 _______________ = __ Ú ________________ __ 2 ÷3 2 1 2 __ ___ t + + 2 2
|
x dx _________ Ú __________ 4 ÷ x + x2 + 1
|
214. We have,
211. We have,
______ ÷x 1/2 – 1 +
= 4 × log x1/4 +
|
= 4 log t + ÷t 2 – 1 + c
)
_____
|
dt = – Ú _______________ _____________ 2 1 1 2 __ t + __ – 2 2
÷( ) ( ) = – log ( t + __ | 12 ) + ÷t + t | + c = – log ( cos x + __ | 12 ) + ÷ cos x + cos x | + c _____ 2
___________ 2
213. We have, dx dt ______ = 4 Ú ______ _____ , Let x1/4 = t Ú __________ 3/4 1/2 2 x ÷ x – 1 ÷ t – 1 1 fi __ x–3/4 = dt 4
1.63
cos x
___________ dx Ú ____________ 2 2 ÷sin x – sin a
( ) ___________ – sin a log | sin x + ÷sin x – sin 2a | + c
cos x = – cos a sin–1 _____ cos a
2
215. The given integral is
exdx
ex
______ dx = Ú _________ ________ Ú _______ 2x x 2
÷4 – (e ) dz = Ú ______ _____ , Let z = ex 2 ÷ 4 – z fi dx = ex dt
÷4 – e
( )
z = sin–1 __ + c 2 ex = sin–1 __ + c 2
( )
216. The given integral is sec2x ________ dx Ú ÷__________ 16 + tan x dt = Ú ______ _____ , 2 ÷ t + 4
Let t = tan x fi dt = sec2 x dt
_____
| | ________ = log | tan x + ÷ tan x + 4 | + c = log t + ÷ t 2 + 4 + c 2
1.64 Integral Calculus, 3D Geometry & Vector Booster 217. The given integral is
_______ ÷sec x – 1 dx
Ú
________
÷
1 – cos x = Ú _______ dx cos x
÷
__________________
(1 – cos x) (1 + cos x) = Ú __________________ dx cos x (1 + cos x)
sin x = Ú ______________ _____________ dx + cos x) ÷ cos x (1
dt = – Ú ______ _____ , Let t = cos x 2 ÷ t + t fi dx = – sin x dx dt = – Ú _______________ _____________ 2 1 1 2 __ __ t + – 2 2
÷( ) ( ) = – log ( t + __ | 12 ) + ÷t + t | + c = – log ( cos x + __ | 12 ) + ÷ cos x + cos x | + c _____ 2
___________ 2
218. The given integral is
÷
________
_________
1 – sin x – 1 dx = Ú _______ dx Ú ÷cosecx sin x
÷
__________________
220. The given integral is
÷
_________
sin (x – a) Ú _________ dx sin (x + a)
÷
__________________
sin (x – a) sin (x – a) = Ú __________________ dx sin (x – a) sin (x + a) sin(x – a) ___________ = Ú ____________ dx 2 2x – sin a ÷ sin =
sin x cosa – cos x sina
___________ dx Ú ___________________ 2 2
÷sin x – sin a
sin x dx cos x dx = cosa Ú __________________ _________________ Ú _____________ ____________ 2 2 2 a – sin x – sina ÷sin 2x – sin a ÷ sin sin x dx cos x dx = cosa Ú _____________ ____________ – sina Ú _____________ ____________ 2 2 2 a – cos x ÷sin 2x – sin a ÷ cos
(
)
cos x = – cos a sin–1 _____ cosa ___________
|
(1 – sin x) (1 + sin x) = _________________ dx sin x (1 + sin x)
221. The given integral is
cos x = Ú ______________ ____________ dx + sin x) ÷ sin x (1
dt = Ú _______ _____ , 2 ÷ t + t
Let t = sin x fi dt = cos x dt
dt = Ú ______________ _____________ 1 1 2 t + __ – __ 2 2
÷( ) ( ) = log ( t + __ | 12 ) + ÷ t + t | + c = log ( sin x + __ | 12 ) + ÷ sin x + sin x | + c _____ 2
__________ 2
219. The given integral is
e– x
dx
_______ = Ú ___________ __________ dx Ú ________ 2x 2 ÷1 – e ÷(e – x) – 1
dt = – Ú ______ _____ , 2 ÷ t – 1
Let t = e– x
| = – log | e
fi dt = – e –x dt
_____
|
= – log t + ÷ t 2 – 1 + c – x
_______
|
+ ÷e – 2x – 1 + c
|
2 – sin a log sin x + ÷sin 2x – sin a + c
dx
dx
______ ________ = Ú ____________ Ú __________ 2/3 2/3 2/3 1/3 2
x ÷x
– 4 x ÷(x ) – 4
dt 1 _____ = __ Ú ______ , 3 ÷t 2 – 4
dt 1 _____ = __ Ú ______ 3 ÷t 2 – 4
Let t = x1/3 1 fi dt = __ x–2/3 dx 3
_____
| |
|
1 = __ log t + ÷ t 2 – 4 + c 3 ______ 1 = __ log x1/3 + ÷x 2/3 – 4 + c 3 222. The given integral is ______ x 3 3 dx Ú ______ a –x Let x = a sin2/3q
|
÷
fi dx = a ◊ sin–1/3 q dq
__
1/3 –1/3 ÷a sin q sin q cosq = Ú ___________________ 3/2 a cosq 1 = __ a Ú dq
q = __ a + c
x 3/2 sin–1 __ a = ____________ + c a
{ ( ) }
1.65
Indefinite Integrals
223. The given integral is
(
cosq + sinq
)
_________ dq Ú ___________ ÷ 5 + sin 2q cosq + sinq _____________ = Ú _______________ dq ÷6 – (1 – sin 2q)
cosq + sinq ________________ = Ú _________________ dq 2 ÷ 6 – (sinq – cosq)
Put (sinq – cosq) = t fi (cosq + sinq)dq = dt
dt _____ = Ú _______ – t2 ÷ 6 t = sin–1 ___ __ + c ÷6
( )
(
)
sinq – __cos q = sin–1 __________ + c ÷6
Let
(x2 – 3x + 2) = t2
fi
(2x + 3) dx = 2t dt
| (
_________
sinq – cosq ______________ = Ú _______________ dq ÷ 3 – (1 + sin 2q)
sinq – cosq ________________ = Ú _________________ dq 2 ÷3 – (sinq + cosq)
dt = – Ú ______ _____ , Let t = (sinq + cosq) 3 – t2 ÷ dt = (cosq – sinq) dq
( )
t = cos–1 ___ __ + c 3 ÷
sinq + cosq __ = cos–1 __________ + c 3 ÷
(
3x + 4
÷x + 5x + 2
(
)
3 15 __ (2x + 5) + 4 – ___ 2 2 _________ = Ú ___________________ dx 2 + 2 ÷ x + 5x
(2x + 5)dx 3 dx 7 __________ __________ = __ Ú ___________ – __ Ú ___________ 2 2 ÷x 2 + 5x 2 + 2 + 2 ÷ x + 5x dx 3 2t dt __ 7 = __ Ú ____ – Ú ___________________ __________________ t 2 2 2 5 25 __ x + + 2 – ___ 2 4
÷( ) (
)
Let (x2 + 5x + 2) = t2
fi (2x + 5) dx = 2t dt dx 7 ________________ = 3 Ú dt + __ Ú _________________ ___ 2 ÷17 2 5 2 __ ____ x + – 2 2
÷( ) ( )
| ( | (
_________
)
|
5 7 = 3t + __ log x + __ + ÷ x 2 + 5x + 2 + c 2 2
)
)
_________
|
x–1 __________ dx Ú ___________ 2 + 2 ÷ x – 3x
227. We have,
÷x + 5x + 6
÷x + 5x + 6
(2x + 5) 1 1 1 __________ __________ = __ Ú ___________ dx – __ Ú ___________ dx 2 2 ÷x 2 + 5x 2 + 6 ÷x + 5x + 6
– 3) + 1 1 (2x __________ = __ Ú ___________ dx 2 ÷x 2 – 3x + 2
(2x – 3) dx __________ 1 1 __________ __________ = __ Ú ___________ dx + __ Ú ___________ = x 2 + 5x + 6 ÷ 2 2 ÷x 2 – 3x 2 + 2 ÷x – 3x + 2
(2x + 5) 1 1 1 __________ = __ Ú ___________ dx – __ Ú _______________ ______________ dx 2 ÷x 2 + 5x 2 5 2 1 2 + 6 __ x + __ – 2 2
dx 1 2t dt 1 ________________ = __ Ú ____ dx + __ Ú _________________ t 2 2 9 3 2 x – __ + 2 – __ 2 2
÷( ) ( )
(2x + 5) – 1
x+2
1 __________ __________ dx = __ Ú ___________ dx Ú ___________ 2 2 2
1 2 x – 2 __________ = __ Ú ___________ dx 2 ÷x 2 – 3x + 2
)
5 7 = 3 ÷x 2 + 5x + 2 + __ log x + __ + ÷ x 2 + 5x + 2 + c 2 2
225. We have,
|
_________
__________ dx Ú ___________ 2
_________
| (
c
226. We have,
1.65
|
__________ + 2 + ÷ x 2 – 3x
1 1 = ÷x 2 – 3x + 2 + __ log x – __ + ÷x 2 – 3x + 2 + c 2 2
sinq – cosq
________ dq Ú ___________ ÷2 – sin 2q
)
1 1 = t + __ log x – __ + 2 2
224. The given integral is
÷( ) ( )
| (
)
_________
|
5 1 – __ log x + __ + ÷ x 2 + 6x + 6 + c 2 2
1.66 Integral Calculus, 3D Geometry & Vector Booster 231. We have,
6x – 5 __________ 228. Ú ___________ dx 2 – 5x + 1 ÷ 3x
Let 3x 2 – 5x + 1
fi (6x – 5) dx = 2t dt
2t dt = Ú ____ t
= 2 Ú dt
= 2t + c
= 2÷3x 2 – 5x + 1 + c
_____
____________
÷
÷
a______ – x = Ú _______ dx 2 ÷ a – x2
÷ 4 1–t = __ Ú ____ dt 3 ÷1 + t
–1
= a sin
÷
______
( )
x __ a + ÷ a 2 – x2 + c
______
a2 – x2 230. Ú x ______ dx a2 + x2
4 – 4t 4 = __ Ú ______ dt 3 4 + 4t
Let x2 = a2t 2
fi
a2(1 – t) a2 fi __ Ú ________ dt 2 a2(1 + t)
2x dx = a dt
÷ Ú ÷
– t) 4 (1_____ = __ Ú ______ dt 3 ÷ 1 – t2
dt t 4 = __ Ú _______ _____ – Ú ______ _____ dt 3 ÷1 – t2 ÷1 – t2
4 = __ ( sin–1(t) + ÷ 1 – t2 ) + c 3
x3 x3 2 4 = __ sin–1 __ + 1 – __ + c 3 4 4
(
______
(1 – t) a2 = __ Ú ______ dt 2 (1 + t)
(1 – t) 1____ a2 –t = __ ______ × dt 2 (1 + t) 1 – t
÷
( ( ) ÷ ( ) )
232. We have, sec2x dx
dx
= Ú ______________ Ú _________ 3 + 4sin2x 3 sec2x + 4 tan2x
=
________
_____________
__
÷
– t) a2 (1_____ = __ Ú ______ dt 2 ÷ 1 – t2
__
7 1 ___ = ____ × tan–1 tan x __ + c 3 21 ÷
233. We have,
(
)
dt t a2 = __ Ú ______ _____ – Ú ______ _____ dt 2 2 ÷1 – t ÷1 – t2
_______
÷ ( )
)
a2 x2 x 2 = __ sin–1 __ 2 + 1 – __ a + c 2 a
sec2x dx
dx
2 = Ú _________ Ú _____________ 3sin x + 4cos2x 3tan2x + 4 =
a2 = __ ( sin–1(t) + ÷ 1 – t2 ) + c 2
( ( )
( ÷ ) ( ÷ )
7 7 1 ÷ = __ × ___ tan–1 t __ + c 7 3 3 __
_____
( ÷ )
______
sec2x dx
Ú __________ 3 + 7 tan2x
sec2x dx 1 = __ Ú ________ 7 3 tan2x + __ 7 dt 1 __ = __ Ú _________ 7 2 3 2 t + __ 2
(1 – t)2 a2 = __ Ú ______ dt 2 (1 – t2)
)
________
4 x2dx = __ dt 3
_____
dx x = a Ú _______ ______ – Ú ________ _______ dx 2 2 2 ÷a – x2 ÷ a – x
fi
–x a – x a_____ –x dx = Ú _____ × dx Ú aa_____ +x a+x a–x
Let x3 = 4t
_____
229. We have,
3
–x dx Ú x 44_____ + x3
______
__________
÷
______
2
dt
2 Ú ______ 3t + 4
Let tan x = t fi sec2 x dx = dt
dt 1 = __ Ú _________ 2 3 2 2 t + ___ __ 3 ÷
( )
Indefinite Integrals
1 = __ × 3
__
( ) ( ) __
3 3 ÷ ÷ ___ tan–1 ___ t + c 2 2
__
3 ÷ 1 = ___ __ tan–1 ___ tanx + c 2 3 ÷
234. We have,
2
dx sec x dx 2 = Ú __________ Ú _____________ (2sinx + cosx) (2tan x + 3)2
dt = Ú _______ (2t + 3)2
1 = – ________ + c 2(2t + 3)
1 = – __________ + c 2(2tanx + 3)
sec2x dx 1 = __ Ú _____________________ 4 1 2 1 2 __ ___ tan x + + 1 – 4 16
[ (
) (
1 = __ tan–1 (tan2x) + c 2 2
dx sec x = Ú _________ dx Ú _____________ (sin x + 2 cos x)2 (tan x + 2)2 =
dt Ú ______ (t + 2)2
sec2x dx 1 = __ Ú _________________ 4 15 2 tan x + 1 2 ___ _______ + 4 16
[ (
) ] ÷15 1 Let ( tan x + __ ) = ____ tanq 4 4
15 ___ dq 16 2 1 ÷ = __ × ____ × Ú _____ 4 4 15 sec2q
1 = – _________ + c (tan x + 2)
( )
8 = ______ 3/2 Ú (2cos2q)dq (15) 8 = ______ 3/2 Ú (1 + cos2q)dq (15) sin 2q 8 = ______ 3/2 q + _____ + c 2 (15)
(
)
8 = ______ 3/2 (q + sinq + cosq) + c, (15)
where
4 1 q = tan–1 ____ ___ tan x + __ 4 ÷15
[ (
)]
237. We have, dx 1 __ Ú _____________ 2 4 sin x ____ + sec x 2
)
2
sin x dx 2 sin x cosx = Ú ___________ dx Ú ___________ sin4x + cos4x sin4x + cos4x
Divide the numerator and the denominator by cos4x we get
(
( )
fi sec2 x dx = dt
1 = – ______ + c (t + 2)
dx = Ú _____________ (sinx + 2 sec x)2
÷15 2 fi sec x dx = ____ sec q dq 4 2
238. The given integral is
Let tan x = t
)]
÷15 2 ____ sec q 1 4 = __ Ú _________ sec4q 4 15 2 ___ 16
236. We have,
)
___
dt 1 Let tan2x = t = __ Ú _____ 2 t2 + 1 fi 2 tan x sec2 x dx = dt 1 = __ tan–1 (t) + c 2
(
___
sec2x dx 1 = __ Ú ________________ 2 tan x 4 tan2x + ____ + 1 2
1 2tan x sec2x dx = __ Ú __________ 2 tan4x + 1
)
___
tan x sec2x dx = Ú _________ tan4x + 1
(
sin x cos x dx Ú ___________ sin4x + cos4x
sec2x dx 1 = __ Ú ___________ 2 tan x 4 ________ 2 + sec2x
___
235. We have,
1.67
2
sec x __________ dx Ú 2 tan x tan4x + 1 Let tan2x = t fi 2 tan x sec2x dx = dt
dt = Ú __2 + 1 t = tan–1 (t) + c
= tan–1 (tan2x) + c
1.68 Integral Calculus, 3D Geometry & Vector Booster 239. The given integral is
dx Ú ______________ (2sinx + 3cosx)2
Divide the numerator and the denominator by cos2x, we get sec2x dx Ú ___________ (2 tan x + 3)2 Let 2 tan x + 3 = t fi 2 sec2 x dx = dt 1 dt = __ Ú __2 2 t 1 = – __ + c 2t 1 = – __ (2 tan x + 3) + c 2
2
2
dx sec x dx sec x dx 2 = Ú _________ = Ú _________ Ú ________ 2 2 + cos x 1 + 2 sec x 2tan2x + 3
Let tan x = t fi sec2 x dx = dt
dt = Ú ______ 2 2t + 3
÷ 2 = __ ÷ 3 tan __
( ÷ ) ( ÷ ) ( ÷ )
__
__
2 __ tan x + c 3
–1
sinx sin x dx = Ú _____________ dx Ú _____ sin 3x 3 sin x – 4 sin3x dx = Ú _________ 3 – 4sin2x
sec2x dx = Ú _____________ 3 sec2x – 4 tan2x
sec2x dx = Ú ________ 3 – tan2x
dt = – Ú _____ t2 – 3
Let tan x = t sec2 x dx = dt
| | __
sinx = Ú _____________ dx 3 sin x – 4 sin3x
dx = Ú _________ 3 – 4 sin2x
sec2x dx = Ú _____________ 3 sec2x – 4 tan2x
sec2x dx = Ú ________________ 3 + 3 tan2x – 4 tan2x
sec2x dx = Ú ________ 3 – tan2x Let tan x = t fi sec2 x dx = dt
dt = – Ú _____ 2 t –3
t–÷ 3 – 1 = ____ __ log ________ + c 2÷3 t+÷ 3
t+÷ 3 1 = ____ __ log ________ 2÷3 t – ÷3
| | | | | | __
__
241. We have,
3
cosec x sinx dx = Ú _____ dx Ú _______ cosec x sin3x
tan x + ÷ 3 1 = ____ __ log ___________ + c 2÷3 tan x – ÷3 243. The given integral is
__
2 2 = __ tan–1 __ t + c 3 3
|
__
dt __ = Ú _________ 3 2 2 t + __ 2
240. The given integral is
|
__
tan x – ÷3 1__ = – ____ log ___________ + c 2÷3 tan x + ÷ 3 242. The given integral is
t – ÷3 1__ = – ____ log ________ + c 2÷3 t+÷ 3
sec3x cosx dx = Ú _____ dx Ú _____ secx cos3x
cos x = Ú _____________ dx 4 cos3x – 3 cos x dx = Ú _________ 4 cos2x – 3 Divide the numerator and the denominator by cos2x we get, sec2x dx = Ú _________ 4 – 3 sec2x
sec2x dx = Ú _____________ 4 – 3 – 3 tan2x
sec2x dx = Ú __________ 1 – 3 tan2x
Let tan x = t fi sec2 x dx = dt dt = Ú ______ 1 – 3t 2
Indefinite Integrals
| | | |
dt 1 = __ Ú __________ __ 3 t2 – ( 1/÷3 )2
1 t – ___ __ 3 ÷ 1 = ____ __ log ______ + c 1 2÷3 t + ___ __ 3 ÷ __
244. We have,
( ) ( )
3 – dt 1 1 = __ Ú ________ fi __ sec2 __ dx = dt 2 t2 – 2t – 2 2 2
dt 1 = – __ Ú _____________ 2 2 (t – 1)2 – ( ÷__ 3 )
(t – 1) – ÷ 3 1__ = – ____ log ___________ __ + c 2÷3 (t – 1) + ÷ 3
__ x tan __ – 1 – ÷3 2 1__ = – ____ log ________________ __ + c x 2÷3 tan __ – 1 + ÷ 3 2
)
( )
( ) ( )
2dt = Ú _________ 2 t + 4t + 1 dt = 2 Ú _____________ __ (t + 2)2 – ( ÷ 3 )2
|
__
|
(t + 2) – ÷ 3 2 = ____ __ log __________ __ + c 2÷3 (t + 2) + ÷ 3
|
( (
( ) ) ( ) )
__
245. We have,
dx Ú _____________ sin x + cos x + 1
(
) (
dx = Ú _____________________________ x x __ 2 tan 1 – tan2 __ 2 2 __________ + __________ x x + 1 1 + tan2 __ 1 + tan2 __ 2 2
( ) ( ) ( ) ( ) x sec ( __ ) dx 2 _____________________ = Ú x x ( 2 tan ( __ ) – tan ( __ ) + 2 ) 2 2 dt x 1 = __ Ú _________ Let ( __ ) = t 2 2t – t + 2 2
)
2
2
( (
|
( ) ) ( ) )
dx Ú _________ 1 + 2 sin x dx = Ú _________________ 2tan(x/2) 1 + 2 ___________ 1 + tan2 (x/2)
(
)
sec2 (x/2)dx = Ú ____________________ tan2 (x/2) + 4tan(x/2) + 1
Let tan(x/2) = t
x 1 fi __ sec2 __ dx = dt 2 2
2dt = Ú _________ t2 + 4t + 1 dt = 2 Ú _____________ __ (t + 2)2 – ( ÷ 3 )2
| |
__
( )
|
(t + 2) – ÷ 3 ___________ __ + c (t + 2) + ÷ 3
1 = ___ __ log 3 ÷
(tan(x/2) + 2) – ÷ 3 1 = ___ __ log _________________ __ + c 3 (tan(x/2) + 2) + ÷ 3 ÷
__
|
245. The given integral is dx dx = __________________ Ú ________ 3cosx + 4 Ú 1 – tan2 (x/2) 3 ___________ + 4 1 + tan2 (x/2)
(
)
(1 + tan2 (x/2)) = Ú __________________________ dx 4 + 4tan2 (x/2) + 3 – 3tan2 (x/2)
sec2 (x/2) = Ú ___________ dx tan2 (x/2) + 7
2
|
|
(1 + tan2 (x/2))dx = Ú ____________________ 1 + tan2 (x/2) + 4tan(x/2)
|
__
x tan __ + 2 – ÷3 2 1 = ___ __ log ________________ __ + c x 3 ÷ tan __ + 2 + ÷3 2
|
244. The given integral is
x Let tan __ = t 2 x fi sec2 __ dx = 2dt 2
( ) ( )
2 tan 2 1 + 2 __________ x 2 __ 1 + tan 2 x 2 __ sec 2 = Ú ___________________ x x dx 2 __ 1 + tan + 4 tan __ 2 2
(
dx dx = ________________ Ú _________ x 1 + 2 sin x Ú __
( )
3 tan x – 1 ÷ 1 __ = ____ __ log __________ + c 2÷3 ÷3 tan x + 1
1.69
Let t = tan(x/2)
fi
( )
x 2dt = sec2 __ dx 2
1.70 Integral Calculus, 3D Geometry & Vector Booster
(
)
2dt = Ú _____ 2 t +7
1 1 1 1 = __ Ú _____ – ______ dt + __ Ú sec2 (x/2)dx 2 t2 – 1 (t + 1)2 2
t 2 = ___ __ tan–1 ___ __ + c 7 ÷7 ÷
)
t–1 1 __ 1 1 = __ log ____ + ______ + tan(x/2) + c 2 2 t+1 (t + 1)
tan(x/2) 2 __ = ___ __ tan–1 ______ + c 7 ÷7 ÷
cos x – 1 1 1 1 = __ __ log ________ + _________ 2 2 cos x + 1 (cosx + 1)
( )
(
246. The given integral is dx Ú _____________ sinx + cosx + 1
dx = Ú ___________________________ 2tan(x/2) 1___________ – tan2 (x2) ___________ + + 1 1 + tan2 (x/2) 1 + tan2 (x/2)
( ) ( ) ( )
x Let tan __ = t 2 x 1 fi __ sec2 __ dx = dt 2 2 x fi sec2 __ dx = 2dt 2
2dt = Ú _____ 2t + 2
(
= Ú dt/t + 1
x = log tan __ + 1 + c 2
|
( )
(1 + tan2 (x/2))dx = Ú ____________________ tan2 (x/2) – 4tan(x/2) + 1
sec2 (x/2)dx = Ú ____________________ tan2 (x/2) – 4tan(x/2) + 1
Let tan(x/2) = t x fi sec2 __ dx = 2dt 2
dt = 2 Ú _________ 2 t – 4t + 1 dt = 2 Ú _____________ __ (t – 2)2 – ( ÷ 3 )2
|
|
__
)
|
249. The given integral is
1 + sinx dx Ú ____________ sinx(1 + cosx)
dx dx = Ú ____________ + _________ sinx(1 + cosx) Ú (1 + cosx)
sin x dx = Ú _________________ + (1 – cos2x)(1 + cosx)
dt = Ú ___________ + 2 (t – 1)(1 + t)
( )
x 1 __ Ú sec2 __ dx 2 2
( )
x 1 __ Ú sec2 __ dx 2 2
Let t = cos x
fi dt = – sin x dx
__
(tan(x/2) – 2) – ÷ 3 = log ________________ __ + c (tan(x/2) – 2) + ÷ 3
247. The given integral is
(
( )
(t – 2) – ÷ 3 1 = 2 × __ log __________ __ + c 2 (t – 2) + ÷ 3
= log |t + 1| + c
)
)
dx = Ú _________________ 2tan(x/2) 1 – 2 ___________ 1 + tan2 (x/2)
(1 + tan2 (x/2))dx = Ú _________________________________ 1 + tan2 (x/2) + 2tan(x/2) + 1 – tan2 (x/2) sec (x/2)dx = Ú ___________ 2tan(x/2) + 2
| | ) | |
+ tan(x/2) + c 248. The given integral is dx Ú ________ 1 – 2 sin x
2
( (
dt = Ú ___________ + (t – 1)(t + 1)2
( )
x 1 __ Ú sec2 __ dx 2 2
dx Ú ________________ 3 sin x + 4 cos x + 5
dx = Ú _________________________________ 2tan(x/2) 1___________ – tan2 (x/2) ___________ 3 + 4 + 5 1 + tan2 (x/2) 1 + tan2 (x/2)
(
) (
)
(1 + tan2 (x/2))dx = Ú ____________________ tan2 (x/2) + 6tan(x/2) + 9 2dt = Ú _________ t 2 + 6t + 9 2dt = Ú ______ (t + 3)2
Let tan(x/2) = t x fi sec2 __ dx = 2dt 2
( )
Indefinite Integrals
2 = – ______ + c (t + 3)
dx 1 = ___ __ Ú _______________ 1__ 1 2 ___ ÷ sinx + ___ __ cosx 2 ÷2 ÷
2 = – ___________ + c (tan(x/2) + 3)
dx 1 = ___ __ Ú _________ p 2 sin x + __ ÷ 4 p 1__ ___ = Ú cosec x + __ dx 4 2 ÷
(
250. The given integral is
dx Ú ___________ cos x + cos a dx = Ú _________________ , where k = cos a 2 1 – tan (x/2) ___________ + k 1 + tan2 (x/2) [1 + tan2 (x/2)]dx = Ú __________________________ 1 – tan2 (x/2) + k[1 + tan2 (x/2)]
(
)
sec2 (x/2)dx = Ú __________________________ 1 – tan2 (x/2) + k[1 + tan2 (x/2)]
( ) ( )
2dt = Ú ________________ 2 2 (1 – t ) + k(1 + t )
x Let tan __ = t 2 x fi sec2 __ dx 2 = 2 dt
2dt = Ú _______________ (1 + k) + (k – 1)t2
2dt 1 = ______ __________ (1 – k) Ú (1 + k) ______ – t2 (1 – k)
2dt 1 ______ = ______ ___________ – t2 (1 – k) Ú (1 + k) 2 ______ (1 – k)
( ÷
)
| | | ÷ ÷
______
(1 + k) ______ +t (1 – k) 1 1 ______ + c = ______ × ________ ______ log __________ (1 – k) (1 + k) (1 + k) 2 ______ ______ –t (1 – k) (1 – k)
1 = ____ log sina
|
÷
cot(a/2) + tan(x/2) ________________ + c cot(a/2) – tan(x/2)
251. The given integral is
)
( ) x p 1 = ___ log tan ( __ + __ ) + c 2 8 | | ÷2 __
253. We have,
1.71
dx __ Ú ____________ ÷3 sinx + cosx dx 1 __ = __ Ú ______________ 2 ___ ÷3 1 __ sinx + cosx 2 2 dx 1 _________ __ = Ú p 2 sin x + __ 6 p 1 = __ Ú cosec x + __ dx 2 6 p x ___ 1 __ __ = log tan + + c 2 2 12
(
|
)
( ) ( ) |
257. We have, 2 sin x + cos x = l (3 sin x + 2 cos x) + m(3 cos x – 2 sin x) = (3l – 2m) sin x + (2l + 3m) cos x Comparing the co-efficients of sin x and cos x, we get 3l – 2 m = 2 and 2l + 3 m = 1 On solving, we get, l = 4/5 and m = – 3/ 5 4 3 sin x + 2 cos x \ __ Ú _____________ dx 5 3 sin x + 2 cos x 3 3 cos x – 2 sin x + __ Ú _____________ dx 5 3 sin x + 2 cos x
(
)
3 3 cos x – 2 sin x 4 = __ Ú dx – __ Ú _____________ dx 5 5 3 sin x + 2 cos x
3 4 = __ x – __ log |3 sin x + 2 cos x| + c 5 5
258. We have,
dx Ú ______________ tanx + 4cotx + 4
4sin x cos x = Ú _______________________ dx sin2x + 4cos2x + 4sin x cos x
2 sin x 1 = __ Ú __________ dx 2 sin x + cos x
1 (sin x + cos x) + (sin x – cos x) = __ Ú _________________________ dx 2 sin x + cos x
252. We have,
dx Ú __________ sinx + cosx
sin x dx Ú __________ sin x + cos x
1.72 Integral Calculus, 3D Geometry & Vector Booster
1 (1 + cos x – sin x = __ Ú _______________ dx 2 sin x + cos x)
1 1 = __ x + __ log |sin x + cos x| + c 2 2
259. The given integral is + 3 cos x _____________ dx Ú 2 sin x 3 sin x + 4cosx
= (3l – 4m)sin x + (4l + 3m) cos x Comparing the co-efficients of sin x and cos x, we get (3l – 4m) = 2, (4l + 3m) = 3
On solving, we get 18 1 l = ___ , m = ___ 25 25
The given integral reduces to 18 1 ___ (3 sin x + 4 cos x) + ___ (3 cos x – 4 sin x) 25 25 Ú ____________________________________ dx (3 sin x + 4 cos x)
(
cos x 1 dx = Ú __________ dx Ú _______ 1 + tan x sin x + cos x 2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
1 = __ (x + log |sin x + cos x|) + c 2 263. The given integral is
= l(3 sin x + 4 cos x) + m(3 cos x – 4 sin x)
we have 2 sin x + 3 cos x
1 = __ (x + log |sin x + cos x|) + c 2 262. The given integral is
)
18 1 (3cosx – 4sinx) = Ú ___ + ___ _____________ dx 25 25 (3sinx + 4cosx) 18 1 = ___ x + ___ log |3 sin x + 4 cos x| + C 25 25
1 dx Ú _______ 1 – tan x
cos x = Ú __________ dx cos x – sin x
2 cos x 1 = __ Ú __________ dx 2 cos x – sin x
1 (cos x – sin x) + (cos x + sin x) = __ Ú _________________________ dx 2 cos x – sin x
(cosx + sinx) 1 = __ Ú 1 + ___________ dx 2 cosx – sinx
1 = __ (x – log |cos x – sin x|) + c 2
(
)
264. The given integral is
260. The given integral is
1 dx Ú _______ 1 + cot x
sin x dx Ú __________ sin x – cos x
cos x = Ú __________ dx sin x + cos x
2 sin x 1 = __ Ú __________ dx 2 sin x – cos x
2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
1 (sin x – cos x) + (sin x + cos x) = __ Ú __________________________ dx 2 sin x – cos x
1 (cos x + sin x) + (cos x – sin x) = __ Ú __________________________ dx 2 sin x + cos x
cos x + sin x 1 = __ Ú 1 + __________ dx 2 sin x – cos x
(cos x – sin x) 1 = __ Ú 1 + ___________ dx 2 sin x + cos x
1 = __ (x + log |sin x + cos x|) + c 2
cos x dx Ú __________ sin x + cos x
265. We have, 3 sin x + 2 cos x + 4 = l(3 cos x + 4 sinx+ 5)
2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
+ m(– 3 sin x + 4 cos x) + n
1 (cos x + sin x) + (cos x – sin x) = __ Ú __________________________ dx 2 sin x + cos x
(cosx – sinx) 1 = __ Ú 1 + ___________ dx 2 sinx + cosx
+ (5 l + n) Comparing the co-efficients of sin x and cos x we get 4 l – 3 m = 3, 3 l + 4 m = 2 and 5 l + n = 4
(
)
1 1 = __ x + __ log |sin x – cos x| + c 2 2 261. The given integral is
(
)
(
)
= (4 l – 3 m) sin x + (3 l + 4 m) cos x
Solving, we get
18 1 1 l = ___ , m = – ___ , n = __ 25 25 5
Thus,
+ 2 cos x + 4 ________________ dx Ú 3 sin x 3 cos x + 4 sin x + 5
(
(
))
(
)
x – tan–1 (4/3) 2 + ___ tan–1 ____________ + c. 2 25
269. We have,
|
271. We have,
dx 2 + __ Ú ________________ 5 4 sin x + 3 cos x + 5 18 1 = ___ x – ___ log |4 sin x + 3 cos x + 5| 25 25
|
1.73
49(sinx – cosx) – 5 1 = – ___ log _________________ + c 40 4(sinx – cosx) + 5
18 4 cos x – 3 sin x 1 = Ú ___ + ___ ________________ dx 25 25 4 sin x + 3 cos x + 5
Indefinite Integrals
Ú (cos x – sin x)(2 + 3 sin x 2x)dx
dx Ú ____________ cos x + cosec x dx = Ú __________ 1 cos x + ____ sin x sin x dx = Ú ___________ cos x sin x + 1 2 sin x dx = Ú _____________ 2 cos x sin x + 2
(sin x + cos x) + (sin x – cos x) = Ú __________________________ 2cos x sin x + 2
(sinx + cosx) = Ú ______________ dx 2 cos x sin x + 2
= Ú (cos x – sin x)(2 + 3(–1 + 1 + sin 2x))dx
= Ú (cos x – sin x)(–1 + 3(sin x + cos x)2)dx
(sin x – cos x) + Ú _____________ dx 2 cos x sin x + 2 (sin x + cos x) (sin x – cos x) = Ú ___________ dx + Ú ___________ dx 2 + sin 2 x 2 + sin 2 x
= Ú (3t 2 – 1) dt Let t = sin x + cos x
fi dt = (cos x – sin x)dx
3
= (t – t) + c = (sin x + cos x)3 – (sin x + cos x) + c 270. We have,
sin x + cos x dx Ú ___________ 9 + 16 sin 2 x
sin x + cos x = Ú ___________________ dx 9 + 16(1 – 1 + sin 2x)
sin x + cos x = Ú ___________________ dx 25 – 16(sin x – cos x)2
Let sin x – cos x = t
fi (cos x + sin x)dx = dt
dt = Ú ________ 25 – 16t2 dt 1 = – ___ Ú ________ 2 16 2 5 __ t – 4
( )
| | |
5 t – __ 1 1 4 = – ___ × _____ log _____ + c 16 5 5 2 × __ t + __ 4 4 4t – 5 1 = – ___ log ______ + c 40 4t + 5
|
(sin x + cos x) = Ú _______________ dx 3 – (sin x – cos x)2
(sin x – cos x) + Ú ________________ dx 1 + (sin x + cos x)2
|
__
|
(sin x – cos x) – ÷ 3 1__ __ = – ____ log _________________ 2÷3 (sin x – cos x) + ÷ 3 –1
– tan (sin x + cos x) + c 271. The given integral is dx Ú ____________ cos x + cosec x
sin x cos x = Ú ___________ dx sin x + cos x
2 sin 2x 1 = __ Ú __________ dx 4 sin x + cos x
2 2 1 (sin x + cos x) – (sin x – cos x) = __ Ú ___________________________ dx 4 sin x + cos x
(sin x – cos x) 1 = __ Ú (sin x + cos x) – ___________ dx 4 sin x + cos x
1 1 = __ (cos x – sin x) – __ (sin x – cos x)3 + c 4 3
[
[
]
]
272. The given integral is
Ú (sin x + cos x)(2 + 3 sin 2 x)dx = Ú (sin x + cos x)[2 + 3{1 – (1 – sin 2x)}]dx
1.74 Integral Calculus, 3D Geometry & Vector Booster
= Ú (sin x + cos x)(5 – 3(sin x – cos x)2)dx
Let sin x – cos x = t fi (cos x + sin x)dx = dt
= Ú (5 – 3t2)dt 3
= (5t – t ) + c = (5(sin x – cos x) – (sin x – cos x)3) + c
273. The given integral is
Ú (sin x – cos x)(3 – 4 sin 2 x)dx = Ú (sin x – cos x)(7 – 4(1 + sin 2 x)) = Ú (sin x – cos x)(7 – 4(sin x + cos x)2)dx = Ú (cos x – sin x)(4(sin x + cos x)2 – 7)dx
Let sin x + cos x = t fi (cos x + sin x)dx = dt
= Ú (4t2 – 7)dt
4t3 = ___ – 7t + c 3
(
)
(
)
4(sin x + cos x)3 = _____________ – 7(sin x + cos x) + c 3 274. The given integral is
cos x – sin x dx Ú ( ___________ 3 + 2 sin 2 x )
cos x + sin x = Ú _________________ dx 1 – 4(sin x – cos x)2
Let sin x – cos x = t fi (cos x + sin x)dx = dt
dt = Ú ______ 1 – 4t2
dt 1 = – __ Ú _________ 4 t2 – (1/2)2
2t – 1 1 = – __ log ______ + c 2 2t + 1
1 = __ log 2
| |
|
|
2(sin x – cos x) – 1 ________________ + c 2(sin x – cos x) + 1
276. The given integral is
+ sin x 2 cos x dx Ú (____________ 9 + 16 sin 2x )
3 1 __ (cos x + sin x) + __ (cos x – sin x) 2____________________________ 2 = Ú dx 9 + 6 sin 2x 3 (cos x + sin x) 1 (cos x – sin x) = __ Ú ___________ dx + __ Ú ___________ dx 2 2 (9 + 6 sin 2x) 9 + 6 sin 2x (cos x + sin x) 3 = __ Ú __________________ dx 2 15 – 6(sin x – cos x)2 (cos x – sin x) 1 + __ Ú __________________ dx 2 (3 + 6(sin x + cos x)2)
(cos x – sin x) = Ú ______________ dx 1 + 2(1 + sin 2x)
Let sin x – cos x = t and sin x + cos x = z fi (cos x + sin x) dx = dt and (cos x – sin x) dx = dz
(cos x – sin x) = Ú _________________ dx 1 + 2(sin x + cos x)2
3 dt dz 1 = __ Ú _______ + __ ______ 2 15 – 6t2 2 Ú 6z2 + 3
dt dz 1 1 = __ Ú ______ + __ ______ 2 5 – 2t2 6 Ú 2z2 + 1
dt dz 1 1 __ + __ = – __ Ú __________ Ú ______ 4 t2 – ( ÷5 / 2 )2 6 2z2 + 1
__ t – ( ÷5 /2 ) 1 1 __ + ____ = – __ log __________ __ tan–1 ( z÷2 )+ C 4 t + ( ÷5 /2 ) 6÷2
where
sin x – cos x = t and sin x + cos x = z
Let sin x + cos x = t fi (cos x + sin x)dx = dt
dt = Ú ______ 1 + 2t2 dt 1 = __ Ú _________ 2 ___ 1__ 2 2 + t 2 ÷ 1__ –1 ( __ ) ___ = tan t ÷2 + c 2 ÷ 1__ –1 __ = ___ tan ( ÷2 ( sin x + cos x) ) + c 2 ÷ 275. The given integral is
( )
+ sin x __________ dx Ú ( cos x 5 – 4 sin 2x )
(
)
cos x + sin x = Ú ___________________ dx 5 – 4(1 – (1 – sin 2x))
|
__
277. We have,
2
x + 1 4 dx Ú ______ x + 1
(
)
1 1 + __ 2 x = Ú _______ dx 1 2 x + __ 2 x
|
Indefinite Integrals
(
)
1 1 + __ 2 x = Ú ______________ dx 1 2 1 x – __ x + 2 ◊ x ◊ __ x
(
)
(
)
1 Let x – __ x = t
1 fi 1 + __ 2 dx = dt x
(
dt __ Ú _________ 2 ( )2
=
t 1 = ___ __ tan–1 ___ __ + c 2 ÷2 ÷
1 1 1 = ___ __ tan–1 ___ __ x – __ x + c 2 ÷2 ÷
(
))
(
)
Ú tan–1 ( x – __ 1x ) – __ 23 tan–1 (x3) + c
2
– 3x + 1 dx Ú x__________ 4 2 x + x + 1
(x2 – x + 1) – 2x
x + __ 2 x
=
dx – Ú _________ 2x dx Ú ________ 2 4 2
=
dx dt Let x2 = t __ , Ú ________ Ú ______________ 2 2 3 2
( )
1 1 – __ 2 x = Ú ______________ dx 2 1 1 x + __ x – 2 ◊ x ◊ __ x
(
)
(
)
1 Let x + __ x = t 1 fi 1 – __2 dx = dt x
( )
t – ÷ 2 1 __ = ____ __ log ______ + c 2÷2 t + ÷2
__ 1 __ x + – 2 ÷ x 1 = ____ __ log | ____________ __ + 1 2÷2 x + __ x + ÷ 2
| | __
| | ( (
279. We have,
) )
(x2 – x + 1) x + x + 1
x +x+1
x +x +1
x +x +1
( ) ( )
t + t + 1
fi 2x dx = dt
( ) ( )
1 1 x + __ t + __ 2__ –1 _____ 2 2 2 –1 _____ ___ ___ = tan __ + __ tan __ + c 3 ___ ÷3 ___ ÷ ÷3 ÷3 2 2
2x __ +1 2 = ___ __ tan–1 ______ + 3 ÷3 ÷
2x +__ 1 2x2 __ +1 2 2 = ___ __ tan–1 ______ + ___ __ tan–1 ______ + c 3 ÷3 ÷3 ÷3 ÷
( ) ( )
(
)
2t +__ 1 2 ___ __ tan–1 _____ + c 3 ÷3 ÷
(
)
281. We have,
4
(x2 + 1)2 – 2x2 = Ú _____________ dx 6 x + 1 (x2 + 1)2 – 2x2 = Ú _________________ dx (x2 + 1)(x4 – x2 + 1) (x + 1)
2
(x – x + 1)
x +1
dx – 2 Ú _____ x dx Ú ___________ 4 2 6
( ) ÷ 2
dx – x ___________ __________ Ú 1 1 _____ 2 4 + x2 1 + x + x
( ) ) ÷ ( ( ) ) ÷(
1 – x2 1 – __ 2 dx x _____________ = Ú _____________________ 1 1 2 2 __ __ x x + x x x + 2 + 1 x
1 1 – __ 2 dx x ___________ = – Ú __________________ 1 1 2 __ __ x + x x + 2 + 1 x
x + 1
2
x + x + 1
÷ 1 x + __ + ___ 2 2
x + 1 dx Ú ______ 6
=
(
dx Ú _________ 2x dx Ú __________ 4 2 4 2
1 1 + __ 2 2 2 x __________ dx – __ Ú _____ 3x dx 2 6 3 x +1 1 x – __ x + 1
=
= Ú _________ dt __ t2 – ( ÷ 2 )2
=
( ) ) ( Ú
( )
)
dx Ú _______________ 4 2
=
x +1
=
1 2 1 – __ x = Ú ________ dx 1 2
(
2 – 1 dx Ú x______ 4 x + 1
1 1 + __ 2 2 x 2 = Ú ___________ dx – __ Ú _____ 3x dx 6 3 x +1 1 x2 + __ 2 – 1 x
(x – x + 1)
280. We have,
278. We have,
=
t + ÷2
( )
2
2 dx – __ Ú _____ 3x dx Ú ___________ 4 2 6 3
)
(x2 + 1)
1.75
(
(
)
)
1.76 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 dt = – Ú _______ _____ Let x + __ x = t 2 t ÷ t – 1 1 fi 1 – __ 2 dx = dt x = cosec–1 (t) + c 1 = cosec–1 x + __ x + c x = sin–1 _____ 2 + c x +1 282. We have,
( )
(
(
)
÷x + x + x
( ( Ú (
)
=
)
)
)
÷(
)
) ÷(
)
)
( ) ( )
(
2 tdt Ú _______ 2
) ÷( ) 1 Let ( x + __ x + 1 ) = t
( )
=
_____ tdt Ú ________ 2 2
=
v dv Ú ________ 2
2
= 2 tan–1 t + c
1/2 1 = 2 tan–1 x + __ x + 1 + c
2
÷x + 1
x +1
dx x2 – 1 = Ú _______ × ___________ __________ 1 1 __ 2 x x + x x x2 + __2 x 2 dx x – 1 ___________ _________ = Ú × __________ 1 1 x2 x + __ x x2 x2 + __2 x
÷ ( ) ( ) ÷ ( )
(
)
fi 2t dt = 2v dv
dv Ú _______ 2
(v + 2) v 1 = ___ __ tan–1 ___ __ + c 2 ÷2 ÷
( )
( ) ( ÷ ( ) )
(t + 1)t
dx – 1 × _______ _____ Ú x_____ 2 4
Let t2 – 2 = v2
(v + 2)v
1 1 1 __ = ___ __ tan–1 ___ __ x2 + 2 + c 2 ÷2 ÷ x
= Ú ______ 2 dt (t2 + 1)
t ÷t – 2
=
283. We have,
( )
2 ÷ t 2 –__ 1 = ___ __ tan–1 ______ + c 2 ÷2 ÷
)
)
284. We have,
1 fi 1 – __2 dx = dt x
(
=
(
1 Let x + __ x = t 1 fi 1 – __ 2 dx = dt x
___________
1 1 – __ 2 x dx = Ú __________ × ____________ ___________ 1 1 __ __ x + x + 2 x + x + 1
t÷t – 2
)
_____
dx x2 – 1 ____________ 2 × ____________ ___________ 1 x(x + 2x + 1) x + __ x + 1
( (
dt ______ Ú _______ 2
dx x2 – 1 = Ú ____________ × ____________ ___________ 1 1 x2 x + __ x + 2 x + __ x + 1
=
)
÷(
) ÷(
(
dx x2 – 1 = Ú _______ × _____________ ____________ 1 (x + 1)2 2 x x + __ x + 1 dx x2 – 1 = Ú ________ × ____________ ___________ 2 1 x(x + 1) __ x + x + 1
÷ (
)
dx x x +– 11 ) × ___________ __________ Ú (_____ 3 2
( )
1 1 – __ 2 dx x = Ú _______ × ____________ ___________ 1 1 2 __ x + x x + __ x – 2
2
x – 1 ___________ Ú ______________ 3 4 2
dx x ÷2x – 2x + 1 2
x – 1 dx _____________ = Ú ______________ 2 1 3 __ x 2 – __ 2 + 4 x x
) ÷ ( Ú ) ÷( ) ( Ú ) ÷( ( (
x2 – 1 dx ___________ = ______________ 2 1 __ x5 2 – __ 2 + 4 x x 1 1 __ 3 – __ 5 x x ___________ = ____________ dx 2 1 __ __ 2 – 2 + 4 x x
)
2 1 Let 2 – __2 + __ 4 = t2 x x 4 4 fi __ 3 – __5 dx = 2t dt x x
1 2t dt = __ Ú ____ t 4 1 = __ × t + c 2 1 2 1 1/2 = __ × 2 – __ 2 – __ 4 + c 2 x x
(
)
)
Indefinite Integrals
285. We have,
(
4
x –1 _________ Ú ____________ 2 4 2
=
)
dx x ÷x + x + 1
( ( ÷
4
x –1 _____________ Ú ________________ 1
)
dx x x x + __ 2 + 1 x 2
2
2
( ( )) ( ( ( ) ) )
)
( )
÷
÷
= Ú ___ t dt t
=
=t+c
1/2 1 = x + __ 2 + 1 + c x
Ú dt
(
)
2
– 1 dx x – 3x Ú __________ 4 2 x + x + 1
3 x2 – 1 = Ú _________ dx – __ Ú _________ 2x dx 4 2 x4 + x2 + 1 x + x2 + 1
( )
1 1 – __ 2 x 3 2x = Ú __________ dx – __ Ú __________ dx 1 2 x4 + x2 + 1 2 __ x + 2 + 1 x 1 1 – __ 2 x 3 = Ú ___________ dx – __ Ú _________ 2x dx 2 x4 + x2 + 1 1 2 __ 1 – x – 1
(
( ) )
(
)
1 Put x + __ x = t and x2 = z
1 fi 1 + __ x dx = dt and 2x dx = dz 2
(
)
3 dt – __ Ú ________ Ú _____ 2 2 2 dz
t –1
z +z+1
(
( ) ( )
÷ (
))
÷(
))
1 x2 x + x2 1 + __ 2 x
)
) | (
| |
dx _________ Ú _________________
(
(
)
(
1 x + __ x – 1 __ 2(x2) + 1 1 __ __________ __ = log – ÷ 3 tan–1 ________ + c 2 1 ÷3 x + __ x + 1
dx ________________ _______ 1 2 x x + x 1 + __ 2 x 1 = 1 + __ 2 = t2 x =
|
2z + 1 3 t–1 1 2 __ = __ log ____ – __ × ___ __ tan– 1 ______ + c 2 t+1 2 ÷ 3 ÷3
x ( x + ÷ 1 + x )
( Ú (
= log |t + 1| – t + c 1 1/2 1 1/2 = log 1 + 1 + __ 2 – 1 + __ 2 + c x x
1 – 1 ) dt Ú ( ____ t+1
| |
)
2
dx _____ Ú _____________ 2 2 =
=
286. We have,
=
( ) ( ) ( )
1 Let x2 + __ 2 + 1 = t2 x 2 fi 2x – __ 3 dx = 2t dt x 1 fi x – __ 3 dx = t dt x
=
287. We have,
1 x – __ 3 x ___________ = Ú ____________ dx 1 2 + 1 x2 + __ x
– t dt Ú ____ 1+t
x4 – 1 _____ x3 ____________ = Ú ___________ dx 1 2 __ x + 2 + 1 x
1.77
(
288. The given integral is dx 4 = Ú ___________ sin x + cos4x
=
4
sec x 4 dx Ú ________ tan x + 1
Ú
(1 + tan2x) sec2x ______________ dx tan4x + 1
)
1 Put 1 + __ 2 = t2 x 2 fi – __3 dx = 2t dt x dx ___ fi 3 = – t dt x
Put tan x = t fi sec2 x dx = dt =
and then solve it. 289. The given integral is
)
2
+1 dt Ú t______ 4 t +1
dx 1 2 4 = __ Ú _______ dx Ú _____ x + 1 2 (x4 + 1)
1.78 Integral Calculus, 3D Geometry & Vector Booster
2 2 1 (x + 1) – (x – 1) = __ Ú _______________ dx 2 (x4 + 1)
1 = __ Ú 2
(x2 + 1) 1 _______ dx – __ Ú 4 2 (x + 1)
x2 – 1 _______ dx (x4 + 1)
and then you do it. 290. The given integral is 2
2
x dx 2x 1 4 = __ Ú _____ 4 dx Ú _____ 2 x +1 x +1
2 2 1 (x + 1) + (x – 1) = __ Ú ________________ dx 2 x4 + 1
2
1 (x + 1) 1 = __ Ú _______ dx + __ Ú 2 2 x4 + 1
( ) 4
=
2
x ÷x + x + 1 4
=
x –1 ______________ dx Ú ________________ 1
=
x –1 ____________ dx Ú ______________ 1
=
Ú
÷ ( ÷(
2
4
)
x3 x2 + __ 2 + 1 x 1 x – __ 3 x ____________ ____________ dx 1 2 x + __ 2 + 1 x
÷(
) ( ) ( )
1 Put x2 + __ 2 + 1 = t2 x
1 x – __3 dx = t dt x
= Ú dt
=t+c
(x2 + 1)2 – 2x2 __________________ dx (x2 + 1) (x4 – x2 + 1)
Ú
=
3x 2 dx – __ Ú _______ 3 2 dx Ú ___________ 4 2 3 (x – x + 1) (x ) + 1
(x2 + 1) 2 ___________ dx – __ tan– 1 (x3) + c 3 (x4 – x2 + 1)
=
(x2 + 1)
÷
__________
2
dx dx 2 = Ú ___________ Ú __________ 1 3/4 5 __ x (1 + x4)3/4
(
dx dx 2 = Ú ___________ Ú __________ 5 4/5 1 4/5 6 __ x (1 + x )
(
) ( )
x 1 + 4 x
(
= – Ú dt
= – t + c
1 1/5 = – 1 + __ 5 + c x
(
296. The given integral is
–1 ___5 dx = t3 dt x = – Ú dt
x ÷1 + x
)
1 Put 1 + __ 5 = t5 x 5 fi – __6 dx = 5t4 dt x 1 fi – __6 dx = t4 dt x
1 Let 1 + __ 4 = t4 x
= – t + c
)
x 1 + 5 x
and then you do it. 293. The given integral is
)
x2 x2 x2 + __ 2 + 1 x
1 = x2 + __ 2 + 1 + c x 295. The given integral is
(x + 1) – 2x
)
4
=
(
dx Ú _____________ (x6 + 1)
Ú
)
x –1 __________ dx Ú _____________ 2 4 2
(x – 1) _______ dx x4 + 1
x +1 6 dx Ú ______ x +1 2
294. The given integral is
2
and then you do it. 291. The given integral is 2 (x_______________ + 1) – (x2 – 1) 2dx _____ = dx Ú x4 + 1 Ú x4 + 1 2 (x2 + 1) (x – 1) _______ = Ú _______ dx – dx Ú 4 4 x +1 x +1 and then you do it. 292. The given integral is
(
1 1/4 = – 1 + __ 4 + c x
(
)
2
x _____ –1 dx Ú ________ 4
=
(x2 – 1)
_______ dx Ú __________ 1 2 2
÷
x x + __ 2 x
)
Indefinite Integrals
Ú
=
( )
1 1 – __ 2 x _______ ____________ dx 1 2 x + __ x – 2
÷(
)
(
)
1 Let x + __ x = t 1 fi 1 – __2 dx = t4 dt x
(
Ú
dt ______ _____ 2 ÷ t – 2
)
=
= log t + ÷t 2 – 2 + c
_____
|
| (
)
|
÷
______
|
1 1 = log x + __ x + x2 + __ 2 + c x
Ú
=
÷
2 __ 2 x 1 = __ Ú __________ __________ dx 2 1 2 __ x + 2 + 3 x
) ( )
1 1 1 + __ 2 – 1 – __ 2 x x 1 _________ = __ Ú _________________ dx 2 1 2 __ x + 2 + 3 x 1 = __ Ú 2
(
)
( )
1 1 1 – __ 2 1 + __ 2 x x 1 __________ __________ dx __________ dx – __ Ú ___________ 2 1 1 2 __ x2 + __ 2 + 3 x + 2 + 3 x x
÷
(
÷
)
( )
1 1 1 + __ 2 1 – __ 2 x x 1 1 ___________ ___________ = __ Ú ____________ dx – __ Ú _____________ dx 2 2 2 1 2 __ x + 1x + 1 x – x + 5
÷( ) 1 1 = __ log ( x – __ x ) + x 2 ÷
|
÷(
__________ 2
|
1 + __ 2 + 3 x
298. The given integral is
)
( )
)
1 fi 1 – __2 dx = dt x
(
dt = – Ú _______ _____ 2 t÷t + 3
t dt = – Ú ________ _____ 2 2 t÷ t + 3
)
Let t2 + 3 = v2 fi 2t dt = 2v dv
v dv = – Ú ________ 2 (v – 3)v
dv = – Ú _______ 2 (v – 3)
v– ÷ 3 1__ __ = – ____ log _______ + C 2÷3 v+÷ 3
÷t 2 + 3 – ÷ 3 1__ _____ = – ____ log ___________ __ + c 2 2÷3 ÷t + 3 + ÷ 3
__ 1 x2 + __ 2 + 1 – ÷3 x 1__ ____ ________________ = – log _________ __ + c 2÷3 1 2 __ x + 2 + 1 + ÷ 3 x
| |
(
)
__
_________
|
|
dx –1 ) × ___________ ___________ Ú ( x_____ x 4 2 2
÷x + 3x + 1
=
=
( ) ÷ Ú ( ) (
dx 1 2 × __________ __________ Ú 1 – __ 1 x 2
x + __ 2 + 3 x dx 1 1 – __ 2 × ____________ _______ 1 2 x x + __ + 1 x
| (
÷
)
÷
)
__________
|
1 1 = log x + __ x + x2 + __ 2 + 3 + c x
300. The given integral is
2
÷x + x + 1
|
_____
÷ ÷
dx +1 × ___________ __________ Ú x______ 2 4 2 1–x
|
299. The given integral is
1 1 – __ tan–1 x + __ x + c 2
(
1 Let x – __ x = t
__
x x + __ 2 + 3 x
÷
)
dx __________ Ú ____________ 1 2 2
(
) ÷(
(
dx ____________ __________ 4 x ÷x + 3x2 + 1
÷
)
1 1 + __ 2 x dx = – Ú _______ × ____________ __________ 1 1 2 __ x – x x – __ x + 3
297. The given integral is
(
1.79
xx(x2x + 1) (ln x + 1)
dx Ú _________________ (x4x + 1)
Let xx = t
fi xx(1 + ln x) dx = dt
1.80 Integral Calculus, 3D Geometry & Vector Booster
=
=
2
t +1 4 dt Ú ______ t +1
(Ú )
1 1 + __ 2 t _______ dt 1 2 t + __ 2 t
Ú
=
x dx Ú __________ (1 + x2)1006
=
x dx Ú ______________ 1 1006 2012 __
( )
1 1 + __ 2 t __________ dt 1 2 __ t – + 2 t
=
1 1 1 = ___ __ tan– 1 ___ __ t – __ + c t 2 ÷2 ÷
1 1 = ___ __ tan– 1 ___ __ (x x – x– x) + c ÷2 2 ÷
( )
=
( ( ) ) ( )
2
=
1 1 __ 3 – __ 5 x x ____________ ___________ dx 2 1 __ __ 2 – 2 + 4 x x
1 = __ Ú 2 1 = __ Ú 2 1 = __ t 2
dt 1 = – __ Ú ____ 2 t1006
t– 1005 1 = – __ × ______ +C 2 – 1005
1 = __________ + C 2010 × t1005
÷
__________
(
2009
(2x + 1) dx
=
(2x + 1) dx
Ú _______________ 4 __ 1 3/2 3 __
(
)
)
x 1 + x + 2 x
2x4 – 2x2 + 1 = ____________ +c 2x2 x dx Ú __________ (1 + x2)1006
)
Ú _____________ (x2 + 4x + 1)3/2
4 __ 4 dx = 2t dt x
302. The given integral is
(
303. We have,
1 2 1 __ = __ 2 – __ 2 + 4 + c 2 x x
÷
1 = _________________ + C 1 1005 __ 2010 × 1 + 2 x
=
Ú
1 2 __ 2 + __ 3 x x dx _____________ 4 __ 1 3/2 __ 1 + x + 2 x
(
)
(
)
(
1 __ 3 dx = – t dt x
4 1 Let 1 + __ x + __ 2 = t2 x 4 2 fi – __2 – __ 3 dx = 2t dt x x
+c
( )
1 Let 1 + __2 = t x 2 __ fi – 3 dx = dt x dx ___ dt ___ fi 3 = – 2 x
dt
)
tdt ___ t
____________
1 __ 4 = t2 x
1 1 1 fi __ 3 – __ 5 dx = __ tdt 2 x x
) ÷( ) ( Ú ) ÷( ( ) ( ) ( ) __ x5 2 – __ 2 + 4 x x
4 fi __ 2 – x
(x – 1)
____________ dx Ú ______________ 2 1
2 Let 2 – __ 2 + x
(
x2 – 1 ____________ dx Ú _______________ x3 ÷2x 4 – 2x2 + 1 =
Ú
)
301. The given integral is
(
2009
1 + 2 x dx ____________ 1 1006 3 x 1 + __ 2 x x
2009
2 fi __ 2 + x t dt = – Ú ___ t3 dt = – Ú __ 2 t 1 = __ + c t
)
Indefinite Integrals
304. We have,
Ú
1 = __________ __________ + c 4 __ 2 1 + __ x + 2 x
÷
(
____
____
)
cot x – ÷tan x ÷ ____________ dx 1 + 3 sin 2x
=
1 – tan x 1 ____ dx Ú _____ ( _____________ 1 + 6 sin x cos x ) ÷tan x
( (
)
=
1 – tan x 1 ____ _____________ sec2x dx Ú _____ ÷tan x sec2x + 6 tan x
=
1 – tan x 1 ____ ________________ sec2x dx Ú _____ ÷tan x tan2x + 6 tan x + 1
=
1–t 2t dt Ú __ 1t _________ t4 6t2 + 1
( Ú (
)
2
2
)
( )
1 __ 2 – 1 t = 2 Ú _________ dt 1 2 t + __ 2 + 6 t 1 __ 2 – 1 t _______ = 2 Ú + 4 dt 12 t + __ t dz 1 = 2 Ú _____ 2 Let z = t + __ t z +4 1 2 fi dx = 1 – __ 2 dx t z 1 = 2 × __ tan– 1 __ + c 2 2
( ( ) ) ( )
( )
(
305. We have,
=
2t dt 4 Ú _____ t +1
=
Ú _________________ t4 + 1
=
+ Ú _________ 4 Ú ________ t4 + 1 t +1
2
(t2 – 1) + (t2 + 1) dt (t2 – 1) dt
(t2 + 1) dt
=
(Ú )
( )
=
(Ú )
) ( Ú
1 1 1 – __ 2 dt 1 + __ 2 dt t t _________ + Ú _________ 1 1 2 __ t + 2 t2 + __ 2 t t
1 1 – __ 2 dt t ___________ + 1 2 __ t + – 2 t
1 1 + __ 2 dt t __________ 1 2 __ t – + 2 t
( )
( )
| |
( ( ) )
( ) ( )
__ 1 1 t + __ – ÷2 t – __ t t 1 1 – 1 ___ ______ __ = ____ __ log ___________ + c __ + __ tan 1 2÷2 2 2 ÷ ÷ __ t + + ÷2 t
| | ( (
) )
__ 1 x2 + __ 2 – ÷2 x 1 = ____ __ log _____________ __ 1 2 2÷2 __ x + 2 + ÷ 2 x
( ) ) (
1 x2 – __ 2 x 1 __ tan–1 ________ __ + ___ + c 2 ÷2 ÷
306. We have,
= tan–1
( )
( ) 1 1 ( __ ( t + __ ) ) + c t 2
t ◊ 2t dt 4 Ú ______ t +1
)
1–t = 2 __________ dt t4 + 6t2 + 1
=
1.81
)
____ 1 ____ = tan–1 __ ÷ tan x + ÷ cot x + c 2
____
dx Ú ÷tan x
Put tan x = t2
fi
fi
____
____
– ÷cot x ) dx Ú (÷tan x
( ) tan x – 1 = Ú ( _______ dx tan x ) =
____
1 – _____ ____ dx Ú ÷tan x ÷tan x ____ ÷
Put
tan x = t2
sec2x dx = 2t dt
fi
sec2x dx = 2t dt
2t dt dx = _____ 2 sec x
fi
2t dt dx = _____ 2 sec x
2t dt = ________ 1 + tan2x
2t dt = ________ 1 + tan2x
2t dt = _____ 1 + t4
2t dt = _____ 1 + t4
| |
1.82 Integral Calculus, 3D Geometry & Vector Booster
=
= =
=
=
( ) Ú ( ) Ú ( ) 2
t –1 4 2t dt Ú ______ t +1
Ú
Ú
( )
1 1 – __ 2 t __________ dt 2 1 __ t + – 2 t
=
1 1 – __ 2 t _____ dt 1 2 t + __ 2 t
311.
|( )|
Ú
(
____ ÷cot x –
)
1 ____ _____ dx = cot x ÷
__
____
2t dx = – ______ dt 1 + t4
|
= 2x + 2
2
(
t2 + 1 = 2x + 2 Ú ______ 4 dt Let tan x = t2 t + 1 fi sec2 dx = 2t dt
2
– cosec x dx = 2t dt
____
+1 ________ ____ + 2 dx Ú tan x ÷tan x
)
Let cot x = t
=
____
2t 4 dt Ú ( t 2 – __ 1t ) × ______ 1+t
( ) 2
t –1 = 2 Ú ______ dt t4 + 1
( )
1 1 – __ 2 t = 2 Ú _____ dt 1 t2 + __ 2 t
( ( ) )
1 1 – __ 2 t = 2 Ú ___________ dt 1 2 __ t + – 2 t
| | ( )
__ 1 t + __ – ÷2 t 1 = 2 × ___ __ log ___________ __ + c 1 ÷2 t + __ + ÷ 2 t
( )
2t fi dx = _____ 4 dt 1+t
Ú
1 1 + __ 2 t _____ dt 1 t2 + __ 2 t
Ú
1 1 + __ 2 t __________ dt 1 2 __ t – + 2 t
cot x – 1 ____ dx Ú ________ ÷cot x
____
+ ÷cot x + 2) dx Ú (÷tan x
= 2x + 2
(
|
( )
(÷tan x + ÷cot x ) – ÷ 2 1 __ + c = ____ __ log ___________________ ____ ____ 2÷2 (÷tan x + ÷ cot x ) + ÷ 2
307. Do yourself
__
____ ÷
|
____
) +1 = 2x + Ú (________ tan x dx tan x )
( ) ( ) ____
____
____
4 + ÷ cot x ) 2 dx Ú (4÷tan x
=
( ( ) )
|
__
309. Do yourself 310. Do yourself
__ 1 t + __ – ÷2 t 1 = ____ __ log ___________ __ + c 1 2÷2 t + __ + ÷ 2 t
308.
(÷cot x + ÷tan x ) – ÷ 2 __ + c = ÷2 log ___________________ ____ ____ (÷cot x + ÷ tan x ) + ÷ 2
2t dt t2 – 1 ____ ______ 4 t t +1 t2 – 1 ______ dt t4 + 1
( )
__ 1 t + __ – ÷2 t = ÷2 log ___________ __ + c 1 t + __ + ÷ 2 t __
( ( ) )
( ( ) )
1 1 1 = 2x + 2 × ___ __ tan–1 ___ __ t – __ + c t 2 ÷2 ÷
(
)
__ 1 = 2x + ÷2 tan–1 ___ __ (tan x – cot x) + c 2 ÷
312. We have,
2
dx sec x ____ 4 = Ú ___________ ____ dx ____ Ú ______________ (÷sin x + ÷ cot x ) (÷tan x + 1)4
Put tan x = t2
fi dt Ú ______ (t + 1)4
=
4 = – _______ +c (t + 1)3
4 = – ___________ + c ____ (÷tan x + 1)3
sec2 x dx = 2t dt
Indefinite Integrals
Integration by Parts 313. We have,
Ú x ex dx = x Ú ex dx – Ú (1 ◊ ex) dx
1 = x log (x2 + 1) – 2 Ú 1 – _____ 2 dx 1+x
= x log (x2 + 1) – 2(x – tan–1 x) + c
Ú x sin x dx = x Ú sin x dx – Ú (1 – cos x)dx
= x Ú sin x dx + Ú cos x dx
= – x cos x + sin x + c
x = tan–1 (x) Ú dx – Ú _____ dx 1 + x2 1 = x tan–1 (x) – __ log |1 + x2| + c 2
320. We have,
Ú x2 sin x dx = x2 Ú sin x dx – Ú (2x – cos x) dx
= – x2cos x + 2 Ú cos x dx
= – x cos x + 2x Ú cos x dx – 2 Ú (1 sin x) dx
= – x2cos x + 2x sin x + 2 cos x + c
2
–x dx Ú cos–1 1______ 1 + x2
= Ú (2 tan–1 x) dx
= 2 Ú tan–1 x dx
1 = 2 tan–1 x Ú dx Ú ______ 2 ◊ x dx 1+x
2x 1 = 2 x tan–1 x – __ ______ dx + c 2 1 + x2
1 = 2 x tan–1 x – __ log |x2 + 1| + c 2
( (
2
316. We have,
Ú log x dx = Ú (log x ◊ 1) dx 1 = log x Ú dx – Ú __ x ◊ x dx
= log x Ú dx – Ú 1 ◊ dx
= x log x – x + c
)
__
Ú e x dx ÷
Let x = t2 fi dx = 2 t dt
Ú (log x)2 dx = Ú {(log x)2 ◊ 1} dx
(
)
= 2 Ú t et dt
= 2 t Ú et dt – Ú (1 ◊ et) dt
1 = (log x)2 Ú dx – Ú 2 log x ◊ __ x ◊ x dx
= 2 (t et – et) + c
= x (log x) – 2 Ú log x dx + c
÷x = 2 (÷x e – e ÷ x ) + c
= x (log x)2 – 2 (x log x – x) + c
2
Ú log (x2 + 1) dx = Ú (log (x2 + 1) ◊ 1) dx 1 = log (x2 + 1) Ú dx – Ú _____ 2 ◊ 2x ◊ x dx x +1
(
( )
x2 = x log (x2 + 1) – 2 Ú _____ 2 dx x +1
)
__
__
__
322. We have,
318. We have,
) ) ) )
(
317. We have,
( Ú (
321. We have,
( )
( )
315. We have,
Ú tan–1 x dx = Ú {tan–1 x ◊ 1}dx
314. We have,
)
319. We have,
= xex – ex + c
(
1.83
x + sin x dx Ú ( ________ 1 + cos x )
x sin x = Ú ________ dx + Ú ________ dx 1 + cos x 1 + cos x x x 2 sin __ cos __ 2 2 x ________ ______________ = Ú x dx + Ú dx x 2 cos2 __ 2 cos2 __ 2 2
( ) ( ) ( ) ( ) x x 1 = __ Ú x sec ( __ ) dx + Ú tan ( __ ) dx 2 2 2 2
1.84 Integral Calculus, 3D Geometry & Vector Booster
[
( ( ) ) ]
( )
x x 1 = __ x Ú sec2 __ dx – Ú 2 tan __ dx 2 2 2
( )
x + Ú tan __ dx 2 x x x 1 = __ 2x tan __ – Ú tan __ dx + Ú tan __ dx + c 2 2 2 2
[
( ) ]
( )
( )
( )
x = x tan __ + c 2 323. We have,
( ÷ )
1–x dx Ú tan _____ 1+x
fi dx = – sin q dq
( ÷
_________
)
1 – cos q = – Ú tan _________ sin q dq 1 + cos q –1
( ÷
__________
(
) ] 4 1 = __ p [ q Ú sin 2q dq + __ Ú cos 2q dq ] – x + c 2 4 – q cos 2q = __ p [ ( ________ ) + __ 14 sin 2q ] – x + c 2 4 __ 1 1 __ = __ p [ – 2 (q (1 – 2 sin q)) + 2 sin q cos q ]– x + c 4 __ 1 1 __ = __ p [ – 2 (sin x (1 – 2x )) + 2 x ÷1 – x ]– x + c 2
–1
_____
2
2
325. We have,
Let x = cos q
______
–1
[
cos 2q 4 = __ p q Ú sin 2q dq – Ú 1 ◊ – ______ dq – x + c 2
)
2
x dx Ú ______________ (x sin x + cos x)2
x cos x = Ú ____________ ◊ x sec x dx x sin x + cos x x cos x = x sec x Ú ____________ dx x sin x + cos x
2 sin2 (q/2) = – Ú tan–1 _________ sin q dq 2 cos2 (q/2)
q = – Ú tan–1 tan __ sin q dq 2
(sec x + x sec x tan x) – Ú – _________________ dx x sin x + cos x
1 = – __ Ú q sin q dq 2
x sec x = ___________ + sec x dx + c x sin x + cos x Ú
1 = – __ q Ú sin q q – Ú (1 – cos q) dq 2
x sec x = ___________ + log |sec x + tan x| + c x sin x + cos x
1 = – __ [– q cos q + sin q] + c 2
326. We have,
( ( ) )
[
(
]
1 = __ [q cos q – sin q] + c 2
1 = __ [ x cos–1 x – ÷1 – x2 ] + c 2
_____
324. We have, __
(
(
)
)
Let x = sin2q
fi dx = sin 2q dq
–1
x tan x dx Ú _________ (1 + x2)3/2
Let x = tan q
fi dx = 2 sec2q dq
tan q a tan–1 (tan q) = Ú _______________ ◊ sec2q dq sec3q
= Ú q cos q dq
= q Ú cos q dq – Ú (1 . sin q) dq
= q sin q + cos q + c
x 1 _____ = _______ ◊ tan–1 x + ______ ______ + c. 2 – x2 ÷ 1 + x ÷ 1
__
–1 sin–1 ÷x – cos ÷x dx Ú ________________ –1 __ –1 __ sin ÷x – cos ÷x __ p –1 __ __ sin–1 ÷x – – sin ÷x 2 = Ú _____________________ dx p __ 2 p 2 –1 __ __ __ = p Ú 2 sin ÷x – dx 2 __ 4 = __ p Ú sin–1 ÷x dx – Ú dx
4 = __ p Ú sin–1 (sin q) sin 2q dq – x + c 4 = __ p Ú q sin 2q dq – x + c
)
327. We have,
_____
x dx Ú sin–1 ( ÷_____ a + x )
( ÷
Put x = a tan2q fi dx = 2a tan q sec2q dq
__________
=
Ú sin
)
a tan2q __________ 2a tan q sec2q dq a + a tan2q
–1
1.85
Indefinite Integrals
= 2a Ú sin–1 (sin q) tan q sec2q dq
Let
x = t2 dx = 2t dt
= 2a Ú q tan q sec2q dq
= 2 Ú t sin t dt
= 2 t Ú sin t dt + Ú cos t dt + c
= 2 [– t cos t + sin t] + c
= 2 [– ÷x cos ÷x + sin ÷x ] + c
Let tan q = t
fi sec2q dq = dt
= 2a Ú t tan–1 t dt
t2 1 __ = 2a tan–1 t Ú t dt – Ú _____ 2 dt 1+t 2
[ [
2
( ) ] Ú ( )]
t 1 1 = 2a __ tan–1 t – __ 1 – _____ 2 dt 2 2 1+t
= a [t2 tan–1 t – t + tan–1 t] + c
= a [(t2 + 1) tan–1 t – t] + c
= a [(tan2q + 1)q – tan q] + c
[ (
__
÷ ÷ ]
x2 1 = log (1 + x) Ú x dx – __ Ú ______ dx 2 (x + 1)
2 x2 1 (x – 1) + 1 = __ log (1 + x) – __ Ú __________ dx 2 2 (x + 1)
x2 1 1 = __ log (1 + x) – __ Ú x – 1 + _____ dx 2 2 x+1
x2 1 x2 = __ log (1 + x) – __ __ – x + log |x + 1| + c 2 2 2
x = log (1 + x) Ú dx – Ú _____ dx x+1
(
)
= x log (1 + x) – (x – log|x + 1|) + c
x – sin x dx Ú 1________ – cos x x dx sin x = Ú ________ – _______ dx 1 – cos x Ú 1 – cos x 2 sin (x/2) cos (x/2)
x dx – Ú ________________ Ú _________ 2 sin2 (x/2) 2 sin2 (x/2) dx
=
x 1 = __ Ú x cosec2 (x/2) dx – Ú cot __ dx 2 2
1 = __ x cosec2 (x/2)dx + 2 Ú cot (x/2) dx 2 Ú
( )
[
– Ú cot (x/2) dx + c 1 = __ (– 2x cot (x/2)) + Ú cot (x/2) dx 2
– Ú cot (x/2) dx + c
__
)
Ú (sin–1 x)2 dx –1 2x sin x ______ = (sin–1 x)2 Ú dx – Ú ________ dx ÷ 1 – x2
–1 x sin x ______ = x (sin–1 x)2 – 2 Ú _______ dx ÷ 1 – x2
= x (sin–1 x)2 – 2 Ú q sin q dq, Let sin–1 x = q
= x (sin–1 x)2 – 2 Ú q sin q dq
= x (sin–1 x)2 – 2 [q sin q dq + q cos q dq] + c
= x (sin–1 x)2 – 2 [– q cos q dq + cos q]
= x (sin–1 x)2 – 2 [– ÷1 – x2 sin–1 x + x] + c
= x (sin–1 x)2 + 2 Î÷1 – x2 sin–1 x – x˚ + c
______
333. The given integral is x – sin x dx Ú ________ 1 – cos x x sin x = Ú _______ dx – Ú ________ dx 1 – cos x 1 – cos x
= – cot (x/2) + c
Ú sin ÷x dx
(
______
]
330. The given integral is
)
x _____ ______ dx = dq ÷ 1 – x2
(
329. The given integral is
Ú x log (1 + x) dx
332. The given integral is
Ú log (1 + x) dx 1 = x log(1 + x) – Ú 1 – _____ x+1
__
__
__
331. The given integral is
x x x = a __ a + 1 tan–1 __ a – __ a + c 328. The given integral is
)
__
x = Ú ________ x dx – 2 sin2 __ 2
( )
Ú
( ) ( ) ( )
x x 2 sin __ cos __ 2 2 ______________ dx x 2 sin2 __ 2
1.86 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) x x 1 = __ [ x Ú cosec ( __ ) dx – 1 ◊ ( – 2 cot ( __ ) ) dx ] 2 2 2 x – Ú cot ( __ ) dx 2 x x 1 = __ [ x ( – 2 cot ( __ ) ) + Ú ( 2 cot ( __ ) ) dx ] 2 2 2 x – Ú cot ( __ ) dx + c 2 x = – x cot ( __ ) + c 2
x x 1 = __ Ú x cosec2 __ dx – Ú cot __ dx 2 2 2 2
334. The given integral is
Ú x sin3x dx cos3x = x Ú sin3x dx – Ú _____ – cos x dx 3
( ) )
( (
)
( (
= Ú (2 tan–1 x) dx
= 2 Ú tan–1 x dx
x = 2 tan–1x Ú dx – Ú _____ dx 1 + x2
]
[
]
1 = 2 x tan–1x – __ log | 1 + x2 | + c 2 338. The given integral is
) )
[
x3 sin–1 (x2)
______ dx Ú _________ 4
÷1 – x
–1 2 (x ) × x2 1 2x sin______ = __ Ú ______________ dx 2 ÷ 1 – x4
Let sin–1 (x2) = t 2x ______ fi ______ dx = dt ÷ 1 – x4
cos3x cos3x = x _____ – cos x – Ú _____ – cos x dx 3 3
cos3x sin3x 1 = x _____ – cos x – __ sin x – _____ 3 3 3
1 = __ Ú t sin t dt 2
= Ú [t Ú sin t dt + Ú cos t dt] + c
+ sin x + c 335. The given integral is
(
)
sec 2x – 1 dx Ú x _________ sec 2x + 1 1 – cos 2 x = Ú x _________ dx 1 + cos 2 x
(
)
= Ú x tan x dx
=
= x Ú sec2x dx – Ú tan x dx – x + c
= x tan x – log |sec x| – x + c
336. The given integral is
( )
2x 2 dx Ú sin–1 1_____ +x
= Ú (2 tan–1 x) dx
= 2 Ú tan–1x dx
= 2 tan–1x Ú dx – Ú _____ x dx 1 + x2
[ [
]
]
1 = 2 x tan–1x – __ log | 1 + x2 | + c 2 337. The given integral is 2x 2 dx Ú tan–1 1_____ –x
( )
1 = __ [– t cos t + sin t] + c 2 __________ 1 = __ – sin–1 (– x2) ÷1 – x4 + x2 + c 2
[
]
339. The given integral is
2
Ú x(sec2x – 1) dx = Ú x sec2x dx – Ú dx
sec x (2 + sec x)
dx Ú ______________ (1 + 2 sec x)2
2 cos x + 1 = Ú __________ (cos x + 2)2
cos x (cos x + 2) + sin x = Ú ___________________ (cos x + 2)2
cos x sin x = Ú __________ dx + Ú __________ (cos x + 2) (cos x + 2)2
sin x 1 = _________ cos dx – Ú __________ (cos x + 2) Ú (cos x + 2)2
sin x + Ú __________ (cos x + 2)2 sin x = ________ + c cos x + 2 340. The given integral is
1–x dx. Ú e_____ x +x (1 + ex) – (x + ex) = Ú _______________ dx (ex + x)
+c
Indefinite Integrals
(
)
(1 + ex) = Ú _______ – 1 dx (ex + x)
= log |(ex + x)| – x + c
341. We have,
Here, f (x) = sin x fi f ¢(x) = cos x Thus,
Ú e (sin x + cos x) dx = e sin x + c x
x
342. We have,
x
x e dx Ú (x_______ + 1)2
Ú
ex (x + 1 – 1) ____________ dx (x + 1)2
=
x+1 1 = Ú e x _______ – _____ dx 2 x +1 (x + 1)
(
343. We have,
)
(
(
e x x = _________ + c (1 + x2)1/2
(
)
2
x +1 dx Ú ex _______ (x + 1)2
( ( (
)
x2 – 1 + 2 = Ú e x _________ dx (x + 1)2
(x + 1)(x – 1) + 2 = Ú e x ________________ dx (x + 1)2
(x – 1) _______ 2 = Ú e x ______ + dx (x + 1) (x + 1)2
(x – 1) = e x ______ + c (x + 1)
(
) )
)
)
( ( ) )
}
ex = __ x + c
347. The given integral is x dx Ú ex (x_______ + 1)2
(
)
(
)
sin 2 x 2 = Ú e x _________ + _________ dx 1 + cos 2 x 1 + cos 2 x
(
{
1 1 1 2 dx = Ú ex __ x + – __2 dx Ú ex __ 1x – __ x x
2 + sin 2 x dx Ú e x _________ 1 + cos 2 x
346. The given integral is
1 1 = – Ú e x _____ – _______ dx x + 1 (x + 1)2 ex = – _____ + c x+1
x 1 = Ú e x _________ + _________ dx (1 + x2)1/2 (1 + x2)3/2
Ú ex [ f (x) + f ¢(x)] dx = ex f (x) + c
)
345. We have,
Ú ex (sin x + cos x) dx
We know that,
(
1.87
)
(
)
)
(x + 1) – 1 = Ú ex __________ dx (x + 1)2
1 1 = Ú ex ______ – _______ dx (x + 1) (x + 1)2
ex = Ú ______ + c (x + 1)
(
)
2 sin x cos x 2 = Ú e x ______ 2 + _________ dx 2 cos x 2 cos2x
= Ú e x (sec2 x + tan x) dx
= Ú e x (tan x + sec2 x) dx
sin x 1 = Ú ex _______ – _______ dx 1 – cos x 1 – cos x
2 sin (x/2) cos (x/2) 1 = Ú ex _________ – ________________ dx 2 2 sin (x/2) 2 sin2 (x/2)
x
= e tan x + c
344. We have,
348. The given integral is
(
)
+ x + x3 dx Ú ex 1_________ (1 + x2)3/2
(
)
1 + x (1 + x2) = Ú e x ____________ dx (1 + x2)3/2
x 1 = Ú e x _________ + _________ dx (1 + x2)3/2 (1 + x2)1/2
(
)
1 – sin x dx Ú ex { ________ 1 – cos x }
(
)
( (
)
)
1 = Ú ex __ cosec2 (x/2) – cot (x/2) dx 2 1 = Ú ex (cot (x/2) + – __ cosec2 (x/2) dx 2 x x __ = – e cot + c 2
( )
349. The given integral is
(
)
1.88 Integral Calculus, 3D Geometry & Vector Booster
2 + sin 2 x dx Ú ex { _________ 1 + cos 2 x }
(
)
ex = _______ + c (x + 1)2
sin 2 x 2 = Ú e x _________ + __________ dx 1 + cos 2 x 1 + cos 2 x 2 sin x cos x 2 = Ú e x ______ 2 + _________ dx 2 cos x 2 cos2 x
354. The given integral is 1 1 – ______ dx Ú ____ log x (log x)2
= Ú e x (sec2x + tan x) dx
= Ú e (tan x + sec x) dx
= e x tan x + c
(
)
x
(
(
)
)
2 – 1) + 2 x2 + 1 x _______ x (x __________ e d x = e dx Ú (x + 1)2 Ú (x + 1) 2
(
)
(
)
(x – 1) = e ______ + c (x + 1) x
351. The given integral is
log x __________ dx Ú (1 + log x)2 t
t e = Ú ______ dt (t + 1)2
Let log x = t dx fi dt = ____ log x e
[(t + 1) – 1]et = Ú ____________ dt (t + 1)2
1 1 = Ú et ____ + – ______ dt t+1 (t + 1)2
[
(
e = ____ + c t+1 x = ________ + c log x + 1
Ú e x ( log x + __ 1x ) dx = e x log x + c
(
)
x–1 dx Ú e (x_______ + 1)3
(
)
(x + 1) – 2 = Ú ex __________ dx (x + 1)3
1 2 = Ú ex _______ + – _______ dx (x + 1)2 (x + 1)3
[
(
)]
(
)
1–x dx Ú ex ______ 1 + x2 2
( (
)
(1 – x)2 = Ú ex ________ dx (1 + x2)2
(1 + x2 – 2x) = Ú ex ___________ dx (1 + x2)2
– 2x 1 = Ú ex _______ + _______ dx (1 + x2) (1 + x2)2
ex = _______ + c (1 + x2)
[
(
)
)]
356. The given integral is
–x dx Ú ex ( 11 _____ + x) 2
( ( (
)
(1 – x)2 = Ú ex _______2 dx (1 + x)
(1 + x2) – 2x = Ú ex ___________ dx (1 + x)2
x2 – 1 – 2 (x – 1) = Ú ex _______________ dx (x + 1)2
2 (x – 1) x2 – 1 = Ú e x _______ – ________ dx 2 (x + 1) (x + 1)2
2 (x – 1) x2 – 1 = Ú e x _______ – ________ dx 2 (x + 1) (x + 1)2
2(x – 1) x–1 = Ú ex _____ – _______ dx x+1 (x + 1)2
353. The given integral is x
dx fi dt = ____ log x e
t
352. The given integral is
)]
Let t = log x
355. The given integral is
(x – 1) _______ 2 = Ú ex ______ + dx (x + 1) (x + 1)2
( )
}
1 1 = Ú e t __ – __ dt, t t2 et = __ + c t x = ____ + c log x
2
350. The given integral is
{
( ( ( ( ( (
)
)
) ) )
) 2 (x + 1 – 2) x – 1 = Ú e _____ ) – ___________ dx x+1 (x + 1) ) 4 = Ú e ( _____ dx ( xx +– 31 ) – (x_______ + 1) ) x
2
x
2
Indefinite Integrals
(
)
x–3 = Ú ex _____ + c x+1 357. The given integral is
(
x +1 _______ dx Ú ex (1 + x)2
( (
x – 1 + 2 = Ú ex _________ dx (x + 1)2
x2 – 1 2 = Ú e _______ + _______ dx (x + 1)2 (x + 1)2
x–1 2 = Ú e _____ + _______ dx x+1 (1 + x)2
x–1 = ex _____ + c x+1
x
x
Ú ecos
( ( )
(
)
( ( (
) (
)
)
= esin x (x + sec x) + c
(
)
1 362. Ú e x log x + __ 2 dx x _____
)
Let cos–1 x = t
dx ______ fi ______ = – dt ÷ 1 – x2
)
{ (
(
)
)}
1 1 __ 1 __ = Ú ex log x – __ x + x + x2 dx 1 = e x log x – __ x + c 1 363. Ú log (log x) + ______ 2 dx (log x)
+ 1) + ÷1 – x2 ) x (x ______ _______________ dx (x + 1)2 ÷ 1 – x2
( (
))
1 = et sin–1 t + ______ _____ + c ÷ 1 – t2
358. The given integral is
(
t 1 1 = Ú et sin–1 t + ______ _____ – ______ _____ + ________ dt 2 2 (1 – t2)3/2 ÷1 – t ÷ 1 – t
)
( )
–1
fi cos x dx = dt
t = Ú et sin–1 t – ________ dt (1 – t2)3/2
)
2
)
x – tan x = Ú esin x _______ cos x dx cos2x Let sin x = t
)
2
(
(
{
)
(
}
)
1 = Ú et log t + __ 2 dt, t
Let
t = log x
fi
dx dt = ____ logx . e
1 + cos t + sin t = – Ú et _____________ dt (1 + cos t)2
sin t 1 = – Ú et _________ + _________ dt (1 + cos t) (1 + cos t)2
1 = Ú et log t – __ + t
et = – _______ + c 1 + cos t
1 = et log t – __ + c t
e cos x = – _____ + c 1+x
1 = x log (log x) – ____ + c log x
)
–1
359. We have,
Ú e (2sec x – 1) tan x dx 2
x
= Ú e x (2 sec2x tan x – tan x) dx = Ú e x [(sec2x – tan x) + (2 sec2x tan x – sec2x)] dx = Ú e x (sec2x – tan x) + c 360. We have,
Ú ex (log (sec x + tan x) + sec x) dx
= ex log (sec x + tan x) + c 361. We have,
(
3
)
x cos x – sin x Ú e sinx ____________ dx cos2x
1.89
= Ú esin x (x cos x – sec x tan x) dx
(
[ (
)
(
(
) ( __ 1t + __ t1 ) ] dt 2
)
)
x4 + 2 364. Ú ex _________ dx (1 + x2)5/2
(
)
(1 + x2)2 + 1 – 2x2 = Ú e x ________________ dx (1 + x2)5/2 x 1 = Ú ex _______ _____ – ________ 2 (1 – x2)3/2 ÷1 + x
( {
{
}
})
x 1 – 2x2 + _________ + _________ dx 2 3/2 (1 + x2)5/2 (1 + x )
(
)
x 1 = ex _______ ______ + ________ + c 2 3/2 2 ÷ 1 + x (1 – x ) esin x (x cos3x – sin x) 365. Ú _________________ dx cos2x
= Ú esin x (x cos x – sec x tan x) dx
1.90 Integral Calculus, 3D Geometry & Vector Booster
Let sin x = t fi cos x dx = dt dt dt dx = ____ cos x = ______ _____ – t2 ÷ 1
fi
(
)
t = Ú e sin t – ________ dt (1 – t2)3/2 t
( (
–1
)
)
1 _____ = et sin–1 t – ______ + c – t2 ÷ 1
( (
(
)
3
x – x – 2 2 dx Ú ex _________ (x + 1)2
)
( {
}
{
})
2
1–x – 2x + _______ + _______ 2 2 dx 2 2 (x + 1) (x + 1)
( (
)
x 1 = Ú ex _______ 2 + _______ + c (x + 1) (x2 + 1)
)
x+1 = ex _______ + c (x2 + 1)
368. We have, e2x (sec2x + 2 tan x) dx = Ú e2x (2 tan x + sec2x) dx
= Ú e2x [2 cos x + (– sin x)] dx
= e2x 2 cos x + c
= e 2 tan x + c
(
( ) sin 2 x 1 = Ú e ( _________ + _________ dx 1 + cos 2 x 1 + cos 2 x ) 1 + sin 2 x = Ú e2x __________ dx 1 + cos 2 x 2x
(
)
2 sin x cos x 1 = Ú e2x __________ + _________ dx 2 1 + 2cos x 2 cos2x
1 = Ú e2x __ sec2x + tan x dx 2
1 = __ Ú e2x (sec2x + 2 tan x) dx 2
1 = __ Ú e2x (2 tan x + sec2x) dx 2
e2x 2 tan x = ________ + c 2
(
)
)
(
________
)
x – __ ÷1 – sin x e 2 _________ dx
Ú
1 + cos x
(
)
x – __ cos(x/2) – sin (x/2) = Ú e 2 ________________ dx 2 cos2 (x/2)
x – __ 1 x x x 1 = Ú e 2 __ sec __ – __ sec __ tan __ dx 2 2 2 2 2
( ( ) ( ) ( ) ) x x x 1 1 = Ú e ( – __ sec ( __ ) + __ sec ( __ ) tan ( __ ) ) dx 2 2 2 2 2 x – __ 2
( )
x – __ x = Ú e 2 sec __ + c 2
2 sin 4 x – 4 dx Ú e2x ___________ 1 – cos 4 x
(
1 + sin 2 x dx Ú e2x ( 1_________ + cos 2 x )
2x
369. We have,
Ú e2x (– sin x + 2 cos x) dx
Ú e3x (3 sin x + cos x) dx = e3x sin x + c
372. We have,
367. We have,
= e2x cot 2x + c
)
x 1 = Ú ex _______ + _______ 2 2 (x + 1) (x + 1)
x (x2 + 1) + (x2 + 1) + (1 – 2x2) – 2x = Ú e ________________________________ dx (x2 + 1)2
= Ú e2x (2 cot 2 x + (– 2 cosec22x)) dx
(x3 + x) + (x2 + 1) + (1 – 2x2) – 2x = Ú ex _______________________________ dx (x2 + 1)2 x
371. We have,
= esin x (x – sec x) + c
)
4 sin 2 x cos 2 x _______ 4 = Ú e2x ____________ – 2 dx 2 sin22x 2sin 2x
370. We have,
t 1 1 = Ú et sin–1 t + ______ _____ – ______ _____ – ________ dt 2 2 (1 – t2)3/2 ÷ 1 – t ÷1 – t
366. We have,
(
373. We have,
)
2 sin 4 x 2 = Ú e x _________ – _________ dx 1 – cos 4 x 1 – cos 4 x
Ú e2 x [2 × log (sec x + tan x) + sec x] dx = e2 x (log (sec x + tan x)) + c
1.91
Indefinite Integrals
= Ú e– 2t sin t dt,
374. We have,
Ú e sin 3 x dx x
ex = ______ 2 2 (1 ◊ sin 3 x – 3 cos 3 x) + c 1 +3 ex = ___ (sin 3 x – 3 cos 3 x) + c 10
375. We have,
Ú e4x cos 3 x dx e4x = ______ 2 2 (4 cos 3 x + 3 sin 3 x) + c 4 + 3
4x
e = ___ (4 cos 3 x + 3 sin 3 x) + c 25
Ú e2x sin 3 x dx
e2x = ___ (2 sin 3 x – 3 cos 3 x) + c 13
e– 2 log x = ______ [2 sin (log x) + cos (log x)] + c 5 1 = – ___ 2 [2 sin (log x) + cos (log x)] + c 5x
x x = __ ÷ 4 – x2 + 2 sin–1 __ + c 2 2
382. We have,
1 dx = 3 Ú ( __ dx Ú ÷1 – 9x ÷ 3 ) – x2 2
( ÷
( ) x ( ___ ) ) + c 1/3
( ÷
x 1 1 = 3 __ __ – x2 – ___ sin–1 2 9 18
e = ______ (– cos x + sin x) + c (1 + 1)
383. We have, _____
– x
e = ___ (– cos x + sin x) + c 2
)
1 __ ______ 9 x __ x 1 2 __ __ = 3 – x – sin–1 ___ + c 2 9 2 1/3 ______
– x
______
_____
| + c Ú ÷x 2 + 1 dx = __ 2x ÷ x2 + 1 + __ 12 log | x + ÷x 2 + 1
384. We have, _______
Ú e– x cos (3x + 4) dx
e2x = _______ [2 cos (3x + 4) + sin (3x + 4)] + c 3(4 + 1)
379. We have,
2 + 1 dx Ú ÷3x
÷ x 1 1/3 =÷ 3 ( __ x + __ + ___ log x + x + __ 2÷ 3 2 | ÷ 13 | )+ c ______
__ 1 = ÷3 Ú x2 + __ dx 3
Ú e
x
=
=
=
=
380. We have, 1 sin (log x) dx Ú __ x3
______ 2
385. We have,
1 __ Ú ex (2cos2x) dx 2 1 __ Ú ex (1 + cos 2 x) dx 2 1 __ Ú (ex + ex cos 2 x) dx 2 ex 1 __ ex + __ (cos 2 x + 2 sin 2 x) + c 2 5
[
2
2
cos x dx
______
__
( )
________
2
378. We have,
______
x x 4 – x2 dx = __ ÷ 4 – x2 + __ sin–1 ( __ ) + c Ú ÷4 2 2 2
Ú e– x cos x dx
_____
_______
377. We have,
e– 2t = ____ (– 2 sin t – cos t) + c 5 e– 2t = – ____ (– 2 sin t + cos t) + c 5
______
e2x = ______ (2 sin 3 x – 3 cos 3 x) + c (4 + 9)
= Ú e– 2t sin t dt
381. We have,
376. We have,
Let t = log x dx fi dt = ____ logx e
]
______
dx Ú ÷x 2 – 9 ______
______
|
|
9 x = __ ÷x 2 – 9 – __ log x + ÷ x 2 – 9 + c 2 2
386. We have,
_______
2 – 1 dx Ú ÷4x
________
÷ ( )
1 2 = 2 Ú x2 – __ dx x
x 1/4 1 ___ 1 = 2 __ x2 – __ – log x + x2 – __ + c 2 4 2 4
( ÷
______
|
÷
______
|)
1.92 Integral Calculus, 3D Geometry & Vector Booster
( ÷
_____
1 = x x – __ – 4
2
|
÷
_____
1 __ log x + x2 – 4
|)
1 __ + c 4
387. We have,
______________
= Ú ÷– {(x – a)2 – a2} dx
= Ú ÷a 2 – (x – a)2 dx
(x – a) ________2 __ a2 x – a = ______ ÷2ax – x + sin–1 _____ a + c 2 2
___________
__________
+ 3 dx Ú ÷x 2 + 2x _______________ (x2 + 2x + 1) + 2 dx
= Ú ÷
=
=
________
Ú ÷
|
c
__________
|
_______________
= Ú ÷– (x 2 + 4x – 3) dx = Ú ÷
= Ú ÷7 – (x + 2)2 dx
_______
___________
=
dx Ú ÷x – 4x2
(x + 2) __________ ______ ÷3 – 4x – x2 + 2
( )
x+2 7 __ sin–1 _____ __ + c 2 ÷7
________________
Ú
=
2
2
fi 2dx = dt
÷( ) t 1 1 = __ ___ – t + ___ sin (4t) + c 4 ÷16 32 1 ( 2x – __ ) 4 ___ 1 1 = _______ – ( 2x – __ ) ÷16 4 4 1 1 + ___ sin [ 4 ( 2x – __ ) ] + c 32 4
( 2 )
__________ x+1 2 _____ ÷ x + 2x + 3
|
1 1 2 = __ Ú __ – t2 dt 2 4 _______
__________
2
|
388. We have,
–1
_____________ 2
__________ ÷3 – 4x – x2 dx
Ú
= Ú ÷– (x 2 + 4x – 3) dx
= Ú ÷– {(x + 2)2 – 7} dx
–1
____________
_____________
392. We have,
__________ 7 – (x + 2)2 dx
__________
= Ú ÷
(x + 2) _________2 __ x +__ 2 7 = ______ ÷3 – 4x – x + sin–1 _____ + c 2 2 ÷7
( )
389. We have, ________
– x2 dx Ú ÷2ax __________
2
________
1 + __ log (x + 1) + ÷ x 2 + 2x + 3 + c 2
2
______________ __
= Ú ÷(x + 1)2 + ( ) 2 dx ÷2
1 2 1 2 __ = Ú __ – 4 x – dx 4 8
________
__________ ÷x 2 + 2x + 3 dx
÷( ) ( ) 1 1 = Ú ( __ ÷ 4 ) – 4 ( 2x – __ 4 ) dx 1 1 1 = __ Ú ( __ – t dt, Let ( 2x – __ ) = t 2 ÷ 4) 4 _______________
387. We have,
|
391. We have,
______________ – {(x2 + 2)2 – 7} dx
___________
= Ú ÷(x + a)2 – a2 dx
x2 dx Ú ÷3 – 4x –
+ x2 dx Ú ÷2ax
(x + a) ________ = ______ ÷ x 2 + 2ax 2 ________ a2 + __ log (x + a) + ÷ x 2 + 2ax + c 2
388. We have,
)
390. We have,
______________ __ (x + 1)2 + ( ) 2 dx ÷2 __________ log (x + 1) + ÷ x 2 + 2x + 3 +
|
(
= Ú ÷– (x 2 – 2ax) dx
+ 4 dx Ú (2x + 1) ÷x 2 + 3x __________ = Ú ((2x + 3) – 2) ÷ x 2 + 3x + 4 dx __________ = Ú ((2x + 3)) ÷ x 2 + 3x + 4 dx __________ – 2 Ú ÷x 2 + 3x + 4 dx
Let x2 + 3x + 4 = t2 fi (2x + 3) dx = 2t dt
÷(
___________________
) ( )
9 3 2 = Ú (2t) t ◊ dt – 2 Ú x + __ + 4 – __ dx 2 4
Indefinite Integrals
÷( 2 ) ( 2 )
= 2 Ú t dt – 2 Ú 2
__________________ __ ÷7 2 3 2 __ ___ x + + dx
( ) [ ( )
_________
__________
_________ 2
__________
)
|]
3 7 + __ log x + __ + ÷x 2 + 3x + 4 + c 4 2
[ ( )
395. We have,
__________ 3 __ ÷ x2 + 3x + 4
1 – 2 __ x + 2 2
| (
_________
__________
)
|]
392. We have, __________
__________
2
__________
= Ú (2x + 3) ÷ x 2 + 3x + 4 dx – 2 Ú x2 + 3x + 4 dx 2
Let x + 3x + 4 = t
396. We have,
÷( 2 ) ( 2 )
_______________ __ ÷7 2 3 2 ___ x + __ + dx
( ) 7 – [ __ log ( x + __ 4 | 32 ) + ÷x
_________
|]
__________ 2 + 3x + 4 +
c
_________
1 = __ Ú (2x – 2 ______ 1 = __ Ú ((2x + 1) – 11) ÷ x 2 + x dx 2 ______ 1 = __ Ú (2x + 1) ÷ x 2 + x dx 2
÷( ) ( )
_____________
11 1 2 1 2 – ___ Ú x + __ – __ dx 2 2 2 and then you do. 394. We have,
÷( ) ( )
÷5 2 1 1 2 __ + __ Ú ___ – x – dx 2 2 2 and then you do it.
Partial Fractions 397. We have,
dx Ú (3x – 2) ÷x 2 + x + 1
( ) 3 4 = __ Ú ( 2x – __ ) ÷ x 2 3
_________
2 = 3 Ú x – __ ÷ x2 + x + 1 dx 3 _________ 2 + x + 1 dx
(2x + 1)dx
Ú ____________ (x + 2)(x + 3)
(2x + 1) A B fi ____________ = ______ + ______ (x + 2)(x + 3) (x + 2) (x + 3)
_________
1 = – __ Ú (1 – 2x) ÷1 + x – x2 dx 2
_______________ __
______ 10) ÷x 2 + x dx
1 = – __ Ú (– 2x) ÷1 + x – x2 dx 2 _________ 1 = – __ Ú ((– 2x) – 1) ÷1 + x – x2 dx 2
______
1 = __ Ú (2x) ÷1 + x – x2 dx 2
dx Ú (x – 5) ÷x 2 + x
dx Ú x ÷1 + x – x2 _________
393. We have,
_________
__________ 2 (x2 + 3x + 4)3/2 3 = _______________ – x + __ ÷x 2 + 3x + 4 3 2
2
and then you do it.
2
fi (2x + 3) dx = 2t dt
= 2 Ú t2 dt – 2 Ú
_________ 2
____________
= Ú (2x + 3 – 2) ÷x 2 + 3x + 4 dx
( ) 3 = 2 Ú ( (2x – 1) + __ ) ÷ x – x – 2 dx 2 3 = 2 Ú ( (2x – 1) + __ ) ÷ x – x – 2 dx 2 3 + 3 Ú (x ÷ – 1) – ( __ 2 ) dx 1 = 2 Ú 2x + __ ÷x 2 – x – 2 dx 2
_________ 2
+ 4 dx Ú (2x + 1) ÷x 2 + 3x
dx Ú (4x + 1) ÷x 2 – x – 2
_________
3 7 + __ log x + __ + ÷ x 2 + 3x + 4 + c 2 4
__________________ __ 2 ÷ 2
and then you do.
2 = __ (x2 + 3x + 4)3/2 3
( ( ) ) 3 7 4 = __ Ú ( ( 2x + __ ) – __ ) ÷ x + x + 1 dx 2 3 3 3 7 1 + __ Ú ( x + __ ) + ( ___ ) dx 2 ÷ 2 2 3 7 4 = __ Ú 2x + __ – __ ÷ x2 + x + 1 dx 2 3 3
3 t3 1 = 2 __ – 2 __ x + __ ÷x 2 + 3x + 4 3 2 2
| (
1.93
fi
(2x + 1) A(x + 2) + B(x + 3) ____________ = _________________ (x + 2)(x + 3) (x + 2) (x + 3)
fi
2x + 1 = A(x + 3) + B (x + 2)
When
x = – 2, A = – 4 + 1 = – 3
1.94 Integral Calculus, 3D Geometry & Vector Booster and
x = – 3, B = – 6 + 1 = – 5
Thus,
Ú ____________ (x + 2)(x + 3)
dx dx = – 3 Ú _____ – 5 Ú _____ x+2 x+3
= – 3 log |x + 2| – 5 log |x + 3| + c
dx dx 1 1 + __ Ú ____________ – __ ____________ 6 (x + 2) (x + 3) 6 Ú (x + 1) (x + 2)
(2x + 1)dx
(
)
1 1 = Ú _____ – _____ dx x – 2 x – 1
= log |x – 2| – log |x – 1| + c
x–2 = log _____ + c x–1
| |
dx Ú _________________ (x + 1)(x + 2)(x + 3) ((x + 3) – (x + 1)) 1 = __ Ú __________________ 2 (x + 1)(x + 2)(x + 3) dx 1 = __ Ú ____________ – 2 (x + 1)(x + 2)
dx 1 __ Ú ____________ 2 (x + 2)(x + 3)
)
(
)
| | | | | ( ) ( ) |
x+1 x+2 1 1 = __ log _____ – __ log _____ + c 2 x+2 2 x+3
) (
)|
400. We have, dx Ú _______________________ (x + 1)(x + 2)(x + 3)(x + 4)
((x + 4) – (x + 1)) dx 1 = __ Ú _______________________ 3 (x + 1)(x + 2)(x + 3)(x + 4)
dx 1 = __ Ú __________________ 3 (x + 1)(x + 2)(x + 3)
(x + 1) – 2 = Ú ____________ dx (x + 1) (x – 2)
dx 2 = Ú ______ – ____________ dx (x – 2) Ú (x + 1) (x – 2)
dx 1 1 = Ú ______ – 6 Ú _____ – _____ dx (x – 2) x–2 x+1
= log |x – 2| – 6 log |x – 2| + 6 log |x + 1| + c
= 6 log |x + 1| – 5 log |x – 2| + c
1 ((x + 4) – (x + 2)) dx = – __ Ú __________________ 6 (x + 2)(x + 3)(x + 4)
)
2x – 1
dx Ú __________________ (x – 1) (x + 2) (x – 3)
(2x – 1) Now __________________ (x – 1) (x + 2) (x – 3)
(2x – 1) = A (x + 2) (x – 3)
+ B (x – 1) (x – 3) + C (x – 1) (x + 2)
When
x = 1,
x = – 2, then B = –1/3
and
x = 3,
then A = – 1/6 then C = 1/2
dx dx dx 1 1 1 – __ Ú ______ + __ ______ + __ ______ 6 (x – 1) 3 Ú (x + 2) 2 Ú (x + 3) 1 1 1 = – __ log |x – 1| + __ log |x + 2| + __ log |x + 3| + c 3 2 6 403. We have,
)
x k 1 _________ n – 1 + 2(n – 1) Ú _________ 2 n –1 – _______ dx 2 2 (x + k) (x + k) (x + k)n dx 1 = __ Ú ____________ – 6 (x + 2) (x + 3)
(
Thus, the given integral reduces to
dx 1 – __ Ú __________________ 3 (x + 1)(x + 2)(x + 3)
(
x–1
C A B ______ + ______ + ______ (x – 1) (x + 2) (x – 3)
x+3 x + 1 _____ 1 = __ log _____ 2 x+2 x+2
402. We have,
1 1 1 1 1 1 = __ Ú _____ – _____ dx – __ Ú _____ – _____ dx 2 x+1 x+2 2 x+2 x+3
| (
dx Ú ____________ (x + 1) (x – 2)
399. We have,
(
)|
401. We have,
dx Ú ____________ (x – 1)(x – 2)
) (
x + 2 _____ x+2 1 + __ log _____ + c x+3 x+1 6
398. We have,
| (
x + 2 _____ x+4 1 = __ log _____ x+3 x+3 6
1 dx __ Ú ____________ 6 (x + 3)(x + 4)
x3
dx Ú ____________ (x – 1) (x – 2) x3 = Ú __________ 2 dx x – 3x + 2
( Ú (
)
Indefinite Integrals
| | | |
7x – 6 = Ú x – 3 + __________ dx x2 – 3x + 2
t+2 = log _____ + c t+3
7x – 6 = x – 3 + ____________ dx (x – 1)(x – 2)
) )
sinq + 2 = log ________ + c sinq + 3
(
2
406. We have,
7x – 6 x = __ – 3x Ú ____________ dx 2 (x – 1) (x – 2)
1.95
(1 – cos x)
dx Ú _____________ cos x (1 + cos x)
7x – 6 A B Now, ____________ = ______ + ______ (x – 1) (x – 2) (x – 1) (x – 2)
(1 + cos x – 2 cos x) = Ú ________________ dx cos x (1 + cos x)
(7x – 6) = A (x – 2) + B (x – 1)
When
x = 1, then A = –1
dx dx = Ú ____ cos x – 2 Ú _________ (1 + cos x)
and
x = 2, then B = 8
1 – cos x = Ú sec x dx – 2 Ú ________ dx sin2x
= Ú sec x dx – 2 Ú (1 – cosec2 x – cosec x cost x) dx
= log |sec x + tan x| – 2(cosec x – cost x) + c
(
Thus, the given integral reduces to
(
) ( ( )
)
8 x2 –1 + ______ dx __ – 3x + Ú ______ 2 (x – 1) (x – 2) x2 = __ – 3x – log |(x – 1)| + 8 log |x – 2| + c 2
2x
2 dx Ú ______________ (x + 1) (x2 + 2)
Let x2 = t fi 2x dx = dt Thus, the given integral reduces to
dt 2 Ú ___________ (t + 1) (t + 2)
(
407. We have,
404. We have,
)
dx
Ú ___________ sin x – sin 2x
dx = Ú _______________ sin x – 2sin x cos x
dx = Ú ______________ sin x (1 – 2cos x)
sin x dx = Ú ______________ 2 sin x (1 – 2cos x)
sin x dx = Ú ___________________ (1 – cos2x) (1 – 2cos x)
1 1 – ____ dt = 2 Ú ____ t+1 t+2
)
(t + 1) = 2 log ______ + c (t + 2)
dt = Ú _____________ , where cos x = t (1 – t2) (1 – 2t)
dt = – Ú __________________ (t – 1) (t + 1) (2t – 1)
(x2 + 1) = 2 log _______ + c (x2 + 2)
1 Now, __________________ (t – 1) (t + 1) (2t – 1)
405. We have,
| |
| |
c A B ______ + ______ + _______ (t – 1) (t + 1) (2t – 1)
cosq dq Ú __________________ (2 + sinq ) (3 + sinq)
– 1 = A (t + 1) (2t – 1) + B (t – 1) (2t – 1) + c (t 2 – 1)
Let sinq = t
fi cosq dq = dt
Put
t =1,
t = –1, then B = 1/10
t =1/2, then C = 4/3
Thus, the given integral reduces to
dt Ú ___________ (t + 2) (t + 3)
(
)
1 1 = Ú _____ – _____ dt t+2 t+3
then A = –1/2
The given integral reduces to
dt 1 – __ Ú ______ + 2 (t – 1)
dt 1 ___ Ú ____ + 10 t + 1
dt 4 __ Ú _______ 3 (2t – 1)
1.96 Integral Calculus, 3D Geometry & Vector Booster 1 1 = – __ log |t – 1| + ___ log |t + 1| 2 10 4 + __ log |2t – 1| + c 3 1 1 = – __ log |cos x – 1| + ___ log |cos x + 1| 2 10
When Put and
x = – 2, A = – 3 x = 3, C = 7/5 x = 0, 9A – 6B + 2C = 1
fi
6B = 14/5 – 27 – 1
fi
B = – 21/5
4 + __ log |2cos x – 1| + c 3
2x + 1 Thus, _____________ dx (x + 2)(x – 3)2
408. We have,
dx = – 3 Ú _____ – x+2
dx
Ú _____________ (x + 1) (x + 1)2
dx dx 7 21 ___ Ú _____ + __ _______ 5 x – 3 5 Ú (x – 3)2
7 21 = – 3 log |x + 2| – ___ log |x – 3| – _______ + c 5 5 (x – 3) C 1 A B Now, ____________ = ______ + ______ + _______ (x + 1) (x – 1)2 (x + 1) (x – 1) (x – 1)2 415. We have, (2x + 3) dx 1 Ú (x_____________ _____________ fi 2 + 1) (x2 + 4) (x + 1) (x – 1) (2x + 3) Bx + C A Now, _____________ = ______ + _______ A (x – 1)2 + B (x2 – 1) + C (x – 1) 2 2 (x + 1) = _____________________________ (x + 1)(x + 4) (x + 4) (x + 1) (x – 1)2 fi 2x + 3 = A(x2 + 4) + Bx (x + 1) + C (x + 1) fi 1 = A (x – 1)2 + B (x2 – 1) + C (x + 1) When x = –1, A = 1/5 When x = –1, A = 1/4 and x = 0, 4 A + C = 3 x = 1, C = 1/2 fi C = 3 – 4 A = 3 – 4/5 = 11/5 and x = 0, A – B + C = 1 Also x = 1, 5A + 2B + 2 C = 5 fi B = 3/4 – 1 = –1/4 fi 2B = 5 – 5A – 2C = 5 – 1 – 22/5 = – 2/5 dx Thus, 2 Ú _____________ fi B = – 1/5 (x + 1) (x + 1) dx dx dx 1 1 1 = __ Ú ______ – __ ______ + __ _______ 4 (x + 1) 4 Ú (x – 1) 2 Ú (x – 1)2 1 1 1 = __ log |x + 1| – __ log |x – 1| – _______ + c 4 4 2 (x – 1) 409. We have,
2x + 1 dx Ú _____________ (x + 2)(x – 3)2
2x + 1 _____________ (x + 2)(x – 3)2
C A B = ______ + ______ + _______ (x + 2) (x – 3) (x – 3)2
2x + 1 _____________ (x + 2) (x – 3)2 fi
(2x + 3) dx Thus, Ú _____________ (x + 1)(x2 + 4) dx xdx dx 1 1 11 = __ Ú _____ – __ _____ + ___ Ú _____ 5 x + 1 5 Ú x2 + 4 5 x2 + 4 1 1 1 = __ log |x + 1| – ___ |x2 + 1| + __ tan–1 (x) + c 10 5 5 416. We have,
(3x – 2) dx
Ú _____________ (x – 1)(x2 + 9)
(3x – 2) Bx + C A Now, ____________ = ______ + _______ (x – 1)(x2 + 9) (x – 1) (x2 + 9) fi
(3x – 2) = A (x2 + 9) + Bx (x – 1) + C (x – 1)
A (x – 3)2 + B (x2 – 9) + C (x + 2) = _____________________________ (x + 2) (x – 3)2
When
x = 1, A = 1/10
and
x = 0, 9 A – C = – 2
2x + 1 = A (x – 3)2 + B (x2 – 9) + C (x + 2)
fi
C = 2 + 9 A = 2 +9/10 = 29/10
Indefinite Integrals
Also
x = –1, 10A + 2B – 2 C = – 5
421. We have,
fi
2 B = 2 C – 10A – 5 = 29/5 –1 – 5
fi
2B = –1/5
x dx Ú _____________ (x2 – 1)(x2 + 1)
fi
B = –1/10
Put
x2 = t
Thus,
(3x – 2) dx Ú _____________ (x – 1)(x2 + 9)
2
t A Now, ___________ = _______________ (t – 1)(t + 1) (t – 1) + B (t + 1) A(t + 1) + B (t – 1) t fi ___________ = _________________ (t – 1)(t + 1) (t – 1)(t + 1)
29 dx xdx dx 1 1 = ___ Ú _____ – ___ _____ + ___ _____ 10 x – 1 10 Ú x2 + 9 10 Ú x2 + 9 fi 1 1 Put = ___ log |x – 1| – ___ log |x2 + 9| 10 20 Put
( )
29 x + ___ tan–1 __ + c 30 3 417. We have, 2x – 1 dx Ú _____________ (x + 1)(x2 + 2)
Bx + C 2x – 1 A = ______ + _______ Now, _____________ (x + 1) (x2 + 2) (x + 1) (x2 + 2)
(2x – 1) = A (x2 + 2) + Bx (x + 1) + C (x + 1)
When
x = –1, then A = –1
x = 0, then C = 1 and x = 1, then B = 1 Thus, the given integral reduces to
t = A (t + 1) + B (t – 1) t = 1, A = 1/2 t = –1, A = –1/2 2
Thus,
x dx Ú _____________ 2 (x – 1)(x2 + 1)
dx 1 1 = __ Ú _____ – __ _____ 2 x2 – 1 2 Ú x2 + 1
x–1 1 1 = __ log _____ – __ tan–1 (x) + c 4 x+1 2
| |
426. We have,
(x2 + 3)(x2 + 1)
dx Ú ______________ (x2 – 1)(x2 + 2)
x2 = t
Put
(t + 3)(t + 1) __________ t2 + 4t + c Thus, ___________ = 2 (t – 1)(t + 2) t +t–2
dx dx x – Ú ______ + _______ dx + Ú _______ (x + 1) Ú (x2 + 2) (x2 + 2)
( )
x 1 1 __ tan–1 ___ __ + c = – log |x + 1| + __ log |x2 + 2| + ___ 2 2 ÷2 ÷ x _____________ dx 418. Ú (x + 1)(x2 + 4)
1.97
(
) (
)
– 3t + 5 – 3t + 5 = 1 + ________ = 1 + ___________ 2 (t + 1)(t – 2) t +t–2 – 3t + 5 A B Now, ___________ = ____ + ____ (t + 1)(t – 2) t – 1 t – 2 fi
(– 3t + 5) = A(t – 2) + B(t – 1)
Bx + C x A Now, Ú ____________ = ______ + _______ (x + 1)(x2 + 4) (x + 1) (x2 + 4)
When
t = 1, then A = –2
and
t = 2, then B = –1
x = A(x2 + 4) + Bx (x + 1) + C(x + 1)
Therefore, the given integral reduces to
Let and Thus, the
dx dx – Ú _____ Ú dx – 2 Ú x_____ 2 –1 x2 – 2
x – ÷2 x–1 1 __ = x – log _____ – ____ __ log ______ + c x+1 2÷2 x + ÷2
x = –1, then A = –1/5 x = 0, then C = 4/5 x = 1, then B = 1/5 given integral reduces to
427. We have,
dx dx x 1 1 4 – __ Ú ______ + __ _____ + __ _____ 5 (x + 1) 5 Ú x2 + 4 5 Ú x2 + 4
( )
x 1 1 2 = – __ log |(x + 1)| + __ log |x2 + 4| + __ tan–1 __ + c 2 5 5 5 419. Do yourself 420. Do yourself
| |
| |
(x2 + 1)(x2 + 2)
dx Ú ______________ (x2 + 3)(x2 + 4)
x2 = t (t + 1)(t + 2) Now, ___________ = (t + 3)(t + 4) Put
t2 + 3t + 2 __________ t2 + 7t + 12
__
1.98 Integral Calculus, 3D Geometry & Vector Booster
(
) (
)
(
)
4t + 10 4t + 10 = 1 – __________ = 1 – ___________ 2 (t + 3)(t + 4) t + 7t + 12
Thus,
6 2 2 – ______ dt Ú 1 + ______ (t + 2) (t2 + 4)
4t + 10 A B Again, ___________ = ______ + ______ (t + 3)(t + 4) (t + 3) (t + 4)
__ t t = t + ÷ 2 tan–1 ___ __ – 3 tan–1 __ + c 2 2 ÷
(4t + 10) = A (t + 4) + B (t + 3)
When
t = – 4, then B = 6
and
t = – 3, then A = – 2
Hints
and
2
( )
( )
1 – __ (sec x ◊ tan x + log |sec x + tan x|) + c 2
2I = sec3 x tan x 1 – __ (sec x tan x + log |sec x + tan x|) + c 2 1 I = __ (sec3 x tan x) 2 1 – __ (sec x tan x + log |sec x + tan x|) + c 4
( ) )
solutions
fi
Ú tan x ◊ sec x dx 4
( )
__ x2 x2 = (x2 + ÷2 tan–1 ___ __ – 3 tan–1 __ + c 2 2 ÷ 428. Do yourself.
(
Let tan x = t fi sec2 x dx = dt = Ú t4dt
fi
t5 = __ + c 5 5 (tan x) = ______ + c 5
2. Let I = Ú sec x dx 3
4. Ú tan2 x sec4 x dx = Ú tan4 x (1 + tan2 x) sec2 x dx
Let tan x = t
fi sec2xdx = dt
= Ú t2 (1 + t2) dt
= Ú (t4 + t2) dt
= sec x ◊ tan x – Ú (sec x ◊ tan x) dx
t 5 t3 + c = __ + __ 3 5
= sec x ◊ tan x – Ú (sec x (sec2 x – 1)) dx
tan5 x _____ tan3 x + + c = _____ 3 5
= sec x ◊ tan x – Ú sec3x dx + Ú sec x dx
= sec x ◊ tan x – I + log |sec x + tan x| + c
= Ú tan3 x (1 + tan2 x)2 sec2 x dx
fi 2I = sec x ◊ tan x + log |sec x + tan x| + c 1 fi __ [sec x ◊ tan x + log |sec x + tan x|] + c 2
= Ú t3 (1 + t2) 2 dt,
= Ú sec2 x ◊ sec x dx = sec x Ú sec x dx – Ú (sec x ◊ tan x) tan x dx 2
2
3. Let I = Ú sec5 x dx
( (
)
)
5. Ú tan3 x sec6 x dx
Let tan x = t fi sec2x dx = dt
= Ú t3 (t4 + 2t2 + 1) dt
= Ú sec3 x ◊ sec2 x dx
= Ú (t7 + 2t5 + t3) dt
= sec3 x Ú sec2 x dx – Ú (3 sec3 x ◊ tan x) ◊ tan x dx
t8 t3 __ t4 = __ + __ + + c 8 3 4
= sec x ◊ tan x – 3 Ú sec x (sec x – 1) dx
= sec3 x ◊ tan x – 3 Ú sec5 x dx + 3 Ú sec3 x dx
tan8 x _____ tan3 x _____ tan4 x = _____ + + + c 8 3 4
= sec3 x ◊ tan x – I
3
3
2
( (
6. Ú tan3 x sec5 x dx
)
)
Indefinite Integrals
1.99
= Ú sec4 x ◊ tan2 x sec x ◊ tan x dx
– Ú (– cot x) (– cosec x ◊ cot x) dx
= Ú sec4 x (sec2 x – 1) sec x ◊ tan x dx
= – cosec x ◊ cot x – Ú (cosec2 x – 1) cosec x dx
= – cosec x ◊ cot x – I – Ú cosec3 x dx + Ú cose x dx
= – cosec x ◊ cot x – I + log |cosec x – cot x| + c
Let sec x = t
fi secx ◊ tan xdx = dt
= Ú t4 (t2 – 1) dt,
= Ú (t6 – t4) dt
fi
2I = – cosec x cot x + log |cosec x – cot x| + c
t7 t5 = __ – __ + c 7 5
fi
1 1 I = – __ cosec x cot x + __ log |cosec x – cot x| + c 2 2
sec7 x _____ sec5 x = _____ – + c 7 5
( (
7. Ú sec x dx
8. Ú sec9 x dx
9. We have,
)
12. Ú cot2 x cosec4 x dx
)
4
Ú cosec2 x ◊ cot2 x dx
Let cot x = t 2
fi cosec x dx = – dt
= – Ú t dx 3
t = – __ + c 3 3 (cot x) = – ______ + c 3 10. Let I = Ú cot3 x cosec3 x dx Let cosec x = t fi – cosec x ◊ cot x dx = dt fi cosec x ◊ cot x dx = – dt
2
= Ú cot3 x ◊ cosec3 x dx
= Ú cot2 x (1 + cot2 x) cosec2 x dx = – Ú t2 (1 + t2) dt, = – Ú (t4 + t2) dt
(
)
t5 t3 = – __ + __ + c 3 5 x x 5 __ = – cot + cot3 __ + c 3 5
(
)
13. Ú tan– 5 x ◊ sec6 x dx
sec6 x = Ú _____ dx tan5 x
(1 + tan2 x)2 sec2 x = Ú _______________ dx tan5 x
(1 + t2)2 = Ú _______ dt t5
t4 + 2t2 + 1 = Ú __________ dt t5
( (
)
= Ú cot2 x ◊ cot x ◊ cosec3 x dx
= Ú (cosec x – 1) ◊ cosec x (cosec x ◊ cot x) dx
= – Ú (t2 – 1) t2 dt
= Ú (t4 – t2) dt
t5 t3 = __ – __ + c 5 3
14. Ú cot3 x ◊ cosec– 8 x dx
cosec5 x _______ cosec3 x = _______ – + c 3 5
2
( (
)
11. Let I = Ú cosec3 x dx
= Ú cosec2 x ◊ cosec x dx = cosec x Ú cosec2 x dx
2
)
Let cot x = t fi cosec2x dx = dt
( ( (
)
)
1 2 __ 1 = Ú __ + __ 3 + 5 dt t t t 1 1 = log |t| – __ 2 – ___ 4 + c t 4t 1 1 = log |tan x| – _____ 2 – ______ 4 + c tan x 4 tan x
)
cot3 x = Ú _______ dx cosec8 x
cos3 x = Ú _____ dx sin5 x
(1 – sin2 x) cos x = Ú _____________ dx sin5 x
)
1.100 Integral Calculus, 3D Geometry & Vector Booster
( )
1 – t2 = Ú _____ dt, Let t = cos x t5 dt = – sin xdx
( ( (
)
1 1 = Ú __ 5 – __ 3 dt t t 1 1 = ___ 2 – ___ 4 + c 2t 4t 1 1 = ______ 2 – ______ + c 2 sin x 4 sin4 x
| |
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
| | | |
x3 1 dx 20. Ú ________ = __ log _____ 3 + c x +1 x (x4 + 1) 3
)
dx dx 21. Ú __________ = Ú ___________ 1 6/7 8 x2 (x7 + 1)6/7 x 1 + __ 7 x
( )
)
15. Ú cosec5 x dx
16. Ú cosec7 x dx
17. We have,
dx dx = Ú _________ Ú ________ 7 8 – 7
x (x + 1)
x (1 + x )
Put
1 + x– 7 = t x– 8
fi
dx 1 fi ___8 = – __ dt 7 x
– 7 dx = dt
1 dt = – __ Ú __ 7 t 1 = – __ log |t| + c 7 1 __ = – log |1 + x– 7| + c 7
t6 dt = – Ú ____ t6 = – t + c
(
dx dx = Ú ___________ Ú x__________ 2 3/4 5 (1 + x) x (1 + x– 4)3/4
Let 1 + x– 4 = t4
fi – 4x– 5 dx = 4t3 dt dx fi ___5 = – t3 dt x
dt = – Ú t3 __3 t = – Ú dt
)
(
)
1 1/3 dx dx 1 + __ 22. Ú ___________ = Ú ___ 3 3 3 1/3 4 x x (1 + x ) x
(
t2 dt = – Ú ____ t
= – Ú t dt t2 = – __ + c 2 1 1 2/3 = – __ 1 + __ 3 + c 2 x
(
)
23. We have,
= – t + c = – (1 + x– 4)1/4 + c
dt 1 = __ Ú ______ , Let x4 = t 4 t (t – 1) 4x3dx = dt
(
)
1 Put 1 + __ 3 = t3 x 3 fi – __4 dx = 3t2 dt x dx fi ___4 = – t2 dt x
dx x4 _________ 19. Ú ________ = dx Ú x (x5 + 1) x5 (x5 + 1)
)
1 1/7 = – 1 + __ 7 + c x
18. We have,
(
1 Let 1 + __ 7 = t7 x 7 fi – __6 dx = 7t6 dt x dx ___ fi 6 = – t6 dt x
)
1 1 ____ 1 = __ Ú __ – dt 4 t t+1
2
x dx Ú _______ (x + 3)2
Let x + 3 = t fi dx = dt (t – 3)2 = Ú ______ dt t2 t2 – 6t + 9 = Ú _________ dt t2 6 9 = Ú 1 – __ + __ 2 dt t t
(
)
1.101
Indefinite Integrals
(
)
9 = t – 6 log |t| – __ + c t
9 = (x + 3) – 6 log |x + 3| – ______ + c (x + 3)
(
)
24. We have,
3
x dx Ú ________ (2x + 3)2
Let 2x + 3 = t
fi
2 dx = dt
fi
dx = 1/2 dt
fi
( )
3–t x = ____ 2
(
2
3
)
27 1 27 ___ = ___ Ú ___ – + 9 – t dt t 16 t2
t2 27 1 = ___ – ___ – 27 log |t| + 9t – __ + c t 2 16
( Ú (
)
)
4 1 (t + 2) = ____ Ú ______ dt 243 t3
t4 + 8t2 + 24t2 + 32t + 15 1 = ____ Ú ______________________ dt 243 t3
x2 25. Ú _______ dx (x + 2)3
Let (x + 2) = t
=
=
=
fi dx = dt (t______ – 2)2 Ú t3 dt t2 – 4t + 4 dt Ú _________ t3 Ú __ 1t – __ t42 + __ t43 dt 4 2 log |t| + __ – __ + c t t2 4 2 log |(x – 2)| + ______ – _______ + c (x + 2) (x + 2)2
(
26. Ú x2 (ax + b)2 dx
)
)
(
)
(
)
15 1 24 32 ___ = ____ Ú t + 8 + ___ + ___ 2 + 3 dt t 243 t t 2 32 15 1 t = ____ __ + 8t + 24 log |t| – ___ – ___2 + c t 243 2 2t
(
)
where t = (3x – 2) 29. We have,
dx dx = Ú __________ Ú __________ +2 3 5 3x ______ x3 (3x + 2)3
(
)
x x
Let ax + b = t 1 fi dx = __ a dt b fi x = t – __ a
Let 3x – 2 = t dt fi dx = __ 3 t____ +2 and x = 3
where t = 2x + 3
(
dt 1 = __ Ú ________ where (x3 + 1) = t 3 t2 (t3 – 1)
=
)
x4 28. Ú ________ dx (3x – 2)3
1 (27 – 27t + 9t – t ) = ___ Ú _________________ dt 16 t2
(
b2 1 = __ 2 (ax + b) – 2 log |(ax + b)| – _______ + c (ax + b) a
2x = 3 – t
=
)
and then use partial fractions.
3 1 (3 – t) = ___ Ú ______ 16 t2 dt
(
dx x2 27. Ú _________ 3 2 = Ú _________ dx x (1 + x ) x3 (1 + x3)2
3–t 3 ____ 2 1 = __ Ú ______ dt 2 t2
2 1 (t – b) = __ 2 Ú ______ dt a t2 2 2 1 (t – 2bt + b = __ 2 Ú ____________ dt a t2 1 = __ 2 Ú (1 – 2b/t + b2/t2) dt a b2 1 = __ 2 t – 2b log |t| – __ + c t a
( (
)
2 Let 3x + __ x = t 2 fi 3 + __ x = t 2 fi – __2 dx = dt x dx dt fi ___2 = – __ 2 x
)
2 Also, __ x = t – 3 t–3 1 ____ fi __ x = 2
1.102 Integral Calculus, 3D Geometry & Vector Booster
3 1 (t – 3) = ___ Ú ______ dt 16 t3
3 2 1 (t – 3t + 3t – 27) = – ___ Ú _________________ dt 16 t3
3 3 ___ 27 1 = – ___ Ú 1 – __ + __ 2 – 3 dt t 16 t t
3 27 1 = – ___ t – 3 log |t| – __ + ___ 2 + c t 16 2t
( (
(
dt 1 = – __ Ú ________ 2 ____ 2 3 3 t t –1
( )
3 1 (t – 1) = – ___ Ú ______ dt 16 t3
)
)
)
3x + 2 where, ______ x = t
(
(
)
x x
( ) b fi ( a + __ x ) = t
b + ax Let ______ x = t
b fi – __2 dx = dt x dx dt fi – ___2 = __ b x –a 1 t____ Also, __ x = b
3 1 (t – a) = – __4 Ú ______ dt b t3
( (
)
3a 3a2 __ a3 1 = – __4 Ú 1 – ___ + ___ 2 – 3 dt t b t t 3a2 1 = – __4 t – 3a log |t| – ___ + t b b where a + __ x = t
)
a3 ___ 2 + c 2t
)
dx dx 31. Ú _________ = Ú _________ x+2 3 5 _____ x2 (x + 2)3 x x
( )
)
)
dx dx 32. Ú __________ 2 = Ú _________ 3 a + bx 2 5 ______ x (a + bx) x x
(
)
(
)
a + bx Let ______ x = t
fi
a – __2 dx = dt x
dx fi ___2 = – __ a x a Also, x = ____ t–b
dt 1 ________ = – __ a Ú ____ a 3 2 t t–b 3 1 (t – b) = – __4 Ú ______ dt a t2
( Ú (
)
1 t3 – 3t2b + 3t b2 – b2 = – __4 Ú _________________ dt a t2
)
3b2 __ b3 1 = – __4 t – 3b + ___ – 2 dt t a t
(
)
b2 1 t2 = – __4 __ – 3bt + 3b2 log |t| + __ + c t a 2
(
( )
x+2 Put _____ x = t
fi
)
1 __ 3 dt t
)
a + ax where ______ x = t
3 __ 2 – t
( )
3 2 2 3 1 (t – 3t a + 3ta – a ) = – __4 Ú ___________________ dt b t3
( (
3 1 = – ___ Ú 1 – __ + t 16
2 where t = x + __ x
dx dx = Ú __________ Ú __________ b + ax 2 5 ______ x3 (b + ax)2
(
)
3 1 1 = – ___ t – 3 log |t| – __ + ___ 2 + c t 16 2t
30. We have,
(
1 t3 – 3t2 + 3t – 1 = – ___ Ú ______________ dt 16 t3
2 – __2 dx = dt x
dx dt fi ___2 = – __ 2 x 2 Also, 1 + __ x = t 2 fi x = ____ t –1
dx 33. Ú _________ 2 x (1 + x2)3
(
)
dx dx ______ a – bx 2 ___ 34. Ú _________ = Ú x x2 (a – bx)2 x2
(
)
a – bx Let ______ = t x
a fi __2 dx = – dt x
Indefinite Integrals
dx dt fi ___2 = – __ a x a Also, __ x = t + b a fi x = ____ t+b
dt 1 ________ = – __ a Ú ____ a 2 2 t t+b
( )
)
)
)
a – bx where t = ______ x dx dx 35. Ú __________ 2 = Ú __________ 4 2x + 1 3 x (2x + 1) x7 ______ x 2x + 1 Let ______ = t x
(
)
(
)
dx fi ___2 = – dt x 1 Also, 2 + __ x = t 1 fi x = ____ t–2
( )
(
)
80 ___ 2 + t
( )
dt = – Ú __3 × (t – 1)5 t
(t – 1)5 = – Ú ______ dt t3
(t5 – 5t4 + 10t3 – 10t2 + 5t – t5) = – Ú ___________________________ dt t3 10 5 __ 1 = – Ú t2 – 5t + 10 – ___ + __ 2 – 3 dt t t t
(
)
80 ___ – 40 + 10t – t2 dt t
(
( )
)
dx dx _____________ = Ú _______________ Ú _______________ 4 3 5 (x – 1)3/4 (x + 2)5/4 ÷(x – 1) (x + 2)
dx = Ú ______________ 3/4 x – 1 _____ (x + 2)2 x+2
( )
x–1 Let _____ = t x+2
dx fi _______ = dt (x + 2)2
16 ___ 80 t3 = – ___ + + 80 log |t| – 40t + 5t2 – __ + c 2 t 3 t 2x + 1 where t = ______ x
1 dt Thus, __ Ú ___ 3 t3/4
36. We have,
)
5 t3 5t2 1 = – Ú __ – ___ + 10t – 10 log |t| – __ + ___ 2 + c t 3 2 2t x–1 where, t = _____ x–2
(t – 2)5 dt = – Ú ________ t3 (2 – t)5 dt = Ú ________ t3 32 – 80t + 80t2 – 40t3 + 10t4 – t5 = Ú ____________________________ dt t3
(
dx = Ú _____________ x–1 2 _____ (x – 2)7 x–2
37. We have,
dt = – Ú ______ 5 t3 1 ____ t–2
32 = Ú ___ 3 – t
b2 1 = – __3 t + 2b log |t| – __ + c t a
(
(x – 2) ◊ 1 (x – 1) ◊ 1 fi _______________ dx = dt (x – 2)2 – dx fi _______ = dt (x – 2)2
2t – 1 Also, x = _____ t–1 1 fi (x – 2) = ____ t–1 dx Thus, Ú _____________ (x – 1)3 (x – 2)4
2b b2 1 = – __3 Ú 1 + ___ + __ 2 dt t a t
(
( )
x–1 Let _____ = t x–2
2 1 (t + b) dt = – __3 Ú ________ a t2 2 2 1 (t + 2bt + b ) = – __3 Ú ____________ dt a t2
(
1.103
dx Ú _____________ (x – 1)3 (x – 2)4
1 = __ Ú t– 3/4 dt 3 1 __ = (4t1/4) + c 3
( )
4 x – 1 1/4 = __ _____ + c 3 x+2
1.104 Integral Calculus, 3D Geometry & Vector Booster dx 38. Ú _____________ (x – 1)3 (x – 2)2 dx = Ú _____________ x+2 2 (x – 1)5 _____ x–1
dx dx ______________ 42. Ú ________________ = Ú _______________ 4/5 5 4 6 (x + 1) (x + 3)6/5 + 1) (x + 3) ÷ (x
( ) x+2 Put ( _____ = t x – 1)
(x – 1) ◊ 1 – (x + 2) ◊ 1 fi __________________ dx = dt (x – 1)2
fi
dx fi _______ = (x – 1)2
x+2 Also, _____ = t x–1
fi
fi
fi
3 – _______ dx = dt (x – 1)2
(
dt ___ – 3
)
x + 2 = tx – t t+2 x = _____ t–1 3 t+2 x – 1 = _____ – 1 = ______ t–1 (t – 1)
dt 1 = – __ Ú ________ 3 ____ 3 3 2 t t–1
( )
3 1 (t – 1) dt = – __4 Ú ________ 3 t2 3 2 1 (t – 3t + 3t – 1) dt = – __4 Ú ________________ 3 t2
( (
)
3 1 1 = – __4 Ú – 3 + __ – __ dt t t2 3 1 t2 1 = – __4 __ – 3t + 3 log |t| + __ + c t 2 3 x + 2 where t = _____ x–1
)
dx dx 39. Ú _____________ 5 = Ú _____________ 4 x–3 4 (x – 3) (x – 2) (x – 2)9 _____ x–2
( )
dx = Ú ______________ x + 1 4/5 2 _____ (x + 3) x+3
( )
x–3 Put _____ = t and then you do it x–2
( )
(
(
)
)
x+1 Put _____ = t5 and then you do it. x+3 43. We have,
2
dx x dx 3 = Ú __________ Ú _________ 3 x (2 + 3x ) x (2 + 3x3)
3x2 dx 1 = __ Ú __________ 3 x3 (2 + 3x3)
dt – __2 t 1 = __ Ú ________ 3 __ 3 1 2 + __ t t
dt 1 = – __ Ú _____ 3 2t + 3
1 = – __ Ú log |2t + 3| + c 6
1 2 = – __ log __ 3 + 3 + c 6 x
(
| |
44. We have,
)
4
5x dx dx 1 = __ Ú __________ Ú _________ 5 5 5 x (3 + 5x ) x (3 + 5x5)
1 Let x5 = __ t
dt fi 5x4 dx = – __2 t
dt 1 = – __ Ú _______ 5 (3t + 5) 1 = – ___ log |3t + 5| + c 15 3 1 = – ___ log __ 5 + 5 + c 15 x
| |
dx x dx 45. Ú _________ 2 = Ú __________ 2 x (2 + 3x ) x (2 + 3x2)
x – 3 3/2 dx 40. Ú _______________ = Ú dx _____ (x – 2)5 3/2 7/2 x–2 (x – 3) (x – 2)
Let
x2 = t
Put (x – 3/x – 2) = t2 and then you do it.
fi
1 x dx = __ dt 2
dx dx _____________ 41. Ú _______________ = Ú ______________ 5 7 x – 2 5/2 6 – 2) (x – 5) (x – 5) _____ ÷ (x x–6 x–2 Put _____ = t2 and then you do it. x–6
(
( )
)
dt 1 = __ Ú ________ 2 t (2 + 3t)
3 1 1 ______ = __ Ú __ – dt 4 t 2 + 3t
(
)
Indefinite Integrals
1 = __ (log |t| – log |2 + 3t|) + c 4 t 1 = __ log ______ + c 4 2 + 3t
( |
dx 4x3 dx 1 49. Ú _________ = __ Ú __________ 4 4 x (3x + 1) 4 x (3x4 + 1)
|)
x2 1 = __ log ______ + c 4 2 + 3x2
3x2 dx dx 1 46. Ú _________ 3 = __ Ú __________ x (3 + 4x ) 3 x3 (3 + 4x3) Let x3 = t
fi
2
3x dx = dt
dt 1 = __ Ú ________ 3 t (3 + 4t)
(
)
1 1 __ 4 = __ Ú __ – + 4t dt 9 t 3 1 = __ (log |t| –log |3 + 4t|) + c 9 t 1 = __ log ______ + c 9 3 + 4t
( |
( |
|)
|)
x3 1 = __ log _______ + c 9 3 + 4x3
dt 1 = __ Ú ________ 4 t (3t + 1)
3 1 1 _____ = __ Ú __ – dt 4 t 3t + 1
t 1 = __ log _____ + c 4 3t + 1
x4 1 = __ log ______ 4 + c 4 3x + 1
Let x3 = t
fi
Let
(x2 – 1) = t
fi
2x dx = dt
Also, x2 = t + 1
3x2 dx = dt
2 1 (t – 1) dt = __ Ú ______ 2 t3
1 t2 – 2t + 1 dt = __ Ú __________ 2 t3
1 1 __ 2 1 – + __ dt = __ Ú __ 2 t t2 t3
dt 1 = – __ Ú _______ 3 t (5t – 2)
5 1 1 __ = __ Ú __ – – 2 dt 6 t 5t
( |
|)
x3 1 = __ log ______ 3 + c 6 5x – 2
dx 4 3 dx 1 48. Ú _________ 4 = __ Ú _________ x (1 – 4x ) 4 x4 (1 – 4x4)
)
)
1 2 – = __ log |t| + __ t 2
(
)
1 ___ 2 + c 2t
)
(x2)4 4x dx
9
x 1 = __ Ú _________ Ú ___________ 2 5 (2x + 3) dx 4 (2x2 + 3)5
Let 2x2 + 3 = t
Thus,
fi 4x dx = dt
dt 1 = – __ Ú _______ 4 t (4t – 1)
1 1 _____ 4 = __ Ú __ – dt 4 t 4t – 1
t 1 = __ log _____ + c 4 4 –1
| |
( (
dt 1 = __ Ú _______ 4 t (1 – 4t)
)
(
1 2 1 2 = __ log |x2 – 1| + _______ – _________ + c 2 (x – 1) 2 (x2 – 1)2 51. We have,
(
| |
Ú x5 (x2 – 1)4 dx
dt 1 = __ Ú _______ 3 t (2 – 5t)
)
| |
50. We have,
(
(x2)2 ◊ 2x x5 1 _______ __ dx = dx Thus, Ú ________ Ú 2 (x2 – 1)4 (x2 – 1)4
3x2 dx dx 47. Ú _________ 3 = Ú _________ 3 x (2 – 5x ) x (2 – 5x3)
( ) 1 = __ ( log _____ 6 | 5t t– 2 | ) + c
x3 1 = __ log ______ 3 + c 4 4x – 1
|)
( |
| |
1.105
( )
t –3 4 ____ dt – 3)4 dt 2 1 1 (t________ ___ __ Ú ________ = Ú 4 64 t5 t5
4 3 2 1 (t – 12t + 54t – 108t + 81 = ___ Ú ________________________ dt 64 t5
(
1 1 = ___ Ú __ – 64 t
12 ___ 2 + t
54 ___ 3 – t
108 ____ 4 + t
)
81 ___ 5 dt t
1.106 Integral Calculus, 3D Geometry & Vector Booster
(
)
36 81 1 12 27 ___ = ___ log |t| + ___ – ___ 2 + 3 – ___ 4 + c t 64 t t 4t
where 2x2 + 3 = t x3 52. Ú ________ dx 2 (x + 1)4
Let (x2 + 1) = t fi 2x dx = dt
2 1 x (2x) dx = __ Ú ________ 2 (x2 + 1)4 1 (t – 1) = __ Ú ______ 2 t4 dt
(
)
1 1 1 = __ – ___2 + ___ 3 + c 2 2t 3t
1 __ 4 dt t
1 1 1 = __ – _________ + _________ + c 2 2 (x2 + 1)2 3 (x2 + 1)3
)
x5 53. Ú ________ dx 2 (x – 3)4
Let
(x2 – 3) = t
fi
2xdx = dt
2 2 1 (x ) (2x) dx = __ Ú __________ 2 (x2 – 3)4 2 1 (t + 3) dt = __ Ú ________ 2 t4
(
)
)
3 3 __ 2 – __ 3 + c t t 2 3 (x – 3) 3 3 1 = __ _______ – ________ 2 2 – ________ 2 3 + c 2 3 (x – 3) (x – 3) )
(
)
x7 54. Ú _________ dx 2 (3x – 2)4
2 3 1 (x ) (6x) dx = __ Ú __________ 6 (3x2 – 2)4
( )
1 1 6 ___ 12 8 = ____ Ú __ + __ 2 + 3 + __ 4 dt 162 t t t t
6 6 __ 1 2 = ____ log |t| – __ – __ – + c t t2 t3 162
(
(
) )
2 4 4 (x ) (6x) dx = __ Ú __________ 3 (3x2 – 2)5
( )
t+2 4 ____ dt 3 4 = __ Ú ________ 3 t5
Let 3x2 – 2 = t fi 6xdx = dt
+ 2)4 dt 4 (t________ = ____ Ú 243 t5
t4 + 8t3 + 24t2 + 32t + 16 4 = ____ Ú _______________________ dt 243 t5
8 24 ___ 32 16 4 = ____ t + __ 2 + ___ 3 + 4 + ___ 5 dt 243 Ú t t t t
( (
(
2
)
)
)
8 12 ___ 32 4 t 4 __ = ____ – __ – ___ 2 – 2 – __ 4 + c t 243 2 t 2t t
where t = 3x2 – 2 56. Do yorself. 57. We have,
9 6 1 = __ Ú t2 + __ 3 + __ 3 dt 2 t t 1 t3 = __ __ – 2 3
2 1 (t + 6t + 9) dt = __ Ú ____________ 2 t4
3 + 6t2 + 12t + 8) dt 1 (t__________________ = ____ Ú 162 t 4
)
(
8x6 55. Ú ________ dx (3x2 – 2)5
1 1 = __ Ú __ – 2 t3
(
3 1 (t + 2) dt = ____ Ú ________ 162 t4
where t = 3x2 – 2
(
t+2 3 _____ dt 3 1 = __ Ú _________ Let 3x2 – 2 = t 4 6 t fi 6xdx = dt
2 sin x + 3 dx Ú ___________ (3 sin x + 2)2
2 sec x tan x + 3 sec2 x = Ú __________________ dx (3 tan x + 2 sec x)2
Let 2 sec x + 3 tan x = t
fi (2 sec x tan x + 3 sec2 x) dx = dt
dt = Ú __2 t 1 = – __ + c t 1 = – ______________ + c (2 sec x + 3 tan x)
58. Do yourself 59. Do yourself
1.107
Indefinite Integrals
60. We have,
(
)
4 cos x + 3 dx Ú ___________ (3 cos x + 4)2
(
)
(4 cosec x cot x + 3 cosec2 x = Ú ______________________ dx (3 cot x + 4 cosec x)2
dx dt fi ___________ = _____ (2 + 3 cos x)2 5 sin x
3 – 2t Also, cos x = ______ 3t – 2
__ _____
Let 3 cot x + 4 cosec x = t fi – (3 cosec2 x + 4 cosec x cot x) dx = dt dt = – Ú __2 t 1 __ = + c t 1 = ________________ + c (3 cot x + 4 cosec x)
1 dt = __ Ú ____ 5 sin x
– 2) 1 (3t _____ = ____ __ _______ dt Ú 2 5 ÷5 ÷ t – 1
1 __ = __ ÷ 5 3÷t 2 – 1 – 2 log t + ÷t 2 – 1 + c 5
(
_____
)
fi 4ex + 6e– x = ex (9l + 9m) + e– x (– 6l + 6m)
fi
__ _____ 2 ÷________ 7 ÷ t – 1
cos x =
(3 – 4t)
(
dx 1 dt = – __ Ú ____ Ú ___________ 7 cos x (3 + 4 sin x)2 3 – 4t 1__ ______ = – ____ _____ dt Ú 7÷7 ÷t 2 – 1 _____ _____ 1__ = – ____ (3 log (t + ÷ t 2 – 1 ) – 4 ÷t 2 – 1 )+c 7÷7
4 + 3 sin x where, _________ x=t 3 + 4 sin 64. Do yorself. 65. Do yorself.
5 13 9ex + 6e– x = – ___ Ú dx + ___ Ú __________ dx 18 18 (9ex – 6e– x)
5 13 = – ___ x + ___ log |(9ex – 6e– x)| + c 18 18
70. Do yorself. 71. Do yorself. 72. We have,
(
)
)
4ex + 6e– x Thus, Ú _________ x dx 9e – 4e– x
Thus,
dx 66. Ú __________ = t (2 + 3 cos x)2
|)
68. Do yorself. 69. Let 4ex + 6e– x = l (9ex – 6– x) + m (9ex + 6e– x)
3t – 4 Also, sin x = ______ 3 – 4t
|
4 + 3 sin x Let _________ = t fi 9 (l + m ) = 4 and – 6 ( l – m ) = 6 3 + 4 sin x dx dt 5 13 fi ___________ = – ____ cos x fi l = – ___ , m = ___ 18 18 (3 + 4 sin x)2
_____
(
2 cos x + 3 where, t = _________ 2 + 3 cos x 67. Do yorself.
dx Ú ___________ (3 + 4 sin x)2
2 ÷5 ÷ t – 1 sin x = ________ (3t – 2)
fi
61. Do yorself. 62. Do yorself. 63. We have,
dx Let __________ = t (2 + 3 cos x)2
5 sin x fi ___________ dx = dt (2 + 3 cos x)2
dx ______________ Ú _______________ 5 3
÷(x + 2) (x + 1)
dx ______________ = Ú ________________ x+1 3 8 _____ (x + 2) x+2 dx = Ú ______________ x + 1 3/2 4 _____ (x + 2) x+2
÷
(
)
( ) x +1 Put ( _____ = t x + 2)
2
dx fi ______ = 2t dt (x + 2)2
1.108 Integral Calculus, 3D Geometry & Vector Booster
Also,
2t2 – 1 x = ______ 1 – t2
fi
2t2 – 1 1 x + 2 = ______ + 2 = _____ 2 1 – t2 1 –t
Thus,
dx Ú _______________ + 1 3/2 4 x _____
( )
(
)
8t8 – 12t6 + 6t4 – t2 = – 2 Ú _________________ dt (3t2 – 2) and then you do it
(x + 1) 74. Ú ______________ dx (x + 2) (x + 3)3/2
(x + 2) x+2
(1 – t2)4 = Ú _______ × 2 t dt t3
(x + 2) – 1 = Ú ______________ dx (x + 2) (x + 3)3/2
1 – 4t2 + 6t4 – 4t6 + t8) = 2 Ú ____________________ dt t2
dx dx = Ú ________ – Ú _____________ (x + 3)3/2 (x + 2) (x + 3)3/2
1 = 2 Ú __ 2 – 4 + 6t2 – 4t4 + t6 dt t
t7 1 4 = 2 – __ – 4t + 2t3 – ___ 5 + __ + c t 7 5t
(
)
(
(
÷
_____
)
)
x+1 where t = _____ x+2
(
)
dx x + 2 1/2 ___ 73. Ú ______ ◊ x 2x + 3
dx dx = Ú ________ 3/2 – Ú ____________ x + 2 _____ (x + 3) (x + 3)2 x+3
( ) x+2 Put ( _____ = t x + 3)
(
)
(x + 3) ◊1 – (x + 2) ◊ 1 fi __________________ ◊ dx = dt (x + 3)2
dx fi _______ = dt (x + 3)2
x+2 Put ______ = t2 2x + 3
dt 2 _____ = – ______ – Ú __ t x + 3 ÷
(2x + 3 – 2x – 4) fi _______________ dx = 2t dt (2x + 3)2
x+2 2 _____ = – ______ – log _____ + c x+3 x + 3 ÷
dx fi ________ = – 2t dt (2x + 3)2
dx 2–x 2 75. Ú _______ 2 3 _____ dx = 2 Ú ____________________ 5/3 2 + x (2 – x) (2 – x) (2 + x/2 – x)1/3
– dx fi ________ = – 2t dt (1 – 2t2)2
dx fi _________ = 2t dt (2x2 – 1)2
÷
_____
3t2 – 2 Also, x = ______2 1 – 2t
( )
(2 – x) ◊1 + (2 + x) ◊1 fi __________________ dx = dt (2 – x)2
dx dt fi _______ 2 = __ 4 (2 – x)
1 – 2t2 = Ú t ◊ ______ 2t (2t2 – 1)2 dt 2 3t – 2
t2 (1 – 2t2)3 = 2 Ú _________ dt 3t2 – 2)
t2 (1 – 6t2 + 12t4 – 8t6) = 2 Ú ___________________ dt (3t2 – 2)
4
6
4
(t – 6t + 12t – 8t ) = 2 Ú __________________ dt (3t2 – 2)
( )
( )
2
dx = 2 Ú _______________ 2 + x 1/3 (2 – x)2 _____ 2–x 2+x Put _____ = t 2–x
dx = 2t (2t2 – 1)2 dt
fi
| |
1 dt = __ Ú ___ 2 t1/3 1 = __ t2/3 + c 3
(
)
1 2 + x 2/3 = __ _____ + c 3 2–x
dx 76. Ú _______________ _____ _____ 3 x + 1 + ÷ x + 1 ÷
Let x + 1 = t6
fi
dx = 6t5 dt
1.109
Indefinite Integrals
6t5 dt = Ú _____ t3 + t2 3
dx 79. Ú ________ x1/2 + x1/3
Let x = t6 fi dx = 6t5 dt
t = 6 Ú ____ dt t+1
(t3 + 1) – 1 = 6 Ú __________ dt t+1
t5 = 6 Ú ______ dt 3 t + t2
(t3 + 1) dt = 6 Ú ______ dt – 6 Ú ____ t+1 t+1
t3 = 6 Ú _____ dt t + 1
(t3 + 1) – 1 = 6 Ú _________ dt t+1
1 = 6 Ú (t2 – t + 1) – ____ dt t+1
t3 t2 = 6 __ – __ + t – log |t + 1| + c 3 2
(t + 1) (t2 – t + 1) dt = 6 Ú ________________ dt – 6 Ú ____ t + 1 t+1 1 = 6 Ú ((t2 – t + 1) – ____ ) dt t+1
(
t3 = 6 __ – 3
)
t2 __ + t – log |t + 1| + c 2
_____
(
)
(
)
(
)
where t = 6÷ x + 1
where t = (x)1/6
dx 77. Ú _________________ _____ x + 1 – (x + 1)1/4 ÷
x ÷ __ 80. Ú _______ dx 4 3 ÷ x + 1
Put x = t4
fi dx = 4t3 dt
__
Let (x + 1) = t4
fi dx = 4t3 dt
2 × t3 = 4 Ú t______ dt 3 t + 1 t5 = 4 Ú _____ dt t3 + 1
4t3 dt = Ú _____ t2 – t
t2 dt = 4 Ú ____ t–1
(
t2 = 4 Ú t2 – _____ 3 dt t +1
t3 1 = 4 Ú __ – __ log |t3 + 1| + c 3 3
)
1 = 4 Ú t + 1 + ____ dt t–1
(
)
2
t = 4 __ + t + log |t – 1| + c 2
(x + 1)2 = 4 _______ + (x + 1) + log |x| + c 2
(
( Ú (
3
)
__
÷x 81. Ú ________ dx __ 3 __ x + ÷ x ÷
6
Let x = t fi dx = 6t5 dt 2
(
)
where t = x1/4
)
1 + x1/2 – x1/3 78. Ú ____________ dx 1 + x1/3
(
) )
t3 × t5 = 6 Ú ________ t3 + t2
t6 dt = 6 Ú ____ t+1
Let x = t6 fi dx = 6t5 dt
1+t –t 5 = 6 Ú _________ t dt 1 + t2
t8 – t7 + t5 = 6 __________ dt 1 + t2
t+1 = 6 Ú t – t – t + t – t – t – 1 + _____ dt t2 + 1
(t6 – 1) + 1 = 6 Ú _________ dt t+1
t7 t6 __ t5 t4 __ t3 t2 = 6 __ – __ – + __ – – __ – t 7 6 5 4 3 2 1 + __ log |t2 + 1| + tan– 1 (t) + c 2 __ where t = 6÷ x
1 = 6 Ú 1 + t + t2 + t3 + t4 + t5 + ____ dt t +1
t2 t3 __ t4 t5 __ t6 = 6 t + __ + __ + + __ + + log |t + 1| + c 2 3 4 5 6
( [
6
5
4
3
)
2
]
(
(
(
where t = x1/6
)
)
)
1.110 Integral Calculus, 3D Geometry & Vector Booster dx_____ 82. Ú ____________ (x – 1) ÷x + 3
x – 3 = t2
Let
fi dx = 2t dt Also, x – 1 = t2 – 3 – 1 = t2 – 4
= Ú _______ 2t dt (t2 – 4) t
dt = 2 Ú _______ (t2 – 22)
t–2 1 = __ log ____ + c 2 t+2
x + 3 – 2 ÷ 1 = __ log __________ _____ + c 2 ÷x + 3 + 2
|
_____
Let x + 2 = t2
fi dx = 2t dt
= 2 tan– 1 (t) + c
= 2 tan– 1 (÷x + 2 )+c
_____
t 2__ – 1 ___ = ___ tan __ + c 3 ÷3 ÷
x2 2 = ___ __ tan– 1 ___ __ + c 3 ÷3 ÷
Let x2 = t fi 2xdx = dt
( (
( (
)
)
)
)
x = 2 __ – x2 + log |x2 + 1| + c 2
t dt = 2 Ú ________ (t2 + 2) t = 2 Ú dt (t2 + 2)
Let (x – 2) = t2 fi 2xdx = dt
x2 = t
fi
2xdx = dt
( )
( )
t dt = Ú ______ t 2 t +5 _____ 2
( )
2x + 1 = t2 dx = t dt
t2 – 1 Also, x = ____ 2 t2 + 5 t2 – 1 fi x + 3 = ____ + 3 = _____ 2 2 dt = 2 Ú ______ 2 (t + 5) t 2__ – 1 ___ = ___ tan __ + c ÷5 ÷5
( )
( ÷
______
t2 = 2 __ – t + log |t + 1| + c 2 4
t2 dt = 2 Ú ____ t+1 2 (t – 1) + 1 = 2 Ú _________ dt t+1 1 = 2 Ú (t – 1) + ____ dt t +1
Let
dx______ 87. Ú _____________ (x + 3) ÷2x + 1
dx _____ 85. Ú _______ x ÷x – 2
__
÷x 84. Ú _____ dx x +1
dt = 2 Ú ______ (t2 + 3)
Let
t dt = 2 Ú _______ (t2 + 1) t dt = 2 Ú ______ (t2 + 1)
x–2 = ÷2 tan _____ + c 2 dx 86. Ú _________ __ (x + 3) ÷x
– 1
t dt = 2 Ú _______ 2 (t + 3) t
dx_____ 83. Ú ____________ (x + 3) ÷x + 2
______
__
| |
|
( ) ( ÷( ) )
__ 1 = ÷2 tan– 1 ___ __ + c 2 ÷
)
2x + 1 2 = ___ __ tan– 1 ______ + c 5 ÷5
dx 88. Ú ________ _____ x2 ÷x – 1
Let
x – 1 = t2
fi
dx = 2t dt
Also, x = t2 + 1
2t dt = Ú ________ (t2 + 1) 2t
dt = 2 Ú _______ 2 (t + 1)2
dt Now, Ú ______ t2 + 1
Indefinite Integrals
1.111
1 1 = _____ 2 Ú dt – Ú ________ × (– 2t) ◊ t dt t +1 (t2 + 1)2
x dx x dx 91. Ú __________________ _____ = Ú _________________ _____ (x2 + 2x + 2) ÷x + 1 ((x + 1)2 + 1 ÷x + 1
(t2 + 1) – 1 t = _____ 2 + 2 Ú _________ dt t +1 (t2 + 1)2
dt dt t = _____ 2 + 2 Ú _______ – 2 Ú _______ t +1 (t2 + 1) (t2 + 1)2
(t2 – 1)t dt = 2 Ú _________ (t4 + 1)t
t2 – 1 = 2 Ú ______ 4 dt t +1
1 – (1/t2) = 2 Ú _________ 2 dt t + (1/t)2
1–1 _____ t2 __________ dt = 2 Ú 1 2 t + __ – 2 t
__ 1 t + __ – ÷2 t 1 = ___ __ log ___________ __ + c 1 ÷2 t + __ + ÷ 2 t
dt t fi 2 Ú ________ = ______ + 2 tan– 1 t + c (t2 + 1)2 t2 + 1 _____
_____ x – 1 ÷ = ______ + 2 tan– 1 (÷x – 1 )+c x
dx 89. Ú _____________ _____ 2 (x – 4) ÷x + 1
Let (x + 1) = t2
fi dx = 2tdt
Let (x + 1) = t2 fi dx = 2tdt
( )
(
)
( ( ) )
| | ( (
t dt = 2 Ú ____________ ((x + 1)2 – 4) t
dt = 2 Ú __________ 2 (t + 2t – 3)
where t = ÷x + 1
_____
dt = 2 Ú ___________ (t + 3) (t – 1)
1 1 1 = __ Ú ____ – ____ dt 2 t–1 t+3
t– 1 1 = __ log ____ + c 2 t+3
x + 1 – 1 ÷ 1 _____ = __ log __________ + c 2 ÷x + 1 + 3
(
)
| |
dx 90. Ú _________ __ 2 (x + 1) ÷x
|
Let
x = t2
fi
dx = 2tdt
t dt = 2 Ú _______ 4 (t + 1) t
dt = 2 Ú ______ 3 (t + 1)
2 dt = Ú ______ (t4 + 1)
(t2 + 1) – (t2 – 1) = Ú _______________ dt (t4 + 1)
( Ú (
dx 92. Ú _________ __ 2 (x – 1) ÷x
|
_____
) )
Let x2 = t fi 2xdx = dt
t dt = 2 Ú _______ (t4 – 1)t dt = 2 Ú ______ 4 (t – 1) dt = 2 Ú ____________ (t2 – 1) (t2 + 1) 1 1 = _____ 2 – _____ dt t – 1 t2 + 1
(
| | | |
)
t–1 1 = __ log ____ – tan– 1 (t) + c 2 t+1
__ x – 1 ÷ 1 = __ log ______ – tan– 1 (÷x ) + c __ 2 ÷x + 1
__
x dx 93. Ú ________________ _____ 2 (x – 2x + 2) ÷x – 1
) )
2 (t2 + 1) (t – 1) ______ = ______ – dt 4 4 (t + 1) (t + 1) and then you do it.
x dx = Ú __________________ _____ 2 ( (x – 1) + 1 ) ÷ x – 1 Let (x – 1) = t2
fi dx = 2t dt
2
(t + 1) 2t dt = Ú __________ (t4 + 1) t
1.112 Integral Calculus, 3D Geometry & Vector Booster
(t2 + 1) dt = 2 Ú ________ t4 + 1)
1 1 + __ 2 dt t = 2 Ú _________ 1 2 __ t + 2 t
1 where t = _____ x+1
( ) ( ) (Ú )
dx __________ 95. Ú _________________ (x + 2) ÷x 2 + 2x + 2
1 1 + __ 2 dt t = 2 __________ 1 2 __ t – + 2 t
( )
( )
dz 1 = 2 Ú _____ Let z = t – __ 2 t z +2 __ z __ + c =÷ 2 tan– 1 ___ 2 ÷
( )
( ( ) ) (
__ 1 1 = ÷2 tan– 1 ___ __ t – __ + c t 2 ÷
__ 1 _____ 1 = ÷2 tan– 1 ___ __ (÷ x – 1 – ______ _____ + c 2 ÷x – 1 ÷
)
dx_____ 94. Ú _____________ (x + 1) ÷x 2 + 1 1 Let (x + 1) = __ t
1 fi dx = – __2 dt t 1–t 1 Also, x = __ – 1 = ____ t t
1 – __2 dt t ___________ = Ú ______________ 1 1 2 __ 1 – __ + 1 t t
÷(
dt ____________ = – Ú _____________ 2 – 2t + 1) ÷ (2t dt 1__ ___________ = – ___ Ú __________ 2 ÷ 1 2 __ t – t + 2
)
dt 1__ ________________ = – ___ Ú _______________ 2 ÷ 1 2 1 2 t – __ + __ 2 2
÷ ( ( ) ( ) ) 1 1 1 = – ___ log ( t – __ ) + t – t + __ + c ÷ 2 2| | 2 1 1 1 = – ___ log ( t – __ ) + t – t + __ + c ÷ 2 2| 2 | _________
__ ÷ __
÷
1 Put (x + 1) = __ t
fi
2
dt ______ = – Ú _______ 2 ÷ t + 1
= – log t + ÷t 2 + 1 + c
1 1 2 + 1 + c = – log __ x + __ x
÷
_____
|
|
÷
|
______
|
dx _____ 96. Ú ________ x ÷x 2 + 4 1 Put x = __ t
fi
1 – __2 dt t _____ = Ú ________ 1 1 __ __ 2 + 4 t t
dx ______ = – Ú _______ 2 + 1 ÷ 4t
dt 1 _________ = – __ Ú __________ 2 t 2 + (1/2)2
1 1 = – __ log t + t2 + __ + c 2 4
1 1 1 1 = – __ log __ x + __ 2 + __ + c 2 4 x
÷
|
|
÷
______
÷
|
______
|
1 Put (x – 1) = __ t
fi
1 dx = – __2 dt t
÷
dx_____ 97. Ú ____________ (x – 1) ÷x 2 + 4
_________ 2
1 dx = – __2 dt t
1 – __2 dt t = Ú________ _____ 1 1 __ __ + 1 t t2
)
dt = – Ú ____________ ___________ – 1)2 + t2 ÷ (t
÷(
1 __ 2 dt t = Ú – ______________ ____________ 2 1 1 __ __ + 1 + 4 t t
÷(
)
1 dx = – __2 dt t
Indefinite Integrals
dx ___________ = – Ú _____________ + t)2 + 4t2 ÷ (1
dt __________ = – Ú ___________ 2 + 2t + 1 ÷ 5t
÷
dt 1__ ______________ = – ___ Ú _____________ ÷5 1 2 2 2 t + __ + __ 5 5
÷( ) ( ) 1 = – ___ log ( t + __ | 15 ) + ÷ t __ 25 t + __ 15 | + c ÷5 ________
2
__
1 where t = ______ (x – 1) dx _____ 98. Ú ______________ (2x – 1) ÷x 2 + 1
(
1 (2x – 1) = __ t dt dx = – ___2 2t
dt _____________ = – Ú ______________ ÷(1 – 2t)2 – 36t2
dt ___________ = – Ú ____________ 32t2 ÷ 1 – 4t –
fi
(
)
1 1 + 1 Also, x = __ __ 2 t
÷ (
)
dt 1__ _______________ = – ____ Ú ______________ 4÷2 t ___ 1 2 – t – __ – 8 32
÷ {
}
dt 1__ _____________________ = – ____ Ú ____________________ 4÷2 3 2 1 2 ___ – t – – ___ 16 16
÷ { (
) ( ) }
dt 1__ ________________ = – ____ _______________ Ú 2 4÷2 3 1 2 ___ ___ – t – 16 16 1 t – ___ 16 1__ – 1 _______ ____ = – sin + c 3 4÷2 ___ 16
÷( ) ( ) ( )
( )
dt ___ 2 2t ____________ = – Ú _______________ 2 1 __ 1 __ 1 __ + 1 + 1 t 4 t
dt ___________ = – 2 Ú _____________ + t)2 + 4t2 ÷ (1
dt __________ = – 2 Ú ___________ 2 + 2t + 1 ÷ 5t
dx ______ 100. Ú _________ 2 x ÷x 2 – 1
dt 2__ ___________ = – ___ Ú __________ 2t __ 1 ÷5 2 __ t + + 5 5
Put
1 x = __ t
fi
1 dx – __2 dt t
÷ (
)
÷
dt 2__ ______________ = – ___ Ú _____________ 2 ÷5 1 2 2 __ t + + __ 5 5
÷( ) ( ) 2 = – ___ log ( t + __ | 15 ) + ÷ t + __ 25 t + __ 1t | + c ÷5 __________
2
__
1 where t = _______ (2x – 1) dx _____ 99. Ú ______________ (3x + 2) ÷x 2 – 4
)
1 1 Also, x = __ __ – 2 3 t
Put
fi
dt ___ 2 3t ____________ = – Ú _______________ 2 1 1 __ 1 __ __ – 2 – 4 t 9 t
1 (3x + 2) = __ t dt ___ dx = – 2 3t
Put
dx 1__ ___________ = – ___ Ú __________ 2 1 ÷5 2 __ t + t + __ 5 5
1.113
(
)
–1 1__ – 1 16t = – ____ sin ______ + c 3 4÷2
1 – t2 1 Also, x2 – 1 = __ 2 – 1 = _____ t t2 1 – __2 dt dx t _____ _____ Thus, Ú ________ = Ú ________ 2 2 – t2 x ÷x – 1 1 1_____ __ 2 2 t t
÷
– t dt _____ = Ú ______ ÷ 1 – t2
= ÷1 – t2 + c
_____
1.114 Integral Calculus, 3D Geometry & Vector Booster _______
÷ ( )
t dt______ = – Ú _____________ 2 (1 – t ) ÷1 + 2t2
1 2 = 1 – __ x + c
÷ 1 – x2 = ______ + c x
v dv 1 = __ Ú ___________ Let (1 + 2t2) = v2 2 4 _____ v –1 – 1 v 2 fi 4tdt = 2v dv
_____
dx_____ 101. Ú ____________ 2 (x + 1 ÷x 2 + 2 1 Put x = __ t dt fi dx = __ 2 t
dt – __2 t ______ = Ú ______________ 1 1 __ 2 + 1 __ 2 + 2 t t t dt______ = – Ú ______________ 2 (t + 1) ÷1 + 2t2
( ) ÷
dv______ 1 = – __ Ú _____________ , where t2 = v 2 (v + 1) ÷1 + 2v fi 2tdt = dx
Let
dv = z dz 2
z –1 Also, v = _____ 2
(1 + 2v) = z2
z dz 1 = – __ Ú ___________ 2 z2 – 1 _____ + 1 z 2
(
= cot– 1 (z) + c
= cot– 1 (÷1 + 2v) +c
)
dv 1 = __ Ú _______ 2 (v2 – 3)
v –÷ 3 1__ __ = ____ log ______ + c 4 ÷3 v+÷ 3
÷1 + 2t2 – ÷ 3 1 ______ = ____ __ log ____________ __ + c 2 4 ÷3 ÷1 + 2t + ÷3
| |
|
__
______
__
| | ÷ ÷
______
|
__ 1+2 _____ – ÷3 2 x 1__ ____ ___________ = log ______ + c __ 4 ÷3 1+2 _____ + 3 ÷ x2 dx ______ 103. Ú _____________ 2 (1 + x ) ÷1 – x2 1 Put x = __ t
dt fi dx = – __2 t
dt __ 2 t ______ = – Ú _____________ 1 1 1 + __ 2 1 – __ 2 t t
t dt _____ = – Ú _____________ 2 (1 + t ) ÷t 2 – 1
)
dz = – Ú _______ (z2 + 1)
(
( ) ÷
v dv = – Ú ________ , 2 (v + 2) v
______
_______
Let (t 2 – 1) = v fi 2tdt = dx
= cot– 1 (÷1 + 2t2) +c
2 = cot– 1 1 + __ 2 + c x
dv = – Ú _______ (v2 + 2)
v 1__ – 1 ___ = – ___ cot __ + c 2 ÷2 ÷
÷t 2 – 1 1 __ = ___ __ cot– 1 ______ + c 2 ÷2 ÷
1 1 1 = ___ __ cot– 1 ___ __ __ – 1 + c 2 ÷2 x2 ÷
( ÷ ) ______
dx______ 102. Ú _____________ 2 (x – 1) ÷x 2 + 2
fi
dt __ 2 t ______ = – Ú _____________ 1 1 __2 – 1 __ 2 + 2 t t
( ) ÷
( ) _____
1 Let x = __ t
( )
dt dx = – __2 t
( ÷ ) _____
x dx_____ 104. Ú _____________ 4 (x – 1) ÷x 4 + 3
Let x2 = t fi
2xdx = dt
1.115
Indefinite Integrals
dt ______ 1 = __ Ú _____________ 2 (t2 – 1) ÷t 2 + 3
dy ___ 2 y 1 1 ______ = – __ Ú ______________ Let t = __ y 2 __ 1 1 2 – 1 __ 2 + 3 dy y y fi dt = – ___ 2 y
( ) ÷
(
)
t fi _______ _____ dx = dt 2 ÷ x + 1
1 1 2 Also, x2 + 1 = __ t – __ + 1 t 4
fi ÷ x2
y dy 1 ______ = – __ Ú _______________ 2 (1 – y2) 1 + 3y2
v dv 1 = – __ Ú ________ Let (3y2 + 1) = v2 2 4 (4 – v ) v fi 3ydy = 2vdu
2
2
(
_____
)
( ( ) )
1 1 __ t + __ t 2 ________ x = dt t
dv 1 = __ Ú _______ 4 (v2 – 4)
fi
d
v–2 1 = ___ log _____ + c v+2 16
fi
1 t2 + 1 dx = __ ______ 2 dt 2 t
=
|
|
_______ 3y2 + 1 – 2 1 ___ ______ + log ___________ 16 3y2 + 1 +2
÷ ÷
| | ÷ ÷
______
( )
t2 + 1 1 1 __ Ú t10 _____ dt = __ Ú t8 (t2 + 1) dt 2 2 t2
dx __________ 105. Ú __________________ 2 (x – 1) ÷x 2 + 4x + 5
1 Let x = __ t
fi
dt dx = – __2 t
dt – __2 t __________ = Ú _________________ 1 1 __ 4 __ __ 2 – 1 2 + + 5 t t t
( ) ÷
t dt __________ = Ú __________________ 2 (t – 1) ÷5t 2 + 4t + 1
_____
1 = __ Ú (t10 + t8) dt 2
1 t11 = __ ___ + 2 11
______
(x + ÷ x 2 + 1 )=t
Let
1 _____ × 2x fi 1 + ________ dx = dt 2 ÷x 2 + 1
x+÷ x 2 + 1 fi __________ _____ dx = dt ÷ x 2 + 1
(
(
)
t9 __ + c, 9
______
5 107. Ú (x – ÷x 2 + 4) dx
)
_____
)
_____
Put (x – ÷x 2 + 4 )=t
1 _____ × 2x fi 1 – ________ dx = dt 2 ÷x 2 + 4
x fi 1 – _______ _____ dx = dt 2 ÷ x + 4
10
(
where (x + ÷x 2 + 1 )=t
106. Ú (x + ÷x + 1 ) dx 2
_____
( )
Thus, the given integral reduces to
c
3 __ 2 + 1 – 2 x 1 ______ = ___ log ___________ + c 16 3 __ 2 + 1 + 2 x
÷ x 2 + 1 Thus, dx = _______ dt t
| |
_______
÷
___________
______
( ) 1 1 + 1 = __ ( t – __ ) + 4 t 2÷ 1 1 = __ ( t + __ ) t 2÷ 1 1 = __ ( t + __ ) t 2
(
)
(
(
)
_____
)
2 + 4 ÷ x _____ fi x – _______ dx = – dt 2 ÷ x + 4 t fi _______ _____ dx = – dt 2 ÷ x + 4
(
)
( ( ) )
t fi _________ dx = – dt t __ 2 __ – + t 2 fi
(
1 t dx = __ __ + t 2
)
2 __ dt t
1.116 Integral Calculus, 3D Geometry & Vector Booster fi
( )
1 2 dx = __ + __ dt 2 t2
{ ( ) } Ú ( ) ( )
3 = __ ( t– 8/3 – t– 2/3 ) + c 4 _____
where t = (x + ÷x 2 – 4 )
1 2 = Ú t5 __ + __ dt 2 t2
110. Do yourself.
5
t __ + 2t3 dt 2
=
t6 t4 = ___ + __ + c 12 2
dx_____ dx ______ 111. Ú ____________ = Ú _____________ 2 2 9 3 x (x – ÷ x + 9 x 1 – 1 + __ 2 x
( ÷ )
_____
where t = (x – ÷x 2 + 4 ) 108. Ú (x +
______ )n dx ÷ 1 + x2
______
Put (x + ÷ x 2 + 1) =t
( ) 2
1 t +1 dx = __ _____ dt 2 t2
fi
( )
t2 + 1 1 = __ Ú tn ______ 2 dt 2 t
1 = __ Ú tn– 2 (t2 + 1) dt 2
1 = __ Ú (tn + tn– 2) dt 2
tn– 1 1 tn + 1 = __ _____ + _____ + c 2 n+1 n–1
(
t 1 = __ Ú ____ dt 9 t–1
1 (t – 1) + 1 = __ Ú _________ dt 9 t–1
( (
)
)
1 1 = __ Ú 1 + ____ dt 9 t–2 1 = __ (t + log |t – 1|) + c 9
÷
9 where t = 1 + __ 2 x
______
Put (x + ÷ x 2 – 4) =t
1 ______ × 2x fi 1 + ________ dx = dt 2 ÷x 2 – 4
(
)
( (
)
(
fi
1 t __ 2 dx = __ __ – dt t 2 t
fi
1 2 dx = __ – __ dt 2 t2
( )
Let 2 + 3x = t x
fi
2 x = ____ t –3
fi
– 2 dt dx – ______ (t – 3)2
( )
)
t – 5/3 = Ú ____ – 2t– 11/3 dt 2
_______ – 2 dt (t – 3)2 _____________ = Ú 2 1/2 3/2 3/2 ____ t x t–3
– 2 dt _______ (t – 3)2 _________________ = Ú 3/2 2 1/2 3/2 ____ 2 ____ t t –3 t –3
= – Ú t– 3/2 dt
2 = __ _ + c t ÷
)
1 2 __ – __ 2 t2 _______ = Ú 5/3 dt t
)
x fi 1 + _______ ______ dx = dt 2 ÷ x – 4 t fi ______ _____ dx = dt 2 ÷ x – 4
(
dx 112. Ú ____________ x1/2 (2 + 3x)3/2
dx _____ 109. Ú _____________ (x + ÷ x 2 – 4 )5/3
)
______
where t = (x + ÷ x + 1 )
t 1 = – __ Ú ____ dt 9 1 – t
2
)
_____
(
9 Put 1 + __ 2 = t2 x dx t ___3 = – __ dt 9 x
( )
( ) ( )
( )
1.117
Indefinite Integrals
2 = _______ ______ + c 2______ + 3x x
÷
115. Ú dx/x1/3 (2x + 1)5/3
dx 113. Ú ____________ x2/3 (2 + 3x)4/3
Put (2x + 1) = t x
Let (2 + 3x) = t x
1 fi x = ____ t–2
fi x (t – 3) = 2
dt fi dx = – ______ (t – 2)2
2 fi x = ______ (t – 3) – 2 fi dx = ______ dt (t – 3)2
2 ______ 2 (t – 3) __________________ = – Ú dt 2 2/3 4/3 ____ 2 4/3 ____ t t–3 t–3
( )
2 = – Ú __ × t– 4/3 dt 4
1 = – __ Ú t– 4/3 dt 2
3 = __ (t – 4/3) + c 2
( )
( (
3 = __ (t– 2/3) + c 2
)
Put (3x – 1) = t x 1 fi x = – ______ (t – 3) 1 fi dx = ______ 2 dt (t – 3)
1 ______ (t – 3)2 = Ú __________________ dt _________________ 3/4 5/4 1 5/4 1 ____ t ____ t –3 t –3
( ) ( )
= Ú t – 5/4 dt
= – 4t– 1/4 + c
4 = – ___ + c 4 _ t ÷
dx Ú _____________ (x – 1)3 (x + 2)4 dx = Ú _______________ 3 (x – 1) ______ (x + 2)7 (x + 2)
(
)
( )
x–1 Put _____ = t x+2
3 fi _______ dx = dt (x + 2)2
2t + 1 Also, x = _____ 1–t
fi
3 x + 2 = ____ 1–t
dt 1 = __ 6 Ú ________ 1 5 3 t3 ____ t –1
( )
5 1 (t – 1) dt = __ 6 Ú ________ t3 3 5 – 5t4 + 10t3 – 10t2 + 5t – 1) dt 1 t____________________________ __ = Ú t3 36
4 _______ = – ________ + c 1 4 3 – __ x
÷(
( ( ) )
3 1 = __ _________ + c 2 1 2/3 2 + __ x
dt = Ú ___ t5/4
( )
117. Do yourself. 118. We have,
= – Ú t– 5/3 dt
( )
116. Do yourself
dx 114. Ú ____________ 3/4 x (3x – 1)5/4
3 1 = __ _________ 1/3 + c 2 2 3 + __ x
)
dt ______ 2 (t – 2) ___________________ = – Ú 1 1/3 5/3 ____ 1 5/3 ____ (t) t –2 t –2
)
(
10 1 = __ 6 Ú t2 – 5t + 10 – ___ + t 3
5 __ 2 – t
)
1 __ 3 dt t
1.118 Integral Calculus, 3D Geometry & Vector Booster
(
)
5 5 1 t3 __ 1 = __ 6 __ – t2 + 10t – 10 log |t| – __ + ___ 2 + c t 2t 3 3 2 x–1 where t = _____ x+2
( )
dx dx 119. Ú _____________ 2 = Ú ______________ 3 x–2 2 5 _____ (x – 1) (x – 2) (x – 1) x–1 x–2 Put _____ = t x–1
( )
(
1 dt = __ Ú ___ 3 t3/4 4 1 = __ ___ _ + c 3 4÷t
( )
( ( ) ) ÷
4 _______ 1 = __ _________ + c 3 4 _____ x–1 x–2
)
(x – 1) ◊ 1 – (x – 2) ◊ 1 ___________________ dx = dt (x – 1)2 dx _______ = dt (x – 1)2 t–2 Also, x = _____ t–1 t– 2 1 (x – 1) = ____ – 1 = – ______ t–1 (t – 1) dt = Ú ________ 1 3 2 ____ t t–1
( )
dx dx 122. Ú _________ = Ú ______________ x 2 _____ x2 (x + 5)4 (x + 5)6 x+5
(
)
( )
x Put _____ = t x+5
5 dx fi _______ = dt (x + 5)2
dx fi _______ = __ 5 (x + 5)2
dt 1 = __ 5 Ú ________ 1 4 2 5 t ____ 1–t
( )
(t – 1)3 dt = Ú _________ t2
4 1 (t – 1) dt = __ 5 Ú _________ t2 5
(t3 – 3t2 + 3t – 1) = Ú _______________ dt t2
4 3 2 1 (t – 4t + 6t – 4t + 1) dt = __ 5 Ú ______________________ t2 5
3 1 = Ú t – 3 + __ – __ dt t t2
t2 1 = __ – 3t + 3 log |t| + __ + c t 2
1 4 = __ 5 Ú t2 – 4t + 6 – __ + t 5
1 t3 1 = __ 5 __ – 2t2 + 6t – 4 log |t| – __ + c t 3 5
( (
(
)
)
)
x–2 where t = _____ x–1
(
dx dx ______________ 121. Ú _______________ = Ú ________________ 3/4 4 3 5 (x – 1) (x + 2)5/4 ÷(x – 1) (x + 2)
( )
)
1 __ 2 dt t
)
)
x where t = _____ x+5
120. Do yourself.
dx = Ú _______________ x – 1 3/4 _____ (x + 2)2 x+2
( (
dx 123. Ú _______________ 3/2 (x – 1) (x + 1)5/2
dx = Ú _______________ x – 2 3/2 _____ (x + 1)2 x+1
( )
( )
x – 1 Put _____ = t x+2
x–1 Let _____ = t x+1
(x + 2) ◊ 1 – (x – 1) ◊ 1 fi __________________ dx = dt (x + 2)2
(x + 1) ◊ 1 – (x – 1) ◊ 1 fi ___________________ dx = dt (x + 1)2
3 fi _______ dx = dt (x + 2)2
2 fi _______ dx = dt (x + 1)2
dx dt fi _______ = __ 3 (x + 2)2
dx dt fi _______ 2 = __ 2 (x + 1)
(
)
Indefinite Integrals
1 dt = __ Ú ___ 2 t3/2
1 = __ Ú t– 3/2 dt 2 1 = – ___ + c t ÷ 1 _______ = – ________ + c x –1 _____ x+1
÷( ) x+1 = – _____ ÷ x – 1 + c
Let x3 = t
fi
Again let
3x2 dx = dt
dt 1 ______ = __ Ú ________ 3 t ÷3t + 4
Also,
v2 – 4 t = ______ 3
(
)
dv 2 = __ Ú _______ 3 (v2 – 4)
v–2 2 1 = __ × __ log _____ + c 3 2 v+2
fi
– 30x5 dx = 2t dt
fi
t dt x5 dx = – ___ 15
t dt 1 = – ___ Ú _______ 15 2_____ – t2 t 5 1 = __ Ú dt (t2 – 2) 3
( )
| |
|
_______
|
3x 3 + 4 – 2 ÷ _______ 1 = __ log ____________ + c 3 3 ÷3x + 4 + 2
|
__
|
– ÷2 ÷ 2 – 5x6 1 = __ log _____________ _______ __ + c 3 ÷2 – 5x6 + ÷2
3t + 4 – 2 ÷ 1 ______ = __ log ___________ + c 3 ÷3t + 4 + 2
Let 3x – 2 = t2 fi 3 dx = 2t dt
t dt 2 = __ Ú ________ 2 3 _____ t +2 t 3
( )
dt = 2 Ú ______ 2 t +2
__ t =÷ 2 tan– 1 ___ __ + c 2 ÷
__ 3x – 2 =÷ 2 tan– 1 ______ + c 2
( )
( ÷ ) _____
3
dx x _______ _______ 125. Ú _________ = Ú __________ dx 4 4 x ÷5x + 3 x ÷5x 4 + 3
Let 5x4 + 3 = t2
fi
20x3 dx = 2t dt
fi
t dt x dx = ___ 10
Put (2 – 5x6) = t2
|
| | __
dx ______ 127. Ú ________ x ÷3x – 2
( )
_______
t – ÷3 1 __ = ____ __ log ______ + c 4 ÷3 t + ÷3
__
v dv 2 = __ Ú _________ 2 9 v_____ –4 ◊ v 3
|
dt 1 = __ Ú _______ 2 (t2 – 3)
t – ÷2 1 __ = __ log ______ + c 3 t + ÷2
3 dt = 2v dv
______
3t + 4 = v2
fi
| |
)
dx x5 _______ _______ 126. Ú _________ = Ú __________ dx x ÷2 – 5x6 x6 ÷2 – 5x6
(
_______
2 dx x_______ dx _______ 124. We have, Ú _________ = Ú __________ 3 3 x ÷3x + 4 x ÷3x 3 + 4
t dt 1 = ___ Ú ________ 10 2 __ 3 t – t 5
where t = ÷ 5x 4 + 3
_____
1.119
dx x8 _______ _______ 128. Ú _________ = Ú __________ dx x ÷3x 9 – 2 x9 ÷3x 9 – 2
Let 3x9 – 2 = t2
fi 27x9 dx = 2t dt
2t dt fi x8 dx = ____ 27
3
1.120 Integral Calculus, 3D Geometry & Vector Booster
t dt 2 = ___ Ú _______ 27 _____ t2 + 2 t 3
dx dx 131. Ú __________ = Ú ___________ c 3 (c + dx2)3/2 x d + __ 2 3/2 x
dt 2 = __ Ú ______ 9 (t2 + 2)
(
( )
__
( ) ( ______ ÷3x 2– 2 ) + c
2 ÷ t = ___ tan– 1 ___ __ + c 9 ÷2
2 ÷ = ___ tan– 1 9
__
______
Let 2x10 – 3 = t2
fi
10x9 dx = 2t dt
fi
t dt x9 dx = ___ 5
t dt 1 = __ Ú _______ 2 5 _____ t +3 t 10 dt = 2 Ú ______ (t2 + 3)
( )
t 2 = ___ __ tan– 1 ___ __ + c 3 ÷3 ÷
2 = ___ __ tan– 1 3 ÷
10
t dt = – Ú _________ 2 (2t + 3)3/2
1 Let x = __ t 1 fi dx = – __2 dt t
Again let (2t2 + 3) = v2
fi
1 = __ c × 1 = __ c ×
132. Do yourself
1 __ + c t 1 _________ ________ + c c d + __ 2 x
÷ (
)
1 v dv = – __ Ú ____ 2 v3 1 dv = – __ Ú ___2 2 v 1 = ___ + c 2v 1 _________ = _______ + c 2 ÷2t 2 + 3 x = _________ _______ + c 2 ÷2 + 3x2
2t dt = v dv
( ) ( )
2 Let __ 4 – 5 = t2 x 8 fi – __5 dx = 2t dt x dx t dt fi ___5 = – ___ 4 x
dt 1 __ = – __ c Ú t2
( )
dx 130. Ú __________ (2 + 3x2)3/2
x dx x dx dx 134. Ú _________ = Ú __________ = Ú ___________ 3/2 2 3/2 2 6 _____ 5 __ (2 – 5x4)3/2 x 4 x 4 – 5 x –5 x
_______
t dt 1 ___ = – __ c Ú t3
133. Do yourself
( ) –3 ( ________ ) + c ÷ 2x 3
)
2c fi – ___3 dx = 2t dt x dx t dt fi ___3 = – ___ c x
9
(
c Let d + __ 2 = t2 x
dx x9 _______ _______ 129. Ú __________ = Ú ___________ dx x ÷2x 10 – 3 x10 ÷2x 10 – 3
)
1 t dt = – __ Ú ___ 4 t3
1 dt = – __ Ú __2 4 t
1 = __ + c 4t
1 2 = __ __ – 5 + c 4 x4
÷(
_______
)
dx dx 135. Ú x2 _________ = Ú ___________ 3/2 1 7 __ (1 – 4x6)3/2 x 6 – 4 x
( )
( )
1 Let __ 6 – 4 = t2 x 6 fi – __7 dx = 2t dt x dx t dt fi ___7 = – ___ 3 x
1 t dt = – __ Ú ___ 3 t3
1 dt = – __ Ú __2 3 t 1 = __ + c 3t
÷
1 Put x – 2 = __ t 1 fi dx = – __2 dt t Thus, the given integral reduces to 1 – __2 t t ______ Ú ________ ______ dt = – Ú ________ dt 2 1 __ 1 __ 1 + 3t ÷ 2 2 + 3 t t
Again, let 1 + 3t2 = v2
fi
3t dt = v dv
v dv Now, the integral is – Ú ____ v
= – v + c
= ÷1 + 3t2 + c
_______
÷
3 = 1 + _______ + c (x – 2)2
1 Let (x + 1) = __ t fi
1 __ 2 dt t_____ = – Ú _________ 1 __ 1 __ 3 2 + 3 t t
t2 dt ______ = – Ú ________ ÷ 1 + 3t2
÷
1 dx = – __2 dt t
÷
_____
|
÷
_____
|)
|
1 1 + ____ __ log t + t2 + __ + c 3 3÷3
dx 1 __________ = __ Ú __________________ 2 (x – 2)3÷x 2 – 4x + 5 dx 1 __________ = __ Ú __________________ 2 (x – 2)3 (x – 2)2 + 1
÷
1 Let (x – 2) = __ t
fi
dt __ 2 t______ = – Ú __________ 1 1 __ 3 __ 2 + 1 t t
t2 _____ = – Ú _______ dt ÷ t 2 + 1
(t2 + 1) – 1 ______ = – Ú __________ dt ÷ t 2 + 1
dx dx __________ ___________ 137. Ú __________________ = Ú ___________________ 3 2 3 (x + 1) ÷ x + 2x + 4 (x + 1) ÷(x + 1)2 + 3
|
__ t2 1 __ 1 1 = – ÷3 __ t2 + __ + log t + t2 + __ 2 3 6 3
dx _____________ 138. Ú _____________________ 3 (x – 2) ÷ 4x 2 – 16x + 20
__________
( ÷
÷ ( )
1 where t = ______ (x + 1)
÷
÷( ( ) )
__ dt 1 2 1__ ___________ = – ÷3 Ú t2 + ___ __ dt + ____ _________ Ú 3 3÷3 ÷ 1__ 2 2 ___ t + 3 ÷
= – Ú dv
)
________
______
dx __________ = Ú __________________ 2 (x – 2) ÷(x – 2)2 + 3
1 1 = – __ Ú ÷(1 + 3t2) – ________ ______ dt 3 ÷1 + 3t2
dx __________ Ú __________________ 2 2 (x – 2) ÷ x – 4x + 7
____________
1 1 = __ __ – 4 + c 3 x6
136. We have,
)
+ 3t2) –1 1 (1 ______ = – __ Ú __________ dt 3 ÷ 1 + 3t2
______
(
(
1.121
Indefinite Integrals
1 dx = – __2 dt t
( )÷
( (
)
_____
)
1 = – Ú ÷t 2 + 1 – ______ _____ dt 2 ÷ t + 1 _____ _____ t 2 = – __ ÷ t + 1 + log t + ÷ t 2 + 1 2
[
|
|
|
|]
– log t + ÷ t 2 + 1 + c
1 where t = ______ (x – 2)
dx _____________ 139. Ú _________________________ 2 (x – 6x + 9) ÷4x 2 – 24x + 20
_____
dx 1 __________ = __ Ú _________________ 2 (x – 3)2 ÷x 2 – 6x + 5
1.122 Integral Calculus, 3D Geometry & Vector Booster
dx 1 ___________ = __ Ú ___________________ 2 (x – 3)2 (x – 4 ÷ – 3)2 1 Let (x – 3) = __ t dt fi dx = __ 2 t
dt __ 2 t______ = –Ú ___________ 3 1 1 __ __ 2 – 4 t t
1 = – ___ Ú dv 12 v = – ___ + c 12 _______
2 + 1 ÷ 6t = – _______ +c 12
÷
____________
( ) ÷
t2 ______ = – Ú _______ dt ÷ 1 – 4t2
1 1 = __ Ú (1 – 4t2) – _______ ______ dt 4 ÷ 1 – 4t2
1 1 = __ 1 – 4t2 – _______ ______ dt 4 ÷1 – 4t2
6 ________ + 1 (2x + 1)2 _____________ = +c 12
dx dx __________ __________ 141. Ú ___________________ = Ú __________________ 3 2 3 (x + 1) ÷ x + 2x – 4 (x + 1) ÷(x + 1)2 – 5
( Ú ( ÷
) )
______
Let
fi
1 (x + 1) = __ t 1 dx = – __2 dt t
dt __ 2 t______ = – Ú ___________ 3 1 1 __ __2 – 5 t t
t2 dt ______ = – Ú ________ ÷ 1 – 5t2
1 where t = ______ (x – 3)
5t2) – 1 1 1 –______ dt = __ Ú ___________ 5 ÷1 – 5t2
dx ___________ 140. Ú ________________________ 2 (4x + 4x + 1)÷4x 2 + 4x + 7
1 1 – _______ ______ dt = __ Ú ÷1 – 5t2 5 ÷1 – 5t2
÷( )
________
dt 1 1 2 1 ________ = __ Ú __ – t2 dt – __ Ú _________ 2 2 4 1 2 2 __ – t 2
÷( )
[ ÷
_____
]
1 t __ 1 1 1 = __ __ – t2 + __ sin– 1 (2t) + __ sin– 1 (2t) + c 2 2 4 8 4
dt ___ 2 2t_____ = – Ú ___________ 2 1 __ 1 __ 2 + 6 t t
t dt 1 _______ = – __ Ú ________ 2 ÷6t 2 + 1
1 Let (2x + 1) = __ t dt ___ fi dx = – 2 2t
______
÷( )
)
__ dt 1 2 1 __________ dt – ____ __ =÷ 5 Ú ___ __ – t2 Ú _________ 2 ÷5 5÷5 1 ___ __ – t2 ÷5 __
[ ÷
__
÷( )
]
__ t 1 1 = ÷5 __ __ – t2 + ___ sin– 1 (t ÷5 ) 2 5 10 __ 1__ – ____ (sin– 1 t ÷5 ) + c 5÷5
(2x + 3) __________ 142. Ú __________________ dx (3x + 4)÷x 2 + 2x + 4
Again, let 6t2 + 1 = v2
fi
12t dt = 2v dv
fi
1 t dt = __ v dv 6
)
1 where t = ______ (x + 1)
1 v dv = – ___ Ú ____ 12 v
(
( ) ÷
(
_________
dx ___________ = Ú ____________________ 2 (2x + 1) ÷(2x + 1)2 + 6
( ) ÷
Let 2x + 3 = A (3x + 4 ) + B Comparing the co-efficients of x and the constant term, we get fi
A = 2/3, B = 1/3
Indefinite Integrals
dx dx 2 __________ 1 __________ = __ Ú ____________ + __ Ú __________________ 2 3 (x 3 + 1) + 3 (3x + 4) (x + 3 ÷ ÷ + 1)2
|
2 = __ log (x + 1) + ÷ x 2 + 2x + 4 3 3 ____________ dt ___ + – ____ , Ú ___________ 2 28 ÷28t ÷ – 2t + 1
where 3x + 4 = 1/t fi
dt 3dx = __ 2 t
|
2 = __ log (x + 1) + 3
__________ ÷ x 2 + 2x + 4
| (
|
) ÷
___________
|
3 t 1 1 ___ ___ + – ____ log t – ___ + t2 – ___ + + c 28 14 28 28 ÷
1 where t = ______ 3x + 4
|
dt __ 2 t_____ 1 – Ú _________ , Let (x + 1) = __ t 1 1 __ __ + 8 t t2 dt dx = – __2 t
( )÷
|
__________ ÷ x 2 + 2x + 9 –
|
__________ dt 1__ ____________ __________ ÷ x 2 + 2x + 9 ____ 2÷2 2 1__ 2 ____ t + 2÷2
= 2 log (x + 1) +
= 2 log (x + 1) +
(4x + 7) __________ 144. Ú _________________ dx (x + 2)÷x 2 + 4x + 7 4 (x + 2) – 1 ___________ = Ú __________________ dx (x + 2)÷(x + 2)2 + 3 dx dx ___________ ___________ = 4 Ú ____________ – Ú _________________ 2 ÷(x + 2) + 3 (x + 2)÷(x + 2)2 + 3 __________
|
|
= 4 log (x + 2) + ÷ x 2 + 3x + 7 dt __ 2 t ______ – Ú __________ , Let (x + 2) = 1 1 __ __ 2 + 3 t t dt fi dx = __ 2 t
( )÷
|
__________
|
__________
1 __ t
|
dt ______ = 4 log (x + 2) + ÷ x 2 + 3x + 7 – Ú _______ 2 + 1 ÷ 3t
dx dx __________ __________ = 2 Ú ____________ + Ú _________________ 2 ÷(x + 1) + 8 (x + 1)÷(x + 1)2 + 8
|
|
1 where t = ______ (x + 1)
|
2 (x + 1) + 1 ___________ = Ú _________________ dx (x + 1)÷(x + 1)2 + 8
= 2 log (x + 1) +
÷
|
dt 1__ ___________ = 4 log (x + 2) + ÷ x 2 + 3x + 7 – ___ Ú _________ 3 2 ÷ 1 2 t + ___ __ 3 ÷ __________ = 4 log (x + 2) + ÷ x 2 + 3x + 7
(2x + 3) __________ 143. Ú _________________ dx (x + 1)÷x 2 + 2x + 9
__________ +9 ÷ x 2 + 2x
|
______
1__ 1 – ____ log t + t2 + __ + c 8 2÷2
dx dx 2 __________ 1 __________ = __ Ú ___________ + __ Ú __________________ 3 ÷x 2 + 2x 3 + 4 (3x + 4)÷x 2 + 2x + 4
|
|
= 2 log (x + 1) + ÷ x 2 + 2x + 9
2 1 __ (3x + 4) + __ 3 3 ____________________ ____________ 2 (3x + 4)÷x + 2x + 4 dx
__________
__________
|
Thus, the given integral is reduces to
1.123
|
dt ______ Ú _______ 2 + 1 ÷ 8t
|
Ú
÷ ( )
|
÷
______
|
÷ ( )
|
1__ 1 – ___ log t + t2 + __ + c 3 3 ÷
1 where t = ______ (x + 2) x2 + __________ 4x + 2 145. Ú _________________ dx (x + 1)÷x 2 + 2x + 3 Put x2 + 4x + 2
= l (x + 1) (2x + 2) + m (x + 1) + n
= l (2x2 + 4x + 2) + m (x + 1) + n
= 2 l x2 + (4l + m) x + (2l + m + n)
Comparing the coefficients of the like terms of both the sides, we get L = 1/2, M = 2, N = – 1. Thus, the given integral reduces to (x + 1) (2x + 2) 1 __________ __ Ú _________________ dx 2 (x + 1)÷x 2 + 2x + 2
1.124 Integral Calculus, 3D Geometry & Vector Booster (x + 1) __________ + 2 Ú _________________ dx (x + 1)÷x 2 + 2x + 2
__
__
3 3/2 = Ú 3÷ x (1 + ÷ x + 3x + x ) dx
= Ú (x1/3 + 3x5/6 + 3x4/3 + x11/6) dx dx _________________ __________ – Ú (x + 1)÷x 2 + 2x + 2 9 3 18 6 = __ x4/3 + ___ x11/6 + __ x7/3 + ___ x17/6 + c 7 4 11 17 + 2) dx dx 1 (2x ___ __________ _________ = __ Ú ___________ + 2 Ú ___________ 2 2 ÷x 2 + 2x 149. Ú 3÷ x 2 (3 + x– 2/3)– 2 dx + 2 + 2 ÷ x + 2x
(
dx __________ – Ú _________________ Here, a = – 2 fi a negative integer (x + 1)÷x 2 + 2x + 2 Let x = t3
__________
=÷ x 2 + 2x + 2
|
)
+ 2 log (x + 1) +
fi dx = 3t2 dt.
__________ ÷ x 2 + 2x + 2
|
– sec– 1 (x + 1) + c
x2 + __________ 5x + 6 146. Ú _________________ dx (x + 2)÷x 2 + 5x + 4
= Ú t2 (3 + t– 2)– 2 ◊ 3t2 dt
t4 = 3 Ú ________ dt (3 + t– 2)2 dt = 3 Ú _________ 2 1 2 3t + __ 3
(
1 1 __ (x + 2) (2x + 5) + __ (x + 2) 2 2 ________________________ __________ = Ú dx 2 (x + 2)÷(x + 2) + 1
+ 5) dx dx 1 (2x 1 __________ __________ = __ Ú ___________ + __ Ú ___________ 2 ÷x 2 + 2x + 2 5 + 2)2 +1 ÷ (x
dx Now, I = Ú _______ x2 + 1/3
__________
_________
|
|
1 = ÷x 2 + 2x + 5 + __ log (x + 2) + ÷ x 2 + 4x + 5 + c 2
dt 1 = __ Ú ________ 2 3 2 __ 1 t + 3
(
)
– 2x ◊ x 1 = _________ 2 ◊ x – Ú _________ dx (x + 1/3) (x2 + 1/3)2 (x2 + 1.3) – 1/3 1 = _________ 2 ◊ x + 2 Ú _______________ dx (x + 1/3) (x2 + 1/3)2
x2 +__________ 10x + 6 147. Ú _________________ dx 2 (x + 2)÷x + 4x + 9
(
)
dx dx x 2 = ________ 2 + 2 Ú _________ – __ Ú _________ + c (x + 1/3) (x2 + 1/3) 3 (x2 + 1/3)2
1 __ (x + 2) (2x + 4) + 6 (x + 2) – 10 2 ____________ = Ú _____________________________ dx (x + 2)÷x 2 + 4x + 10)
dx x 2 = _________ 2 + 6 tan– 1 (3x) – __ Ú _________ + c 3 (x2 + 1/3)2 (x + 1/3)
(2x + 4) dx 1 ___________ ___________ = __ Ú ____________ dx + 6 Ú ____________ 2 2 ÷x 2 + 4x + 10 10 ÷ x + 4x +
x 2 = _________ 2 + 6 tan– 1 (3x) – __ I + c 3 (x + 1/3)
10 dx 5 x __________ – Ú _________________ fi __ I = _________ 2 + 6 tan– 1 (3x) + c 3 (x + 2)÷(x + 2)2 + 6 (x + 1/3)
___________
|
___________
|
= ÷ x 2 + 4x + 10 + 6 log (x + 2) + ÷ x 2 + 4x + 10
|
______
__
|
÷t 2 + 6 – ÷6 1__ ______ – ____ log ____________ __ + c 2 2 ÷6 + ÷ 6 ÷t + 6
where t = (x + 2) 148. We have,
)
__
__
Ú 3÷x (1 + ÷ x )3 dx
3x 18 fi I = __________ 2 + ___ tan– 1 (3x) + c 5 5 (x + 1/3)
(
_______ __
)
x ÷ 1 + __4÷ 150. Ú ________ dx 3 4 ÷ x 3 1 – __ + 1 __ b + 1 4 4 Here, _____ = _______ = __ = 1 g 1 1 __ __ 4 4
Indefinite Integrals 1 __
Let (1 + x 4 ) = t2
1 – __3 fi __ x 4 = 2t dt 4
Thus, Ú x
(1 + x ) dx
= 8 Ú t dt
8t3 = ___ + c 3
1 3/2 __ 8 = __ 1 + x 4 + c 3
– 3/4
1/4 1/2
)
fi
x4 (1 – t2) = 1
fi
1 x4 = ______ 2 (1 – t )
dx ______ Thus, Ú _________ 7 x÷ 1 + x4
( )
1 1 + x4 = __ _____ – 2 x4
x = t6
fi
dx = 6t5 dt
6t5 = Ú _________ dt 2 t (2 + 3t2)
(
)
(2 + 3t2) – 2 = 2 Ú __________ dt (2 + 3t2)
(
)
2 = 2 Ú 1 – ________ dt (2 + 3t2)
( ÷ ) ÷ ( ÷ ) ( ÷ ) __
__
3 3 __ tan– 1 __ t + c 2 2 __
__
2÷2 3 = 2 (x1/6) – ____ __ tan– 1 __ x 1/6 + c 2 3 ÷
1 1 + x4 3/2 __ _____ + c 3 x4
Here, a = Р10 ΠI
Let x = t4
fi
dx = 4t3 dt
_________ __
÷
7 3 4 154. Ú 3÷x × ( 1 + ÷ x ) dx
__
dx 152. Ú ____________ = Ú x– 1/2 (1 + x1/4)– 10 dx __ 4 __ 10 x ( ÷ x + 1) ÷
4t3 dt = Ú _________ 2 t (1 + t)10
Let
4 = 2t – __ × 3
( ) ( ÷ ( ) ) ______
dt 4 __ = 2 Ú dt – __ Ú _________ 3 2 2 2 __ t + 3
t 1 __ _______ 2 (t2 – 1)2 _________ = Ú dt 1 3 _____ 2 t 1–t
t3 1 = __ t – __ + c 2 3
)
Here, a = Р2 ΠI
Let 1 + x4 = x4 t2
)
153. Ú x– 1/2 (2 + 3x1/3)– 2 dx
1 = __ Ú (1 – t2) dt 2
)
9 8 = 4 _________ 1/4 9 – _________ 1/4 8 + c (x + 1) (x + 1)
– 3 1 = ___ – __ = – 2 = Integer. 2 2
)
9 8 = 4 _______ – _______ + c (t + 1)9 (t + 1)8
b+1 – 7 + 1 __ 1 Here, _____ + p = _______ – g 4 2
(
(t + 1) – 1 = 4 Ú __________ dt (t + 1)10
( ( (
dx ______ 151. Ú _________ 72 x ÷ 1 + x4
t dt = 4 Ú _______ (t + 1)10
1 1 = 4 Ú _______ – _______ dt (t + 1)9 (t + 1)10
2
(
1.125
1 __ + 1 b_____ +1 3 Here, = _____ = 1 ΠI g 4 __ 3
Let (1 + x4/3) = t7 4 fi __ x1/3 dx = 7t6 dt 3 21 fi x1/3 dx = ___ t6 dt 4 21 = ___ Ú (t6 ◊ t) dt 4
1.126 Integral Calculus, 3D Geometry & Vector Booster
21 t8 = ___ × __ + c 4 8
4/3 8/7 21 (1 + x ) = ___ × __________ +c 4 8
1 fi x4 = ______ 2 (t – 1)
2t fi 4x3 dx = – _______ dt (t2 – 1)2
155. Ú x– 6 (1 + 2x3)2/3 dx
t dt fi x3 dx = – ________ 2 2 (t – 1)2
b +1 Р6 + 1 __ 2 Here, _____ + a = ______ + = Р1 ΠI g 3 3
dx ______ = Ú __________ 11 2 x ÷ 1 + x4
1 = – __ Ú (t2 – 1)2 dt 2
1 = – __ Ú (t4 – 2t2 + 1) dt 2
Let 1 + 2x3 = x3t3
fi x3 (t2 – 2) = 1
1 fi x3 = ______ 3 (t – 2)
3t2 dt fi 3x2 dx = – _______ (t3 – 2)2
t2 dt + x4 fi x2 dx = – _______ 2 where t = 1_____ 3 (t – 2) x4
(
÷
______
(
2 3× __
3
_______ __
)
(tx) 3 = Ú ______ dx x6
÷1 + 4÷ x __ 158. Ú ________ dx 4 3 ÷ x
t2x2 dx = Ú ______ x6
t2 dt t2 = Ú ________ × – _______ 3 1 (t – 2)2 ______ 3 2 t – 2)
(
)
= – Ú t4 dt
t5 = – __ + c 5
( )
(
)
1 1 + 2x3 5/3 = – __ ______ + c 5 x3
156. Do yourself. dx _____ 157. Ú __________ 11 2 x ÷ 1 + x4
)
2t3 1 t5 ___ = – __ __ – + t + c 2 5 3
Let (1 + x1/4) = t3 1 fi __ x– 3/4 dx = 3t2 dt 4 dx fi ___ = 12t2 dt x3/4
= 12 Ú t × t2 dt
= 12 Ú t3 dt
t4 = 12 __ + c 4
= 3t4 + c
= 3 (1 + x1/4)4/3 + c
( )
dx _________ 159. Ú ______________ 2 (1 + ÷ x + x + 1 )
= Ú x– 11 (1 + x4)– 1/2 dx
Here a = 1 > 0, therefore we put the Euler first substitution, i.e.
= Ú x– 11 (1 + x4)– 1/2 dx
_________
÷ x2 + x + 1 = t – x
b+ 1 Р11 + 1 __ 2 Here, _____ + a = _______ + = Р2 ΠI g 4 4
fi (x2 + x + 1) = (t – x)2
Put (1 + x4) = x4t2
fi (x2 + x + 1) = t2 – 2tx + x2
fi x4 (t2 – 1) = 1
fi (1 + 2t) x = t2 – 1
Indefinite Integrals
( )
t2 – 1 fi x = _____ 2t + 1
dt = – 2 Ú _____ t2 – 1
2 (t2 + t + 1 fi dx = ___________ dt (2t + 1)2
t – 1 1 = – 2 × __ log _____ + c 2 t+1
t – 1 = log _____ + c t+1
_________
Now, 1 = ÷ x 2 + x + 1 = 1 + t – x
( )
t2 – 1 = 1 + t – _____ 2t + 1 dx _________ Thus, Ú _______________ 2 ( 1 + ÷ x + x + 1 )
(
= – (x – 2) (x – 5) = (x – 2) (5 – x), so we can apply Euler third
2 (t + t + 1) = Ú _________________ dt 2 (t + 3t + 2) (2t + 1)
substitution, i.e. ___________
÷ 7x – 10 – x2 = (x – 2) t
) dt dt 2 – __ ( Ú ____ – 2 _____ 3 t + 2 Ú 2t + 1 )
8 dt dt dt 4 = 2 – 2 Ú ____ + __ ____ + __ _____ t + 1 3 Ú t + 2 3 Ú 2t + 1
_____________
fi ÷– (x – 2) (x – 5) = (x – 2) t fi (5 – x) = (x – 2) t2
)
8 2 = 2 – 2 log |t + 1| + __ log |t + 2| + __ log |2t + 1| 3 3 2 – __ (log |t + 2| – log |2t + 1|) + c 3 _________
where
t = x + ÷x 2 + x + 1
dx _________ 160. Ú _____________ 2 x+÷ x – + 1 Here, c = 1 > 0, therefor we put the Euler second substitution, i.e. _________ ÷ x2 – x + 1 = t x
(x2 – x + 1) = t2x2 – 2t x + 1
fi
2t – 1 x = ______ t2 –1
fi
– 2 (t2 – t + 1) dx = ____________ dt t2 – 1)2
dx _________ Thus, Ú _____________ 2 x+÷ x – x + 1
5 + 2t2 fi x = _______ 1 + t2 6t dt fi dx = – _______ 2 (t + 1)2
(
– 2 (t2 – t + 1) ____________ 2 2 (t – 1) = Ú _____________ dt t2 – t + 1 _________ 2 (t – 1)
)
5 + 2t2 3t Now, (x – 2) t = _______ – 2 t = _____ 2 2 t +1 t +1 dx ___________ Thus, Ú x ____________ – 10 – x2 ÷ 7x
– 6t x = Ú _______ × _______ dt (x – 2) t (1 + t2)2
5 + 2t2 _______ 2 (1 + t ) – 6t = Ú ________ × _______ dt 2 2 3t (1 + t ) _______ (1 + t2)
5 + 2t2 = – 2 Ú _______ dt (1 + t2)2
2 (1 + t2) + 3 = – 2 Ú ___________ dt (1 + t2)2
dt dt = – 4 Ú _______ – 6 Ú _______ (1 + t2) (1 + t2)2
–1
fi
)
Here, 7x – 10 – x2 = – (x2 – 7x + 10)
2
(
_________
x dx ___________ 161. Ú ____________ – 10 – x2 ÷ 7x
( )
| |
1 + ÷x 2 – x + 1 where, t = _____________ x
2(t2 + t + 1) ___________ 2 (2t + 1) = Ú ____________ dt t2 – 1 ______ 1 + t – 2t + 1
(
| |
1.127
1.128 Integral Calculus, 3D Geometry & Vector Booster
(
)
fi
(x – 1) (x – 2) = (x – 1)2 t2
fi
(x – 2) = (x – 1) t2
fi
x (1 – t2) = 2 – t2
– 10 – x2 ÷ 7x where, t = ____________ (x – 2)
fi
dx _________ 162. Ú _____________ 2 x–÷ x – x + 2
t2 – 2 x = ______ 2 t –1
fi
(t2 – 1) (2t) – (t2 – 2) 2t dx = ____________________ dt (t2 – 1)2
fi
2t (t2 – 1 – t2 + 2) dx = ________________ dt (t2 – 1)2
fi
2t dt dx = _______ 2 (t – 1)2
t 1 ______ 1 = – 4 tan– 1 (t) – 6 __ + __ tan– 1 t + c 2 t2 + 1 2
3t = – 7 tan (t) – ______ 2 + c t +1 – 1
___________
_________
Put ÷x 2 – x + 2 = t + x fi (x2 – x + 2) = (t + x)2 fi (x2 – x + 2) = t2 + 2tx + x2 fi (2t + 1) x = 2 – t2
( )
(2x + 1) (– 2t) – (2 – t2) 2 fi dx = ______________________ dt (2t + 1)2
2t dt _______ 2 (t – 1)2 _________________ = Ú dt t2 – 2 1 _____ 2 × – _______ t –1 (t2 – 1)
– 4t2 – 2t + 2t2 – 4 fi dx = _________________ dt (2t + 1)2
2t dt = – Ú _______ 2 (t – 2)
= – log|(t2 –2)| + c
– 2t2 – 2t – 4 fi dx = ____________ dt (2t + 1)2
+ 2 ÷ x 2 – 3x where t = ___________ (x – 1)
( )
2 – t2 fi x = _____ 2t + 1
__________
dx ______ 165. Ú __________ x+÷ x 2 – 1
1 2t2 + 2t + 4 = Ú __ × ___________ dt t (2t + 1)2
]
)
(x – ÷ x 2 – 1 = Ú ___________ dx (x2 – x2 + 1)
= Ú (x – ÷x 2 – 1 ) dx
x2 x ÷x 2 – 1 __ 1 = __ – ________ – log x + ÷x 2 – 1 + c 2 2 2
1 t +t+2 = 2 Ú __ × _________ dt t (2t + 1)2
9 dt 7 _______ dt dt = 2 2 Ú __ – __ Ú – __ ________ t 2 (2t + 1) 2 Ú (2t + 1)2
9 7 = 2 2 log |t| – __ log |2t + 1| – ________ + c 4 4 (2t + 1)
[
(
______
______
_________
where t = ( ÷x 2 – x + 2 – x ) 163. Do yourself. dx __________ 164. Ú _____________ 2 x ÷x – 3x + 2
fi
2
(
______
|
______
Here, x2 – 3x + 2 = (x – 1) (x – 2)
______
(x – ÷x 2 – 1 ) ______ = Ú ______________________ ______ dx (x + ÷ x 2 – 1 ) (x – ÷ x 2 – 1 )
2
__________ Let ÷ x2 – 3x +2
( )
x dx ______ 166. Ú __________ = Ú x (x – ÷x 2 – 1 ) dx 2 x+÷ x – 1 ______
= Ú (x2 – x ÷x 2 – 1 ) dx
x3 = __ – Ú x÷x 2 – 1 dx 3
= (x – 1) t 2 2
(x – 3x + 2) = (x – 1) t
________
______
|
)
Indefinite Integrals
8 cos6 x ◊ sin x ___ = __________ + cos4 x ◊ sin x 9 63
Let x2 – 1 = t2
xdx = tdt
x = __ – Ú t2 dt 3
170. We have,
x3 = __ – 3
Ú cos8 x dx
2 3/2 x3 (x – 1) = __ – _________ +c 3 3
3
t3 __ + c 3
(
Ú sin x dx 5
cos x sin4 x __ 4 cos x sin2 x __ 2 = – _________ – – _________ + I1 3 3 5 5
(
9 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x 10 80 63 cos3 x ◊ sin x __ 5 + ___ __________ + I2 80 6 6
)
(
4
cos x sin x ___ 4 = – _________ – (cos x sin2 x + 2 cos x) + c 5 15
Ú sin x dx
6
(
)
63 cos3 x ◊ sin x ___ 5 + ___ __________ + (x + sin x ◊ cos x) + c 80 12 6
171. We have, Ú tan5 x dx
5 cos x sin5 x __ = – _________ + I4 6 6
( (
)
tan4 x = _____ – I3 4
5 cos x ◊ sin x __ 3 cos x sin x __ = _________ + – __________ + Ú sin2 x dx 4 4 6 6
tan4 x _____ tan2 x = _____ – + I1 4 2
5 cos x sin5 x ___ = – _________ – cos x ◊ sin3 x 24 6
tan4 x _____ tan2 x = _____ – + Ú tan x dx 4 2
(
)
5 cos x ◊ sin3 x __ 3 cos x sin5 x __ = – _________ + – __________ + I2 4 4 6 6 5
(
)
3
3 + __ (x – sin x cos x) + c 4
169. We have,
8 cos6 x ◊ sin x __ = __________ + I5 9 9
(
)
8 48 cos6 x ◊ sin x ___ = __________ + cos4 x ◊ sin x + ___ I3 9 63 63 6
8 cos x ◊ sin x ___ = __________ + cos4 x ◊ sin x 9 63
tan4 x _____ tan2 x = _____ – + log |sec x| + c 4 2 172. We have,
8 cos4 x ◊ sin x __ 6 cos6 x ◊ sin x __ = __________ + __________ + I3 7 7 9 9
)
Ú tan6 x dx
Ú cos7 x dx
)
9 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x 10 80
168. We have,
)
9 63 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x + ___ I4 10 80 80
cos x sin4 x __ 4 = – _________ + I3 5 5
)
9 cos5 x ◊ sin x __ cos7 x ◊ sin x ___ 7 = __________ + __________ + I4 10 10 8 8
(
48 cos2 x ◊ sin x __ 4 + ___ __________ + sin x + c 63 5 5
9 cos7 x ◊ sin x ___ = __________ + I6 10 10
167. We have,
1.129
( (
)
tan5 x = _____ – I4 5
)
tan5 x _____ tan3 x = _____ – + I2 3 5 tan5 x _____ tan3 x = _____ – + Ú tan2 x dx 3 5
(
tan5 x _____ tan3 x = _____ – + tan x – x + c 3 5 2 48 cos x ◊ sin x __ 4 + ___ __________ + I1 173. We have, 63 5 5 Ú cot7 x dx.
(
)
)
tan5 x _____ tan3 x = _____ – + Ú (sec2 x – 1) dx 3 5
1.130 Integral Calculus, 3D Geometry & Vector Booster
cot6 x = – _____ – I5 + c 6
cot6 x cot4 x = – _____ – – _____ – I3 + c 4 6
(
6
)
5 sec5 x ◊ tan x __ = __________ + I5 6 6
5 sec3 x ◊ tan x __ 3 sec5 x ◊ tan x __ = __________ + __________ + I3 4 4 6 6
5 15 sec5 x ◊ tan x ___ = __________ + sec3 x ◊ tan x + ___ I3 24 24 6
5 sec5 x ◊ tan x ___ = __________ + sec3 x ◊ tan x 24 6
4
cot x _____ cot x = – _____ + + I3 + c 4 6
cot6 x _____ cot4 x cot2 x = – _____ + + – _____ – I1 + c 4 2 6
cot6 x _____ cot4 x _____ cot2 x = – _____ + – – I1 + c 4 2 6
cot6 x _____ cot4 x _____ cot2 x = – _____ + – – log |sin x| + c 4 2 6
(
)
(
(
cot5 x – _____ – I4 + C 5 cot5 x _____ cot3 x = – _____ + + Ú cot2 x dx + c 3 5
179. We have, sin 3x Ú _____ dx sin x 2 sin (3 – 1) x = ___________ + I3–2 (3 – 1)
= sin 2x + I1
cot5 x _____ cot3 x = – _____ + + Ú (cosec2 x – 1) dx + c 3 5
= sin 2x Ú dx
cot5 x _____ cot3 x = – _____ + – cot x – x + c 3 5
= sin 2x + x + c
180. We have, Ú_____ sin 5x dx sin x 2 sin (5 – 1)x = ___________ + I5–2 (5 – 1)
175. We have, Ú sec2 x dx
sec x ◊ tan x __ 1 = _________ + I1 2 2
sec x ◊ tan x __ 1 = _________ + log |sec x + tan x| + C 2 2
176. We have, Ú sec4 x dx
2
sec x tan x __ 2 = _________ + I2 3 3
sec2 x tan x __ 2 = _________ + tan x + c 3 3
3 sec x ◊ tan x __ = __________ + I3 4 4
2 sin (3 – 1)x 1 = __ sin 4x + ___________ + I1 2 (3 – 1)
2 = __ sin 4x + sin 2x + x + c 3
3
sec x ◊ tan x = __________ 4
(
3
1 = __ sin 4x + I3 2
sin 6x Ú _____ dx sin x 2 sin (6 – 1)x = ___________ + I6–2 (6 – 1)
Ú sec5 x dx
181. We have,
177. We have,
)
3 x 1 + __ sec x ◊ tan __ + __ log |sec x + tan x| + c 4 2 2
Ú sec7 x dx
2 = __ sin 5x + I4 5 2 2 = __ sin 5x + __ sin 3x + I2 3 5
2 2 = __ sin 5x + __ sin 3x + Ú _____ sin 2x dx 3 5 sin x
2 2 = __ sin 5x + __ sin 3x + 2 Ú cos x dx 3 5
2 2 = __ sin 5x + __ sin 3x + 2 sin x + c 3 5
178. We have,
)
15 sec x ◊ tan x __ 1 + ___ _________ + log |sec x + tan x| + c 24 2 2
174. We have,
)
Indefinite Integrals
182. We have, Ú _____ sin 8x sin x 2 sin (8 – 1) x = ___________ + I6 (8 – 1)
2 2 = __ sin 7x + __ sin 5x + I4 7 5 2 2 2 = __ sin 7x + __ sin 5x + __ sin 3x + I2 7 3 5 2 2 2 = __ sin 7x + __ sin 5x + __ sin 3x + 2 sin x + c 3 5 5
183. We have Ú _______ dx (x2 + 2)2
(
x 1 2.2 – 3 + __ _______ I2–1 = _________ 2.2 (x2 + 2) 2 2.2 – 2
x 1 Ú _______ dx = __ (x2 + 2) + __ 4 4 (x2 + 2)
)
x 1 + __ I2 = __________ 12 (x2 + 3)2 4
x x 1 11 + __ __________ + __ __ I1 = __________ 12 (x2 + 3)2 4 2.3 (x2 + 3) 3 2
)
x x + _________ = __________ 12 (x2 + 3)2 24 (x2 + 3)
( )
x 1 + _____ __ tan– 1 ___ __ + c 24 ÷2 ÷2
185. We have, x + 1 Ú ____________ dx (x2 + 3x + 2)2
3 1 x + __ – __ 2 2 1 ____ ___________ + log + C 1 3 1 __ __ 2 ◊ x + + __ 2 2 2
3 x + __ 2 1 1 = – ____________ – __ ___________ 2 2 2 3 1 __ 2 (x + 3x + 2) x + – __ 2 4
( ) ]
x+1 + log _____ + c x+2
x3 1 = (log x) ◊ __ – __ I 3 3 2,0
x3 = (log x) ◊ __ – 3
x3 x3 = (log x) ◊ __ – __ + C 3 9
1 __ Ú x2 dx 3
187. We have,
(
[ ( ( ) ) ( (( )) )] [ ( ( ) )
Ú x2 log x dx
Ú _______ dx (x2 + 3)2 x 1 2.3 – 3 + __ _______ I3–1 = ________________ 2 (3 – 1) 3 (x2 + 3)2 3 22.3 – 2
3 x + __ 2 1 2 1 __________ __ __ = – (x + 3x + 2) – 2 2 3 2 __ 1 __ x + – 2 4
[ ( ) ( ) ]
186. We have,
( )
(
dx 1 1 = – ____________ – __ Ú _______________ 2 2 (x + 3x + 2) 2 3 2 1 22 __ x + – __ 2 2
x x 1 __ tan– 1 ___ __ + c = __ (x2 + 2) + ____ 4 4 ÷2 ÷2 184. We have,
)
(2x + 2) 1 = __ Ú ____________ dx 2 2 (x + 3x + 2)2 1 (2x + 1) – 1 = __ Ú ____________ dx 2 (x2 + 3x + 2)2 (2x + 3) 1 1 dx = __ Ú ___________ dx – __ Ú ___________ 2 (x2 + 3x +2)2 2 (x2 + 3x + 2)2
1.131
Ú x2 (log x)2 dx
x3 2 = (log x)2 __ – __ I 3 3 2,1
x3 = (log x)2 __ – 3
x3 2 x3 x3 = (log x)2 __ – __ log x) __ – __ + c 3 3 3 9
(
)
2 __ Ú x2 log x dx 3
(
)
188. We have, Ú x3 (log x)2 dx
x4 2 = (log x)2 __ – __ I 4 4 3,1
x4 1 x4 1 = (log x)2 __ – __ (log x) __ – __ I 4 2 4 4 3,0
x4 = (log x)2 __ – 4
x4 = (log)2 __ – 4
( ( (
)
x4 1 __ (log x) __ – 2 4
x4 1 __ (log x) __ – 2 4
)
1 __ Ú x3 dx 4
)
x4 ___ + c 16
1.132 Integral Calculus, 3D Geometry & Vector Booster
[ [ [
189. We have, Ú x2 (1 – x)3 dx 3
x3 (1 – x)3 __ 1 = _________ + I2,2 2 6
(
3
3
2
)
Hints
1. We have,
cos 2x – cos x dx Ú ___________ 1 – cos x
and
solutions
( Ú (
)
1 – 3sin2x cos2x = Ú ______________ dx sin2x cos2x
sin2x + cos2x – 3sin2x cos2x = _______________________ dx sin2x cos2x
= Ú (sec2x + cosec2x – 3)dx
= (tan x – cot x – x) + c
2cos2x – cos x – 1 = Ú _______________ d x 1 – cos x
(cos x – 1)(2cos x + 1) = Ú ___________________ dx 1 – cos x
= – Ú (2cos x + 1) dx
cos 2x Ú _________ dx cos2x sin2x
= – (2sin x + x) + c
cos2x – sin2x = Ú ___________ dx cos2x sin2x
= Ú (cosec2x – sec2x)dx
= – (cot x + tan x) + c
cos 5x + cos 4x Ú _____________ d x 1 – 2 cos 3x
sin3x (cos 5x + cos 4x) = Ú ___________________ dx sin 3x – sin 6x
9x 3x 3x x 2sin ___ cos ___ 2cos ___ cos __ 2 2 2 2 = Ú ____________________________ dx 9x 3x – 2cos ___ sin ___ 2 2
( ) ( ) ( ) ( ) ( ) ( ) 3x x = – Ú 2cos( ___ )cos ( __ ) dx 2 2
= – Ú (cos 2x + cos x) dx
sin 2x = – _____ + sin x + c 2
4. We have,
2. We have,
(
)
3. We have,
(
)]
Note: Q-190 to 197, do yourself.
(
)
sin6x + cos6x Ú ___________ 2 sin x cos2xdx
)]
3 2 x3 (1 – x)3 __ x3 1 x (1 – x) 2 x3 ___ = _________ + _________ + __ __ + + c 2 12 6 5 5 4
x (1 – x) 1 x (1 – x) 2 = _________ + __ _________ + __ I2,1 2 6 5 5
)]
3 2 x3 (1 – x)3 __ 1 x (1 – x) 2 x3 __ 1 = _________ + _________ + __ __ + Ú x2 dx 2 4 6 5 5 4
3
x (1 – x) 3 = _________ + _________ I (2 + 3 + 1) (2 + 3 + 1) 2,2
3
( (
3 2 x3(1 – x)3 __ 1 x (1 – x) 2 x3 __ 1 = ________ + _________ + __ __ + I2,0 2 4 6 5 5 4
5. We have,
sin 2x Ú __________ dx sin 5x sin 3x
sin (5x – 3x) = Ú ___________ dx sin 5x sin 3x
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 5x sin 3x
= Ú (cot 3x – cot 5x)dx
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
6. We have,
Ú___________ sin x dx sinx + cos x
)
Indefinite Integrals
2sin x 1 = __ Ú ___________ dx 2 sin x + cos x
b = – __ a log |t| + k
1 (sin x + cos x) + (sin x – cos x) = __ Ú _________________________ dx 2 sinx + cos x
b – x = – __ a log|a e + c| + k
sin x – cos x 1 = __ Ú 1 + __________ dx 2 sin x + cos x
1 = __ (x – log|sin x + cos x|) + c 2
(
)
10. We have, dx Ú ______ 1 + ex
7. We have,
dx Ú _____________ sec x + cosec x
exdx = Ú _________ x x e (e + 1)
dt = Ú ______ , t(t + 1)
sin x cos x = Ú ___________ dx sin x + cos x
1 2 sin x cos x = __ Ú ___________ dx 2 sin x + cos x
1 (1 + sin2x) – 1 = __ Ú _____________ dx 2 sin x + cos x
2 dx 1 (sin x + cos x) 1 = __ Ú_____________ dx – __ Ú __________ 2 sin x + cos x 2 sin x + cos x
fi exdx = dt
(
)
1 1 = Ú __ – ____ dt t t+1 t = log ____ + c t+1
| |
| |
ex = log _____ x + c e +1
11. We have, x+9 Ú _______ dx x3 + 9x
dx 1 1 = __ Ú (sin x + cos x)dx – __ Ú __________ 2 2 sin x + cos x
x + 9 + x2 – x2 = Ú _____________ dx x(x2 + 9)
dx 1 1__ _________ = __ Ú (sin x + cos x)dx – ____ Ú p 2 2÷2 sin x + __ 4 p x __ 1 1__ __ ____ __ = (sin x – cos x) – log tan + + c 2 2 8 2÷2
x + (9 + x2) – x2 = Ú _______________ dx x(x2 + 9)
dx dx x _______ = Ú _____ + Ú ___ dx x – Ú (x2 + 9) x2 + 9
x 1 1 = __ tan– 1 __ + log|x| – __ log|x2 + 9| + c 3 3 2
|
( (
) )|
8. We have,
Ú tan 3x tan 2x tan x dx = Ú (tan 3x – tan 2x – tan x) dx
1 1 = __ log |sec 3x| – __ log |sec 2x| 3 2
Ú sin x – cos x/ex + sin xdx
– log|sec x| + c
(ex + sin x) – (ex + cos x) = Ú _____________________ dx ex + sin x)
1 – (ex + cos x) = Ú _____________ x dx (e + sin x)
= (x – log|ex + sin x|) + c
9. We have,
b Ú _______ dx a + c ex
be– x = Ú ________ dx c e– x + c
dt b __ = – __ a Ú t ,
( )
12. We have
Let ex = t
1.133
(
)
13. We have, cos x – sin x + 1 – x Ú _________________ d x |ex + sin x + x Let a e– x + c = t x
fi – a e dx = dt
(cos x – 1) – (sin x + x) = Ú ____________________ dx ex + sin x + x
1.134 Integral Calculus, 3D Geometry & Vector Booster
[
cos(x – a)cos(x – b) 1 = _________ __________________ dx cos(b – a) Ú sin(x – a)cos(x – b)
(ex + cos x – 1) – (ex + sin x + x) = Ú ___________________________ dx ex + sin x + x = log|ex + sin x + x| – x + c
]
sin(x – a)sin(x – b) + _________________ dx sin(x – a)cos(x – b)
1 = _________ (cot(x – a) + tan(x – b))dx cos(b – a) Ú
14. We have, sin(x + a) Ú _________ dx sin(x + b)
1 = _________ [log |sin(x – a)| – log(sec{x – b})] + c cos(b – a)
sin(t – b + a) = Ú ____________ dt sin t
1 = _________ [log |sin(x – a)cos(x – b)|] + c cos(b – a)
sin(t + (b – a)) = Ú _____________ dt sin t
17. We have, ____
sin t.cos(a – b) + cos t.sin(a – b) = Ú ____________________________ dt sin t
÷tan x Ú________ dx sin x cos x
= cos(a – b) Ú dt + sin(a – b) Ú cot dt
sec2x ÷tan x = Ú __________ dx tan x
2 sec x ____ = Ú _____ dx tan x ÷
= (x + b)cos(a – b) + sin(a – b)log |sin (x + b)| + c
2t dt = Ú ____ , t
dx 15. We have Ú ________________ sin(x – a)sin(x – b)
= t cos(a – b) + sin(a – b)log |sin t | + c
sin(b – a) 1 = ________ ________________ dx sin(b – a) Ú sin(x – a)sin(x – b)
= 2t + c
= 2÷tan x + c
]
cos(x – a)sin(x – b) – _________________ dx sin(x – a)sin(x – b)
1 = ________ [cot(x – b) – cot(x – a)]dx sin(b – a) Ú 1 = ________ [log|sin(x – b)| – log|sin(x – a)|] + c sin(b – a)
( |
|)
sin(x – b) 1 = ________ log _________ + c sin(b – a) sin(x – a) 16. We have, dx Ú ________________ sin(x – a)cos(x – b) cos(b – a) 1 = _________ Ú _________________ dx cos(b – a) sin(x – a)cos(x – b) cos ( (x – a) – (x – b) ) 1 = _________ ___________________ dx cos(b – a) Ú sin(x – a)cos(x – b)
fi sec2xdx = 2tdt ____
dx _____ Ú _______ x÷x 4 – 1
[
Let tan x = t2
18. We have,
sin ( (x – a) – (x – b) ) 1 = ________ ___________________ dx sin(b – a) Ú sin(x – a)sin(x – b) sin(x – a)cos(x – b) 1 = ________ __________________ dx sin(b – a) Ú sin(x – a)sin(x – b)
____
3 x_____ dx = Ú ________ 4 4 x÷ x – 1
dt 1 = __ Ú t _______ 2 (t2 + 1)t
1 = __ Ú______ dt 2 (t2 + 1)
1 = __ tan– 1(t) + c 2
1 ) + c = __ tan– 1 ( ÷x 4 – 1 2
_____
19. Let dx Ú _______ 2 (x + 1)
(
)
2x◊x 1 = _______ 2 Ú dx – Ú – ________ dx (x + 1) (x2 + 1)2
(x2 + 1) – 1 1 = _______ 2 Ú dx – 2Ú __________ dx (x + 1) (x2 + 1)2
(
)
Indefinite Integrals
(
)
2t + 1 1 = ___ __ tan– 1 ______ __ + c 3 ÷3 ÷
dx dx x = _______ 2 + 2 Ú _______ – 2 Ú _______ (x + 1) (x2 + 1) (x2 + 1)2
1.135
(
)
2x2 + 1 1 = ___ __ tan– 1 _______ __ + c 3 ÷3 ÷
dx x fi 2 Ú ________ 2 = _______ 2 + 2 tan– 1x – tan– 1x + c 2 (x + 1) (x + 1)
dx x 1 fi Ú ________ = _______ + __ tan– 1x + c (x2 + 1)2 (x2 + 1) 2
22. Let I = Ú sec3x dx
20. We have,
= Ú (sec2x ◊ sec x)dx
= sec x Ú sec2x dx – Ú (sec x ◊ tan x ◊ tan x) dx
= sec x Ú sec2x dx – Ú (sec x ◊ tan2 x)dx
= sec ◊ x tan x – Ú (sec x (sec2x – 1))dx
= sec ◊ x tan x – Ú sec3x dx + Ú secx dx
fi
2I = sec x ◊ tan x + Ú secx dx
fi
2I = sec x ◊ tan x – log|sec x + tan x| + c
fi
1 I = __ (sec x ◊ tan x + log|sec x + tan x|) + c 2
÷ Ú ÷
_________
sin(x – a) Ú _________ dx sin(x + a) ____________________
sin(x – a) _________ sin(x – a) = _________ × dx sin(x + a) sin(x – a) sin(x – a) ____________ = Ú _____________ dx 2 2x – sin a ÷ sin sin x cos a – cos x sina ___________ = Ú ____________________ dx 2 2 x – sin a ÷ sin sin x ____________ = cos a Ú _____________ dx 2 2 x – sin a ÷ sin
cos x ____________ – sin a Ú _____________ dx 2 2 x – sin a ÷ sin
23. Let I = Ú cosec3x dx
sin x ____________ = cos a Ú _____________ dx 2 2 a – cos x ÷ cos
cos x ____________ – sin a Ú _____________ dx 2 2 x – sin a ÷ sin
(
= Ú cosec3x ◊ cosec x dx
= cosec x Ú cosec2x dx –
Ú (– cosec x ◊ cot x ◊ – cot x)dx
)
cos x = – cos a sin – 1 _____ cosa
= – cosec x ◊ cot x – Ú cosec x ◊ cot2 x dx
= – cosec x ◊ cot x – Ú cosec x(cosec2x – 1)dx
21. We have,
= – cosec x ◊ cot x – Ú cosec3x dx + Ú cosec x dx
x Ú __________ dx x4 + x2 + 1
fi 2I = – cosec x ◊ cot x + Ú cosec x dx
|
– sin a log sin a +
____________ 2 2x – sin a + ÷ sin
1 dx = __ Ú ___ + t + 1, 2 t2
Let x2 = t
fi 2xdx = dt
( ) __
|
3 2 ÷ dx 1 = __ Ú _______ 2 + ___ 2 2 1 t + __ 2 1 t + __ 1 2 2 – 1 __ = __ × ___ __ tan _____ + c 2 ÷ 3 3 ÷ ___ 2
( )
( )
c
fi 2I = – cosec x ◊ cot x + log|cosec x – cot x| + c 1 fi I = __ (– cosec x ◊ cot x + log|cosec x – cot x| + c 2 24. We have, x2 – 1 ___________ Ú _____________ dx x÷x 4 + 3x2 + 1
x2 – 1 ________ = Ú ______________ dx 1 2 4 x x + __ 2 + 3 x
÷(
)
1.136 Integral Calculus, 3D Geometry & Vector Booster
( )÷(
________
)
1 1 2 = Ú 1 – __ 2 x + __ x – 2 + 3 x
dt ______ = Ú _______ , 2 ÷ t + 1
1 Let x + __ x = t
( )
1 fi 1 – __2 dx = dt x
|
_____ ÷t 2 + 1 +
|
= log t +
1 1 2 __ = log x + __ 3 + c x + x + x2 +
| (
25. We have,
c
÷(
)|
dx Ú _____________ x4 + 18x2 + 81 = Ú _______ dx (x2 + 9)2
Now, Ú ________ dx (x2 + 9)2
(
( )
27. We have, Ú sin 4x etan x dx 2
____________
)
( ) ( ) x x = __ ( 2tan ( __ ) ) + c 2 2 x = x tan ( __ ) + c 2
x x x x = __ Ú sec2 __ dx – Ú tan __ dx + Ú tan __ dx 2 2 2 2
)
= Ú 2sin2x . cos2x.etan x dx
1 – tan2x tan2 x 2 tan x = Ú 2 ________ ◊ ________ ◊ e dx 2 1 + tan x 1 + tan2x
1 – tan x = 2Ú __________ ◊ 2tan x ◊ sec2x ◊ etan x dx (1 + tan2x)3
2
2
2
Let tan2x = t
fi
et(1 – t) = 2 Ú ________ dt (1 + t)3
2tan x sec2x dx = dt
– 2x ◊ x 1 = _______ 2 Ú dx – Ú _______ dx (x + 9) (x2 + 9)2
(x2 + 9) – 9 x = _______ 2 + 2 Ú __________ dx (x + 9) (x2 + 9)2
(t + 1) – 2 = – 2 Ú et _________ dt (t + 1)3
x = _______ 2 + 2 Ú _______ dx – 18 Ú ________ dx 2 (x + 9) (x + 9) (x2 + 9)2
1 2 = – 2 Ú et _______ – ______ dt (t + 1)2 (t + 1)3
t = – 2 Ú _______ e + c (t + 1)2
(
)
dx fi 19Ú ________ (x2 + 9)2
( ) x ( __ ) + c 3
( )
(
)
(
)
x x x 2 1 = _______ 2 + __ tan– 1 __ – __ tan– 1 __ + c 3 3 3 (x + 9) 3
28. We have,
x 1 = _______ 2 + __ tan– 1 (x + 9) 3
2x + 2 ____________ Ú sin– 1 _____________ dx 2 + 8x + 13 ÷ 4x
fi
x x 1 dx = _________ + ___ tan– 1 ( __ ) + c Ú ________ 2 2 3 19(x2 + 9) 57
(x + 9)
26. We have,
(
)
x + sin x Ú ________ dx 1 + cos x
(
)
x sin x = Ú ________ + ________ dx 1 + cos x 1 + cos x
2sin (x/2)cos(x/2) x = Ú _________ + _______________ dx 2 2cos (x/2) 2cos2(x/2)
(
(
)
( )
( ) )
x x x = Ú __ sec2 __ + tan __ dx 2 2 2
(
)
(
)
2x + 2 ____________ = Ú sin1 ______________ dx + 8)2 + 32 ÷ (2x
Let (2x + 2) = 3tanq
fi 2dx = 3sec2q dq
(
)
3tanq 3 = __ Ú sin– 1 ______ sec2q dq 2 3secq
3 = __ Ú q sec2q dq 2
3 = __ (q Ú sec2q dq – Ú tanq dq ) 2
3 = __ (q tanq – log|secq |) + c 2
Indefinite Integrals
(
( ) ( ÷ ( ) ) ) 3 2x + 2 __ ( ______ – log (4x + 8x + 13) + c 3 ) 4 ___________
3 2x + 2 2x + 2 2x + 2 2 = __ ______ tan – 1 ______ – log 1 + ______ + c 2 3 3 3
= (x + 1) tan– 1
2
29. We have,
2 = Ú _________ cos x dx + Ú __________ sin x dx (2 + cos x) (2 + cos x)2
2 1 = Ú _________ Ú cos x dx – Ú __________ sin x dx (2 + cos x) (2 + cos x)2
sin2x + Ú __________ dx (2 + cos x)2
x2 Ú _____________ dx (x sinx + cos x)2
sin x = _________ + c (2 + cos x)
(x cosx)(x sec x) = Ú _____________ dx (x sin x + cos x)2
(x cosx) = x sec x Ú ______________ dx (x sin x + cos x)2
32. We have,
x sec x = – ____________ + sec2x dx x sin x + cos x Ú
( ) 1 + sin2q = Ú cos(2q ) log ( ________ dq cos2q ) 1 + sin 2q = log (_________ cos 2q ) Ú cos (2q ) dq
x sec x = – ______ + cos x + tanx + c x sin x
sin (2q ) ______ 2 – Ú _______ ◊ dq cos 2q 2
cosq + sinq Ú cos(2q ) × log ___________ dq cosq – sinq
(sec x + x sec x ◊ tan x) – Ú – _________________ dx x sin x + cos x
(
30. We have,
x sec2x + tan x = Ú x _____________ dx (x tan x + 1)2
x sec2x + tan x = x2 _____________ dx (x tan x + 1)2
( Ú (
) )
2
(
33. We have,
(
(
)
)
x x cos x = – _________ + 2 Ú _____________ dx (tan x + 1) (x sin x + cos x) x2 = – _________ + 2log|x sin x + cos x| + c (tan x + 1)
31. We have, sec x (2 + sec x)
dx Ú ______________ 2
1 2cos x + = Ú __________ dx (2 + cos x)2
2cos x + cos2x + sin2x = Ú ___________________ dx (2 + cos x)2
cos x(2 + cos x) + sin2x = Ú ____________________ dx (2 + cos x)2
(1 + 2 sec x)
)
x4 + 2 Ú ex _________ dx (1 + x2)5/2
1 – Ú 2x. – __________ dx (x tan x + 1) 2
)
(2q ) __ 1 + sin 2q 1 = log _________ × sin ____ – log |sec 2q | + c 2 2 cos 2q
x2(x sec2x + tan x) dx Ú ________________ (x tan x + 1)2
1.137
4 (x + 1) + 1 = Ú ex __________ dx (1 + x2)5/2
(x2 + 1)2 + (1 – 2x2) = Ú ex __________________ dx (1 + x2)5/2
(x2 + 1)2 (1 – 2x2) _________ = Ú ex _________ + dx (1 – x2)5/2 (1 + x2)5/2
(1 – 2x2) 1 _________ = ex _________ + dx (1 + x2)1/2 (1 + x2)5/2
x 1 = Ú ex _________ + _________ dx (1 + x2)1/2 (1 + x2)5/2
( Ú ( (
)
)
)
(
)
(1 – 2x2) x __________ + Ú ex _________ + dx (1 + x2)3/2 (1 + x2)5/2
(
)
x 1 = ex _______ ______ + _________ + c 2 (1 + x2)3/2 ÷ 1 + x
1.138 Integral Calculus, 3D Geometry & Vector Booster 34. We have,
36. We have,
(
)
Ú cot– 1(1 + x + x2) dx
x cos3x – sin x Ú esin x ____________ dx cos2x = Ú e
(x cosx – sec x ◊ tan x) dx
= Ú e
sin x ◊ cos x x cosx – _________ dx cos3x
sin x
sin x
(
(
)
)
t = Ú et sin– 1t – ________ dt, (1 – t2)3/2
Let t = sin x
fi dt = cosx dx
[
(
)]
t 1 1 = Ú et sin– 1t – ______ _____ + _______ _____ – ________ dt 2 2 (1 – t2)3/2 ÷ 1 – t ÷1 – t
(
)
1 _____ = e sin t – _______ + c ÷ 1 – t2 t
– 1
dx Ú ____________ cosx + cosec x
= Ú tan– 1(1 + x)dx + Ú tan– 1x dx
)
Let 1 + x = t fi dx = dt
= Ú tan– 1t dx +
Ú tan– 1x dx
1 = t tan– 1t – __ log |t2 + 1| + 2 1 x tan– 1x – __ log |x2 + 1| + c 2 1 = (x + 1)tan– 1(x + 1) – __ log|(x + 1)2 + 1| 2 1 x tan– 1x – __ log |x2 + 1| + c 2
37. We have,
= Ú ____________ sin x dx sin x ◊ cos x + 1
= Ú ______________ 2sin x dx 2sin x ◊ cos x + 2
= Ú _________ 2sin x dx sin 2x + 2
(sin x + cos x) + (sin x – cos x) = Ú __________________________ dx sin 2x + 2
Ú tan– 1 (1 + x + x2) dx
(
p = Ú __ – tan– 1(1 + x + x2) dx 2
(sin x + cos x) (sin x – cos x) = Ú ____________ dx + Ú _____________ dx sin 2x + 2 2 + sin 2x + 2x
Let sin x – cos x = t and sin x + cos x = v fi (cos x + sin x)dx = dt and (cos x – sin x)dx = dy
= Ú _________ dt – Ú _________ dv 1 + t2 + 2 v2 – 1 + 2
dv dt = Ú _____ – Ú ______ 2 2 v +1 t –3
| | __
(1 + x) – x = Ú tan– 1 __________ dx 1 + x(1 + x)
35. We have,
)
1 = Ú tan– 1 ________ dx 1 + x + x2
= esin x(x – sec x) + c
( (
t – ÷3 1__ __ = – ____ log ______ – tan– 1(v) + c 2÷3 t +÷ 3 __
(sin x – cos x) – ÷ 3 1__ __ = – ____ log| ________________ 2÷3 (sin x – cos x) + ÷ 3 – tan– 1(sin x + cos x) + c
)
p 1 = __ x – (x + 1)tan– 1(x + 1) – __ log|(x + 1)2 + 1| 2 2 38. We have,
1 – x tan– 1x – __ log |x2 + 1| + c 2
sin x Ú _____ dx sin 4x sin x = Ú ____________ dx 2 sin 2x cos 2x 1 dx = __ Ú _______________ 4 cos x (2cos2x – 1) 1 cos x dx = __ Ú __________________ 4 (1 – sin2x)(1 – 2sin2x) 1 dt = __ Ú _____________ , 4 (1 – t2)(1 – 2t2)
Let t = sin x
fi
1 dt = __ Ú ______________ 4 (t2 – 1)(2t2 – 1)
dt = cos x dx
Indefinite Integrals
(
) ( )
1 1 2 = __ Ú _____ – _______ dt 4 t2 – 1 2t2 – 1
dv dv 3 3 = __ Ú __________ – __ _______ 2 (v2 – v + 1) 2 Ú (v3 + 1)
| | | |
1 1 dt ___ 1 2 = __ Ú _____ dt – __ Ú __ __ 2 2 4 t – 1 4 t ÷2
2
| |
( ) ( )
(
)
2v – 1 1 1 = ___ __ tan– 1 ______ __ – __ log |v3 + 1| 2 3 ÷3 ÷
__
v–1 3 + __ Ú _________ dv 2 v2 – v + 1
t ÷2 – 1 t – 1 ____ 1 1 __ = __ log _____ – __ log _______ + c 8 t + 1 4÷2 t ÷2 + 1
(
)
2v __ –1 1 1 = ___ __ tan– 1 ______ – __ log|v3 + 1| 2 3 ÷3 ÷
where t = sin x. 39. We have, Ú (x
3m
2m
+x
2m
m
+ x )(2x
+ 3x + 6)
dx
= Ú (x3m – 1 + x2m – 1 + xm – 1) × x × (2x2m + 3xm + 6)1/m dx
= Ú (x3m – 1 + x2m – 1 + xm – 1)(2x3m + 3x2m + 6xm)1/m dx 3m
2m
+ 3x
3 (2v – 1) – 1 + __ Ú __________ dv 2 v2 – v + 1
1/m
m
2
dv 3 3 v – (v – 1) __ = __ Ú _______________ – __ Ú ___________ dv 2 2 2 2 3 ÷ 1 (v3 + 1) __ ___ v – + 2 2
1 t – ___ __ 2 ÷ t–1 1 1 1__ = __ × __ log _____ – ____ log ______ + c 4 2 t + 1 4÷2 1__ ___ t + 2 ÷
| |
(
Let 2x
+ 6x = t
fi 6m(x3m – 1 + x2m – 1 + xm – 1)dx = dt
)
1__ – 1 1 1 = ___ tan 2v – ___ __ – __ log |v3 + 1| 3 3 2 ÷ ÷
3 1 = __ log|v2 – v + 1| – __ log |v3 + 1| + c 2 2
1 = ___ Ú t1/m dt 6m 1 = ________ t m + 1 + c 6(m + 1)
where v = t2 = (tan x)2/3
1 = ________ (2x3m + 3x2m + 6xm) m – 1 + c 6(m + 1)
ex = Ú _________ dx ex(ex – 1)2
40. We have,
dt = Ú _______ Let ex = t t(t – 1)2
dx 41. We have Ú _______ x 2 (e – 1)
_____
dx Ú 3÷tan x
fi exdx = dt
3
Let tan x = t
fi sec 2x dx = 3t2 dt
( )
2v – 1 3 1__ – 1 ______ + __ log|v2 – v + 1| – ___ tan __ + c 2 ÷3 ÷3
m
2
3 = Ú _____ 3t dt t6 + 1
(
)
1 1 1 = Ú __ – ____ + ______ dt t t – 1 (t – 1)2
t 1 = log ____ – ____ + C t–1 t–1
ex 1 = log ______ x – ______ x + C e –1 e –1
2
3t 3t fi dx = _____ 2 dt = _____ 2 dt sec x 1+t
| |
| |
42. We have
3 t 2.2t = __ Ú _____ dt 2 t6 + 1 v 3 = __ Ú _____ d v, 2 v3 + 1
Let t2 = v
fi
(v + 1) – 1 3 = __ Ú ________________ dv 2 (v + 1) (v2 – v + 1)
1.139
2t dt = dx
tan– 1x Ú ______ dx x4
[
( ) ]
dx 1 1 = tan– 1x Ú ___4 – Ú ______ 2 × – ___ 3 dx x 1+x 3x
x dx tan– 1x __ 1 = ______ + Ú ________ 3 4 3 (– 3x ) x (1 + x2)
1.140 Integral Calculus, 3D Geometry & Vector Booster
dt tan– 1x __ 1 = ______ + Ú _______ , 3 6 (– 3x ) t(t – 1)2
Let x2 + 1 = t fi
(
2x dx = dt
)
tan– 1x __ 1 1 _____ 1 1 = ______ + Ú __ – + _______ dt 3 t t – 1 6 (– 3x ) (t – 1)2 – 1
( | |
)
tan x __ t 1 1 = ______ + log ____ – ____ + c t–1 t–1 (– 3x3) 6 – 1
( | | )
3 = 4 log|t3 + 1| – __ log|t2 – t + 1| 2
( )
__ 2t – 1 1 + ___ __ tan– 1 ______ __ + c, Let t = 3÷x 3 ÷3 ÷
45. We have, dx Ú ___________ 2sin x + sec x
cos x = Ú _________ dx sin 2x + 1
2
tan x __ x + 1 __ 1 1 = ______ + log ______ 2 – 2 + c 3 6 (– 3x ) x x
43. We have,
2cos x 1 = __ Ú _________ dx 2 sin 2x + 1
dx Ú _______________ 4 ____ _____ (÷cos x + ÷ sin x )
1 (sin x + cos x) – (sin x – cos x) = __ Ú _________________________ dx 2 sin 2x + 1
sec2x = Ú ___________ 4 dx ____ (1 + ÷ tan x )
= Ú _______ 2t dt , (1 + t)4
1 (sin x + cos x) 1 (sin x – cos x) = __ Ú ____________ dx – __ Ú ____________ dx 2 2 sin 2x + 1 sin 2x + 1 Let tan x = t2 fi
sec2xdx = dt
(t + 1) – 1 = 2 Ú _________ dt (1 + t)4
(
( ) x p 1 1 = ____ log tan ( __ + __ ) – __ (sin x + cos x) + c 2 8 2 | | 2 ÷2
)
1 1 = 2 Ú _______ – _______ dt (1 + t)3 (t + 1)4
1 1 = – 2 Ú ________ – _______ + c 3(1 + t)2 4(t + 1)3
1 1 = – 2 ____________ 2 – ____________ ____ ____ 3 + c 3(1 + ÷ tan x ) 4(÷tan x + 1)
44.
– 2/3
dx 1 1 (sin x – cos x) = __ Ú ____________ – __ Ú ____________ dx 2 (sin x + cos x) 2 (sin x + cos x)2 p 1 1 (sin x – cos x) = ____ __ cosec x + __ dx – __ Ú ____________ dx Ú 4 2 (sin x + cos x)2 2÷2
( Ú(
1 (sin x + cos x) 1 (sin x – cos x) = __ Ú ____________ dx – __ Ú ____________ dx 2 2 (sin x + cos x) 2 (sin x + cos x)2
)
__
)
( ) – 2
+x 1+t 2 dx 3 Ú ______ t dt, Ú 1_______ 1+x 1 + t3 Let x = t3 fi dx = 3t2 dt
dt t2 = 3 Ú _____ 3 dt + 3 Ú _____ 1+t 1 + t3 t2 – (t2 – 1) = log|t3 + 1| + 3 Ú __________ dt t3 + 1 (t – 1) = 4 log|t3 + 1| – 3 Ú ________ dt 2 t – t + 1 3 (2t – 1) – 1 = 4 log|t3 + 1| – __ Ú __________ dt 2 t2 – t + 1 3 3 dt = 4 log|t3 + 1| – __ log|t2 – t + 1| + __ Ú ________ 2 2 t2 – t + 1
46. We have, 4 + 1 ______ Ú x dx 6 x +1
(x2 + 1)2 – 2x2 = Ú _____________ dx (x6 + 1) (x2 + 1)2 x2 = Ú _______ dx – 2 Ú _______ dx 6 6 (x + 1) (x + 1) (x2 + 1) x2 = Ú ___________ dx – 2 Ú _______ 6 dx 4 2 (x + x + 1) (x + 1)
(
)
(
)
1 1 + __ 2 x 3x2 2 = Ú ___________ dx – __ Ú _________ dx 3 ( (x3)2 + 1 ) 1 x2 + __ 2 – 1 x
(
)
1 1 + __ 2 x 3x2 2 = Ú ____________ dx – __ Ú _________ dx 2 3 3 ( (x )2 + 1 ) 1 x + __ x + 1
( (
)
)
Indefinite Integrals
(
)
1 2 – 1 3 __ = tan–1 x – __ x – 3 tan (x ) + c 47. We have,
Ú dx(e
2
x
– 1)
= Ú_______ dt , t(t – 1)2
Let t = ex
fi dt = ex dx
(t2 – 1) + 1 t = ______ 2 + 2Ú _________ dt (t – 1) (t2 – 1)2
dt dt t = ______ 2 + 2Ú ______ + 2Ú _______ 2 2 (t – 1) (t – 1) (t – 1)2
)
| |
| |
t t–1 1 = – ______ – __ log ____ + c t+1 (t2 – 1) 2
_____
where t = ÷ x + 1 50. We have,
1 1 1 = Ú __ – ______ + ______ dt, t (t – 1) (t – 1)2
x4 – 1 _________ Ú ____________ dx x2÷ x 4 + x2 + 1
1 1 = log ____ – ____ + C t –1 t– 1
| |
x4 – 1 ________ = Ú ______________ dx 1 3 2 x x + __ 2 +1 x 1 x – __ 3 dx x ________ = Ú ____________ 1 x2 + __ 2 +1 x
÷( ) ( ) ÷( )
x
e 1 = log _____ x – ______ + C e –1 ex – 1 48. We have, dx_____ Ú ____________ (x – 1)÷x + 2
= Ú _______ 2t dt , (t2 – 3)t
Let x + 2 = t2
fi dx = 2t dt
1 Let x2 + __ 2 + 1 = t2 x 1 fi 2x – 2 __3 dx = 2tdt x
t dt = Ú ___ , t
(
dt = 2Ú __________ __ 2 ( t – (÷3 )2 )
= Ú t dt
t–÷ 3 2 __ = ____ __ log ______ + c 2÷3 t + ÷3
=t+c
|
1 = x2 + __ 2 + 1 + c x
| | __
|
__
x2 – 1 ____________ Ú ______________ dx x3÷ 2x 4 – 2x2 + 1
dx 2 _____ Ú ________ x ÷x + 1 2t dt = Ú ________ , 2 (t – 1)2t dt = 2Ú _______ (t2 – 1)2
dt Now, Ú _______ (t2 – 1)
Let x + 1 = t2
fi dx = dtdt
÷
51. We have,
49. We have,
)
__________
(x + 2) – ÷ 3 1 __ + c = ___ __ log ___________ ÷3 (x + 2) + ÷ 3
)
dt dt t fi 2Ú _______ = – ______ – Ú ______ (t2 – 1)2 (t2 – 1) (t2 – 1)
exdx = Ú _________ x x 2 e (e – 1)
(
(
1.141
(
)
2t ◊ t 1 = ______ 2 Ú dt – Ú – _______ dt (t – 1) (t2 – 1)2
x2 – 1 __________ = Ú ____________ dx 2 1 __ x5 2 – __ 2 + 4 x x
(
÷
)
1 1 5 dx __ 3 – __ x x __________ = Ú ___________ 2 1 __ __ 2 – 2 + 4 x x
1 = __ Ú 2
÷
t dt ___ , t
2 Let 2 – __2 + x
1 __ 4 = t2 x
2 fi ___ – 25
)
(
4 ___ dx = 2tdt 25
1.142 Integral Calculus, 3D Geometry & Vector Booster
1 = __ t + c 2
1 2 1 __ = __ 2 – __ 2 + 4 + c 2 x x
÷
__________
52. We have,
Ú
fi
Also, t2 = 1 – sin 2q
fi
=
dt ________ = Ú __________ ÷ 5 + 1 + t2
dt _________ = Ú ___________ __ ( )2 – t2 ÷ ÷6
Ú
÷
t = sin– 1 ___ __ + c ÷6
sinq –__cosq = sin– 1 __________ + c ÷6
______
cos2x _____ 2 d x sin x
÷
____________ 2
2
=
x – sin x ___________ dx Ú cos sin2x
=
Ú
________
÷cot 2x – 1 dx
( )
(
se2x Ú _____________ dx (sec x + tan x)9/2
__________ 2 ÷cosec x – 2 dx
Ú
=
2 cosec x–2 __________ = Ú ___________ dx 2 x – 2 ÷ cosec
cosec2x 2 __________ __________ = Ú ___________ dx – Ú ___________ dx 2 2 x – 2 ÷cosec x – 2 ÷ cosec
cosec2x 2sin x ________ _________ = Ú _________ dx – Ú __________ dx 2 x – 1 ÷1 – 2sin2x ÷ cot
2
cosec x 2sin x ________ _________ = Ú _________ dx – Ú __________ dx 2 x – 1 ÷2cos 2x – 1 ÷ cot
sec x.sec x = Ú _____________ dx (sec x + tan x)9/2
fi cosec2xdx = dt and – sin xdx = dx.
...(i)
Let
sec x + tan x = t
fi
sec x (sec x + tan x) dx = dt
fi
dt sec xdx = __ t
1 1 Also sec x – tan x = ___________ = __ (sec x + tan x) t
1 1 Thus, sec x = __ t + __ dt t 2
( )
Let cot x = t and cos x = v
)
54. We have
sin 2q = 1 – t2
_____
cos2x ÷ ______ dx sin x
(cosq + sinq)dq = dt
The given intergral (i) reduces to
dv dt = – Ú _______ _____ + 2 Ú ________ _______ 2 ÷2 v 2 – 1 ÷ t – 1
( )
__ dv dt _____ _________ = – Ú ______ + ÷ 2 Ú ___________ 2 1__ 2 2 ___ ÷ t – 1 v – 2 ÷
÷ ( )
1 1 dt __ t + __ __ t t 2__________ Ú 9/2 t
2 1 (t + 1) = __ Ú ______ dt 2 t13/2
1 1 1 = __ Ú ___ + ____ dt 2 t9/2 t13/2
1 2 2 = __ ____ + ______ + c 2 7t7/2 11t11/2
53. We have,
cosq + sinq __________ Ú ___________ dq ÷ 5 + sin(2q )
1 1 = ____ 7/2 + ______ 11/2 + c 7t 11t
1 1 = ______________ + ________________ + c 7(sec x – tan x)7/2 11(sec x + tan x)11/2
_____
|
________
|
= – log cot x + ÷cot 2x – 1
|
÷
______
|
__ 1 = – log|t + ÷ t 2 – 1 |+÷ 2 log v + v2 – __ + c 2
__
|
÷
_________
|
1 + ÷2 log cos x + cos x – __ + c 2
Let
2
sinq – cosq = 1
( Ú (
(
)
)
)
Indefinite Integrals
55. We have,
tan2q ___________ Ú ____________ dq 6 6 q + cos q ÷ sin
sin2x = tan x log(1 + sin2x) – 2Ú ________ dx 1 + sin2x
sin2q __________ = Ú ________________ dq 3 2 cos2q 1 – __ sin 2q 4
÷
(
2sin2q __________ = Ú ________________ dq cos2q ÷3cos 2 2q – 1
dt ______ = Ú ________ , t÷3t 2 – 1
Let t = cos2q
fi dt = 2sin2q dq
)
1 = tan x log(1 + sin2x) – 2Ú 1 – ________ dx 1 + sin2x
Ú sec2x log(1 + sin2x)dx
2sin x ◊ cos x = log(1 + sin2x) Ú sec2x dx – Ú tan x ◊ __________ dx (1 + sin2x)
2sin2q __________ = Ú ________________ dq 2 cos2q ÷4 – 3sin 2q
log(1 + sin2x) 2. Ú ____________ dx cos2x
=
tan2q ______________ = Ú _______________ dq 2 q ÷ 1 – 3sin2q cos
2 = tan x log(1 + sin2x) – 2(x) + 2Ú _________ sec x dx 2 2tan x + 1
dt = tan x log(1 + sin2x) – 2(x) + 2Ú ______ 2t2 + 1
Let tan x = t
t dt = – Ú _________ ______ 2 t ÷ 3t 2 – 1
__
3.
Ú x 3 ( 1 + x3 )
2 – 1 __
2 – __
1 = – __ tan– 1v + c 9
1 = – __ tan– 1 ( ÷3t 2 – 1 ) + c 9
1 = – __ tan– 1 ( ÷3cos 2 2q – 1 ) + c 9
dx dx = Ú __________ x2/3(1 + x2/3) Let x1/3 = t dx fi ____ = dt 3x2/3
______
__________
x2 + 6 1. Ú _______________ dx (x sin x + 3cos x)6
(x + 6)x = Ú _________________ dx 3 (x sin x + 3x2cos x)6 3
(x + 6)x cos x.(x sec x) = Ú ___________________ dx (x3 sin x + 3x2cos x)6 2
x(x + 6) cos x = (x3sec x)Ú _________________ dx + Ú sec2x dx 3 (x sin x + 3x2cos x)6 – x sec x = _____________ + tan x + c x sin x + 3 cos x
dx fi ___ = 3dt x2/3
dt = 3Ú _____ 2 1+t
= 3tan– 1(t) + c
= 3tan– 1(x1/3) + c
4
2
__
= tan x log(1 + sin2x) – 2(x) + 2÷2 t an– 1(t ÷2 ) + c where t = tan x
dv 1 = – __ Ú _____ , Let 3t2 – 1 = v2 9 v2 + 1 6 tdt = 2x dx
2
sec2x dx = dt
1.143
[
(
) ] [
(
)]
5p p 4. Ú 1 + tan ___ – x 1 + tan – ___ + x dx 16 16
= Ú 2dx
= 2x + c
(
p (1 + tan A)(1 + tan B) = 2, if A + B = __ 4 5. Útan (x – a ) ◊ tan (x + a ) ◊ tan 2x dx
)
= Ú(tan (2x) – tan (x – a ) – tan(x + a ))dx
1 = – __ log |cos (2x)| + log |cos (x – a )| 2
+ log |cos (x – a )| + c
1.144 Integral Calculus, 3D Geometry & Vector Booster
( ( ÷ ) ) _____
dt t Now, I1 = Ú ___________ t2 + 2t – 3 dt
1–x 6. Ú cos 2cot _____ dx 1+ x – 1
Let x = cosq
fi dx = – sinq dq
=
Ú
[ ( ÷
________
=
Ú
[ [ [ (
= – Ú cos (q ) dq
= – sinq + c ______ – ÷1 – x2 +
(
c
x (x + 1)dx
÷ { (
__________________________
(
)
}
)
1 1 (x2 – 1) x2 x2 + __ 2 + 2 x + __ x – 1 x = Ú _________________________________ dx x2(x2 + 2x + 1) ________________________
( )
)
)
}
1 1 1 1 – __ 2 x2 + __ 2 + 2 x + __ x – 1 x x = Ú _________________________________ 1 x + __ x + 2 1 Let x + __ x + 2 = t _________ ÷__________ t 2 + 2t – 3
( )
1 fi 1 – __ 2 dx = dt x
= Ú
t + _________ 2t – 3 dt = Ú ________________ 2 (t + 2)÷t + 2t – 3
dt (t + 2)
2
t(t + 2) – 3 _________ = Ú ________________ dt (t + 2)÷t 2 + 2t – 3
dt t _________ = Ú __________ _________ dt – Ú _______________ 2 (t + 2)÷t 2 + 2t – 3 ÷ t + 2t – 3
= I1 + I2 (say)
)
|
)
__
|
dx __________ Also, I2 = Ú _________________ (x + 2)÷x 2 + 2x – 3
2
(
(
1 where t = x + __ x
___________________ (x – 1)÷x 4 + 3x3 – x2 + 2x + 1 __________________________
( ) ÷{ (
)
– 1
Ú cos (p – q ) dq
7. Ú
(
(t + 1) – ÷2 1 1__ __ + c = __ log|t2 + 2t – 3| – ____ log __________ 2 2÷2 (t + 1) + ÷2
=
) )
(2t + 2) dt 1 = __ _________ dt – Ú ____________ 2 t2 + 2t – 3 (t + 1)2 – (2)2
q cos 2cot tan __ dq 2
=
)]
– 1
)
(2t + 2) dt 1 = __ _________ dt – Ú _________ 2 2 t2 + 2t – 3 t + 2t – 3
1 – cosq cos 2cot– 1 ________ dq 1 + cosq
{ ( ) } ] p q = Ú cos 2cot { cot ( __ – __ ) } ] dq 2 2 p q = Ú cos 2 __ – __ ) ] dq 2 2
( Ú ( Ú (
1 (2t + 2) – 2 = __ Ú __________ dt 2 t2 + 2t – 3
dt – __2 t = Ú _________________________ _____________________ 2 1 1 1 __ __ – 2 + 2 __ – 2 – 3 t t t
( ) ÷(
)
(
)
dt _______________________ = – Ú _________________________ 1 2 2 __ (1 – 2t) + 2t – 2 – 3t2 t
÷
(
)
dt = – Ú ___________________________ _________________________ 2 ÷(1 – 4t + 4t ) + 2t – 4t2 – 3t2 dt = – Ú _____________ ___________ ÷(1 – 2t + 3t2) dt = – Ú ________________ _______________ 2 + 1} ÷ – {3t + 2t dt = – Ú _________________ ________________ 2 1 – 3 t2 + __ t – __ 3 3
÷ {
}
dt = – Ú ___________________ _________________ 1 2 __ 4 – 3 t + __ – 3 9
÷ { ( ) }
dt 1__ _______________ = – ___ Ú _____________ 2 3 ÷ 2 1 2 __ __ – t + 3 3
÷( ) ( ) 3t + 1 1 = – ___ sin ( ______ + c 2 ) ÷3
– 1 __ 1 where t = x + __ x
(
)
(
p tan __ – x 4 __________________ 8. Ú _______________________ dx 2 3 cos x ÷tan x + tan2x + tan2x
)
Indefinite Integrals
(1 – tan x) sec2x dx _________________ = Ú __________________________ (1 + tan x)÷tan 3x + tan2x + tan x
(1 – t)dt _________ = Ú ________________ Let tan x = t (1 + t)÷t 3 + t2 + t fi sec2xdx = dt
(1 – t2 )dt _________ = Ú _________________ (1 + t)2 ÷ t 3 + t2 + t (1 – t2)dx ___________ = Ú _______________________ 1 2 (t + 2t + 1) t2 t + __ + 1 t
÷ (
)
( )
1 1 – __ 2 dt t __________ = Ú _____________________ 1 1 __ t + + 2 t + __ + 1 t t
(
(
)÷(
)
)
1 Let t + __ + 1 = v2 fi sec2xdx = dt t
– t2 )dt q 1 (1 __ ________ = – _______ + Ú 3 3 tan3q t3
q 1 1 = – _______ + __ Ú __ – 3 3 t3 3 tan q
q 1 ___ 1 = – _______ – __ 2 + log|t| + c 3 3 3 tan q 2t
q 1 ______ 1 = – _______ – __ 2 + log |sin q | + c 3 3 3 tan q 2sin q
( ) ( ) (
(1 – x sin x)ecos x = Ú ________________ dx xecos x( 1 – (xecos x)3 )
dt = Ú _______ , t(1 – t3)
=
dv = – 2Ú _______ 2 (v + 1)
dt Ú _______ t(1 – t3)
=
dt 3 Ú _______ t(t – 1)
3t2 1 = __ Ú ________ 3 3 dt 3 t (t – 1)
dv 1 = __ Ú _______ , 3 v(v – 1)
v –1 1 = __ log _____ + c v 3
t3 – 1 1 = __ log ______ 3 + c 3 t
(xex)3 – 1 1 = __ log ________ + c 3 (xex)3
= – 2 tan– 1(v) + c
( ÷
)
1 + c = – 2tan– 1 t + __ + 1 t 1 where t = x + __ x
9.
– 1
tan x 4 dx Ú ______ x Let x = tan q fi dx = sec2xdq
tan– 1(tan q )
Let t = xex
where v = t3
| |
| |
| |
=
sec2 q dq Ú __________ tan4q
=
q sec q 4 dq Ú _______ tan q
sec2q 1 dq = q Ú _____ 4 dq + __ Ú _____ 3 tan3q tan q
(1 + x cos x)esin x = Ú _______________ dx xesin x(1 – (xesin x )2)
q dq 1 = – ______ + __ Ú _____ 3 tan3q 3 tan3q
dt = Ú _______ , t(1 – t2)
q 1 cos3q = – _______ + __ Ú _____ dq 3 3 sin3q 3 tan q
dt = Ú _______ t(1 – t2)
– sin2q)cosq q 1 (1 __ _____________ = – _______ + dq 3Ú 3 tan3q sin3q
2t dt 1 = – __ Ú ________ 2 t2(t2 – 1)
2
)
(1 – x sin x) 10. Ú ____________ dx x(1 – x3e3 cos x)
v dv = – 2Ú ________ 2 (v + 1)v
________
1 __ dt t
where tan q = x
1.145
(1 + x cos x) 11. Ú _____________ dx x(1 – x2e2sin x)
where t = xesin x
1.146 Integral Calculus, 3D Geometry & Vector Booster
dv 1 = – __ Ú _______ , where v = t2 2 v(v – 1)
v 1 = __ log _____ + c 2 v –1
t2 1 = __ log _____ 2 + c 2 t –1
(xesin x)2 1 = __ log __________ sin x 2 + c 2 (xe ) – 1
| |
| |
|
14.
( ) (
( )
)
|
( )
1 – Ú x sec2 __ x dx + c
( ) 1 = x tan ( __ x ) + c
( )
( )
1 1 1 = x3 tan __ x + Ú x sec2 __ x dx – Ú x sec2 __ x dx + c 3
cos x = Ú _________ dx sin 2x + 1
2cos x 1 = __ Ú _________ dx 2 sin 2x + 1
1 (sin x + cos x) + (cos x – sin x) = __ Ú _________________________ dx 2 (sin x + cos x)2
dx cos x – sin x 1 1 = __ Ú ____________ + __ ____________ 2 (sin x + cos x) 2 Ú (sin x + cos x)2
p 1 1 cos x – sin x = ____ __ dx + __ Ú ____________ dx Ú cosec x + __ 4 2 (sin x + cos x)2 2÷2
__________
÷3cos 2x – 1 15. Ú ___________ dx cos x 3cos 2x –1 _________ = Ú ______________ dx cos x ÷cos 2x – 1 3(2cos2x – 1) __________ = Ú _______________ dx cos x ÷3cos 2x – 1 dx cos x _______________ _________ = 6Ú ________________ dx – 4Ú _______________ dx 2 cos x 3cos2x – 1 ÷ – 2sin x) – 1 ÷ 3(1
( ) 1 1 = ____ log tan ( __ + c (sin x + cos x) | 2x + __ p 8 ) | – ___________ 2÷2
__ cos x dx cosx _________ = 3÷2 Ú ____________ __________ dx – 4Ú ___________________ 2 – 3sin x (1 – sin2x)÷1 – 3sin2x ÷ (1
__
dx _________________ 13. Ú ______________________ cos x ÷sin(2x + a) + sin a
Ú ( 3x2tan ( __ 1x ) – x sec2 ( __ 1x ) ) dx
1 1 1 = tan __ x Ú (3x2)dx – Ú sec2 __ x – __2 (x3) dx x
dx 12. Ú___________ 2sin x + sec x
tan x cos a + sin a ______ = _______________ + c 2cos a ÷
__ cos x dx cos x ____________ _________ = ÷6 Ú ______________ dx – 4Ú ___________________ 2 1 2 (1 – sin2x)÷1 – 3sin2x ___ __ – sin x 3 ÷
÷( )
__ __ cos x dx _________ = ÷ 6 s in– 1(÷3 s in x) – 4Ú ___________________ (1 – sin2x)÷1 – 3sin2x
dx _____________________________ = Ú __________________________________ cos x÷sin(2x)cos a + cos(2x)sina + sin a
__ __ dt______ = ÷ 6 sin– 1(÷3 sin x) – 4Ú ______________ 2 (1 – t )÷1 – 3t2
dx ____________________________ = Ú __________________________________ cos x÷sin (2x) cos a + (1 + cos (2x) sin a
dx _________________________ = Ú _______________________________ cos x ÷2sin x cos x cos a + 2cos2x sina
where t = sin x __ __ y dy ______ = ÷ 6 s in– 1(÷3 sin x) + 4Ú _____________ 2 (y – 1)÷y 2 – 3
1 dx = ___ __ Ú _____________________ ________________ 2 2 cos x÷tan x cos a ÷ + sin a
sec2x dx 1 _______________ = ___ __ Ú ________________ 2 ÷tan x cos a ÷ + sin a
Let tan x cos a + sin a = t
where t = 1/y
__ __ v dv = ÷ 6 sin– 1(÷3 sinx) + 4 Ú ________ 2 , v2 = y2 – 3 (v + 2)v 2
__ __ dv = ÷ 6 s in– 1(÷3 s inx) + 4 Ú _______ 2 (v + 2)
( )
2t dt 1 ____ = _______ __ Ú t 2 cos a ÷
__ __ __ v = ÷ 6 s in– 1(÷3 sin x) + 2÷2 tan– 1 ___ __ + c 2 ÷
1 = _______ __ Ú dt 2 cos a ÷
y2 – 3 = ÷ 6 s in– 1(÷3 sin x) + 2÷2 tan– 1 _____ + c 2
( ÷ ) ______
__
__
__
Indefinite Integrals __
__
( ÷
_________
__
)
= – Ú ________ 2t dt (t2 + 1)t3
cos2x – 3 = ÷6 s in (÷3 s inx) + 2÷2 t an ________ + c 2 – 1
– 1
( ) [ ( )
( ) ]
dt = – 2Ú ________ 2 (t + 1)t3
2x 2x 1 16. Ú _____ 8 cos– 1 _____ 2 + tan– 1 _____ 2 dx 1–x 1+x 1–x
(
(
)
(
)
1 = – 2Ú – __ – tan– 1t + c t
p dx = __ Ú _______ 2 (1 – x8)
( ÷
_________
( ÷
[
]
] ]
)
(x2 + 1) – (x2 + 1) p 1 1 = __ Ú ______ 2 + ______ 2 + Ú _______________ dx 8 1–x 1+x x4 + 1
)
(x2 + 1) (x2 + 1) _______ 1 1 ______ 2 + ______ 2 + Ú _______ – 4 dx 1–x 1+x x +1 x4 + 1
4 cos = Ú _______________ x dx sin6x(1 + cot5x)3/5
4 2 x dx = Ú _____________ cos x cosec 5 3/5 (1 + cot x)
and then you do it.
4 = Ú _________ t dt , (1 + t5)3/5
x2 – x3 17. Ú __________________ dx (x + 1)(x3 + x2 + x)3/2
5t4dt 1 = – __ Ú _________ 5 (1 + t5)3/5
x2(1 – x) = Ú ___________________ dx 3/2 1 (x + 1)x3 x + __ x + 1
)
(1 – x) = Ú __________________ dx 3/2 1 x(x + 1) x + __ x + 1
(
)
(1 – x2) = Ú ___________________ dx 3/2 1 x(x + 1)2 x + __ x + 1
(
)
2
(1 – x ) = Ú ________________________ dx 3/2 1 2 x(x + 2x + 1) x + __ x + 1
(
)
(1 – x2) = Ú ________________________ dx 3/2 1 1 2 __ x x + x + 2 x + __ x + 1
(
) (
( )
)
) (
)
Let (1 + t5) = y5
fi
4 1 5y dy = – __ Ú _____ 3 5 y
= – Ú y dy
y2 = – __ + c 2
1 = – __ (1 + t5)2/5 + c 2
1 = – __ (1 + cot5x)2/5 + c 2
(
)
1 Let x + __ x + 2 = t2 1 fi 1 – __ 2 dx = 2t dt x
( )
Let cot x = t
5t4dt = 5y4dy
( )
sin3x dx 19. Ú _________________________________ 4 2 (cos x + 3cos x + 1) tan– 1(sec x + cos x)
1 1 – __ 2 dx x = Ú ______________________ 3/2 1 1 __ x + x + 2 x + __ x + 1
(
)
4 x cos 18. Ú ___________________ 3 5 sin x(sin x + cos5x)3/5
p dx dx = __ Ú _______ 4 + Ú ______ 4 4 (1 – x ) x +1
(
)
1 1 = 2 ________ _______ + tan– 1 x + __ x + 1 + c 1 x + __ x 1
p dx = __ Ú _____________ 2 (1 – x4)(1 + x4)
p = __ 8
)
1 1 = – 2Ú __ 2 – _____ 2 dt t t +1
p dx = Ú _______ 8 __ – 2 tan– 1x + 2 tan– 1x dx (1 – x ) 2
[ ( [ Ú (
1.147
(1 – cos2x) sin x dx = Ú _________________________________ (cos4x + 3cos2x + 1) tan– 1(sec x + cos x) (1 – t2)dt = – Ú ______________________ , where cos x = t 1 4 2 (t + 3t + 1)tan= 1 t + __ t
( )
1.148 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) ( ) ) ( Ú
dy 1 = 2Ú _______ , y = t + __ 2 t 4y – 2
dy = Ú _______ 2 2y – 1
) ( ) dy 1 = Ú ____________ , where ( t + __ ) = y t (y + 1)tan y
dy 1 = __ Ú ___________ __ 2 2 y – (1/÷2 ) 2
y÷2 – 1 1 = ____ __ log _______ __ + c 2÷2 y÷2 + 1
– 1 = Ú ___ dv v , where tan (y) = v
= Ú ___ dv v
1 __ t + __ ÷ 2 – 1 t 1 = ____ __ log ____________ + c 1 __ 2÷2 __ t + ÷ 2 + 1 t
1 1 – __ 2 dt t = Ú _____________________ 1 1 2 __ t + 2 + 3 tan= 1 t + __ t t
1 1 – __ 2 dt t = ______________________ 2 1 1 = 1 __ __ t + + 1 tan t + t t
( ( )
2
1 = log tan– 1 t + __ + c t
= log |tan– 1(cos x + sec x)| + c
|
( ) |
dx 4 + 3sin 2x
)
tan x – 1 = Ú ______________________ dx ____ 2tan x tan x 4 + 3 ________ ÷ 1 + tan2x
( Ú ( Ú (
))
)
(tan x – 1)sec2x = Ú _______________________ dx ____ tan x ( 4 + 4tan2 x + 6tan x ) ÷ 2
x2 = – __________ + 2 Ú ___________ x cos x dx (x tan x + 1) x sin x + cos x
x2 = – __________ + 2log|x sinx + cos x| + c (x tan x + 1)
÷
___________
)
(t – 1)2t dt = ____________ dt, t(4t4 + 6t + 4)
)
where tan x = t
(t2 – 1)dt = 2 _____________ dt t(4t4 + 6t2 + 4)
(
|
(x sec2x + tan x) = x2 Ú _____________ dx + 2Ú ________ x dx x tanx + 1 (x tan x + 1)2 x2 x = – __________ + 2Ú ________ dx (x tan x + 1) x tanx + 1
( (
|
__
tan x) x2(x sec2x + 21. Ú _______________ dx (x tan x + 1)2
____ ____ ÷____________ cot x – ÷ tan x
(
|
( ) ( )
(tan x + cot x)÷2 – 1 1 __ = ____ __ log __________________ + c 2÷2 (tan x + cot x)÷2 + 1
= log |v| + c
Ú
|
– 1
20.
| | __
( ) ( ( ) ) ) ( Ú
)
1 1 – __ 2 dt t = 2Ú ______________ dt 1 4 t2 + __ 2 + 6 t
( ( )) ) ( ( )
1 1 – __ 2 dt t = 2 ___________________ dt 1 2 4 t + __ – 2 + 6 t
{ ( ) }
1 1 – __ 2 dt t = 2Ú _____________ dt 1 2 4 t + __ – 2 t
( ( ) )
sin x – sin3x 22. Ú ___________ dx 1 – sin3x
÷
_____________
sin x(1 – sin2x) = Ú _____________ d x 1 – sin3x
cos x÷sin x = Ú _________ ________ dx 3 ÷ 1 – sin x
cos x÷sin x = Ú ____________ ___________ dx 3/2 2 x) ÷ 1 – (sin
____
____
Let (sin3/2x) = t 3 fi __ sin1/2x cos x dx = dt 2 2 fi sin1/2x cos x dx = __ dt 3
dt 2 _____ = __ Ú ______ 3 ÷ 1 – t2
Indefinite Integrals
x+2 dx Ú ex ( _____ x + 4) 2
2 = __ sin– 1(t) + c 3
25.
2 = __ sin– 1(sin3/2x) + c 3
(x + 2)2 = Ú ex _______ dx (x + 4)2
x2 + 4x + 4 = Ú ex __________ dx (x + 4)2
x2 + 4x 4 = Ú ex _______2 + _______ dx (x + 4) (x + 4)2
x(x + 4) 4 = Ú ex _______ + _______ dx (x + 4)2 (x + 4)2
x
{
}
{1 + (1 – x)}2 ______ = Ú ex _____________ dx (1 – x)÷1 – x2
(
{ Ú { ÷
÷
_____
} }
)
( ( (
2
e (2 – x) ______ 23. Ú ____________ dx (1 – x)÷1 – x2
)
) )
1 + x 1 ______ = Ú ex _____________ + _____ dx 2 (1 – x)÷1 – x 1 – x
1 + x _____________ 1 ______ = ex _____ + dx 1 – x (1 – x)÷1 – x2
x 4 = Ú ex ______ + _______ dx (x + 4) (x + 4)2
1+x = ex _____ + c 1–x
x = ex ______ + c (x + 4)
______
( ÷ ) _____
24.
26. Ú
÷sin(x + a )cos (x – b )
Put (x – b ) = t dx = dt
=
dt _______________________ Ú _________________________ 3 ÷cos t(sin t ◊ cosq + cos t sinq )
=
dt ______________________ ____________________ 4 ÷cos t(tan t ◊ cosq + sinq )
=
Let (tant ◊ cosq + sinq ) = v
fi sec2t cosq dt = 2v
2v fi sec2t dt = _____ cosq
= 2secq ÷cosq tan t + sinq + c
= 2sec(a + b )
t fi 4x7 dx = – ______ dt (t2 – 1)
÷
=
(x4t) = Ú ____ × x7dx x20
t x7 = Ú ___ dx x16
c
8
+x dx Ú 1_____ x13
_____________
______________________________ ÷cos(a + b )tan(x – b ) + sin(a + b ) +
1 fi x8 = ______ 2 (t – 1) 2t 7 fi 8x dx = – ______ dt (t2 – 1)
______
Let (1 + x8) = x8t2
2
dx
x13
= Ú{ x– 13(1 + x8)1/2 }dx
÷
2v = ____ + c cosq
)
sec t dt ________________ Ú __________________ (tan t ◊ cosq + sinq )
v dv 2 = _____ ____ cosq Ú v
(
2
)
b+1 – 13 + 1 __ 1 Here, _____ + a = _______ + = – 1 g 8 2
dt ________________ = Ú __________________ 3 ÷sin(t + b + a )cos t dt = Ú ______________ _____________ , (given) 3 + q )cos t ÷ sin(t
Ú
(
______ _______ ÷1 + x8
dx __________________ Ú ____________________ 3
1.149
t t 1 = – __ Ú _______ × – _______ 2 dt 2 4 _______ 1 (t – 1) 2 2 (t – 1)
1 = – __ Ú t2 dt 4
t3 = – ___ + c 12
1.150 Integral Calculus, 3D Geometry & Vector Booster
( )
ex(x + 1) = Ú ___________ dx xex(1 + xex)2
dx 27. Ú ___________ 3 sin x + cos3x
dt = Ú _______ , t(1 + t)2
dx = Ú ________________________ (sin x + cos x)(1 – sin x cos x)
=
t 1 = log ____ + ______ + c t+1 (t + 1)
1 1 + x8 3/2 = – ___ _____ + c 12 x8
(
)
sin x + cos x 1 2 = __ Ú ____________ – _____________ dx 3 (sin x + cos x) (1 – sin x cos x)
( ) ( )
)|
x p 2 2 = ____ __ log tan __ + __ – __ tan– 1 (sin x – cos x) + c 2 8 3 3÷2 ______
( x + ÷ 1 + x2 ) 28. Ú _____________ dx ______ ÷ 1 + x2 3
dx _________ = Ú __________ 2 + 3cot x ÷
Let 2 + 3cot x = t2
fi – 3cosec2x dx = 2t dt
2t dt fi dx = – _______ 4 3(t + 1)
______
Let (x + ÷ 1 + x2 ) = t
x fi 1 + _______ ______ dx = dt ÷ 1 + x2
______ x+÷ 1 + x2 ___________ fi ______ dx = 2 ÷1 + x
t fi _______ ______ dx = dt ÷ 1 + x2
dx dx fi _______ ______ = ___ t 2 ÷ 1 + x
(
)
( (
(
dt = Út3 × __ t
= Út2dt
t3 = __ + c 3
)
)
)
dt
dt 2 = – __ Ú ______ 3 (t4 + 1)
( Ú ( )
(x + ÷1 + x2 )3 = ____________ +c 3
) Ú ( )
2 2 1 (t + 1) – (t – 1) = – __ Ú _______________ dt 3 (t4 + 1)
2 2 1 (t + 1) 1 (t – 1) = – __ _______ 4 dt + __ ______ dt 3 (t + 1) 3 (t4 + 1)
( ) ( )
( ) ( ) ) ( Ú
( ) ( ) ) ( Ú
1 1 1 + __ 2 1 – __ 2 t t 1 1 = – __ Ú _______ dt + __ Ú _______ dt 3 3 1 1 2 __ t + 2 t2 + __ 2 t t
(
) ( )
1 1 1 + __ 2 1 – __ 2 t t 1 1 = – __ __________ dt + __ _______2 – 2 dt 2 3 3 1 1 __ __ t – + 2 t + t t
( )
_____
(x + 1) 29. Ú _________ dx x(1 + xex)2
÷
sin x 30. Ú ____________ dx 2sin x + 3cosx
t dt 2 = – __ Ú _______ 3 (t4 + 1)t
| |
____________
sin x + cos x 2 ___________ 2 = ____ __ – __ Ú ________________ dx Ú dx p 3 3÷2 (sin x + __ (1 + (sin x – cos x)2) 4 p sin x + cos x 2__ 2 ____ = dx – __ Ú ________________ dx Úcosec x + __ 4 3 (1 + (sin x – cos x)2) 3÷2
(
dt dt dt – Ú ______ – ______ Ú __ t (t + 1) Ú (1 + t)2
where t = xex
dx sin x + cos x 2 1 = __ Ú ____________ – __ _____________ dx 3 (sin x + cos x) 3 Ú (1 – sin x cos x)
|
where xex = t
( ) 1 ( t + __ ) – ÷2 t 1 1 __ ____ ___________ ( t – ) + log + c t 1 6÷2 ( t + __ ) + ÷2 t
| | __
1__ – 1 = – ____ tan 3÷2
_________
where t = ÷ 2 + 3cot x cos 4x + 1 31. Ú __________ dx cot x – tan x
__
__
Indefinite Integrals
2cos2x 2x dx = Ú ___________ cos2x – sin2x ___________ sin x cos x
2cos2x 2x dx = Ú ___________ 2cos 2x _________ 2sin x cos x
= Ú cos 2x sin 2x dx
1 = __ Ú 2cos 2x sin 2x dx 2
1 = __ Ú sin (4x) dx 2
1 = – __ cos (4x) + c 8
cos x + cos x 32. Ú ____________ dx sin2x + sin4x 3
5
cosx (cos2x + cos4x) = Ú _________________ dx (sin2x + sin4x)
(1 – t2) + (1 + t4)2 = Ú _______________ dt, where sin x = t t4 + t2
2 – 3t2 + t4 = Ú __________ 4 dt t + t2
=
– 4t2 – 4 + 6 = 1 + ____________ dt t4 + t2
– 4t2 – 4 + 6 = Ú 1 + ___________ dt t2(t2 + 1)
4(t2 – 1) 6 = Ú 1 – ________ + ________ 2 2 dt 2 2 t (t + 1) t (t + 1)
4 1 1 = Ú 1 – __ 2 + 6 __ 2 – ______ 2 dt t t t +1
2 1 = 1 + __ 2 – 6 ______ 2 dt t t +1
2 = 1 – __ – 6tan– 1(t) + c t
= sin x – 2cosec x – 6tan– 1(sin x) + c
( Ú (
2
)
(
dx 33. Ú ___________ sin6x + cos6x
(1 + tan2x)2sec2x = Ú _______________ dx tan6x + 1
(1 + t2)2 = Ú _______ dt, t6 + 1
(1 + t2)2 = Ú ________________ dt (t2 + 1)(t4 – t2 + 1)
(1 + t2) = Ú ___________ dt (t4 – t2 + 1)
1 1 + __ 2 t = Ú ___________ dt 1 2 __ t + 2 – 1 t
1 1 + __ 2 t = Ú ____________ dt 2 1 t – __ + 1 t
(
(
sec6x dx = Ú ________ tan6x + 1
(
( ) (
where t = tan x
)
)
( ( ) ) 1 = tan ( t – __ ) + c t – 1
= tan– 1(tan x – cot x) + c
34. Let f (x) = ax2 + bx + c
– 4t + 2 dt Ú 1 + ________ t4 + t2
( Ú (
1.151
Given f (0) = f (1) = – 3 = 3 f (2)
)
On solving, we get
)
x2 – x – 3 Thus, I = Ú _______________ dx (x – 1)(x2 + x + 1)
)
( )) ( ) ) )
a = 1, b = – 1, c = – 3
x2 – x – 3 Now, _______________ (x – 1)(x2 + x + 1)
Bx + C A = ______ + __________ (x – 1) (x2 + x + 1)
Solving, we get
A = – 1, B = 2, C = 2
The given integral reduces to
2x + 2 – Ú ______ dx + _________ dx (x – 1) x2 + x + 1
dx dx 2x + 1 = – Ú ______ + _________ dx + Ú _________ 2 (x – 1) Ú x2 + x + 1 x +x+1
1.152 Integral Calculus, 3D Geometry & Vector Booster
|
dx = – log|(x – 1)| + log (x2 + x + 1) + Ú ________ 2 x + x +1
sec4x dx = Ú ____________ sec4x – 3tan2x
(x2 + x + 1) dx __ = log __________ + _______________ (x – 1) 2 3 2 ÷ 1 __ ___ x + + 2 2
(1 + tan2x)sec2x = Ú __________________ dx (1 + tan2x)2 – 3tan2x
(1 + t2)dt = Ú ____________ (1 + t2)2 – 3t2
t2 + 1 = Ú _________ dt t4 – t2 + 1
1 1 + __ 2 t = Ú __________ dt 1 2 __ t + 2 – 1 t
1 1 + __ 2 t = Ú ___________ dt 1 2 t + __ + 1 t
1 = tan– 1 t – __ + c t
= tan– 1(tan x – cot x) + c
| |
| |
|
( ) ( )
(
)
(x2 + x + 1) 2x + 1 2 = log ___________ + ___ __ tan– 1 ______ __ + c (x – 1) ÷3 ÷3
Integer Type Questions
(
( ( ) )
1. The given integral is
+ cos 2x ____________ dx Ú cos x 2cos x – 1
2 x + cos x – 1 dx ________________ = Ú 2cos 2cos x – 1
(2cosx – 1)(2cosx + 1) = Ú ____________________ dx 2cos x – 1
= Ú (cosx + 1)dx
= sin x + x + c
Clearly, A = 1, B = 1
)
(sin x + cos x) _____________ dx (1 – sin x cos x)
)
(sin x + cos x) dx 2 2 = __ Ú ___________ + __ Ú _______________ dx 3 (sin x + cos x) 3 (2 – 2sin x cos x) __
(sin x + cos x) 2 ÷ dx 2 = ___ Ú __________ + __ Ú ________________ 2 dx p 3 sin x + __ 3 (1 + (sinx – cosx) 4 __ 2 ÷ x 3p 2 = ___ log tan __ + ___ + __ tan– 1(sin x + cos x) + c 3 2 8 3
(
|
(
)
)|
Clearly, L = 3, M = 8, N = 2 and P = 2 Hence, the value of L + M – N – P = 7. 3. The given integral is dx Ú ___________ sin6x + cos6x
dx = Ú ______________ 1 – 3sin2x cos2x
4. The given integral is
x+1 Ú _____ dx x3 + x
dx = Ú _________________________ (sin x + cos x)(1 – sin x cos x)
(
( )
Hence, the value of L + M + 4 = 4.
dx Ú ___________ sin3x + cos3x
(
( ( ) )
Clearly, L = 1, M = – 1
Hence, the value of A + B + 3 = 5. 2. The given integral is
1 2 = __ Ú ____________ + 3 (sin x + cos x)
)
x+1 = Ú ________ dx x(x2 + 1) dx dx = Ú _______ + ________ (x2 + 1) x(x2 + 1)
| |
x2 1 = tan– 1x + __ log ______ 2 + c 2 x +1 Clearly, L = 2, M = 2 and N = 2
(
)
L+M + N Hence, the value of __________ = 2 2
5. The given integral is
(
) 2sin x 1 = __ Ú ( ___________ dx 2 sin x + cos x )
sin x Ú ___________ dx sin x + cos x
(
)
1 (sin x + cos x) + (sin x – cos x) = __ Ú ________________________ dx 2 sin x + cos x
Indefinite Integrals
(
)
1.153
(cos x – sin x) 1 = __ Ú 1 – ____________ dx 2 sin x + cos x
= tan– 1(x3 + x2 + 1) + c
1 = __ Ú (x – log|sin x + cos x|) + c 2
Hence, the value of L + M + N + P + Q = 8.
Thus, L = 1, M = 3, N = 2, P = 1 and Q = 1 cos4x 8. Ú ________________ 6 dx sin x(1 + cot5x)3/5
1 1 Clearly, A = __ and B = – __ 2 2
Hence, the value of A + B + 1 is 1
4 2 x dx = Ú _____________ cos x cosec 5 3/5 (1 + cot x)
(
)
x2 6. The given integral is Ú ________ dx (2x + 3)2
Let 2x + 3 = t
fi 2dx = dt
fi dx = 1/2dt
(
3–t fi x = _____ 2
)
x2 dx = Ú ________ (2x + 3)2
3–t 2 _____ 1 2 dt = __ Ú _______ 2 t2
( )
2x = 3 – t
2 1 (3 – t) dt = __ Ú _______ 8 t2
(
2 1 (t – 6t + 9) dt = __ Ú __________ 8 t2 6 9 1 = __ Ú 1 – __ + __2 dt t 8 t
(
)
(
)
9 1 = __ Ú (2x + 3) – log (2x + 3) – _______ + c 8 (2x + 3) (3 + 2x) __ 9 1 = ________ + log |3 + 2x | – ________ + c L M 8(2x + 3) Clearly, L = 8 and M = – 8 Hence, the value of L + M + 4 = 4. 7. We have,
(
2
)
3x + 2x Ú _________________________ dx 6 5 x + 2x + x4 + 2x3 + 2x2 + 2 3x2 + 2x = Ú _____________________________ dx 6 4 1 + ( x + x + 1 + 2(x5 + x3 + x2) ) 3x2 + 2x = Ú _______________ dx 1 + (x3 + x2 + 1)2
where cot x = t
5t4dt 1 = – __ Ú _________ 5 (1 + t5)3/5
Put (1 + t5) = y5
fi 5t4dt = 5y4dy
4 1 5y dy = – __ Ú _____ 5 y3
= – Ú y dy
y2 = – __ + c 2
1 = – __ (1 + t5)2/5 + c 2
1 = – __ (1 + cot5x)2/5 + c 2
1 tan5x + 1 2/5 = – __ ________ + c 2 tan5x
( )
(
)
Thus, L = 2, M = 2 and N = 5
)
9 1 + c = __ Ú t – logt – __ t 8
t4dt = – Ú _________ , (1 + t5)3/5
Hence, the value of L + M + N = 9 13 ___ + 1 b_____ + 1 ______ 2 9. Now, = = 3 g 5 __ 2 Put (1 + x5/2) = t3 5 __ x3/2 = 3t2 dt 2 6 x3/2 = __ t2dt 5 Thus, the given integral reduces to
6 __ Ú (t3 – 1)2 ◊ t3/2 ◊ t2 dt 5
6 = __ Ú (t6 – 2t3 + 1)t7/2 dt 5
6 = __ Ú (t19/2 – 2t13/2 + t7/2) dt 5
1.154 Integral Calculus, 3D Geometry & Vector Booster
(
)
6 t21/2 2t15/2 = __ ______ – ______ + 5 (21/2) (15/2)
t9/2 _____ + c (9/2)
5/2 21/2 2(1 + x5/2)15/2 (1 + x5/2)9/2 6 (1 + x ) = __ __________ – ___________ + __________ + c (21/2) (9/2) 5 (15/2)
(
)
(
)
4 4 = ___ (1 + x5/2)21/2 ___ 8 (1 + x5/2)15/2 + ___ (1 + x5/2)9/2 + c 35 15 25 Clearly, L = 4, M = – 1 and N = 4
Qustions asked in Past IIT-JEE Examinations
(
)
sin x = Ú __________ dx sin x – cos x
2sin x 1 = __ Ú __________ dx 2 sin x – cos x
1 (sin x – cos x) + (sin x + cos x) = __ Ú _________________________ dx 2 sin x – cos x
(sin x + cos x) 1 = __ 1 + ___________ dx 2 sin x – cos x
1 = __ (x + log |sin x – cos x|) + c 2
(
4. We have,
÷
( ) x x = Ú 1 + 2 sin ( __ ) cos ( __ ) dx 4 4 ÷ _________________
)
)
2. We have,
( ( ) ( ) ) x x = 4( – cos ( __ ) + sin ( __ ) ) + c 4 4 x x = Ú sin __ + cos __ dx 4 4
5. We have,
x2 _____ Ú ______ dx ÷1 – x (1 – t2)2 = Ú _______ × – 2 t dt, where 1 – x = t2 t
)
= – 2Ú (1 – t2)2dt = – 2Ú (1 – 2t2 + t4)dt
dt 1 = __ Ú _____ 2 1 + t2
1 1 = __ tan– 1(t) + c = __ tan– 1(x2) + c 2 2
3. We have, 2
(
Put a + bx = t fi
)
t5 2 = – 2 t – __ t3 + __ + c 3 5
(
)
(1 – x)5/2 2 = – 2 (1 – x)1/2 – __ (1 – x)3/2 + ________ + c 3 5
1 dx = __ dt b t____ –a also x = b Thus, the given integral reduces to
)
_________
)
( Ú (
x dx Ú ________ (a + bx)2
)
x __ Ú 1 + sin dx 2
x dx Ú _____ 1 + x4 2x 1 = __ Ú _______ dx 2 1 + (x2)2
(
a2 1 = __3 t – 2a log |t| – __ + c t b a2 1 = __3 (a + bx) – 2a log |(a + bx)| – _______ + c (a + bx) b
dx Ú _______ 1 – cot x
( )
)
(
1. We have,
(
1 t2 – 2ta + a2 = __3 Ú __________ dt b t2
(
10. Do yourself.
2 1 (t – a) = __3 Ú _______ dt b t2
2a a2 1 = __3 Ú 1 – ___ + __ 2 dt t b t
Hence, the value of L + M + N = 7
( )
t–a 2 ____ b 1 _______ × __ dt Ú t2 b
6. We have,
Ú (elog x + sin x) cos x dx
= Ú (elog x◊ sin x + sin x cos x) dx
= Ú (x◊ sin x + sin x cos x) dx
Indefinite Integrals
(
)
sin2x = Ú x◊ sin x + _____ dx 2
cos 2x = x Ú sin x dx + Ú cos x dx – _____ + c 4
cos 2x = – x cos x + sin x – _____ + c 4
) ( (
)
x+1 – 2 = Ú ex ________ dx (x + 1)2
1 –2 = Ú ex _______ 2 + _______ dx (x + 1) (1 + x)3
ex = _______ + c (x + 1)2
(
))
dx dx Ú __________ = Ú ___________ 2 4 3/4 1 3/4 x (x + 1) x5 1 + __ 4 x
( )
t3 = – Ú __3 dt t = – Ú dt
= – t + c
1 1/4 = – 1 + __ 4 + c x
(
)
)
______ __ 1 – ÷x ______ dx __ 1 + ÷x
1 + cos 2q = 2 Ú cos q – _________ dq 2
( ) sin 2q 1 = 2 Ú ( sin q – __ ( q + ______ + c 2 2 ))
= (2sinq – (q + sin 2q cosq )) + c
) ] + c = 2÷1 – x – [ ( cos– 1(÷x ) + ÷ x ÷ 1 – x
_____
__ _____
__
(
__
__
)
– 1 sin– 1÷ x – cos ÷ x Ú ________________ – 1 __ dx – 1 __ sin ÷x + cos ÷x
(
(
)
__ p – 1 __ __ sin– 1÷x – – sin ÷x 2 = Ú _____________________ dx p __ 2 __ p 2 __ = __ p Ú 2sin– 1÷ x – dx 2
(
)
)
(
4 = __ p Ú q Ú sin(2q ) dq +
cos 2q dq – Ú dx Ú ( ______ 2 )
)
where x = sin2q
Let x = cos q
fi ________
1 – cosq = Ú ________ × – sin (2q) dq 1 + cosq
q sin __ 2 = Ú _______ × 2sin (q ) cos (q ) dq q cos __ 2
( ) ( )
2
÷
= 2 Ú (cos q – cos 2q ) dq
4 = __ p Ú (q sin2q )dq – Ú dx
÷
9. We have – Ú
__ 4 = __ p Ú (sin– 1÷x ) dx – Ú dx
(
4 fi – __5 dx = 4t3 dt x 1 fi __5 dx = – t3 dt x
= 2 Ú (1 – cosq ) cos(q )dq
10. We have,
1 Let 1 + __ 4 = t4 x
8. We have,
2
x–1 Ú ex _______ dx (x + 1)3
2
7. We have,
(
( ) ( ) ( ) ( ) q = 4 Ú sin ( __ ) cos (q ) dq 2 q = 2 Ú ( 2sin ( __ ) ) cos(q ) dq 2
q sin __ 2 q q = 4 Ú _______ × sin __ cos __ cos (q )dq 2 2 q cos __ 2
1.155
dx = – sin 2q dq
(
)
q cos 2q ______ sin 2q 4 = __ p Ú – _______ + – x + c 2 4
(
__
__ _____
)
2 sin– 1(÷x ) (1 – 2x ) ÷x ÷1 – x 4 = __ p – ________________ + _________ – x + c 2 2
__ __ _____ 2 2 ) – x + c = __ p ( – sin– 1(÷x ) (1 – 2x ) + ÷ x ÷1 – x _____
cos2x ÷ 11. We have Ú ______ dx sin x
_________
2 x ÷ 1 – 2sin = Ú __________ dx sin x
1.156 Integral Calculus, 3D Geometry & Vector Booster
÷
( )
________
1 = Ú _______ dx 2 sin x – 2
1 1 + __ 2 t = 2Ú ______ dt 1 2 t + __ 2 t 1 1 + __ 2 t = 2Ú __________ dt 1 2 __ t – + 2 t
__________ 2 ÷cosec x – 2 dx
Ú
=
= Ú ÷ cot 2x – 1 dx
( ( ) )
________
Let cot x = secq
fi
– cosec2x dx = secq tanq dq
dz = 2Ú _____ 2 z +2
secq tanq = – _________ dq 1 + cot2x
z 1__ – 1 ___ = 2 ◊ ___ tan __ + c ÷2 ÷2
__ 1 1 = ÷2 tan– 1 ___ __ t – __ + c t 2 ÷
__ 1 = ÷2 tan– 1 ___ __ (tan x – cot x) + c 2 ÷
secq tanq dx = – _________ dq cosec2x
secq tanq = – _________ dq 1 + sec2x
secq = – ________ dq cos2q + 1
( )
( ( ) ) ( )
13. We have
(
)
4ex + 6e– x I = Ú _________ x dx 9e + 4e– x
tanq ◊ sinq = – Ú _________ dq cos2q + 1
sin2q = – Ú ______________ dq cosq (cos2q + 1)
2 = – Ú ______________ sin q ◊ cosq dq cos2q (cos2q + 1)
Comparing the co-efficients of ex and e– x, we get 3 4 m + n = __ and n – m = __ 9 2 Solving, we get,
t2 = – Ú ____________ dt, where sin q = t (1 – t2)(2 – t2)
n = 35/36 and m = – 19/36. The given integral reduces to
t2 = – Ú ____________ dt (t2 – 1)(t2 – 2) 1 = – Ú _____ 2 – _____ dt t2 – 2 t2 – 1
19 35 9ex + 4e– x – ___ Ú dx + ___ Ú _________ dx 36 36 9ex – 4e– x
( Ú ( )
4ex + 6e– x = m(9ex – 4e– x) + n(9ex + 4e– x)
)
( )
(
)
19 35 = – ___ x + ___ log | 9ex – 4e– x | + c 36 36
| |
2 1 = – _____ 2 dt + Ú _____ 2 dt t –2 t –1
19 9e2x – 4 35 = – ___ x + ___ log _______ + c 36 36 ex
t – ÷2 1 __ = – 2◊ ____ __ log ______ + 2÷2 t+÷ 2
19 35 35 = – ___ x + ___ log | 9e2x – 4 | – ___ log |ex| + c 36 36 36
19 = – ___ + 36
3 35 = – __ x + ___ log |9e2x – 4| + c 2 36
(
| |) | |
where t = sinq 12. We have,
____ ____ (÷tan x + ÷cot x ) dx
Ú
( ) t + 1 _____ 2t = Ú ( _____ dt t )t + 1 tan x + 1 = Ú ________ ____ dx tan x ÷ 2
4
( )
t2 + 1 = 2Ú ______ 4 dt t +1
__
t – 1 1 __ log ____ + c 2 t +1
(
(
)
35 35 ___ x + ___ log | 9e2x – 4 | + c 36 36
)
log(1 + x1/6) 1 14. Ú ________ 1/3 1/4 + ___________ __ dx 1/3 x +x x + x ÷ = I1 + I2 (say) dx Now, I1 = Ú _________ x1/3 + x1/4
Indefinite Integrals
(
)
t11 dt = 12 Ú _____ , where x = t12 4 t + t3
cos q + sin q I = cos 2q log ___________ dq cos q – sin q
t8 dt = 12Ú ____ t+1
(
)
8
(t – 1) + 1 = 12Ú __________ dt t+1
(
(
)
)
t8 t7 __ t6 t5 __ t4 t3 __ t2 = 12Ú __ + __ + + __ + + __ + + log|t + 1| + c 7 8 4 3 2 6 5
log (1 + x1/6) Let I2 = Ú ___________ dx x1/2 + x1/3
log(1 + t) 5 = 6Ú _________ t dt, where x = t 6 t3 + t2
t3log(1 + t) = 6 Ú __________ dt t+1
(t3 – 1 + 1)log(1 + t) = 6 Ú __________________ dt t+1
(
)
log (1 + t) = 6 Ú (t2 – t + 1) log (1 + t) – _________ dt (1 + t)
(
)
) ( )
cosq + sinq _____ sin 2q = log __________ 2 cosq – sinq
2 2 1 sin(2q) ◊ 2 (sin q + cos q) = – __ Ú _____________________ ◊ dq 2 cos2q – sin2q
sin (2q) sin 2q cosq + sinq = _____ × log __________ – Ú _______ ◊ dq 2 cos (2q) cosq – sinq
( ) ( ) sin 2q cosq + sinq = ( _____ × log ( __________ 2 ) cosq – sinq )
( (
) (
(
)
3
)
cos3x + cos5x dx Ú ____________ 2 sin x + sin4x
(t – 1)log(1 + t) log(1 + t) = 6Ú _______________ – _________ dt t+1 (1 + t)
(
cosq – sinq cosq + sinq 1 – __ Ú sin (2q) ◊ ___________ + ___________ dq 2 cosq + sinq cosq – sinq
1 __ log | sec (2q) | + c 2 16. We have,
3
[
)
(
cos q + sin q = log ___________ cos (2q) dq cos q – sin q Ú
1 = 12Ú t7 + t6 + t5 + t4 + t3 + t2 + t + ____ dt t+1
1.157
2
)
cos2x + cos4x cos x dx = ____________ 2 sin x + sin4x
]
( Ú ( Ú ( Ú (
)
(1 – sin2x) + (1 – sin2x)2 = Ú _____________________ cos x dx sin2x + sin4x
t t __ – __ + t 3 2 t__2 – t2) + (1 – t2)2 + t – Ú _________ dt = (1 ________________ dt, where t = sin x 1+t 2 t2 + t4 – 6 (log (1 + t))2 + c 4 ((t + t2) + (2 – 4t2)) __________________ 3 2 = dt t t 2 4 = 6 log (1 + t) __ – __ + t – log (1 + t) t + t 3 2 (2 – 4t2) 3 2 ________ 2t – 3t + 6 = 1 + dt 2 4 + Ú ___________ dt t + t t+1
(
)
t3 = 6 log (1 + t) __ – 3
(
)
(
(
)
(
)
1 + Ú 2t2 + 5t + 5 + ____ dt t+1
(
)
)
( Ú (
)
4 2 = Ú 1 – ______ 2 + ________ 2 2 dt t + 1 t (t + 1)
(
))
4 1 1 = 1 – _____ 2 + 2 __2 – ______ dt t +1 t (t2 + 1)
t3 t2 = 6 log (1 + t) __ – __ + t – log (1 + t) 3 2
(
)
)
2 (2t2 – 1) = Ú 1 – _________ 2 2 dt t (t + 1)
t3 t2 = 6 log (1 + t) __ – __ + t – log (1 + t) 3 2
)
)
( (
)
where t = x1/6.
2 = t – 4 tan–1t – __ – 2 tan–1t + c t 2 –1 = t – 6 tan t – __ + c t
15. We have,
= (sin x – 6 tan–1(sin x) – 2cosec x) + c
2t3 5t2 + ___ + ___ + 5t + log (1 + t) + c 3 2
)
1.158 Integral Calculus, 3D Geometry & Vector Booster 1 1 + sin x 1 1 = – __ Ú _______ dx – __ Ú ________________ dx 5 cos2 x 5 __________ 2 tan (x/2) + 4 1 + tan2 (x/2) 1 = – __ Ú ( sec2 x + sec x tan x) dx 5 sec2 (x/2) 1 = – __ Ú _________________________ dx 5 ( 4 tan2 (x/2) + 2 tan (x/2) + 4 )
17. We have,
(
+1 Ú __________ dx x (1 + x ex)2
(x + 1)e x = Ú ___________ dx x ex(1 + x ex)2
dt = Ú _______ 2 , Where t = x ex t (1 + t)
(
1 1 1 = Ú __ – ____ – ______ dt t t + 1 (t + 1)2
1 1 = log ____ + ______ + c t+1 (t + 1)
x ex 1 = log _______ x + ________ x + c x e + 1 (x e + 1)
| |
|
dt 1 1 = – __ (tan x + sec x) – __ Ú _________ 5 5 2t2 + t + 2 x where tan __ = t 2 and then you do it.
)
( )
20. We have,
|
(
dx ____________ Ú ______________ ÷(x – p)3(x – q) dx = Ú _______________ (x – p)3/2(x – q)1/2 dx = Ú _______________ (x – q) 1/2 2 _______ (x – p) (x – p)
(
)
(x – q) Let ______ = t (x – p)
(x – p) ◊1 – (x – q) ◊ 1 fi __________________ dx = dt (x – p)2
dx dt fi _______ = ______ (x – p)2 (q – p)
_ 1 = ______ × 2÷t + c (q – p)
÷
_____
x–p 2 = _______ × 2 _____ x – q + c (q – p)
19. We have,
dx Ú _________________ (sin x + 4) (sin x – 1)
(
= 2 Ú (cos2x – sin2x)dx
= 2 Ú cos (2x)dx
sin 2x = 2 _____ + c 2
( )
= sin 2x + c
÷
)
1 1 = __ Ú________ 1 – ________ dx 5 (sin x – 1) (sin x – 4) 1 1 1 1 = __ Ú ________ dx – __ Ú _________ dx 5 (sin x – 1) 5 (sin x + 4) 1 1 1 1 = – __ Ú ________ dx – __ Ú _________ dx 5 (1 – sin x) 5 (sin x + 4)
______ _ 1 – ÷x dx ______ × ___ x __ 1 + ÷x
Let x = cos22q fi dx = – 2 cos x(2q) sin (2q)
÷
__________
dt 1 = ______ ___ (q – p) Ú t1/2
)
cos x – sin x = 2 Ú __________ × (cos x + sin x)2 dx cos x + sin x
21. Ú
)
The given integral reduces to
(
(
)
cos x – sin x Ú __________ (2 + 2sin 2x)dx cos x + sin x
18. We have,
)
1 – cos (2q) – 2cos (2q) sin (2q) = Ú __________ × ________________ dq 1 + cos (2q) cos2(2q) 2sin (q) cos (q) sinq _____________ = – 2Ú ____ × dq cosq cos(2q) 2sin2(q) = – 2 Ú _______ dq cos(2q)
(cos (2q) – 1) = 2 Ú ____________ dq (cos (2q)
= 2 Ú (1 – sec (2q))dq
1 = 2 q – __ log |sec (2q) + tan (2q)| + c 2
(
(
|
)
______
÷
|)
__ 1 1 1 __ ___ __ = 2 cos–1(÷x ) – log _ + 1 + x + c ÷x 2
22. We have, dx ____________ Ú ____________________ (2x – 7)÷x2 – 7x + 12
Indefinite Integrals
(
)
dx _____________ = 2Ú _____________________ (2x – 7)÷4x 2 – 28x + 48
2x + 2 ____________ Ú sin–1 ______________ dx 2 + 8x + 13 ÷ 4x
2 ◊ dx ___________ = Ú ___________________ (2x – 7) ÷ (2x – 7)2 –1
dt _____ = Ú _______ , where (2x – 7) = t t ÷t 2 – 1
Let (2x + 2) = 3 tanq
= sec–1(t) + c
–1
= sec (2x – 7) + c
x2 + 3x + 2 23. Ú _____________ dx (x2 + 1)2(x + 1)
x2 + 3x + 2) = Ú _____________ dx (x2 + 1)2(x + 1) (x + 1) (x + 2) = Ú _____________ dx (x2 + 1)2(x + 1)
(x + 2) = Ú ________ dx (x2 + 1)2
(x + 2) = Ú ________ dx (x2 + 1)2
dx 2x 1 = __ Ú ________ dx + 2 Ú ________ 2 (x2 + 1)2 (x2 + 1)2
dx 1 = – ________ 2 + 2 Ú ________ (x2 + 1)2 2(x + 1)
x 1 = – ________ + tan–1x + _____ + c 2 (x2 + 1) x2 + 1
)
3 = __ Ú (q sec2q)dq 2
3 = __ Ú (q tanq – Ú tanq dq) 2
3 = __ (q tanq – log |secq |) + c 2
3 = __ q tanq – log ÷1 + tan2q + c 2
|
(
_________
|)
3 3 = __ (q tanq) – __ log (1 + tan2q) + c 2 4 2x + 2 = (x + 1) tan–1 ______ 3 3 – __ log (4x2 + 8x + 13) + c 4
(
)
25. We have Ú (x3m + x2m + xm) (2x 2m + 3x m + 6)1/mdx = Ú (x3m – 1 + x2m – 1 + xm – 1) (2x 2m + 3x m + 6)1/m x dx
)
= Ú (x3m – 1 + x2m – 1 + x m – 1) (2x 3m + 3x 2m + 6xm)1/mdx Put (2x 3m + 3x 2m + 6x m) = t m
dx Let I1 = 2Ú ________ (x2 + 1)2
)
(
(
(
2x + 2 _____________ = Ú sin–1 _______________ ÷(2x + 2)2 + (3)2
(2 x + 2) _____________ Thus, ______________ = sin q + 2)2 + 32 ÷ (2 x
2
2
sec q dq = 2 Ú _______ , where x = tanq sec4q
6m (x 3m –1 + x 2m –1 + x m –1)dx = m t m – 1dt
dq = 2 Ú _____ sec2q
t m – 1 (x3m –1 + x 2m –1 + x m –1)dx = ____ dt 6 1 Thus, the given integral reduces to __ Ú t mdt 6
= Ú (2cos2q) dq
= Ú (1 + cos 2q) dq
1 t m + 1 = __ _____ + c 6 m+1
3m 2m m m + 1 1 (2x + 3x + 6x ) = __ ____________________ + c m+1 6
= (q + sinq + cosq) + c
x = tan–1x + _____ 2 + c x +1
(
24. We have,
)
1.159
( )
(
26. We have, x2 – 1 ____________ Ú ______________ dx x3÷ 2x 4 – 2x2 + 1
)
1.160 Integral Calculus, 3D Geometry & Vector Booster
x2 – 1 _____________ = Ú ________________ dx 2 1 3 4 __ x x 2 – __ 2 + 4 x x
) ÷ ( Ú ) ÷( ) ( Ú ) ÷( ( (
x2 – 1 ____________ = ______________ dx 2 1 5 __ x 2 – __ 2 + 4 x x
1 = ________ t n – 1 + c n (n – 1)
n – 1 ____ 1 = _______ (1 + nx n) n + c n (n – 1)
1 __ 1 = ________ (1 + nx n)( 1 – n ) + c n (n – 1)
28. We have
1 1 __ 3 – __ 5 x x ____________ = ______________ dx 2 1 __ 2 – __ 2 + 4 x x
J – I
)
2 1 Let 2 – __ 2 + __ 4 = t2 x x 4 4 fi __ 3 – __ 5 dx = 2t dt x x
)
1 2t dt = __ Ú ____ t 4
e xdx e– xdx _____________ = Ú ___________ – Ú e 4x + e 2x + 1 e– 4x + e– 2x + 1
(ex – e3x) dx = Ú ___________ e4x + e2x + 1
(1 – e2x) e xdx = Ú ____________ e4x + e2x + 1 Let ex = t fi e xdx = dt
1 = __ × t + c 2
(
)
1 – t2 = Ú _________ dt 4 t + t2 + 1
1 1 – __ 2 t = – Ú _________ dt 1 2 __ t + 2 + 1 t
1 1 – __ 2 t = Ú __________ dt 1 2 __ t + – 1 t
1 t + __ – 1 t 1 __ __________ = – log + c 2 1 t + __ + 1 t
t2 – t + 1 1 = – __ log ________ + c 2 t2 + t + 1
Similarly, we can write, (f0 f0 f 0....0 f )(x)
t2 + t + 1 1 = __ log _________ + c 2 t2 – t + 1
x = __________ . (1 + nx n)1/n
e2x + ex + 1 1 = __ log __________ + c 2 e2x – ex + 1
(
1 2 = __ × 2 – __ 2 + 2 x
)
1 1/2 __ 4 + c x
27. We have, ( f0 f) (x) = f ( f (x)) f (x) = __________ (1 + f (x))1/n x = __________ (1 + 2xn)1/n
Also, ( f0 f0 f ) (x)
= f (( f0 f ) (x))
( f 0 f )(x) = _____________ (1 + ( f 0 f (x))1/n
x = __________ (1 + 3xn)1/n
Thus, I = Ú x
n – 2
g (x)dx
29. We have,
(
( )
( ( ) )
| | ( ) ( )
| |
|
x = Ú x n – 2 __________ dx (1 + nx n)1/n
sec2x Ú ______________ dx (sec x + tan x)9/2
1 n – 2 n n = __ n Ú t dt, Where t = (1 + nx )
)
|
|
|
secx ◊ sec x = Ú ______________ dx (sec x + tan x)9/2
...(i)
Indefinite Integrals
Put sec x + tan x = t fi
sec x (sec x + tan x) dx = dt
dt sec x dx = __ t 1 1 ____________ Also sec x – tan x = = __ t (sec x + tan x) fi
( )
1 1 Thus, sec x = __ t + __ dt t 2 The given intergral (i) reduces to
( )
1 t + __ dt t 1 ________ __ t 2 __________ Ú t9/2
1.161
2 1 (t + 1) = __ Ú _______ dt 2 t13/2
1 1 = __ Ú ___ + 2 t9/2
1 2 2 = – __ ____ + ______ + c 2 7t7/2 11t11/2
1 1 = – ____ 7/2 + ______ 11/2 + c 7t 11t
1 1 = _______________ + ________________ + c 7(sec x + tan x)7/2 11(secx + tan x)11/2
(
( ( (
)
1 ____ 13/2 dt t
)
)
)
Chapter
2
Definite Integrals
Concept Booster 1. What
is
definite Integral?
Let f (x) be a function of x defined in the closed interval [a, b] and j (x) be another function such that j ¢(x) = f (x) for every x in the domain of f (x). Then b
Ú f (x) dx = [j (x) + c] ba = j (b) – j (a) a
is called the definite integral of the function f (x) over the interval [a, b]. Here, a is called the lower limit and b is called the upper limit.
3. Evaluation
of
definite integrals
by
substitution
Rules 1. When the variable in a definite integral is changed, the substitution in terms of new variable should be affected at three places. (a) in the integrand (f (x) will be changed) (b) in the differentials (dx will be changed) (c) in the limits (old limits will be changed)
4. Geometrical interpretation
of
definite integral
Consider the function y = f (x), where f (x) ≥ 0 for all b
x in [a, b]. The integral Ú is numerically equal to the area a
Notes: b
1. Ú is read as ‘the integral of f (x) from a to b’
bounded by the curve y = f (x), the x-axis and the lines x = a and x = b.
a
Y
2. To evaluate the definite integral, there is no need to keep the constant of integration.
C D
b
3. Geometrically, Ú f (x) dx represents the area bounded
by the curve y = f (x), the x-axis and the lines x = a and x = b. 4. Area bounded means, we shall use,
a
b
Ú | f (x)| dx, when the area lies below x-axis. a
O
A x=a
B x=b
X
b
Area of the region ABCD = Ú f (x) dx a
b
2. Evaluation
of
Definite Integrals
ules R 1. Simply find the indefinite integrals of the given function. 2. There is no need to keep the constant of integration. 3. Use the limits of integration.
In general, Ú f (x) dx represents an algebraic sum of the a
area bounded by the graph of the function y = f (x), x-axis and the lines x = a and x = b. The area above the x-axis is taken as the sum with the positive sign, while the area which is below x-axis is taken as the sum with the negative sign. So, the value of the integral may be positive, negative or zero.
2.2 Integral Calculus, 3D Geometry & Vector Booster
6. Evaluation of the limit of Newton-Leibnitz Formula
Y
O a
the
sum
using
From the definition of the definite integral, we have
A1
A3
A2
A5
A4
b
X
b
Ú f (x) dx a
b
Thus, Ú f (x) dx = A1 – A2 + A3 – A4 + A5 a
5. Definite integral
as the
limit
of
h Æ 0
r = 0
(
[ (
)
b – a n b–a = lim _____ n S f a + _____ n r n Æ • r = 0
1 y (x) __r = lim __ n S f n , n Æ •
where
(i)
r = j (x)
( )
 is replaced by Ú symbol
1 (ii) __ n is replaced by dx r (iii) __ n is replaced by x j (x) (iv) a = lim ____ n n Æ •
( ) y (x) and b = lim ( _____ n )
n Æ •
Divide the interval [a, b] into n equal sub-intervals denoted by [a, a + h], [a + h, a + 2h], [a + 2h, a + 3h], ...,
[a + (n – 1)h, a + nh]
7. Properties
of
Definite Integrals
Property I
b
b
a
a
Ú f (x)dx = Ú f (t)dt
Clearly,
b = a + nh
d Proof: Let ___ [g(x) + c] = f (x) dx
fi
nh = b – a
fi
fi
b–a h = _____ n
Now, Ú f (x) dx = g(x) + c|ba
Ú f (x) dx = g(x) + c b
a
b
Thus, Ú f (x) dx
a
= g(b) – g(a)
b
Also, Ú f (t)dt = g(t) + c|ba
= Area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b.
= Area of the region PQRSP between the curve, x-axis and the ordinates x = a and x = b.
Property II
= lim h [f (a) + f (a + h) + f (a + 2h) + ...
d Proof: Let ___ [g(x) + c] = f (x) dx
h Æ 0
+ f (a + (n – 1)h] n – 1
= lim h S f (a + rh) h Æ 0
a
Hence, the result
(
)
where nh = b – a which is known as the first principle of integration.
= g(b) – g(a)
b
a
a
b
Ú f (x)dx = – Ú f (x) dx
fi
r = 0
b – 1 n – 1 = lim _____ S f (a + rh), n n Æ • r = 0
)]
sum
Let f be a continuous real function on [a, b]. Assume that all the values taken by the function are non-negative, the graph of the function is a curve above the x-axis.
n
= lim h S f (a + rh)
Ú f (x)dx = g(x) + c b
Now, Ú f (x) dx = g(x) + c|ba a
a
Also,
= g(b) – g(a)
– Ú f (x) dx = – [g(x) + c]|ab b
Definite Integrals
= – [g(a) – g(b)]
Property V
= g(b) – g(a)
Ú f (x) dx = Ú f (a + b – x) dx
Hence, the result. Property III
b
b
a
a
Proof: We have
b
c
b
a
a
c
b
Ú f (x)dx = Ú f (x)dx + Ú f (x),
Ú f (a + b – x)dx
where
Let a+b–x=t a
a
a 0, (x3 + 16)4 " x Π(0, 1) Thus, f (x) is concave upaward.
(
1
)
331.
0
______ ÷ 1 + x3 dx
Ú 0 1
(
)
x4 1 £ x + __ 4 0
______
1£x£2
fi
1 £ x4 £ 16
fi
2 £ (1 + x4) £ 17
1 1 1 fi ___ £ _______ £ __ 17 (1 + x4) 2 1
dx 7 1 fi ___ £ Ú _____ £ ___ 17 0 1 + x4 24 •
(
0
5 fi 1 £ Ú ÷ 1 + x3 dx £ __ 4 0 334. We have,
0
x fi (1 – 0)f (0) £ Ú ______ 3 dx 0 x + 16
1
______
1
(1 – 0) £ Ú (e ) dx £ e(1 – 0) 1
1
fi Ú 1 dx £ Ú ÷ 1 + x3 dx £ Ú (1 + x3)dx
0
1
fi
2
)
(
)
tan– 1ax – tan– 1x 335. Let I = Ú _____________ dx x 0
1 – (0)(f (0) + f (1) • £ ________________ 2 dI 1.x 1 ___ fi = Ú _________ dx x 17 da 0 (1 + a2x2)x fi 0 £ Ú ______ 3 dx £ ___ 2 0 x + 16 • dx Hence, the result. = Ú ________ _____ 0 (1 + a2x2 3 Let f (x) = ÷ 3 + . • 3x2 dx 1 fi f ¢(x) = ________ _____ = __2 Ú _________ 3 a 0 (1/a2) + x2 2÷3 + x
fi
(
)
3x2 f ¢(x) = ________ ______ > 0, " x Œ (1, 3) 2÷3 + x3
Thus, f (x) increases in (1, 3). 3
______ 3 + x3) dx
So, (3 – 1)f (1) £ Ú (÷ 3
fi
4£
1
_____ (÷ 3 + x3 )dx
Ú 1
£
332. We have for every x in [0, 1], _______
÷ 4 – 2x
2
_________ £ ÷4 – x2 – x3
_____
£ ÷4 – x 2
1 1 1 ______ fi _______ £ ___________ _________ £ _______ ______ 2 2 3 ÷4 – x – x ÷ 4 – 2x2 ÷ 4 – x
)
)
(
)
|
• 1 = __2 (a tan– 1(ax)) 0 a
1 p = __ 2 a__ = 2 a
( )
£ (3 – 1)f (3)
___ 2÷30
( (
p ___ 2a
p I = __ ln a + c 2
fi
When a = 1, I = 0 fi c = 0 p Hence, I = __ ln a 2 p /2
(
)
dx 1 + a sin x ____ 336. Let I = Ú ln _________ , (|a| < 1). 1 – a sin x sin x 0
Definite Integrals
( Ú ( Ú (
p /2
) ) )
dI dx 2 sin x ____ fi ___ = Ú _________ , (|a| < 1) da 0 1 – a2sin2x sin x p /2
dx 2 sin x ____ = _________ 2 2 sin x 0 1 – a sin x
2 = _________ dx 0 1 – a2sin2x
p /2
p /2
(
2
p /2
•
(
xb log x d[I (b)] fi ______ = Ú ______ dx db log x 0
)
1
•
0
(
)
Let t = tan x
(
))
______
• 2 = ________ _______ × tan– 1(t÷1 – a2) 0 – a2) ÷ (1
(
)
p 2 = ________ _______ × __ 2 ÷(1 – a ) 2 p _______ = ________ ÷(1 – a2)
I = p sin– 1a + c
When a = 0, I = 0, then c = 0
(
)
dx 1 + a sin x ____ 336. Given I = Ú ln _________ , (|a| < 1) 1 – a sin x sin x 0 p /2
(
)
dI 1 – a sin x 2 sin x 1 fi ___ = Ú _________ × __________ × ____ d x da 0 1 + a sin x (1 – a sin x)2 sin x p /2
(
(
)
)
)
cot x p /2 –2 ______ = _______ tan– 1 _______ _____ ÷1 – a2 0 ÷ 1 – a2
p p 2 = _______ ______ × __ = _______ _____ 2 2 ÷1 – a2 ÷ 1 – a
Integrating, we get
When b = 0, then c = 0 Thus, I (b) = log(b + 1) 338. We have, x
2sec 2t (f (x))2 = Ú f(t). _______ dt 4 + tan t 0
f (x) . 2 sec2x 2f (x)f ¢(x) = __________ 4 + tan x sec2x fi f ¢(x) = _______ 4 + tan x Integrating, we get f (x) = log|(4 + tan(x))| + c When x = 0, then c = – log(4) f (x) = log|(4 + tan(x))| – log(4) p 5 fi f __ = log|(4 + 1)| – log(4) = log __ 4 4 339. Given, fi
( )
p /2
f (x) = Ú log(1 + x sin2q )/sin2q dx 0 p /2
fi
(
)
sin2q 1 f ¢(x) = Ú ___________ × _____ 2 dq 2 0 (1 + x sin q ) sin q
(
)
dq = Ú ___________ 0 (1 + x sin2q )
cosec2q dq = Ú ______________ 2 0 (cot q + (1 + x))
cot q p /2 1 _____ = – ______ tan– 1 ______ _____ 1 + x ÷1 + x 0 ÷
p _____ = _______ 2÷1 + x
p /2
2 cosec2x = Ú _____________ 2 dx 0 cot x + (1 – a2)
(
I (b) = log(b + 1) + c
p /2
2 cosec2x = Ú __________ dx 0 cosec2x – a2 p /2
)
2 dx = Ú _________ 0 1 – a2sin2x p /2
(
( )
Hence, I = p sin– 1a p /2
xb + 1 = _____ b+ 1 Integrating both sides w.r.t. b, we get
2dt 1 = _______ Ú ______________ 2 1 2 ________ (1 – a2) 0 t + _______ – a2) ÷ (1
= Ú (xb )dx
)
(
( )
1
2dt = Ú ___________ , 0 1 + (1 – a2)t2
fi
1
2 sec2x = Ú ______________ dx 0 1 + (1 – a2)tan2x
I = p sin– 1(a) + c When a = 0, I (a) = 0, then c = 0 Thus, I = p sin– 1(a) xb – 1 337. Given I (b) = Ú _____ dx 0 log x
2 sec x = Ú _____________ 2 dx 0 sec x – a2tan2x
2.91
(
(
(
Integrating, we get _________ f (x) p ÷1 + x + c When x = 0, f (0) = 0 , then c = – p
)
))
2.92 Integral Calculus, 3D Geometry & Vector Booster _____
_____
f (x) p ÷1 + x – p = p (÷1 + x – 1)
(
1
)
x cos a – 1 340. Let I = Ú ________ dx logex 0
(
1
)
x cos a log x dI fi ___ = Ú _________ dx da 0 logx
=
=
=
=
1
= Ú (x cos a )dx
0
(
)
x cos a + 1 1 ________ 1 = ________ = cosa + 1 0 1 + cos a Integrating, we get
I (a ) = log|(1 + cos a )| + c p When a = __ , I (a ) = 0, then c = 0 2 Thus, I (a ) = log|1 + cos a | 341. Do yourself 2
____________
0
÷
÷
___________
___________
2
2
£ Ú (1 + x)dx × Ú (1 + x )dx
0
) ÷(
÷(
4
0
)
x x5 2 = x + __ × x + __ 2 0 5 0 2 2
÷(
________
)
______ 32 = ÷(2 + 2) × 2 + ___ 5 ___ 42 = 2 × ___ 5 Hence, the maximum value of the given integral is
÷
÷
___
42 2 × ___ . 5 343. We have,
Ú ÷
÷
÷(
) ÷(
________
________
)
___
3 5 15 = __ __ = ___ ___ 2 4 8 15 Hence, the maximum value is ___ . 8 344. We have, p a + __ 2 (sin4x a p __ 2
Ú
÷
+ cos4x)dx
= Ú (sin4x + cos4x)dx 0
)
3 = Ú (__ + 4 0
( (
1 __ cos|4x|)dx 4
)
3 sin|4x| p /2 = __ x + ______ 4 16 0 3 p = __ × __ 4 2 3p = ___ 8
)
2p
1 = ___ Ú 2sin2x dx 4p 0
1 = ___ Ú (1 – cos2x) dx 4p 0
sin2x 2p 1 = ___ x – _____ 4p 2 0
2p
(
)
1 = ___ (2p – 0) 4p 1 = __ 2 2
÷ ÷ ÷ __
(
1 – cos4x 1 – _________ d x 4
0
x2 1 x4 1 = x + __ x + __ 2 0 4 0 __
)
2 sin2(2x) 1 – ________ d x 4
1
3 £ Ú (1 + x)dx Ú (1 + x )dx 0
Ú
)
sin2(2x) 1 – _______ dx 2
dx 1 346. Mean value = ______ Ú _____ (2 – 0) 0 ex + 1
÷
1
( Ú ( Ú
2p
1
(1 – 2 sin2x cos2x)dx
1 345. Mean Value = _______ Ú sin2x dx (2p – 0) 0
_____________ (1 + x)(1 + x3)d x 0 __________ ___________
________
________
Ú ÷(1 + x)(1 + x4) dx
Ú
p __ 2
342. We have,
p __ 2 0 p __ 2 0 p __ 2 0 p __ 2 0
2
e– xdx 1 ______ = __ Ú – x 20 e +1
( (
)
2 1 = – __ log|(e– x + 1)| 2 0 1 = – __ log|2(e– 2 + 1)| 2
( | ( ) | )
e2 1 = __ log 2 _____ 2 2 e+1 1
__ 1 347. Mean value = ______ Ú (3÷x ) dx (1 – 0) 0 1
__ 1 = ______ Ú (3÷x ) dx (1 – 0) 0
)
Definite Integrals
( )
1 3 1 = ______ × __ x4/3 (1 – 0) 4 0 3 = __ 4 2p 1 348. Mean value = _______ Ú (sin3x)dx (2p – 0) 0
= G (6 + 1) = 6! = 720 353. We have, 1
1 = ___ Ú (3sin x – sin|3x|)dx 8p 0
=
=
=
349. Mean value =
( (
)
cos|3x| 2p 1 ___ – 3cos x + ______ 8p 3 0 1 1 1 ___ – 3 + __ + 3 – __ 8p 3 3 0 1 dx 1 ______ Ú _____ (1 – 0) 0 1 + x2
)
) = ( tan p = __ 4 1 ex 1 350. Mean value = ______ Ú _____ x dx (1 – 0) 0 e + 1 (x) 10
– 1
( )
= (log|(ex + 1)|)10
( )
e+1 = log _____ 2
351. We have,
(
1
)
= lim (j – 1 – j log(j ))
= lim (j – 1) – lim ( j log(j ) )
•
j Æ 0 j Æ 0
( )
log(j ) = lim (j – 1) – lim ______ j Æ 0 j Æ 0 1 __ j
( )
1 __ j _____ = lim (j – 1) – lim j Æ 0 j Æ 0 1 – ___ 2 j
0
= G (11) = G (10 + 1) = (10)! 354. We have, •
= – 1 352. We have,
0 • 0
= G (7) = (3)! =6 355. We have,
j Æ 0
= Ú (e– tt n –1) dt,
(Let t = – log x)
0 •
= Ú (e– x t n – 1)dx
= G (n)
0
1
______
356. Let I = Ú x4 ÷1 – x2 dx 0 p /2
= Ú sin4q cos2q dq 0
•
1
0 •
= Ú e– xx7 – 1 dx
= G (7)
0
( ( ) )
= Ú (– log|x|)n – 1dx
0
= Ú e– xx4 – 1 dx
•
Ú e– xx6 dx
= Ú e– xx3 dx
= lim (j – 1) + lim (j ) j Æ 0
•
= Ú e– xx11 – 1 dx
0
= lim (x log(x) – x)1j
j Æ 0
0
= – Ú e– tt10 dt
1 n–1 Ú log __ x dx
j Æ 0
0
1
lim Ú log(x)dx j Æ 0 j
( ( ) )
1 10 Ú log __ x dx
2p
2.93
( ) ( ) ( ) ( ) ( )
2+1 4+ 1 G _____ G _____ 2 2 = _________________ 4 +2 +2 2G _______ 2 5 3 G __ G __ 2 2 = _________ 2G (4) 3 1 __ __ 1 __ __ ◊ __ ◊÷p × ÷ p 2 2 2 = ______________ 2×6 p = ___ 32
(Let sinq = x)
2.94 Integral Calculus, 3D Geometry & Vector Booster p /2
357. We have,
I = Ú sin8x cos4x dx
0
( ) ( ) ( ) ( ) ( ) ( )
358. We have,
______
Ú x6÷1 – x2 dx 0
p /2
= Ú sin6q cos2q dq 0
359. We have,
( ) ( ) ( ) ( ) ( )
6+1 2+1 G ____ G _____ 2 2 = ________________ 6+2+2 2G ________ 2 3 7 G __ G __ 2 2 _________ = 2G(5) 5 3 __ 1 __ 1 __ ◊__ ◊ ÷ p × __ p 222 2 = _____________ 2 × 24 5p = ____ 256
p /2
Ú sin4x cos6x dx 0
( ) ( ) ( ) 5 7 G( __ )G( __ ) 2 2 = _________
6+1 4 +1 G ____ G ____ 2 2 = ________________ 4+6+2 2G ________ 2
2G(6)
5 3 __ 3 1 __ __ 1 __ __ ◊ __ ◊ ÷p ◊ __ ◊ ÷ p 2 2 2 2 2 = ________________ 2 × 120 3p = ____ 512 1
360. We have,
______ x10 ÷1 – x2 dx
Ú 0
(Let sinq = x)
(Let sin q = x)
(
) ( ) ( ) ( ) ( )
10 + 1 2 +1 G ______ G _____ 2 2 = _________________ 10 + 2 + 2 2G __________ 2 3 11 G ___ G __ 2 2 __________ = 2G(7)
3 1 __ __ 3 1 __ 7 5 __ __ ◊ __ ◊ ◊ __ ◊ ÷p × ◊ __ ◊ ÷p 2 2 2 2 2 2 = _____________________ 2 ◊ (6)! 7p _____ = 2048
1
0
8+1 4+1 G ____ G ____ 2 2 ________________ = 8+4+2 2G ________ 2 9 5 G __ G __ 2 2 = _________ 14 ___ 2G 2
= Ú sin10q cos2q dq
p /2
9 7 5 3 1 __ 1 __ __ ◊ __ ◊ __ ◊ __ ◊ __ ÷ p ◊ __ ÷ p 2 2 2 2 2 2 = __________________ 2 × (6!) 21p = ____ 512
361. We have, p /2
p /2
____ dx Ú _____ ____ × Ú ÷sin x dx sin x 0 ÷ 0 p /2
p /2
0
0
1 __
= Ú sin–1/2x cos0x dx x × Ú sin2 x cos0x dx x
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 = ( G( __ ) ) 2
3 1 1 1 G __ G __ G __ G __ 4 2 4 2 = _________ × _________ 3 5 __ __ 2G 2G 4 4 3 1 1 1 G __ G __ G __ G __ 4 2 4 2 = _________ × _________ 3 1 1 2◊ ◊G __ 2G __ 4 4 4 2
= p 362. We have, •
Ú e– a x dx 2 2
0
•
1 = ___ Ú e– t t – 1/2dt, 2a 0
1 1 – __ 1 = ___ Ú e– t t 2 dt 2a 0
( )
•
363. We have,
( )
1 G __ 2 _____ = 2a __ p ÷ = ___ 2a p /2
Ú sin7x cos5x dx 0
(
) ( ) ( )
5+ 1 7+1 G _____ G _____ 2 2 = ________________ 7 + 5 + 2 2G _________ 2
(Let (ax)2 = t)
Definite Integrals
364. We have,
G(2)G(3) = ________ 2G(7) 3.2.1.2.1 = ___________ 2.6.5.4.3.2.1 1 = ____ 120 p /2
Ú sin10x dx 0
365. We have,
9 7 5 3 1 p = ___ ◊ __ ◊ __ ◊ __ ◊ __ ◊ __ 10 8 6 4 2 2 63p = ____ 512 p /2 0
366. We have,
6 4 2 = __ ◊ __ ◊ __ 7 5 3 16 = ___ 35
p /4
Ú 8 cos4x sin4x dx 0
p /4
1 = __ Ú 16 cos4x sin4x dx 2 0
1 = __ Ú (sin2x)4dx 2 0
1 = __ Ú (sin4t)dx 2 0
p /4
p /2
367. We have, p /2
1 3 1 p = __ ◊ __ ◊ __ ◊ __ 4 4 2 2 3p = ___ 64
5 3 1 __ 3 1 __ __ ◊ __ ◊ __ ÷ p ◊ __ ◊ __ ◊ ÷p 2 2 2 2 2 = _________________ 2 × (5!) 3p ____ = 512 370. We have, p /2 0
1
= Ú t7 dt,
(Let sin x = t)
0
( )
t8 1 = __ 8 0 1 __ = 8 371. We have, p /2
Ú sin7x cos5x dx
( ) ( ) ( )
0
5+1 7+1 G _____ G _____ 2 2 = _______________ 7+5+2 2G _________ 2 G(4)G(3) = ________ 2G(7) (3!) × (2!) = __________ 2 × (6!) 6×2 1 = _______ = ____ 2 × 720 120 372. We have, p /2
Ú sin x cos x dx 0
( ) ( ) ( ) ( ) ( )
6+1 4+1 G _____ G _____ 2 2 _______________ = 6+4+2 2G ________ 2 5 7 G __ G __ 2 2 = _________ 2G(6)
Ú sin7x cosx dx
Ú cos7x dx
6
5
(5◊3◊1)(4◊2) = ____________ (11◊9◊7◊5◊3◊1) 8 = ____ 693 p /2
368. We have, Ú sin5x cos3x dx 0
(4◊2)(2) = ________ (8◊6◊4◊2)
1 = ___ 24 p /2
369. We have, Ú sin6x cos4x dx 0
2.95
Ú sin11x dx 0
(
)(
)(
)(
)(
)
–9 – 3 11 – 5 11 – 7 11 11 – 1 11 = ______ ______ ______ ______ ______ 11 11 – 2 11 – 4 11 – 6 11 – 8 10 8 6 4 2 = ___ ◊ __ ◊ __ ◊ __ ◊ __ 11 9 7 5 3 256 = ____ 693 p /2 373. We have, Ú cos8x dx 0
( )(
)( )( )
p 8 – 1 _____ 8 – 3 _____ 8 – 5 _____ 8 – 7 __ = _____ ◊ 8 8–2 8–4 8–6 2 7 5 3 1 p = __ ◊ __ ◊ __ ◊ __ ◊ __ 8 6 4 2 3 35p ____ = 256
2.96 Integral Calculus, 3D Geometry & Vector Booster
1. Let I = Ú (|sin x| – |cos x|)dx
1 8 ×9 = __ × _____ 3 2
= 12
0
2p
10p
2p + p /4
= Ú (|sin x| – |cos x|)dx + Ú (|sin x| – |cos x|)dx 0
4. Let I = Ú ([sec– 1x] + [tan– 1x])dx 1
2p
p
sec 1
p /4
= 2Ú (|sin x| – |cos x|)dx + Ú (|sin x| – |cos x|)dx
= Ú ([sec– 1x] + [tan– 1x])dx
1
0
0
p /2
10p
+ Ú ([sec– 1x] + [tan– 1x])dx
p /4
sec 1
= 4Ú (sin x – cos x)dx + Ú (sin x – cos x)dx 0
0
= 4(– cos x –
sin x)p /2 0 +
(– cos x –
__
sec 1
= Ú (0 + 0)dx + Ú (1 + 0)dx
sin x)p /4 0
1
= 4 . 0 + (1 + ÷ 2 )
5p /12
5 Let I = Ú [tan x]dx 0
Hence, the value of the given integral __
= Î(1 + ÷ 2 ) ˚ = – 1 sq.u.
[ [ ( ) ] ] p x = Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2 p x + Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2 p x + Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2
p x 2. Let I = Ú x 1 + cos ___ + 1 dx 2 – 2 – 1
– 2
1
2
0
1
= Ú [tan– 1x]dx + Ú [tan– 1x]dx
1
sec 1
= (10p – sec 1) sq. u.
__
= (1 + ÷ 2 )
10p
3
5p /12
+ Ú [tan– 1x]dx + Ú [tan– 1x]dx 2
3
= 0 + (tan– 12 – tan– 11)◊1 + (tan– 13 – tan– 12) ◊ 2
0
– 1 1
0
– 1
0
1
– 2
– 1
0
= Ú 1.dx + Ú 0.dx + Ú 1.dx
= 1( – 1– ( – 2)) + 0 + 1(1 – 0)
=1+1
= 2 sq.u. 3
5p p = ___ – __ – (tan– 13 + tan– 12) 4 4 = p + p – tan– 1( – 1) 9p = ___ sq.u. 4 p /4
6. Let I = Ú [sin x + {cos x + tan x + (sec x)}]dx 0
p Since 0 < x < __ 4 __ \ 1 < sec x < ÷ 2
( [ ] [ ] )
1 2 3. Let I = Ú [x] + x + __ + x + __ dx 3 3 0
(
1/3
2/3
1
0
1/3
2/3
= Ú 0. dx + Ú 1. dx + Ú 2. dx 4/3
5/3
2
1
4/3
5/3
7/3
8/3
3
2
7/3
8/3
Ú 6. dx + Ú 7. dx + Ú 5. dx
1 = __ (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) 3
fi
[sec x] = 1
fi
[tan x + [sec x]] = [tan x + 1]
+ Ú 3. dx + Ú 4. dx + Ú 5. dx
)
5p + ___ – tan– 13 ◊3 12
= [tan x] + 1 = 0 + 1 = 1
fi
[cos x + [tan x + [sec x]]]
= [cos x + 1] = [cos x] + 1 = 0 + 1 = 1
fi
[sin x + [cos x + [tan x + [sec x]]]]
= [sin x + 1] = [sin x] + 1 = 0 + 1 = 1 p /4
Thus, I = Ú [sin x + [cos x + [tan x + [sec x]]]dx 0
Definite Integrals
(
p m Now, __ p + n p + __ + 4 2 = 50 + 6 + 4
p /4
= Ú 1.dx
0
p = __ 4
p /2
10. Let
0
I = Ú tan(x2 – 6)dx + Ú tan(x2 + 18x + 75)dx
( ) ( ) ( ) ( )
p–1 p–2 p–3 = _____ + _____ + _____ 2 2 2 3p – 6 = ______ sq.u 2 sin(sin– 1b)
8. Let
| |
=
– 1
|
– 2
– 3
– 6
– 7
– 2
– 3
– 6
– 7
7
– 3
3
– 7
7
= Ú tan(x2 – 6)dx + Ú tan(y2 – 6)dx
|
= Ú tan(x2 – 6)dx + Ú tan(x2 – 6)dx – 3
cos(cos x) __________ dx sin(sin– 1x) cos(cos– 1a)
Ú
– 7
= Ú tan(x2 – 6)dx + Ú tan(x + 9)2 – 6)dx
cos(cos– 1x) = __________ dx Ú sin(sin– 1x) cos(cos– 1a) sin(sin– 1b)
)
= 60
7. Let I = Ú [sin– 1(cos x) + cos– 1(sin x)]dx
Graph:
2.97
3
7
7
3
3
= Ú tan(x2 – 6)dx + Ú tan(x2 – 6)dx
b
= 0
a
11. We have,
b x a
= Ú 1dx
=
= (b – a)
2n p
In = Ú |sin x|[sin x]dx – 2n p 2n p
= Ú |sin x|[sin x]dx
– 2n p
9. We have, 100
2n p
m = Ú sec– 1(secp x)dx
= Ú |sin x|(– 1 – [sin x])dx
0
– 2n p
2
= 50 × Ú sec– 1(secp x)dx
0
1 = 50 × __ × 2 × p 2 = 50p
2n p
fi 2In = – Ú |sin x|dx – 2n p
= Ú [sec (secp x)]dx + Ú [sec (secp x)]dx – 1
= – 2Ú |sin x|dx 0 2n p
1/2
2/p
3/p
1/p
2/p
+ Ú [sec– 1(secp x)]dx + Ú [sec– 1(secp x)]dx 1/p
2/p
1/2
( ) ( 6 1 = ( __ p – __ ) 2
1/p
fi In = – Ú |sin x|dx 0
p
3/p
= – 2nÚ |sin x|dx
2/p
= – 2n × 2 = – 4n
= Ú 0. dx + Ú 1. dx + Ú 2. dx + Ú 3. dx 0
2n p
1/p
– 1
1/2
– 2n p
– 2n p
0
0
– 2n p 2n p
3/p
2n p
= – Ú |sin x|dx – In
Also, n = Ú [sec– 1(secp x)]dx 1/2
2n p
= – Ú |sin x|dx – Ú |sin x|[sin x]dx
) (
)
3 2 1 1 2 1 = 0 + 1. __ p – __ + 2. __ p – __ p + 3. __ p – __ p 2
0
100
Now, S (In) n = 1
100
= S (– 4n) n = 1
2.98 Integral Calculus, 3D Geometry & Vector Booster
(
n = 1
= – 4(1 + 2 + 3 + ........... + 100)
( [ ] )
0
|sin x| m = Ú _______ dx x 1 – 2 __ p + __ 2 Let x = – y fi dx = – dy
( [ ] ) ( [ ] ) ( [ ] ) ( [ ] )
0
|sin(– y)| Thus, m = Ú ________ – y × – dy 1 ___ __ 2 p + 2
1 1 3. __ £ x2 + __ £ 1 fi 2 2 1 fi 0 £ x £ ___ __ 2 ÷ 3 1 4. 1 £ x2 + __ £ __ fi 2 2 1__ fi ___ £ x £ 1 2 ÷
2
|sin y| = – Ú _________ dy y 1 0 – __ – __ p 2
( [ ] )
– 1
1
– 1
0
p /2
2
(
1/÷2
__ 1/÷2
p /2
)
1
+ 2 Ú (cos– 1( – 1))dx + Ú __ cos– 1(0)dx 0
1/÷2
( ( ) ) ( ( ( ) ( )
p p 1 = 2 0 + __ 1 – ___ __ + 2 ___ __ + 2 2 2 ÷ ÷ p 1 = p 1 – ___ __ + 2 __ + 2 2 ÷
p ____ __ 2÷2
)
Dividing the numerator and the denominator by sin2 x, we get
)
1
(
8 + 7 cos x 15. Let I = Ú ___________ dx 0 (7 + 8 cos x)2
= 2 Ú (sin– 1(0))dx + Ú __ (sin– 1(1))dx 0
( )
( ( ) )
1
0
(
0
)
p p = – __ 0 – – __ 2 3 2 p = ___ 6
2
2
__
(
p = Ú __ dx – p /3 2
0
1/÷2
0
– 1
p = Ú – p + __ dx 2 – p /3
1
– 1
– p /3
( [ ] [ ] ) 1 1 = 2Ú ( sin [ x + __ ] + cos [ x – __ ] )dx 2 2 1 1 = 2Ú ( sin [ x + __ ] )dx + 2Ú ( cos [ x – __ ]dx ) 2 2 – 1
]
0
1
2
[
)) ) ( ( ( 1 1 = Ú ( – p + tan ( cos x – __ ) + cot ( cos x – __ ) )dx 2 2
1 1 I = Ú sin– 1 x2 + __ + cos– 1 x2 – __ dx 2 2 – 1 – 1
1 Ï ÔÔ0 : 0 £ x £ 2 1 5. x2 + __ = Ì 2 Ô1 : 1 £ x £ 1 ÔÓ 2 0
|sin x| = – Ú _______ dx x 1 0 __ p + __ 2 = – n fi m + n = 0. Hence, the result. 13. Let
1 __ £ x2 £ 1 2
2 1 I = Ú cot– 1 _________ + cot– 1 cos x – __ dx 2 cos x – 1 2 – p /3
|sin y| = – Ú _______ y dy 1 0 __ p + __ 2 2
1 0 £ x2 £ __ 2
14. Let
2
1 1 1 2. __ £ x2 + __ £ 1 + __ 2 2 2
0
|sin(y)| = – Ú ___________ × dy y 1 __ 2 – 1 – p + __ 2
)
Notes: 1. 0 £ x £ 1 fi 0 £ x2 £ 1
100 × 101 = – 4 × _________ 2 = – 20200
12. Given
) (
1 1 = p 1 – ___ __ + p 1 + ____ __ 2 2÷2 ÷ = 2p
100
= – 4S (n)
))
p 1 __ 1 – ___ __ 2 2 ÷
(
)
8 cosec2 x + 7 cosec x cot x I = Ú ______________________ dx 0 (7 cosec x + 8 cot x)2 Let
7 cosec x + 8 cot x = t
fi
(8 cosec2 x + 7 cosec x cot x)dx = – dt 7
( )
Thus, I = – Ú __ 2 • t 7 1 1 = __ = __ t • 7
( )
Definite Integrals
Put
x=0=y
fi
f (0) = 0
Put
y = – x, we get, f (x) – f (– x) = 0
fi
f (x) = f (– x)
\
f (x) is an even function.
– 1
3
Now, a = Ú (x – 1) f (x – 1)dx 2
1 2
( ( ) ) 1 – cos a – cos a = ( tan ( ________ – tan (__________ – 1sin a ) ) sin a ) 1 – cos a t = tan– 1 _____ sin a – 1 – cos a
16. Given f (x + y) = f (x) – f (y)
= Ú t f (t)dt, 2
(Let x – 1 = t)
– 1
( (
( ( ( ) ) (
a p __ a = __ + __ – 2 2 2
)
x3______ sin– 1x 19. Let I = Ú ________ dx 0 ÷1 – x2
1
= Ú y2f (y)dy,
(Let x + 1 = y)
– 2
p /2
0
( ( ( ( (
) )
4 (y – 2)2 + 1 )sin(y – 2) = Ú ___________________ dy 0 2(y – 2)2 – 7
2
)
(t2 + 1)sin t = Ú __________ 2 dt, – 2 2t – 7)
(Let y – 2 = t)
)
(y2 + 1) sin y = Ú __________ dy – 2 2y2 – 7 = 0, 1
( it is an odd function.)
(
(
cos3t = 0 – Ú _____ – cos t dt 3 0
1 = – __ Ú (cos3t)dt + Ú cos t dt 3 0 0
sin3t p /2 1 = – __ sin t – ____ + (sin t)p /2 0 3 3 0
1 1 = – __ 1 – ___ + 1 3 13
2 = 1 – __ 9
7 = __ 9
))
(
)
p /2
p /2
(
p /2
p /2
(
)
)
| |
(
)
)
sin a = Ú _________ 2 dt – 1 – cos a t + sin2a
1
(
)
(
1
Let x – cos a = t fi dx = dt 1 – cos a
( (
)
sin a = Ú ________________ 2 dx 2 – 1 x – cos a) + sin a
)
1 – x2 _______ 1 20. Let I = Ú _____2 × ______ dx 0 1+x ÷1 + x4
sin a 18. Let I = Ú ______________ 2 dx – 1 x – 2x cos a + 1 1
(
cos3t cos3t = t _____ – cos t – Ú _____ – cos t dt 3 3 0 0 p /2
(y2 – 4y + 5)sin(y – 2) 17. Let I = Ú ___________________ dy 0 2y2 – 8y + 1
2
)
= 2a
Thus, the value of 2a – b + 4 = 0 + 4 = 4
p /2
p /2 cos3t = t Ú sin3t dt 0 – Ú _____ – cos t dt 3 0
2
= 2Ú x2f (x)dx
(
– 2
dx Let sin– 1 x = t fi _______ ______ = dt ÷ 1 – x2
0
= Ú y2f (x)dx
4
)
Thus, I = Ú (t sin3t)dt
2
(
2
( ( ) ) )
0 1
= Ú x2f (x)dx
))
a a = tan– 1 tan __ + tan– 1 cot __ 2 2
a = __ 2
– 3
(
)
2sin2(a /2) 2cos2(a /2) = tan– 1 __________ + tan– 1__________ sin a sin a
0 2
Also, b = Ú (x + 1)2f (x + 1)dx
2.99
dx 1 – x2 = Ú ________ × _______ ______ 1 1 0 x2 x + __ x x2 + __ 2 x 1 __ 1 1 – 2 x dx = Ú _______ × ____________ __________ 1 1 2 0 x + __ x x + __ x – 2
(
( )
(
)
÷
) ÷(
)
)
2.100 Integral Calculus, 3D Geometry & Vector Booster
(
( )
)
1 1 Put x + __ x = t fi 1 – __ 2 dx = dt x
p /2
= 2Ú (log secq )dq 0
2
dt = – Ú _______ _____ • t ÷t 2 – 2
p /2
0
2
t dt = – Ú ________ _____ • t2 ÷t 2 – 2
Let t2 – 2 = y2 fi t dt = y dy
= – 2Ú (log cosq )dq
( ( ) )
p 1 = – 2 __ log __ 2 2
= p log 2
p = 8 __ log 2 8 = 8 m
__
(
)
÷2 dy = Ú _____ • y2 + 2
y ÷2 1__ – 1 ___ = – ___ tan __ ÷2 • ÷2
p 1__ __ = – ___ ( – 2 4 ÷ p = ____ __ 4÷2
2x 2x cos– 1 _____ 2 + tan– 1 _____ 2 1 + x 1 –x _________________________ 22. Let I = Ú __ dx ex + 1 – 1/÷3
(
21. Given, 1
(
( ) ) |
__
p __ ) 2
Hence, the result.
)
p /4
= Ú log (1 + tan q )dq
0
p /4
(
(
))
p = Ú log 1 + tan __ – q dq 4 0
( ) 1 + tan q + 1 – tan q = Ú log ( __________________ ) dq 1 + tan q
0
p /4
(
)
)
( ) p dx = __ Ú ( ______ 2 e + 1) p dx = __ Ú __ _____ x 2 – 1/÷3 e +1
...(i)
__
1/÷3
__ – 1/÷3
–x
__ 1/÷3
= p /2Ú __ (ex/ex + 1)dx
p /4
p /4
0
0
__
1/÷3
p = __ log 2 – m 4 p fi 2 m = __ log 2 4 p fi m = __ log 2 8 • log(1 + x2) Also, n = Ú _________ dx 0 1 + x2
(
)
( )
__
1/÷3
p p 2 = __ Ú __ dx = __ × ___ __ 2 – 1/÷3 2 3 ÷
p fi I = ____ __ 2÷3
( Ú ( ( Ú ( (
5p /4
) ) )
sin x + cos x 23. Let I = Ú __________ dx p ( x – __ ) – 3p /4 e 4 + 1 5p /4
2
cos x + sin x = __________ dx p p __ – x– __ – 3p /4 e 2 4 )+ 1
= Ú (log sec2q )dq
cos x + sin x = __________ dx p – x– __ – 3p /4 e 4 )+ 1
= Ú [log(1 + tan q )]dq
0
...(ii)
p 1 + ex fi 2I = __ Ú __ _____ dx 2 – 1/÷3 ex + 1
0 p /2
(
Adding Eqs (i) and (ii), we get
= Ú log(2)dq – Ú log(1 + tan q )dq
)
– 1/÷3
2 = Ú log ________ dq 1 + tan q 0
p /2
( )
__
p /4
( )
)
1/÷3
p /4
1 – tan q = Ú log 1 + ________ dq 1 + tan q 0
( )
p 2x 2x __ – sin– 1 _____ 2 + tan– 1 _____ 2 2 1+x 1–x = Ú __ ____________________________ dx x e + 1 – 1/÷3 __ p __ – 2tan– 1x + 2 tan– 1x 1/÷3 2 = Ú __ __________________ dx ex + 1 – 1/÷3 __
1/÷3
log(1 + x) m = Ú _________ dx 0 1 + x2
( )
( (
__
1/÷3
5p /4
...(i)
(
p x– __
)
1
cos x + sin x)e( 4 ) = Ú ________________ dx p x– __ – 3p /4 e( 4 )+ 1 5p /4
...(ii)
( x– __p ) )
(cos x + sin x) e( 4 ) + 1 2I = Ú _____________________ dx p ( x– __ ) – 3p /4 4 e + 1 5p /4
(
5p /4
= Ú (cos x + sin x)dx
– 3p /4
5p /4
fi 24. Given,
(
( (
1__ I = – ___ cos x + 2 ÷ 1
(
)
5p ___ p 4 __ = 3p 4 – ___ 4
))
0
)
( ) q = Ú ( _________ dq sin q cos q ) 2q = Ú ( __________ dq 2sin q cos q ) 2q = Ú ( _____ dq sin 2q )
0
t = Ú ____ dt, sin t 0
x = Ú ____ dt sin x 0
1 = __ I1 2
fi
(Let 2q = t)
( )
I1 Thus, __ = 2 I2
Now,
Thus,
(
)
p /2
x2 I2 = Ú _________ 3 dx 0 ex (2 – x3)
1
( ) Ú ( )
sin nx dx Ú ( _____ sin x )
)
dt 1 = __ Ú _________ 3 0 e1 – t(t + 1)
et dt 1 = ___ ______ 3e 0 (t + 1)
(
)
sin(n – 2)x = Ú2 cos(n – 1)x dx + Ú _________ dx sin x sin 5x dx Ú _____ sin x sin3x = Ú2 cos 4 x dx + Ú _____ dx sin x
( )
( )
( )
sin 5x Ú _____ dx sin x 0
1 (Let 1 – x3 = t fi x2 dx – __ dt) 3 0 dt 1 __ _________ = – Ú 1 – t dx 3 1 e (t + 1)
(
In + 1 = In = In – 1 = In – 2 = ... = I1 = I0
Hence, In = p, n ≥ 0. 27. We have, sin nx – sin(n – 2)x = 2 cos(n – 1)x sin x
0
( )
2 = _____ (sin(n + 1)x)p0 = 0 n+1
Therefore, I0 = Ú sin x dx = (x)p0 = p
p /4
1
( )
p
0
25. Given,
)
= 2 Ú cos(n + 1)x dx
fi
p /2
)
Thus, In + 1 – In
p /2
(
q = Ú ____ sec2 q dq tan q 0
p /4
)
1 1 sin n + 1 + __ x – sin n + __ x 2 2 ___________________________ = Ú dx x __ 0 sin 2
0
0
( ( ( )) ) ( (
1 sin n + __ x 2 __________ 26. Let In = Ú dx. x __ 0 sin 2 p
x
tan– 1x I2 = Ú ______ x dx 0
p /4
I1 Thus, __ = 3e I2
x
p /4
)
Now, In + 1 – In
__ p = ÷2 Ú sin x + __ d x 4 – 3p /4
(
ex dx 1 = ___ Ú ______ 3e 0 (x + 1) I1 = ___ 3e
Adding Eqs (i) and (ii), we get
2.101
Definite Integrals
p /2
p /2
( )
sin3x = 2Ú cos4 x dx + Ú _____ dx sin x 0 0
3sin x – 4sin3x = 0 + Ú ____________ dx sin x 0
= Ú (3 – 4sin2 x)dx
p /2
(
)
p /2 0 p /2
1
= Ú [3 – 2(1 – cos2 x)]dx 0
2.102 Integral Calculus, 3D Geometry & Vector Booster p
p /2
= Ú (1 + 2 cos 2 x)dx
sin y B = Ú _____ dy y +2 0
sin x = Ú _____ dx x +2 0
p cos x 1 = _____ Úsin x dx – Ú _______ dx 2 x+2 0 0 (x + 2)
0
p
= (x + sin 2x)p /2 0
( )
p = __ 2 28. Given
(
1
m = Ú x50(2 – x)50 dx 0
(Let x = 2t fi dx = 2dt) 1/2
m = 2 Ú 250 t 50 250(1 – t)50dt
fi
1/2
= =
1 2100 0
Ú
Ú x
50
(1 – x) dx = 2
•
0
(
)
(
)
•
( )
( )
3 3 sin3x sin x = __ Ú ____ x dx – __ Ú _____ dx 40 40 3x
siny 3 sin x 1 = __ Ú ____ x dx – __ Ú ____ y dy, (Let 3x = y) 40 40
( )
•
( )
•
)
3 p __ 1 p = __ × __ – × __ 4 2 4 2 p = __ 4 30. We have, p /2
(
)
sin2 x B = Ú ______ dx (x + 1) 0 y dy Let x = __ fi dx = ___ 2 2
(
( )
( )
3 sin x sinx 1 = __ Ú ____ x dx – __ Ú ____ x dx 40 40
(
)
)
p /2
3sin x _____ sin3x 1 = __ Ú _____ x – x dx 40
•
(
p /2
– sin3x 1 3sin x = __ Ú ___________ dx x 40
•
)
32. Let I = Ú log(a2 cos2 x + b 2 sin2x)dx
3
•
)
p p = ___________________ = ___ 2(2 + 3)(3 + 0)(0 + 2) 60
0
•
(
Let a = 2, b = 3 and c = 0, dx Ú ______________ 2 0 (x + 4)(x2 + 9)
( ) 4 sin x 1 = __ Ú ( ______ ) dx x 4 •
1 __ – A 2 1 1 A + B = _____ + __ p+2 2
(
sin3x 29. Let I = Ú _____ x dx 0
)
p = __________________ 2(x + a)(x + b)(x + c)
.n
•
(
)
1 = _____ + p+2
100
m 100 fi __ n = 2 Hence, the result.
•
p
(
x2 Ú ______________________ 2 dx 0 (x + a2)(x2 + b2)(x2 + c2)
– t)50dt
50
)
)
•
0
1 2100 t 50(1 0
(
p
– cos x p cos x = ______ – Ú _______ dx x + 2 0 0 (x + 2)2
31. We have,
= 2100◊2 Ú t 50(1 – t)50dt
( )
fi
0
( )
fi
= Ú log[a2 (cos2 x + k 2 sin2x)]dx 0
(Let k = b/a)
p /2
= p log a + Ú log (cos2 x + k 2 sin2x))dx
= p log a + I1,
0
p /2
Where
I1 = Ú log (cos2 x + k 2 sin2x))dx p /2
(
0
)
dI1 2 k sin2x fi ___ = Ú ______________ dx dk 0 cos2 x + k 2sin2x
( Ú (
)
p /2
tan2x ◊ sec2x = 2k Ú ____________________ dx 0 (1 + k 2 tan2x)(1 + tan2x)
)
p /2
t2 dt = 2k _______________ dx, 2 2 0 (1 + k t )(1 + t 2) (Let t = tan x) • 2k 1 1 = ______ 2 Ú _____ – _______ dt k – 1 0 1 + t 2 1 + k 2 t 2
(
(
)
)
• 2k 1 = ______ 2 tan– 1t – __ tan– 1(tk ) k 0 k – 1
Definite Integrals
(
•
p p 2k __ = ______ 2 – ___ k – 1 2 2k
p k – 1 2k = ______ 2 × __ _____ k – 1 2 k
)
( ) p k – 1 2k = ___ – 1 × __ ( _____ 2 k ) k
fi
2
When k = 1, I1 = 0, then c = – p log 2
| |
| | | | | |
1+k Therefore, I = p log a + p log ____ 2
•
| |
(
p ln2 2 p __ = ___ × ___ __ __ – 2 6 3 2 ÷
) )
(
•
(
)
0
1
ln(1 + x) = – Ú ________ dx x 0
)
(( ) ) ( )
1 ln __ y 1 1 __ ____________ __ = – Ú dy, Let t = y 2 • __ 1 1 1 2 + __ + 1 __ 2 y y y
– ln(y) 1 = – __ Ú __________ 2 dy 2 • (y + y + 1)
ln(y) 1 = __ Ú __________ 2 dy 2 • (y + y + 1)
0
0
0
(
•
(
(
)
)
)
ln(y) 1 = – __ Ú __________ 2 dy 2 0 (y + y + 1)
(
)
(
)
(
)
x2 x3 __ x4 x – __ + __ – + ... 2 3 4 = – Ú __________________ dx x 1
)
)
ln(1 + x) = 0 – Ú ________ dx x 0
0
ln t 1 Now, I2 = __ Ú ________ dt 2 0 (t2 + t + 1
(
(
ln(1 + x) = (ln x ln(1 + x))10 – Ú ________ dx x
1
= I1 + I2 (say)
(
)
( )
1
dt ln t dt ln2 1 = ___ Ú _________ + __ Ú _________ 2 0 (t2 + t + 1) 2 0 (t2 + t + 1)
•
(
1
ln2 + ln t = 2Ú ___________ dt 0 4(t2 + t + 1) •
( ( ) )
ln2 2 2 1 • = ___ × ___ __ tan– 1 ___ __ × t + __ 2 2 0 3 ÷3 ÷
1
)
(
)
)
(Let x = 2t fi dx = 2dt) •
( (
dt ln2 = ___ Ú __________ 2 0 3 1 2 __ __ t + + 2 4
ln x 34. Let I = Ú _____ dx. 1 +x 0
ln x 33. Let I = Ú __________ 2 dx 0 x + 2x + 4
)
a +b = p log a + p log _____ 2a a+b = p log _____ 2
(
p ln 2 = _____ __ 3÷3 p ln 2 p ln 2 Thus, I = 0 + _____ __ = _____ __ 3÷3 3÷3
1 + b/a = p log a + p log _______ 2
•
•
I1 = p log|k + 1| + c
I2 = 0
dt ln2 Also, I1 = ___ Ú __________ 2 0 (t2 + t + 1)
1+k Thus, I1 = p log _____ 2
)
fi 2 I2 = 0
p = _____ k+1 Integrating, we get
(
ln(t) 1 = – __ Ú _________ 2 dt 2 0 (t + t + 1) = I2
(
)
x x2 __ x3 = – Ú 1– __ + __ – + ... dx 2 3 4 0
(
)
1 x2 x3 __ x4 = – x – __2 + __ 2 – 2 + ... 2 3 4 0
(
)
1 1 1 = – 1 – __ 2 + __ 2 – __ 2 + ... 2 3 4
p 2 = – ___ 12 1 ln(1 + x4) 35. Let I = Ú _________ dx x 0
(
(
)
x8 x12 ___ x16 x20 x4 – __ + ___ – + ___ – ... 2 3 4 5 = Ú _________________________ dx x 1
)
0
2.103
2.104 Integral Calculus, 3D Geometry & Vector Booster 1
(
x7 = Ú x3 – __ + 2 0
(
4
8
x11 ___ – 3 12
)
x15 ___ + 4
x19 ___ – ... dx 5
In + 2 – 2In +1 – In
)
3 1 sin n + __ x – sin n + __ x 2 2 _______________________ = Ú dx x __ 0 sin 2
x 2 cos(n + 1)x ◊ sin __ 2 = Ú _________________ dx x 0 sin __ 2
= Ú 2 cos(n + 1)x dx
p
1
16
x x x x = __ – ___ + ___ – ___ + ... 4 16 36 64 0
(
)
1 1 1 1 = __ – ___ + ___ – ___ + ... 4 16 36 64
( (
p
) )
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
(
)
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
)
p 2 p 2 1 ___ = ___ – __ ◊ 24 8 6 p 2 = ___ 48 36. Given,
(
(
)
p
(
p
= In + 2 – In – 1 = I1 – I0.
Thus,
In – In – 1 = I1 – I0 = p – 0 = p
fi
In = p + In – 2 = p + p + In – 2 = 2p + In – 2 = 3p + In – 3 = ... = n p + I0 = n p + 0 = n p
)
)
) ( )
(
p
(
)
1–1 Now, I0 = Ú _______ dx = 0 1 – cos x 0 p
(
3 x 2 sin n + __ x sin __ 2 2 dx = Ú __________________ 1 – cos x 0
Subtracting, we get
3 x p sin n + __ x sin __ 2 2 = Ú ________________ dx x 2 __ 0 sin 2
( ( () ) ( )) ( ( ( ) ) ) ( ( ( ) ) )
3 sin n + __ x 2 __________ = Ú dx x __ 0 sin 2 p
Subtracting Eq. (ii) from Eq. (i), we get
I0 – I1 = p
fi
In = I0 + n p = n p
In = n p p /2
...(i)
( )
p /2
(
...(ii)
) )
1 – cos(2 nq ) = Ú ____________ dq 1 – cos(2q ) 0
1 – cos(n x) 1 = __ Ú ___________ dq = n p 2 0 1 – cos(x)
p
p
In + 2
sin2 nq Now, In = Ú ______ 2 dq 0 sin q
Similarly,
1 sin n + __ x 2 __________ – In = Ú dx x __ 0 sin 2
)
1 – cos x I1 = Ú ________ dx = p 1 – cos x 0
and
)
)
1 – x cos n x In = Ú __________ dx = n p. 1 – cos x 0
fi
cos(n + 1)x – cos(n + 2)x = ______________________ dx 1 – cos x 0
(
In + 2 – In +1 = In + 1 – In
p
)
(
In + 2 – 2In +1 – In = 0
1 – cos(n + 2)x 1 – cos(n + 1)x = Ú _____________ dx – Ú _____________ dx 1 – cos x 1 – cos x 0 0
p
)
=0
)
In + 2 – In + 1
( Ú (
( )
Similarly, In + 2 – In +1 = In + 1 – In
p
p
( )
)
( )
1 – x cos n x In = Ú __________ dx 1 – cos x 0
Now,
( )
)
2 = _____ sin(n + 1)x)p0 n +1
2 1 1 – __2 1 + __ 2 + ___ 2 + ... fi 4 2 12 fi
(
(
0
1 1 1 – 2 __ 2 – __ 2 + ___ 2 + ... 4 8 12
(
)
p
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
( (
(
37. We have
(
)
(12 + 22 + 33 + ... + n2)(13 + 23 + 33 + ... + n3) lim ______________________________________ n Æ • (16 + 26 + 36 + ... + n6)
Definite Integrals
( (
r = 1
)
( )
( )
n
n
S (r2) × S (r3) r = 1 r = 1 = lim _____________ n n Æ • (r6) S
)
1 n r 2 __ 1 n r 3 __ n S __ n × n S __ n r = 1 r = 1 = lim __________________ n Æ • 1 n r 6 __ n S __ n r = 1
( )
( ) ( ) ( Ú ) 1
1
Ú x dx × Ú x dx 0 0 ________________ = 2
3
(( () () ) ) 0
x x4 1 __ __ 3 0 4 0 = __________ x7 1 __ 7 0 1 1 __ × __ 3 4 ___ 7 _____ = = 1 12 __ 7 38. We have, n n – r 4r lim S _____ cos ___ n 2 n Æ • r = 1 n n n – r 1 4r = lim __ S _____ cos ___ n n n Æ • n r = 1
( ) )
( ( ) ( ) ) 1 r 4r = lim __ n S ( ( 1 – __ n )cos ( ___ n ) ) n
r = 1
1
= Ú (1 – x)cos(4x)dx
(
)
1
sin (4x) 1 __ 1 = (1 – x) ______ + Ú sin(4x)dx 4 4 0 0 1
1 = 0 + __ Ú sin(4x)dx 4 0
(
(
)
( Ú (
)
p /2
3 1
0
p
sinq Ú _______________ 2 2 dq , " a, b Œ R +, a > b 2 2 0 a cos q + b sin q p /2
x6 dx
n Æ •
)}
1 1 = lim __ – _____ n Æ • 2 n+2 1 = __ – 0 2 1 = __ 2 40. We have
sinq = 2 Ú _______________ 2 2 dq 0 a cos q + b2 sin2q
1
(
{ ( ( )
2.105
)
sinq = 2 ____________________ 2 2 dq 0 a cos q + b2 – b2 cos2q
(
p /2
)
sinq = 2 Ú _________________ 2 dq 0 b + (a2 – b2)cos2q 0
Let cosq = t
(
)
dt = – 2 Ú ____________ 2 1 b + (a2 – b2)t2 1
(
)
dt = 2 Ú ____________ 2 0 b + (a2 – b2)t2 1
(
)
dt 2 = _______ 2 2 Ú ___________ (a – b ) 0 _______ b2 2 2 + t2 (a – b )
(
______
)
1
÷a 2 – b2 2 1 = _______ 2 2 × ________ tan– 1 _______ t b b 0 (a – b ) ________ ______ 2 2 ÷ a – b
(
______
)
÷a 2 – b2 2 = __________ _______ × tan– 1 _______ 2 2 b b÷(a – b )
)
cos (4x) 1 = – _______ 16 0 1 = ___ [1 – cos (4x)] 16
b
39. We have, n
1
lim S Ú (xr – xr + 1)dx n Æ • r = 1 0 n
(
)
= b sin2q – a(1 – cos2q)
(
)
= b sin2q – a sin2q)
1 1 = lim S _____ – _____ n Æ • r = 1 r + 1 r+2
(
(x – a) = a cos2q + b sin2q – a
Now,
xr +1 xr + 2 1 = lim S _____ – _____ n Æ • r = 1 r + 1 r+2 0 n
÷
______
x–a Ú _____ d x b–x a Let x = a cos2q + b sin2q fi dx = (b – a)sin(2q )
1 = lim __ – n Æ • 2
) ( __ 13 – __ 14 ) + ( __ 14 – __ 15 ) + ... 1 1 + ( _____ – _____ n + 1 n + 2)
1 __ + 3
and
= (b – a)sin2q (b – x) = b – a cos2q – b sin2q = b (1 – sin2q) – a cos2q = (b – a)cos2q
2.106 Integral Calculus, 3D Geometry & Vector Booster a
The given integral reduces to p /2
÷
(b – a)sin2q Ú ___________ × (b – a)sin(2q )dq 0 (b – a)cos2q
0
sinq = Ú ____ × (b – a)sin(2q )dq cosq 0
= (b – a) Ú (2sin2q )dq
p /2
(
)
p /2
p /4
0
p aI + bJ = __ 2
Also,
b cos x – a sin x bI – aJ = Ú _____________ dx a cos x + b sin x 0
...(i)
(
)
= (log|a cos x + b sin x|)p /2 0
b = log __ a
( )
...(ii)
( ) ) ( ) )
a
( )
p Ú log(cot a + tan x)dx, a Œ 0, __ 2 0
(
a
(
)
)
= Ú log[cos(a – x)]dx – Ú log(sin a)dx
a
0
0
0
0
a
0
= Ú log(cos x)dx – Ú log(sin a)dx a
(
{ (
)}
)
p 4 – 3 sin2 2 __ – x 4 1 = __ Ú ______________________ dx p p 4 0 sin __ – x + cos __ – x 4 4 p /4
(
(
(
)
)
)
4 – 3 cos22x 1 = ____ __ Ú __________ dx cos x 4÷2 0
( Ú (
)
p /4
4 – 3(2cos22x – 1)2 1 = ____ __ Ú _________________ dx cos x 4÷2 0
)
1 = ____ __ Ú (sec x + 12cos x – 12cos3x)dx 4÷2 0
– Ú log(cos x)dx a
)
p /4
a
a
(
4 – 3(4cos4x – 4cos2x + 1) 1 = ____ __ ______________________ dx cos x 4÷2 0
cos(a – x) = Ú log _________ dx sin a cos x 0
)
p /4
sin x ____ cos x dx
)
4 – 3 sin22x 1 = __ Ú ___________ dx 4 0 sin x + cos x p /4
( (
cos a = Ú log _____ + sin a 0
( Ú (
sin6x + cos6x I = Ú ____________ dx sin x + cos x 0
p /4
3. We have,
)
p /4
ap b 1 ___ + b log __ a I = ________ (a2 + b2) 2 bp b 1 2 ___ – a log __ a . J = ________ 2 2 (a + b )
a
(
sin6x + cos6x = 2Ú ____________ dx sin x + cos x 0
1 – 3 sin2x cos2x = _____________ dx sin x + cos x 0
Solving Eqs (i) and (ii), we get
)
p /4
fi
p /2
(
...(ii)
sin6x + cos6x 2I = Ú ____________ dx sin x + cos x 0
2. Now, aI + bJ = Ú dx
and
)
Adding Eqs (i) and (ii), we get
p /2
(
...(i)
0
sin(2q ) p /2 = (b – a) q – _______ 2 0 p __ = (b – a) 2
)
p /2
= (b – a) Ú (1 – cos(2q ))dq
(
sin6x 4. Let Ú __________ dx. 0 sin x + cos x
cos6x = Ú __________ dx 0 sin x + cos x
0
= a log(cosec a) p /2
p /2
= – a log(sin a)
p /2
= – Ú log(sin a)dx
___________
– Ú log(cos x)dx 0
1__ = ____ [log|sec x + tan x| + 12sinx 4÷2
[
(
)]
p /4 1 – 12 sin x – __ sin3x 3 0
(
__ __ 1 1 = ____ __ log(÷2 + 1) + 6÷2 – 12 ___ __ – 4÷2 2 ÷ __ __ 1 = ____ __ [log(÷2 + 1) + ÷ 2 ] 4÷2
)]
1 ____ __ 6÷2
Definite Integrals
(
5. We have 1
(
)
dx I = Ú ______________________ 0 (5 + 2x – 2x2)(1 + e(2 – 4x))
...(i)
dx = Ú __________________________________ 2 2 – 4(1 – x) 0 {5 + 2(1 – x) – 2(1 – x) }(1 + e ) dx = Ú _____________________ 2 4x – 2 0 (5 + 2x – 2x )(1 + e )
...(ii)
(
p
{
}
{
}
dx 1 _____ 1 = Ú ____________ 2 , since _____ 2 – 4x + 4x – 2 = 1 0 (5 + 2x – 2x ) e e 1
dx 1 fi I = __ Ú _____________ 2 0 5 2 – 2 x – x – __ 2
(
)
= – p 2 + 2k
|
( ) ( ) ( ) ( ) ___
|)
|
___
1 + ÷11 1 ___ = ____ ___ log _______ 11 ÷10 ÷
(
)
(
4
)
( Ú ( Ú (
fi
p 2k = Ú _______ dx 0 1 + sin x
fi
dx 2k = 2p Ú _______ 1 + sin x 0
fi
dx k = p Ú _____________ p 0 1 + cos __ – x 2
p /2
fi
)
(
1
)
t4(t4 – 4t3 + 6t2 – 4t + 1) = _____________________ dt 0 t2 + 1
)
1
(
(
7
6
4t ___ + t5 6 1 2 = __ – __ + 1 – 7 3 t = __ – 7
(
( ) p p x k = __ Ú ( sec ( __ – __ dx 2 4 2 )) p x k = ( p tan ( __ – __ 4 2 ))
4
2
0
p __ 2 0
k=p
Thus, J – I = – p 2 + 2p = p (2 – p ) 8. We have, p /2
)
4 = Ú t – 4t + 5t – 4t + 4 – _____ 2 dt 0 t +1 5
)
fi
(t8 – 4t7 + 6t6 – 4t 5 + t4) = _____________________ dt 0 t2 + 1 6
)
dx k = p Ú ____________ p x 0 2cos2 __ – __ 4 2 p /2
fi
t4(t – 1)4 1 = Ú ________ 2 dt, Let t = __ x fi 0 t +1
1
(
p /2
(x – 1) Ú _______ dx 1 x8 + x10 1
(p – x) k = Ú _______ dx 0 1 + sin x
p
6. We have, •
fi
p
1
÷11 1 x – __ – ____ 2 2 1 1 ___ = – __ × ______ ___ log ______________ 4 11 ÷ 11 ÷ 1 2◊____ x – __ + ____ 2 2 2 0 (÷11 + 1) 1 1 = __ × ____ ___ log_________ ___ 4 ÷ 11 (÷11 – 1)
p
( ( ) ( ) )
___
)
p
)
|
(
– x2 x = _________ + 2 Ú _______ dx (1 + sin x) 0 0 1 + sin x
1
(
p
p
x where k = Ú _______ dx 1 + sin x 0
dx 1 = – __ Ú _________________ ___ 4 0 2 1 2 ÷11 x – __ – ____ 2 2
)
x2(1 + cos x) and J = Ú ___________ dx 0 (1 + x sin x)2 x2 cos x (J – I) = Ú _________ dx 2 0 (1 + sin x)
dx 1 1 2i Ú ___________ _________ + _________ 2 2 – 4x 0 (5 + 2x – 2x ) (1 + e ) (1 + e4x – 2 )
)
Now,
Adding Eqs (i) and (ii), we get
(
(
2 x 7. Given I = Ú ________ dx 1 + x sin x 0 p
1
1
p
1
1
)
22 = ___ – p 7
2
)
1
3
4t – ___ + 4t – 4 tan– 1t 3 0 4 – 1 __ + 4 – 4 tan (1) 3
)
____
_____
Ú (÷ sin x + ÷cos x ) – 4 dx 0
p /2
dx = Ú _______________ ____ 4 _____ 0 (÷sin x + ÷ cos x )
sec2 x dx = Ú ___________ ____ 0 (÷tan x + 1) 4
p /2
2.107
2.108 Integral Calculus, 3D Geometry & Vector Booster •
dt = Ú _______ 0 (t + 1) 4
• 1 1 = – __ × ______ 3 (t + 1) 0
1 = __ 3
(
10. Given, p /4
)
In = Ú (tan n x)dx 0
P /4
= Ú (tann – 2x◊ tan2x)dx
0
9. We have, p
(
P /4
)
0
p
(
0
))
dx = Ú ________________ 0 1 – tan2(x/2) ___________ a – 1 + tan2(x/2)
p
( (
(
sec2(x/2)dx = Ú _____________________ 0 (a – 1) + (a + 1)tan2(x/2)
(
)
•
(
)
÷ ÷
______
( ÷
_______ •
)
(a + 1) __ p 2 = ______ × _______ × (a + 1) (a – 1) 2 p = _______ _____ 2 ÷ a – 1
fi
1 In + In – 2 = _____ n –1
( )
1 _______ = 4 (I5 + I3 ) 1 _______ = 5 (I6 + I4 ) 1 and _______ = 6 (I7 + I5 ) Clearly, 3, 4, 5 and 6 are in AP. p
11. Let I = Ú q 3log(sinq )dq
...(i)
0 p
p
= Ú (p – q )3log(sin(p – q ))dq 0 p
p dx Thus, Ú _________ = _______ _____ ◊ a > 1 2 (a – cos x) 0 ÷a – 1
= Ú (p – q )3log(sinq )dq 0
Diffeerentiating w.r.t a, we get
Adding Eqs (i) and (ii), we get
p
t n – 1 1 In + In – 2 = _____ n –1 0
1 Thus, _______ = 3 (I2 + I4 )
(a + 1) (a + 1) 2 = ______ × ______ tan– 1 t _______ (a + 1) (a – 1) (a – 1) 0 _______
( )
fi
1 fi _________ = (n – 1) (In + In – 2 )
)
2
dt 2 = ______ Ú __________ (a + 1) 0 (a – 1) 2 ______ + t (a + 1)
0
0
sec (x/2)dx 1 = ______ Ú ________________ (a + 1) 0 (a – 1) ______ + tan2 (x/2) (a + 1)
0
= Ú (tann – 2x◊ sec2x)dx – In – 2
)
P /4
P /4
p
P /4
= Ú (tann – 2x◊ sec2x)dx – Ú (tann – 2x dx
sec2(x/2)dx = Ú __________________________ 2 2 0 a(1 + tan (x/2)) – (1 – tan (x/2)) p
= Ú (tann – 2x(sec2x – 1))dx
dx Ú a_______ – cos x
p a dx – Ú __________ 2 = – _________ 0 (a – cos x) (a2 – 1)3/2 p
p
2I = Ú {(p – q )3 + q 3}log(sinq )dq 0 p
p a dx fi Ú __________ 2 = _________ 2 3/2 0 (a – cos x) (a – 1)
= Ú {p 3– 3p 2q + 3p q 2}log(sinq )dq
= p 3 Ú log(sinq )dq – 3p 2 Ú (q )log(sinq )dq
(Let a = 2)
0
p
0
p
2p dx fi Ú __________ 2 = ____ __ 3÷3 0 (2 – cos x)
p
0 p
+ 3p Ú (q 2)log(sinq )dq 0
= – p 4log2 – 3p 2I1 + 3p I2
...(ii)
Definite Integrals p
(n + 1)In + 1 = In + 2 – G(n + 1)
0 p
(n + 1)In + 1 = In + 2 – G(n + 1)
Also,
n2In = nIn + 1 – nG(n + 1)
p
Thus,
In + 2 – (n + 1)In + 1 = nIn + 1 – n2In
0
fi 13. Given,
In + 2 – (n + 2)In + 1 + n2In = 0
Now, I1 = Ú q log(sinq )dq = Ú (p – q )log(sinq )dq
0
= p Ú log(sinq )dq – I1
p
fi 2I1 = p Ú log(sinq )dq
= – p 2log(2)
p fi I1 = – __ log(2) 2
(
(
)
2
p Now, 2I = – p log2 – 3p – ___ log(2) + 3p I2 2 2
(
)
(
) ( 1 1 f (x) = ( _______ sin ( x – __ x ) (a + b)x)
p
__ 3p p 3 3p = ___ ◊ ___ log(÷2 ) + ___ Ú [q 2log sin(q )]dq 2 6 2 0 p
I = Ú f (x)dx, (a π b) 2
3p p 3p = ___ ◊ ___ log2 + ___ Ú [q 2log sin(q )]dq 2 6 2 0
1 1 1 = ______ Ú __ sin x – __ x dx (a + b) 1/2 x
1 1 1 1 = ______ Ú __ sin y – __ y dy, x = __ y (a + b) 2 y 1/2
= – I
fi fi
2I = 0 I=0 2
0 •
{ (
) }
xn xn = e– xlog x ◊ __ n 0 – Ú e– x ◊ __ 1 – e– xlog x __ n dx x 0 •
1 = 0 – __ n Ú { (e– x ◊ xn – 1 – e– xlog x ◊ xn) }dx 0 •
= Ú f (x)dx = 0
= Ú {(e– xlog x) ◊x n – 1}dx
•
1 1 = __ n Ú e– xlog x ◊ xn dx – __ n Ú e– x◊ xn – 1dx 0 0 In – 1 ____ G(n) = ____ n – n Thus, nIn = In – 1 – G(n) Replacing n by (n + 1), we get
1/2
14. We have, p /6
(
___________
)
÷3cos(2x) – 1 Ú ____________ dx cos(x) 0 p /6
))
( ( ) ) 1 1 1 = – ______ Ú ( __ y sin ( y – __ y ) )dy (a + b) 1/2
In = Ú (e– xx n – 1log x)dx
)
( (
2
Hence, the result. 12. Given, •
•
) ( ) )
1 1 = Ú _______ sin x – __ x dx (a + b)x 1/2
2
p
__ 3p = ___ ◊ Ú q 2log(÷2 s in(q ))dq 2 0
•
( (
p
__ 3p 3p = ___ ◊ Ú q 2log(÷2 ) dq + ___ Ú [q 2 log sin(q )]dq 2 0 2 0
(
)
1/2
p
0
...(ii)
2
p 3p = ___ log2 + ___ Ú [q 2log sin(q )]dq 4 2 0 3
)
b 1 (a2 – b2) f (x) = a – __ x sin x – __ x
Let
p
4
(
fi
p 4 3p = ___ log2 + ___ I2 4 2
)
1 1 af (– x) + bf (x) = __ x sin x – __ x From Eqs (i) and (ii), we get fi
p 4 p 2 3p 3 fi I = – ___ log2 – __ p 2 – ___ log(2) + ___ I2 2 2 2 2
...(i)
1 1 af (– x) + bf (x) = – __ x sin __ x – x
4
( )
1 1 af (x) = bf (– x) = __ x sin x – __ 2 Replacing x by – x, we get
0
2.109
(
_________
)
2x – 4 ÷ 6cos dx = Ú __________ cosx 0 p /6
( ÷
_________
)
6cos2x – 4 = Ú _________ dx 0 cos2x
2 = Ú (÷ 6 – 4sec x) dx
p /6 0
__________
2.110 Integral Calculus, 3D Geometry & Vector Booster p /6
(
( )
)
6__________ – 4sec2x = Ú ___________ dx 2 0 x ÷ 6 – 4 cos
6 sec2x = Ú ___________ _________ dx + 4 Ú ___________ __________ dx 2 2 0 0 x ÷6 – 4sec x ÷ 6 – 4sec
6cos x sec2x = Ú ___________ __________ dx + 4 Ú ___________ _________ dx 2 0 0 ÷6cos 2x – 4 2 – 2tan x
6cos x sec2x = __________ _________ dx + 4 Ú ___________ _________ dx 2 0 0 2 – 6sin2x 2 – 2tan x
p /6
(
p /6
)
( Ú ( ÷
p /6
) )
p /6
p /6
p /6
( ( ÷ ( ÷
) ) )
Let sin x = t and tan x = v 1/2
(
)
__
1/÷3
(
( ÷ ) ( ÷ ) __
6p 1 2 = ____ __ – 4◊__ sin– 1 __ 2 3 3÷6 __ 2p 2 = ___ __ – 2sin– 1 __ 3 ÷6 15. We have
__
÷3
)
dv 6t = Ú ________ _______ dx _______ dx + 4 Ú ________ 2 0 0 ÷ 2 – 2v2 ÷ 2 – 6t
(
__
(
)
x4 + 2x2 + 1 = Ú __ ___________ 2 dx (x + 1)3 – ÷3 __
÷3
(
)
(x2 + 1)2 = Ú __ ________ 2 dx 3 – ÷3 (x + 1) __
÷3
(
)
dx = Ú __ _______ 2 (x + 1) – ÷3 __
÷3
dx = 2Ú _______ 2 0 (x + 1) __
( ) ( )
p /2 0
( )
9 G __ __ 2 = _____ ×÷ p 2G(5)
)
( )
7 5 3 1 __ ◊ __ ◊ __ ◊ G __ __ 2 2 2 2 ____________ = × ÷ p 2 ◊ (4!)
__
__ 105 × ÷p = ________ × ÷p 4 2 × 24
35 × p = _______ 7 2 So, L = 3, M = 5, P = 2 and Q = 7 Hence, the value of L+M+P–Q =3+5+2–7 =3 3. We have,
p I1 = Ú log(sin x)dx = __ log 2 8 0
__
= 2[tan– 1(÷3 ) – 0] 2p = ___ 3
Integer Type Questions 1. As we know that
( ) ( ) ( )
n+1 G _____ __ 2 Ú sinnx dx = _________ × ÷ p n+2 0 2.G _____ 2 11 p /2 G ___ __ 2 So, Ú sin10x dx = ______ ×÷ p 2G(6) 0 p /2
Thus, A = 2 and B = 8 Hence, the value of B – A is 6. 2. As we know that n+1 p /2 G _____ __ 2 Ú cosnx dx = __________ × ÷p n+2 0 2 . G _____ 2
p /2
= ( 2tan– 1(x) )÷0 3
__
__ 63 × 15 × ÷ p = ____________ 5 × ÷p 2 × 120 63 × p = _______ 5 2 ×8 63 × p = ______ 28
So, Ú cos8x dx
x4 + x3 + 2x2 + x tan2x + x2sin3x + tan– 1(x) + 1 Ú __ _______________________________________ dx x6 + 3x4 + 3x2 + 1 – ÷3 ÷3
9 7 5 3 1 __ ◊ __ ◊ __ ◊ __ G __ __ 2_____________ 2 2 2 2 = × ÷ p 2 ◊ (5!)
p /2
p and I2 = Ú log(cos x)dx = __ log 2 8 0 Hence, the value of I1 – I2 + 5 = 5 1
log(1 + x) 4. Let L = Ú _________ dx 0 (1 + x2) Let x = tanq fi d x = sec2q dq
p /4
log(1 + tanq ) 2 = Ú ___________ sec q dq 0 (1 + tan2q )
Definite Integrals •
p /4
= Ú log(1 + tanq )dq 0 p /4
(
...(i)
0
(
))
p = Ú log 1 + tan __ – q dq 4 0
1 + 1 – tan q = Ú log ___________ dq 1 + tan q 0
( ) 2 = Ú log ( ________ dq 1 + tan q )
p /4
0
p /4
p /4
0
0
= Ú log|2|dq – Ú log(1 + tan q )dq p /4
= Ú log|2|dq – I 0
p /4
fi 2I = Ú log 2 dq 0
p = __ log 2 4 p fi I = __ log 2 8
__ 3 = __ × ÷ p 2 3 = __ × (p)1/2 2 Hence, the value of L + M + N is 5 1 1 1 1 1 6. Clearly, M = __ × 1 × __ + __ × 1 × __ = __ 2 2 2 2 2 1 1 and N = 20 × __ × 1 × __ = 5 2 2 Hence, the value of 2 M + N + 2 = 1 + 5 + 2 = 8 7. We have, 2
M = Ú (– 1)[x] dx
p /4
and Let M = Ú log(1 + tan x)dx
(
(
))
p = Ú log 1 + tan __ – x dx 4 0
( ) 1 + tan x + 1 – tan x = Ú log ( _________________ ) dx 1 + tan x 2 = Ú log ( _______ dx 1 + tan x ) p /4
1 – tan x = Ú log 1 + ________ dx 1 + tan x 0
0
p /4
0
p /4
p /4
0
0
= Ú log 2 dx – Ú log(1 + tan x)dx
p = __ log 2 – I 4 p __ fi 2I = log 2 4 p __ fi I = log 2 8 p p p Now, L + M = __ log 2 + __ log 2 = __ log 2 8 8 4
So, A = 4 and B = 2 Hence, the value of A + B + 3 = 9. 5. We have, I = Ú e– xx3/2 dx 0
1
– 2
– 1
0
= Ú (– 1)
0
– 2
2
+ Ú (– 1)[x] dx 1
1
dx + Ú (– 1)
(– 2)
dx + Ú (– 1)(0) dx
(– 1)
– 1
0
2
+ Ú (– 1)(1) dx – 1
0
1
2
– 2
– 1
0
1
1
= Ú 1 ◊ dx + Ú (– 1) ◊ dx + Ú (1) dx + Ú (– 1)dx = (– 1 + 2) – (0 + 1) + (1 – 0) – (2 – 1) = 1 – 1 + 1 – 1 = 0 4
and N = Ú [x + 1]dx 0 4
= Ú ([x] + 1)dx 0
4
4
0
0
= Ú [x]dx + Ú 1◊dx
43 = ___ + 4 = 10 2 Hence, the value of M + N – 1 = 9. 8. We have,
3
L = Ú (|x – 1| + 1)dx – 2
•
0
– 1
p /4
– 2 – 1
= Ú (– 1)[x] dx + Ú (– 1)[x] dx + Ú (– 1)[x] dx
0
p /4
( ) 3 1 = __ G ( __ 2 2) 5 = G __ 2
p /4
= Ú e– xx5/2 – 1 dx
2.111
1 1 = __ (4 + 1) × 3 + __ (1 + 3) × 2 2 2
2.112 Integral Calculus, 3D Geometry & Vector Booster 23 = ___ 2
Thus, the value of 8
Now, M = Ú |x – 2|dx 0
1 = 2 × __ × 2 × 2 2 =4 2 2 23 Thus, __ (L – M) = __ ___ – 4 = 5 3 3 2
(
1
)
Ú x2 d(ln x) – 1 e
9. Let L = Ú [log|x|]dx 1
3
2p
1
__
10p
0
Hence, the value of B __ = 5. A 11. We have, 1 lim ___ __ + n Æ • ÷ n
(
(
1 ____ ___ + 2n ÷
1 ____ ___ + 3n ÷
)
1 1 ____ ___ + ... + ___ 2n 4n ÷
)
1 1 1 1 1 __ + ___ __ + ___ __ + ... + ___ = lim ___ __ 1 + ___ n Æ • ÷n 2n 2 ÷3 ÷ 4 ÷
1 4n 1_ = lim ___ __ S ___ n Æ • ÷ n r = 1 ÷r 1 4n 1__ = lim __ S ___ n Æ • n r = 1 __ r n 4 dx__ = Ú ___ x ÷ 0
÷
__
__
4 = (2÷x ) = (2÷4 – 0) = 4 0
( ) ( )
15.
( )
1 e4 – 1 = __ _____ 2 e2
1
dx Ú __________________ _____ _____ 1 + x + ÷ 1 – x + 2 0 ÷
Put fi
x = sin(2q ) dx = 2cos(2q )dq
p /4
2cos(2q )dq = Ú _____________________________ __________ __________ ÷1 + sin(2q ) + ÷ 1 – sin(2q ) + 2 0
2cos(2q )dq = Ú ___________________________ cosq + sinq + cosq – sinq + 2 0
2cos(2q )dq = Ú __________ 2(cosq + 1) 0
cos(2q ) = Ú _________ dq (cosq + 1) 0
n(n – 1) _______ Ú [x]dx 0 2 _______ n = _______ = (n – 1) n __ 2 Ú {x}dx
1 – 2sin2q ) = Ú __________ dq 0 (cosq + 1)
1 – 2(1 – cos2q ) = Ú _____________ dq (cosq + 1) 0
1 = Ú ________ – 2(1 – cosq ) dq 1 + cosq 0
1 = Ú _________ – 2(1 – cosq ) dq 2 0 2cos (q /2)
p /4
p /4
p /4
0
10
Ú [x]dx 0 Thus, ________ 10 = (10 – 1) = 9. Ú {x}dx
p /4
p /4
0
13. As we know that
)
f (x) b–a Ú ________________ dx = _____ 2 f (a + b + x) + f (x) a
p /4
p /4
n
(
x2 = __ 2 1/e 1 1 = __ e2 – __ 2 2 e
Clearly, m = 4 and n = 2. Hence, the value of (m + n) = 6. We have,
12. As we know that
b
( ) ( ) e
and B = Ú [sin x + cos x]dx = 5p
= Ú x dx 1/e
A = Ú [÷ 2 s in x]dx = p 0
1 = Ú x2 ◊ __ x dx 1/e e
and M = Ú [|log x|]dx and, then find the value L of __ + 5. M 10. We have,
)
8–2 = _____ = 3 2 14. We have
3
(
[x2] Ú ___________________ 2 dx 2 [x – 20x + 100] + [x2]
4
p /4
(
(
( ( )
)
)
)
q 1 = Ú __ sec2 __ – 2(1 – cosq ) dq 2 2 0
Definite Integrals
( ( )
p __
)
q = tan __ – 2(q – sinq ) 4 2 0
( ( ) (
p /2
= Ú 2((cos x + cos 3x + ... + cos(2k – 1)x)cos x dx 0 p /2
))
p p 1 = tan __ – 2 __ – ___ __ 4 ÷2 8 __ __ p = (÷2 – 1) – __ + ÷ 2 2 __ p = 2÷2 – __ – 1 2 Thus, a = 2, b = 2, c = – 1 Hence, the value of (a + b + c) = 3.
(
= Ú [2cos 2x + 2cos 3x cos x + ... 0
)
+ 2cos(2k – 1)x cos x]dx p /2
= Ú [(1 + cos 2x) + (cos 4x + cos 2x) + ... 0
Questions asked in Past Iit-Jee Examinations
+ cos(2k)x + cos(2k – 2)x]dx
(
(
)
1. Ú [ f (x) + f ( – x)](g(x)) – g( – x))dx
p = __ 2
= Ú h (x)dx = 0 – p /2
where h(x) = [ f (x) + f ( – x)](g(x) – g(– x)
0
x
the curve y + Ú f (t)dt = 2
p /2
2
2. Let I = Ú f (sin 2x)sin x dx
...(i)
0
= Ú f (sin 2x)cos x dx
...(ii)
0
Adding Eqs (i) and (ii), we get 2I = Ú f (sin 2x)sin x + cos x )dx
p /2
( (
__ x = ÷2 Ú f (sin 2x) sin x + __ dx 4 0
=
__ p /4 ÷2 f (cos2t)(cos t)dt
Ú
))
x Let __ – x = t 4
– p /4
)
__ p /4
= 2÷2 Ú f (cos2t)(cos t)dt
)
(
(
)
(
(
)
)
(
(
)
)
)
p (x – p)sin(2x)sin __ cos(x) 2 ______________________ = Ú dx (2x – p) 0 p
)
p (p – x)sin(2x)sin __ cos(x) 2 = Ú _______________________ dx p – 2x 0 p
(
(
p (p – x)sin(2p – 2x)sin __ cos(p – x 2 = Ú ______________________________ dx 2(p – x) – p 0 p
p /2 0
(
0
p x sin(2x)sin __ cos x 2 _________________ 5. Let I = Ú dx ...(i) 2x – p 0 p
p /2
x
4. If f is a continuous function with Ú f (t)dt Æ • as |x| Æ •, then show that every line y = mx intersects
fi h(x) is an odd function.
]
sin(2k)x ___________ cos(2k – 2)x p /2 + _______ + 2k (2k – 2) 0
– p /2 p /2
)
sin 2x p /2 sin 4x _____ sin 2x p /2 = 1 + _____ + _____ + 2 0 4 2 0
p /2
2.113
...(ii)
Adding Eqs (i) and (ii), we get
0
=
__ p /4 2÷2
Ú
p
f (cos2x)(cos x)dx
0
__ p /4
fi
2I = 2÷2 Ú f (cos2x)(cos x)dx
fi
I = 2÷2 Ú f (cos2x)(cos x)dx
__
0 p /4 0
p
(
)
sin2 kx = Ú ______ cos x dx sin x 0
))
(
(
))
p
(
(
))
fi
p = Ú sin(x)cos x sin __ cos(x) dx 2 0
fi
4 = – ___2 Ú (t sin t )dt, p p /2
3. Let I = Ú sin(2kx)cot x dx 0 p /2
(
p fi 2 = 2 Ú sin(x)cos x sin __ cos(x) dx 2 0
p /2
(
p 2 = Ú sin(2x)sin __ cos(x) dx 2 0
– p /2
(
p Let __ cos x = t 2
)
2.114 Integral Calculus, 3D Geometry & Vector Booster
( )
– p /2
fi
4 = – ___2 Ú (t sin t )dt p p /2
8 = – ___2 Ú (x sin x )dx p 0
8 = – ___2 (– x cos x + sin x )p /2 0 p
p /2
5
1 = ___ Ú |sin t |dt, 2p 0
1 = ___ Ú |sin x |dx 2p 0
1 2 = ___ × 4 = __ p 2p
)
4 = Ú 1 + _____ 2 dx 3 x –4
1
10. Given integral = Ú e x(x – 1)n dx
( | | ) = ( x + log _____ | xx +– 22 | ) = ( 5 + log __ | 37 | – 3 – log | __ 15 | ) = ( 2 + log ___ | 157 | ) x–2 5 1 = x + 4 ◊ ___ log _____ 2.2 x+2 3
Let I = Ú e x(x – 1)n dx 0
5
1
3
0
= – (– 1)n – n I n – 1
In = – (– 1)n – n I n – 1 1
= Ú f (x)dx
0
– p
1
= 0, since (1 – x2)sin x cos2x is an odd function.
= ( (x – 1)e x )10 – Ú exdx 0
= – (– 1) – (e x)10
= 1 – (e – 1) = 2 – e
– 1
From Eq. (i), we can write
1
= Ú |x – 1|dx
I2 = – 1 – 2I1
– 1
= – 1 – 2(2 – e) = 2e – 5
0
Again, from Eq. (i), we can write
= Ú |t|dx, – 2
(Let (x – 1) = t)
I3 = 1 – 3I2
= 1 – 3(2e – 5) = 16 – 6e
0
= Ú |x| – 2
p /2
= Ú (– x)dx – 2
(
)
1 11. Given integral is Ú ________ dx 0 1 + tan3x
0
...(i)
Now, I1 = Ú e x(x – 1)dx
1
= (– 1)n – n Ú (x – 1)n – 1e x dx 0
8. Given integral = Ú |1 – x|dx
= ( (x – 1)n e x )10 – Ú n(x – 1)n – 1e x dx 1
– p
0
1
7. Given, Ú (1 + x2)sin x cos2x dx
(Let t = 2p x)
2p
p
9. Given integral = Ú |sin(2p x)|dx 2p
p
=2
0
( )
(
1
x2 6. Given Ú _____ 2 dx 3 x –4
– x2 0 = ____ 2 – 2
8 8 = – ___2 (1 – 0) = ___ 2 p p 5
p /2
(
)
cos3x = Ú ___________ 3 dx 0 cos x + sin3x
...(i)
Definite Integrals p /2
(
)
sin3x = Ú ___________ 3 dx 0 sin x + cos3x Adding Eqs (i) and (ii), we get
p /2
(
3
...(ii)
sin x + cos x 2I = Ú ___________ dx 0 sin3x + cos3x
fi
3
(
)
( ( (
3
5
3
4
)
2
)
(2x3(x2 – 1)) + (x2 + 1)2 = Ú _____________________ dx 2 (x2 + 1)2(x2 – 1)
3
) ( ) 3
(2x3) dx = Ú _______ 2 dx + Ú _______ 2 2 2 2 (x + 1) (x – 1)
3
(
)
( | | )
( ) ( | |
p /4
= p (tan j – sec j )3p /4 p /4 __
( ) ( | | )
3 1 1 1 = Ú __ – __ dt + __ log __ t t2 2 2 5
3 1 10 1 = log|t| + __ + __ log __ t 5 2 2
__
= – 2p (÷2 + 1) __
fi I = – p (÷2 + 1) 1
(
| | ) ) ( | | )
) (
3p /4
(
( )
( ) Ú ( ) 1
x2 = – 2 _____ dx 3 – |x| 0
0 V
n p + V
0 V
V n p
0 V
0
= Ú |sin x|dx + Ú |sin x|dx = Ú sin x dx + Ú |sin x|dx p
= Ú sin x dx + n Ú |sin x|dx 0
V
= – (cos x) 0 + n × 2
= 1 – cos V + 2n
= (2n + 1 – cos V)
16. The given integral = 3
)
)
0
)
p – j = Ú _________ dj 1 + sin(j) p /4
1
1
p – j = Ú ____________ dj 1 + sin(p – j) p /4 3p /4
(
x2 = 0 – Ú _____ dx – 1 3 – |x|
)
(
1
sin x x2 = Ú _____ dx – Ú _____ dx – 1 3 – |x| – 1 3 – |x|
j 13. Let I = Ú ________ dj ...(i) 1 + sin j p /4
)
n p + V
( 1 1 = ( __ log|6| – ___ ) 2 10 3p /4
(
sin x – x2 14. Given integral = Ú ________ dx 3 – |x| – 1
15. Given integral = Ú |sin x|dx
3 1 1 = log|2| – ___ + __ log __ 10 2 2
__
= p [( – 1 – ÷ 2 ) – (1 + ÷ 2 ) ]
| | )
Let x 2 + 1 = t
)
3p /4
t–1 1 1 1 1 = Ú ____ 2 dt + __ log __ – __ log __ 2 2 2 3 t 5
)
= p Ú (sec2j – secj tan j)dj
(
( (
3p /4
x2.2x dx x–1 3 1 __ _____ = Ú _______ 2 + l og 2 x+1 2 2 (x + 1)2
10
)
(1 – sin j) = p Ú _________ dj p /4 cos2j
10
(
3p /4
(2x – 2x ) + (x + 2x + 1) = Ú ________________________ dx 2 (x2 + 1)(x4 – 1)
)
(1 – sin j )dj = p Ú ____________ p /4 1 – sin2(j )
2x5 + x4 – 2x3 + 2x2 + 1 12. Given integral = Ú ______________________ dx 2 (x2 + 1)(x4 – 1) 3
(
3p /4
p = Ú (1)dx = __ 2 0 p I = __ 4
3p /4
dj = p Ú _________ 1 + sin(j) p /4
p /2
Adding Eqs (i) and (ii), we get p 2I = Ú _________ dj 1 + sin(j) p /4
)
3
2.115
...(ii)
(
__
)
x ÷ Ú ___________ __ __ dx ÷5 – x + ÷ x 2
3– 2 = _____ 2
2.116 Integral Calculus, 3D Geometry & Vector Booster 1 = __ 2
1__ ___ 3 ÷
17. Given integral = Ú [2sinx]dx p/4
3p/4
p
0
p/4
3p/4
= Ú [2sinx]dx + Ú [2sinx]dx + Ú [2sinx]dx 5p/4
7p/4
2p
p
5p/4
7p/4
+ Ú [2sinx]dx + Ú [2sinx]dx + Ú [2sinx]dx
( ) 7p 5p 7p + ( ___ – ___ ) ◊ (– 2) + ( 2p – ___ ) ◊(–1) 4 4 4
)
3p p 5p = 0 + ___ – __ ◊ (1) + 0 + ___ – p ◊ (– 1) 4 4 4
(
p p p = __ – __ – p – __ 2 4 4 = – p
1__ ___ 3 ÷ x4 = _____ 4 p – 1__ 1 – x ___ – 3 ÷ 1 ___ __ 3 ÷ x4 = p _____ 4 dx 1__ 1 – x – ___ 3 ÷
=
( )
( )
__ p x 1 18. Given f (x) = Asin ___ + B, f ¢ __ = 2÷2 2 2
Ú
( (
Ú
( ( ) )
p x 2A fi Ú Asin ___ + B dx = ___ p 2 0
)
1 Asin(p x/2) 2A fi – __________ + B = ___ p (p/2) 0
(
)
2A 2A fi – ___ p + B = ___ p 4A fi B = ___ p __ 1 Also, f ¢ __ = 2÷2 2
Ú
( )
( )
p p 1 1 f ¢ __ = A __ cos __ × __ 2 2 2 2
(
)
( )
Ap p Ap = ___ cos __ = ____ __ 2 4 2÷2
__ Ap__ fi ____ = 2÷2 2÷2 8 fi A = __ p 32 4 8 ___ fi B = __ p × __ p = 2 p 1 ___ __ 3 ÷ x4 2x = _____ 4 cos–1 _____ 2 dx. 1__ 1 – x 1+x – ___ 3 ÷
(
( ) Ú ( ) Ú ( )
x4 = 2p Ú _____ 4 dx 0 1–x
1__ ___ 3 ÷
x4 I = p _____ 4 dx 0 1–x
fi
1 ___ __ 3 ÷
x4 = – p _____ 4 dx 0 x –1
(
(
))
)
1 = – p Ú 1 + _____ 4 dx 0 x –1
( )
fi
Ú
÷3
x4 2I = p Ú _____ 4 dx 1__ 1 – x – ___
fi
fi
19. Let I
( ( ( )))
x4 2x – Ú _____ 4 cos–1 _____ 2 dx 1 1 – x 1 +x __ – ___
1 ___ __ 3 ÷
p px f ¢(x) = A __ cos ___ 2 2
1 ___ __ 3 ÷
1 ___ __ 3 ÷ x4 4 dx – I p _____ 1__ 1 – x – ___ 3 ÷ 1 ___ __ 3 ÷
( ) ( )
)))
÷3 1 ___ __ 3 ÷
2A Also, Ú f (x) dx = ___ p 0
(
(
2x cos–1 _____ 2 dx 1+x
( )
1
1
))
÷3
0
(
(
x4 – 2x = Ú _____ 4 cos–1 _____ dx 1 1 – x 1 + x2 __ – ___
2p
= – p
1 ___ __ 3 ÷ 1 0
( Ú ( ( Ú
1 ___ __ 3 ÷ 1 0
)
1 + _____________ 2 dx (x – 1)(x2 + 1)
))
1 1 1 + __ _______ – _______ dx 2 (x2 – 1) (x2 + 1)
= – p
__ x–1 1 __ 1 = – p x + __ log _____ – tan–1x ÷ 3 2 2 x+1 0
1–÷ 3 __ p 1 1 1 __ = – p ___ __ + __ __ log ______ – 2 2 6 3 1+÷ 3 ÷
( ( | |
( |
(
20. We have, Un + 2 – Un + 1 p
(
)
p
(
1 ___
))
__
| )) )
1 – cos(n + 2)x 1 – cos(n + 1)x = Ú ______________ dx – Ú ______________ dx 1 – cosx 1 – cosx 0 0
Definite Integrals p
(
p
)
cos(n + 1)x – cos(n + 2)x = Ú ______________________ dx 1 – cosx 0
(
(
)
)
3 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
( )
21. Given
( (
) ( ) ( ) )
(
)
fi Un + 2 – Un + 1
1 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
( )
(
(
( ) x 2cos(n + 1)x sin ( __ ) 2 = Ú __________________ dx x sin ( __ ) 2
(
0
)
fi
))
)
p
= Ú (2cos(n + 1)x) dx 0
(
)
sin(n + 1)x 0 = 2 __________ (n + 1)x p =0
Thus,
Un + 1 + Un = 2Un + 1
[
fi Now,
Thus, fi fi
( ) 1 – cosx = Ú ( _______ dx = p 1 – cosx )
( (
)
( (
)
)
1 1 = _______ 2 2 Ú a __ x – 5 – b(x – 5) dx (a – b ) 1
x2 1 = _______ 2 2 a(log|x| – 5x) – b __ – 5x 3 1 (a – b )
22b 1 = _______ 2 a(log|2|) – ____ + 10b – 5a 2 3 (a – b )
22. Let
x sin2nx I = Ú ____________ 2n dx 0 sin x + cos2nx
2
( (
)
)
(
(
( (
)
(
)
(
)
(2p – x)sin2nx = Ú ____________ dx 0 sin2nx + cos2nx
sin2nx = 2p Ú ____________ 2n dx 0 sin x + cos2nx
( Ú ( Ú ( Ú ( Ú (
) ) )
sin2nx = 2p ____________ 2n dx – I 0 sin x + cos2nx
)
2p
)
fi
sin2nx 2I = 2p ____________ 2n dx 0 sin x + cos2nx
)
fi
sin2nx I = p ____________ 2n dx 0 sin x + cos2nx
(1 – cosnt) 1 = __ Ú _________ dt, 2 0 (1 – cost)
(
)
2p
(1 – cos2nq) = Ú __________ dq 0 (1 – cos2q) p
)
1 a __ x – 5 – b(x – 5) __________________ = Ú dx 1 (a2 – b2) 2
2sin (nq) = Ú ________ 2 dq 0 sin (q) p/2
)
x sin2nx – ____________ 2n dx 0 sin x + cos2nx
sin nq Also, Ú ______ 2 dq 0 sin q
(
2
2p
( )
(
(
2p
2
2
...(ii)
1 (a2 – b2) f (x) = a __ x – 5 – b(x – 5) 1 a __ x – 5 – b(x – 5) __________________ f (x) = (a2 – b2)
1
0
p/2
( )
2p
U1 – U0 = p U1 = U0 + p Un = U0 + np = np] p/2
...(i)
Ú f (x) dx
p
U1
1 __ x – 5
1 a f __ x + b f (x) = x – 5
2p
p
1–1 Now, U0 = Ú _______ dx = 0 1 – cosx 0
( )
Multiplying (i) by a and (ii) by b, and subtracting, we get 1 a2f (x) – b2f (x) = a __ x – 5 – b(x – 5)
3 1 2 sin n + __ x – sin n + __ x 2 2 = Ú __________________________ dx x 0 sin __ 2 p
( ( )
np 1 = __ × np = ___ 2 2
Replacing x by 1/x, we get
Therefore, (Un + 2 – Un + 1) – (Un + 1 – Un) p
)
(1 – cosnx) 1 = __ Ú __________ dx 2 0 (1 – cosx)
1 a f (x) + b f __ x =
3 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
(
2.117
2p
(Let 2q = t)
)
))
2
)
2.118 Integral Calculus, 3D Geometry & Vector Booster
( Ú (
) )
2p
fi
2p
37
sin2nx I = 4p Ú ____________ 2n dx 0 sin x + cos2nx sin2nx = 4p ____________ 2n dx 0 sin x + cos2nx
...(i)
1
p/2
(
)
sin2nx + cos2nx 2I = 4p Ú ____________ dx 0 sin2nx + cos2nx p/2
= 4p Ú dx
fi
p = 4pp __ = 4p 2 2 I = 2p 2 p
23. Let
(
)
2x sin x = 0 + Ú ________ dx – p 1 + cos2x x sin x = 4 Ú ________ dx 2 0 1 + cos x (p – x)sin(p – x) = 4 Ú ______________ dx 0 1 + cos2x (p – x)sinx = 4 Ú _________ dx 2 0 1 + cos x p
p sinx xsinx = 4 Ú ________ dx – 4 Ú ________ dx 2 2 0 1 + cos x 0 1 + cos x p sin x = 4 Ú ________ dx – I 2 0 1 + cos x p
p sinx 2I = 4 Ú ________ dx 2 0 1 + cos x p
psinx I = 2 Ú ________ dx 2 0 1 + cos x
sinx I = 4p Ú ________ dx 2 0 1 + cos x
= – 4p (tan–1 (cosx))2
= – (–1 – 1)
25. Given
=2
p __
(
0
p = – 4p 0 – __ 4 = p2
)
(
2
)
4
(
2
)
2x esin(x ) = Ú _______ dx 2 1
esin t = Ú ____ dt, x2 = t t 1
esin x = Ú ____ x dx 1
= (F(x))|16 1
= F(16) – F(1)
fi
k = 16
26. Given
a + b = 4.
Let
f (a) = Ú g(x) dx + Ú g(x) dx
16
( )
16
( )
b
0
0
a
4 – a
0
0
= Ú g(x) dx + Ú g(x) dx
df (a) fi _____ = g(a) – g(4 – a) da df (a) fi _______ = g(a) – g(4 – a) d(b – a) df (a) fi ________ = g(a) – g(4 – a) d(4 – 2a)
p/2
a
p
p
= – (cos(37 p) – 1)
p
2x 2x sin x = Ú ________ 2 dx + Ú ________ dx – p 1 + cos x – p 1 + cos2x
p
2esin(x ) Now, Ú ______ x dx 1
p
= – (cos t)|37p 0
4
2x(1 + sinx) I = Ú __________ dx – p 1 + cos2x
p
( )
p
(Let p ln x = t)
0
esinx fi Ú ____ x dx = F(x)
p
= Ú sin tdt,
d esinx ___ (F(x)) = ____ x , x > 0 dx
0
)
37p
...(ii)
Adding Eqs (i) and (ii), we get
(
e p sin(p lnx) 24. Given integral = Ú _________ dx x
fi
df (a) – _____ = g(a) – g(4 – a) 2da
df (a) _____________ g(a – 4) – g(a) fi _____ = >0 2 da Thus,
a
b
0
0
f (a) = Ú g(x) dx + Ú g(x) dx
It increases as (b – a) increases.
Definite Integrals
27. Given integral
x
g (x) = Ú cos4t dt 0
Now,
g(x + p)
= Ú cos t dt
0
x
= Ú cos t dt + Ú cos t dt 4
x
0
28. Given
= Ú cos t dt + Ú cos t dt 4
0
= g(x) + g(p)
x
fi
(1 + x) f (x) = 1
fi
1 f (x) = _____ 1+x 1 f (1) = __ 2
1
–1
= Ú (x – [x]) dx 1
1
(
31. Let
–1
= 0 – Ú ([x]) dx
p = __ – 2 Ú tan–1 x dx 2 0 1 p 1 __ = – 2 xtan–1 x – __ log|x2 + 1| 2 2 0 p p 1 = __ – 2 __ – __ log2 2 4 2 p p = __ – __ + log2 2 2 = log 2
( (
0
1
–1
0
0
1
dx I = Ú _______ ...(i) 1 + cosx p/4
–1
0
dx = Ú _______ p/4 1 – cosx 3p/4
3p/4
= Ú dx
(
( )
2 = Ú _____ 2 dx p/4 sin x
= Ú (2cosec2x) dx
3p/4
= (x)|0–1
30. Given,
= [0 – (– 1)] = 1
p/4
3p ___ 4 – (2cot x)|p __ 4
=
1 Ú tan–1 ________ dx 0 1 – x + x2
= – 2(–1 – 1)
1
=4
(
)
)
x – (x – 1) = Ú tan–1 ___________ dx 1 + x(x – 1) 0
...(ii)
fi
)
1 1 2I = Ú _______ + _______ dx 1 + cosx 1 – cosx p/4
–1
)
Adding Eqs (i) and (ii), we get
= – Ú (–1) dx – Ú (0) dx
(
)
3p/4
= – Ú ([x]) dx – Ú ([x]) dx
1
)
3p/4
= Ú (x) dx – Ú ([x]) dx
0
0
p 1 = __ – Ú tan–1 ________ dx 2 0 1 – x + x2
–1
= 2 Ú tan–1 x dx
1
0
1
– 1
1
= Ú tan–1 x dx – Ú tan–1 (x) dx
p = __ – Ú cot–1 (1 – x + x2) dx 2 0
–1
0
1
1
1
0
1
Ú f (x) dx
1
0
f (x) = 1 – x f (x)
0
1
1
1
fi
29. Given,
0
Also, Ú tan–1 (1 – x + x2)
Ú f (t) dt = x + Ú t f (t) dt
fi
1
1
x
0
1
0
4
0
0
= Ú tan–1 x dx – Ú tan–1 (– x) dx
p
x
0
= Ú tan–1 x dx – Ú tan–1 [(1 – x) – 1] dx
x + p
4
1
= Ú tan–1 x dx – Ú tan–1 (x – 1) dx
x + p
4
1
2.119
I=2
2.120 Integral Calculus, 3D Geometry & Vector Booster 32. Given,
2
fi
p/2
Adding Eqs. (i) and (ii), we get
Ú [2sinx]dx
5p/6
p
p/2
5p/6
= Ú [2sinx] dx + Ú [2sinx] dx 7p/6
7p/6
p
3p/2
(
)
(
)
p 7p __ ◊ 1 + (0) + ___ – p ◊ (–1) 2 6
5p = ___ – 6
(
)
3p 7p + ___ – ___ ◊ (– 2) 2 6 2p __ p ___ 4p ___ = – – 6 6 6 p __ = – 2
( Ú ( p
33. Let
...(i) ...(ii)
Adding Eqs (i) and (ii), we get
(
3
– 2
34. Given
)
2
ecosx sinx dx – 2 ____________
3
= + Ú 2dx odd function 2
= 0 + 2(3 – 2)
=2
| |
= Ú tdt, loge x = 1
( )
t2 2 = __ 2 –1
x
1 = 2 – __ 2
0
3 = __ 2
= p I = __ 2
g(x) = Ú f (t) dt
37. We have,
g(2) = Ú f (t) dt 1
2
0
1
lnt f (x) = Ú ____ dt 1 +t 1
Now,
lnt 1 f __ x = Ú ____ dt 1 +t 1
= Ú f (t) dt + Ú f (t) dt
1 Now, __ £ f (t) £ 1 2
( )
...(i)
1/x
1 1 Let t = __ y fi dt = – __2 dy y The given integral reduces to
1
1 fi Ú __ dt £ Ú 1 ◊ dt 02 0 1
1 fi __ £ Ú f (t) dt £ 1 2 0 1 Also, 0 £ f (t) £ __ 2 2
( ) x
0
2
– 2
–1
(x)p0
1
3
= Ú ecos x sinx dx + Ú 2dx
2
= Ú dx
2
2
2
fi
– 2
Ú
0
fi
3
loge x 36. Given Ú _____ x dx e–1
p
2
= Ú f (x) dx + Ú f (x) dx
e2
ecos x + e– cos x 2I = Ú ___________ dx e– cos x + ecos x 0
2
2
e– cos x = ___________ – cos x + ecos x 0 e p
1
35. Given Ú f (x) dx
) )
p
2
3 1 fi __ £ Ú f (t) dt + Ú f (t) dt £ __ 2 0 2 1
ecos x I = Ú ___________ cos x + e– cos x 0 e
1
...(ii)
1 1 __ £ Ú f (t) dt + Ú f (t) dt £ 1 + __ 2 0 2 1
+ Ú [2sinx]dx + Ú [2sinx] dx
1 0 £ Ú f (t) dt £ __ 2 1
3p/2
...(i) 2
1 fi Ú 0 dt £ Ú f (t) dt £ Ú __ dt 1 1 1 2
y
( ) ( )
1 ln __ y dy = Ú _______ × – ___2 1 1 1 + __ y y y
–ln(y) dy = Ú _____ × – ___ y 1 1 + y
Definite Integrals y
ln(y) = Ú _______ dy y(1 + y) 1 t
ln(t) = Ú ______ dt t(1 + t) 1
...(ii)
x
( )
3x2 3 F¢(x2) = ___ + 1 = ___ + 1 2x 2x
fi
3 f (x2) = ___ + 1 2x
Putting x = 2, we get
Adding Eqs (i) and (ii), we get
fi
x
ln(t) ln(t) 1 f (x) + f __ x = Ú ______ dt + Ú ______ dt 1 (1 + t) 1 t(1 + t)
fi
3 f (4) = __ ◊ 2 + 1 = 4 2
40. **
x
x
(1 + t)ln(t) = Ú _________ dt t (1 + t) 1
41. Given
ln(t) = Ú ____ dt t 1
(ln t)2 (ln x)2 = _____ = _____ 2 2 1
|
x
( ) ( )
lne 1 1 f (e) + f __ e = ___ = __ . 2 2 2
( ) Ú ( ) Ú ( ) p
38. Let
2
cos x I = Ú _____ dx x –p 1 + a
...(i)
x2 – f ¢(x) = 0
fi
x2 = ÷2 – x2
fi
x4 = 2 – x2
fi
x4 + x2 – 2 = 0
fi
(x2 + 2)(x2 – 1) = 0
fi
x = ± 1
...(ii)
(
)
p
= Ú (cos2x)dx
= 2 Ú (cos2x)dx
1 = __ Ú f (t)dt, 2 6
1 = __ Ú f (t)dt 2 0
1 = __ Ú f (x)dx 2 0
1 = __ × 6 Ú f (x)dx 2 0
= 3 I
6I
1
–p
p
0
fi
(
)
2
1 + cos x = 2 Ú ________ dx 2 0
( (
2
(Let 2x = t)
6I
(1 + ax) cos2x 2I = Ú ___________ dx (1 + ax) –p
p
_____
6 + 6I
ax cos2x = _______ dx 1 + ax –p p
Also,
3
Adding Eqs (i) and (ii), we get fi
f ¢(x) = ÷ 2 – x2
42. Given Ú f (2x)dx
cos2x = ______ dx – x –p 1 + a p
_____
fi
3 + 3T
p
_____
f (x) = Ú ÷2 – t2 dt 1
x
Thus,
2.121
))
p
sin x 1 I = __ x + _____ 2 2 0 p = __ 2
43. Given, 1/2
(
( ) )
1–x Ú [x] + ln _____ dx 1+x –1/2 1/2
1/2
( ( ))
1–x = Ú ([x])dx + Ú ln _____ dx 1+x –1/2 –1/2
= Ú ([x])dx + 0
x
39. Given
F(x) = Ú f (t)dt
fi
F ¢(x) = f (x)
Also,
F(x2) = x2 (x + 1) = x3 + x2
fi
F¢(x2) ◊ 2x = 3x2 + 2x
0
1/2
–1/2
0
1/2
–1/2
0
= Ú ([x])dx + Ú ([x])dx
2.122 Integral Calculus, 3D Geometry & Vector Booster
0
1/2
fi
f ¢(x) = 2x ( e– (x + 2x + 1) – e – x )
–1/2
0
fi
f ¢(x) = 2xe– x ( e– (2x + 1) – 1 )
= Ú (–1)dx + Ú (0)dx = (–1)(x)|0–1/2
(
)|
2 47. Given Ú x f (x)dx = __ t 5 5 0
I(m, n) = Ú t (1 + t) dt n
m + 1
)|
1
1
t n m + 1 = (1 + t)n _____ – _____ Ú t (1 + t)n – 1 dt m+1 0 m+10 n 2n = _____ – _____ I(m + 1, n – 1) m+1 m+1 I = Ú f (sin2x)sin x dx
...(i)
t 2 f (t 2)(2t) = 2t 4
1 f (t 2) = __ t 5 1 = __ ( ) ____ 2 2 ( __ ) 5
4 f ___ = 25
p/2
= Ú f (sin2x)cos x dx
...(ii)
0
Adding Eqs (i) and (ii), we get
2I =
p __ 2 0
Ú
p __ 2
1–x 48. Given Ú _____ dx 1+x 0
( (
)) p Let ( __ – x ) = t 4
p __
=
4 __ ÷2 f (cos2t)(cost)dt
=
4 __ 2÷2 f (cos2t)(cost)dt
Ú
p – __ 4
Ú
p __
fi
fi
=
Ú
__
0 p __ 4
2I = 2÷2 Ú f (cos2x)(cosx)dx 0
__
p __ 4
I = ÷2 Ú f (cos2x)(cosx)dx 0
x2 + 1
46. Given
f (x) = Ú e dt
fi
( (( ))) ( (( )))
Let x = cosq
q sin __ 2 q q _______ = Ú 2sin __ cos __ dq 2 2 q 0 __ cos 2
q = Ú 2sin2 __ dq 2 0
= Ú (1 – cosq)dq
p/2
p/2
( ) ( )
( )
p/2 0
p __
= (q – sinq)|02
p = __ – 1 2
| |
49. Let f (x) be a differentiable function defined as f : [0, 4] Æ R, show that (i) 8f ¢(a) f (b) = {f (4)}2 – {f (0)}2 when a, b Œ (0, 4)
(ii) Ú f (x)dx = 2[a f (a 2) + b f (b 2)] 0
x2
2
)
4
–t2
2
________
q p/2 sin __ 2 = Ú _______ sinq dq q 0 cos __ 2
0
4 __ 2÷2 f (cos2x)(cosx)dx
( ÷
1 – cosq = – Ú ________ sinq dq, 1 + cosq p/2
p __
÷
______
0
__ p = ÷2 Ú f (sin2x) sin x + __ dx 4 0
1
f (sin2x)(sinx + cosx)dx
2 Let t = __ , then 5
0
+
t2
p/2
45. Let
2
Clearly f (x) increases in (0, •).
m
(
4
4
O
1 = (–1) 0 + __ 2 1 __ = – 2 0
4
–
1
44. Given
4
4
f ¢(x) = e – (x + 1) ◊ 2x – e – x ◊ 2x
when 0 < a, b < 2 [IIT-JEE, 2004]
p/3
50. Let
(
)
2
p + 4x3 I = Ú _____________ dx p – p/3 2 – cos |x| + __ 3 p/3
Definite Integrals
(
(
)
(
)
)
(
p/3
Differentiating both sides w.r.t. x, we get x2
3
)
4x + Ú _____________ dx p – p/3 2 – cos |x| + __ 3 p/3
( (
(
)
)
p/3
)
)
)
(
)
(
)
(
( ( (
)
(
)
cosx + cosx ________ (2x) 1 + sin2x
( )
dy (–1) Now, ___ = 0 + (–1)(2p) ◊ _____ 1+0 dx x = p = 2p 52. The given integral can be written as 0
Ú ((x + 1)3 + 2 + (x + 1) cos (x + 1))dx – 2
Let (x + 1) = t 1
= Ú (t3 + 2 + t cost) dt –1 1
1
= Ú (t3 + t cos t) dt + Ú 2dt –1
–1
= 0 + Ú 2dt
53. Given,
= 2(1 – (–1)) = 4
–1
sinx
)
(Let t/2 = y)
fi
0 – sin2x f (sinx) ◊ cosx = – cosx
1 f (sinx) = _____ 2 sin x 1 fi f (x) = __ 2 x 1 1 __ = _____ fi f ___ 2 =3 1 3 ÷ ___ __ 3 ÷ 54. Given integral is = fi
)
sec2y dy = 4p Ú ____________ sec2y + 2tan2y p/6
sec2y = 4p Ú _________ dy 2 p/6 1 + 3tan y
sec2y 4p = ___ Ú _____________ dy __ 3 p/6 ( 1/÷3 )2 + tan2y
p/3
___ 16
1
p/3
__
Ú t2 f (t)dt = 1 – sinx
dy = 4p Ú _________ 2 , p/6 1 + 2sin y p/3
(
dy cos÷q __ ___ = – sinx Ú _________ dq dx 2 1 + sin2÷q p
1
dt = 2p Ú ____________ p/3 1 + 2sin2 (t/2) p/3
(
(
dt = 2p Ú ____________ 1 + (1 – cos t) p/3 2p/3
)
dt = 2p Ú _______ , (Let t = x + p/3) 2 – cos t p/3 2p/3
(
(
dx = 2p Ú _____________ p 0 2 – cos x + __ 3 2p/3
)
p = 2 Ú _____________ dx p 0 2 – cos |x| + __ 3 p/3
)
p = Ú _____________ dx + 0 p – p/3 2 – cos |x| + __ 3
(
)
cos÷q __ y = cosx Ú _________ dq 2 2 1 + sin ÷q p ___ 16
p = Ú _____________ dx p – p/3 2 – cos |x| + __ 3
(
__
2.123
)
)
__
÷ 3
( ) ( )
p
[ ( )
(
)]
1 1 Ú e|cosx| 2sin __ cosx + 3cos __ cosx sin x dx 2 2 0 p
[ ( )] Ú [ ( )] Ú [ ( ) ]
4p = ___ Ú __ __________ __ 2 3 1/÷3 ) + v2 (1/÷3
__ __ __ 4p __ = ___ × ÷ 3 ( tan–1 ( v÷3 )) ÷ 3 1/÷3 3
1 + e|cosx| 3cos __ cosx sinx dx 2 0
51. Given,
x2
(
__
)
cosx cos ( ÷q ) __ dq y(x) = Ú ___________ 2( 2 ) p /16 1 + sin ÷q
1 = Ú e|cosx| 2sin __ cosx sinx dx 2 0 p
p
1 = 0 + e|cosx| 3cos __ cosx sinx dx 2 0
2.124 Integral Calculus, 3D Geometry & Vector Booster
( Ú
)
2a
f (x) dx = 0, f (2a – x) = – f (x) 0
[ ( ) ] [ ( )] [ ( ) ] [ ( )]
p
1 = Ú e|cosx| 3 cos __ cosx sin x dx 2 0
1 = 2 Ú e|cos x| 3 cos __ cos x sin x dx 2 0
p/2
p/2
Ú (1 – x50)100 dx 0 5050 × _____________ = 5051 1 Ú (1 – x50)101 dx
1 = 2 Ú e|cos x| 3 cos __ cos x sin x dx 2 0
1 = 6 Ú ecos x cos __ cos x sin x dx 2 0
0
56. Let
sec2x
[ ( ) ] (Let t = cos x) t = 6 Ú e [ cos ( __ ) ] dt 2 e t t 1 = 6 _____ __ sin __ + cos ( __ ) ] 1[ 2 ( 2 ) 2 __ 1 + 0
1
p/2
( ) ( )
Thus, the value of
t , = – 6 Ú et cos __ 2 1
Ú f (t)dt 2 ________ L= lim p x Æ __ p2 4 x2 – ___ 16
( )
4
( 0
)
t
( ( ))
0
[ ( )
1
In = Ú (1 – x50)n dx
dx (A) Ú _____ 2 –1 1 + x 1
0
Integrating by parts, we get 1
In = (1 – x50)n ◊ x|10 + Ú x ◊ 50n ◊ (1 – x50)n – 1 ◊ x49 ◊ dx
dx = 2 Ú _____ 2 0 x + 1
= 2(tan–1x)10
p p = 2 __ – 0 = __ 4 4
0
1
= 0 + 50n Ú x ◊ (1 – x ) 50
50 n – 1
◊ dx
= – 50n Ú ((1 – x50) – 1) ◊ (1 – x50)n – 1 ◊ dx 0
(
1
0
0
)
= – 50n Ú (1 – x50)n ◊ dx – Ú (1 – x50)n – 1 ◊ dx = – 50n (In – In – 1) = – 50n In + 50n In – 1 fi
In = – 50n In + 50n In – 1
fi
(1 + 50n)In = 50n In – 1
In 50n fi ____ = ________ In – 1 (1 + 50n) Putting
(
)
1
0
1
1
( )
57.
( ) ]
24 1 1 1 = ___ e __ sin __ + cos __ – 1 2 2 5 2 1
55. Let
1
4
)
p p f sec2 __ . sec2 __ . 1 4 4 ___________________ = p __ 4 8f (2) = _____ p
1
t
(
2 f (sec x) 2sec2x tanx ________________ = lim p 2x x Æ __
dx ______ (B) Ú ________ _____ 0 1 ÷÷ – x2
p = (sin–1 x)10 = __ 2 3 dx (C) Ú _____ 2 2 1 – x
(
| | )
x–1 3 1 = – __ log _____ 2 x+1 2
1 1 1 1 = __ log __ – __ log __ 2 3 2 2
n = 101, then we get
( ) 1 2 = __ log ( __ ) 2 3 2
dx (D) Ú _______ _____ 1 x÷x 2 – 1
I101 50 × 101 5050 fi ___ = _____________ = _____ I100 (1 + 50 × 101) 5051
I100 5051 fi ___ = _____ I101 5050
= (sec–1 x)21
= (sec=12 – sec–11)
( )
(
Definite Integrals
58. Given
(
)
f (x) = f (1 – x)
Putting
x = 1/2, we get
1 1 1 f ¢ __ = – f ¢ 1 – __ = – f ¢ __ 2 2 2
( ) ( ) ( ) 1 fi 2f ¢ ( __ ) = 0 2 1 fi f ¢ ( __ ) = 0 2 1 Also, it is given that f ¢ ( __ ) = 0 4
fi
1 __ 2 1 Therefore, f x + __ sinx dx = 0. 2 1 – __ 2 1/2 Again, f (1 – t) esin(p t)dt 1 0 = – f (y) esin(p(1 – y)) dy, (Let 1/2
60.
sin(p(1 – py))
dy
0
= Ú f (y) e
)
(
)
p p 2 = ___ __ × __ = ____ __ 6 3 3÷3 ÷ p__ Tn > ____ 3÷3
( )
1
n
sin(p (1 – p y))
dy
0
Ú ÷
0
k = 1
Ú
__________ 2 1 – [f ¢(x)] =
fi ÷
0
f (x)
fi
1 – [f ¢(x)]2 = ( f (x))2
fi
dy 2 1 – ___ = y2 dx
( ) ( ) ÷
_____ dy fi ___ = ± 1 – y2 dx
dy _____ fi Ú ______ = ± dx 1 ÷ – y2
1 – t = y)
fi sin–1 y = c ± x fi y = sin(c ± x) When x = 0, y = 0, then c = 0 Thus, the equation of the curve is y = sin(± x) = ± sinx As
f (x) ≥ 0 for 0 £ x £ 1, we get
f (x) = sinx, for 0 £ x £ 1
Since
sin x < x for all x > 0, we get
f (x) < x for 0 < x £ 1
1/2
Thus,
0
and
= Ú f (t) esin(p y) dt, by Property I.
k = 0
x _________ 2 Given 1 – [f ¢(t)] dt = f (t) dt 0
( )
Ú
1/2
(
x
Ú
p 2 p __ = ___ __ __ – 3 3 6 ÷
(
( ) ( ) ( ) 1 Thus, f ( x + __ ) is an even function. 2 1 fi f ( x + __ ) is an odd function. 2
( )
( ) )
n – 1 k k h S f __ n > Ú f (x) dx > h S f __ n
1 Replacing x by x + __ , we get 2 1 1 fi f x + __ = f __ – x 2 2
= Ú f (y) e
))
__ 2 1 = ___ __ (tan–1 ( ÷3 ) – tan–1 ___ __ 3 ÷3 ÷
Now,
Thus, f ¢(x) = 0 vanishes at least twice. Again, f (x) = f (1 – x)
(
3 1 f ¢ __ = – f ¢ __ 4 4
1/2
(
( ) ( )
2x +__ 1 1 2 = ___ __ tan–1 ______ 3 ÷3 0 ÷
( ) ( ) 3 1 f ¢ ( __ ) = – f ¢ ( __ ) = 0 4 4
dx __ = Ú _______________ 3 2 ÷ 0 1 2 __ ___ x + + 2 2
1 x = __ in f ¢(x) = – f ¢(1 – x), we get 4
Ú
( )
1
f ¢(x) = – f ¢(1 – x)
)
1 n ___________ 1 __ 59. We have Sn < lim S = lim S 2 n Æ • n n Æ • n k = 1 k k __ 1 + n + __ n 1 dx = Ú ________ 2 0 1 + x + x
p = __ – 0 3 p = __ 3
fi
Putting
2.125
( ) ( )
1 f __ < 2 1 f __ < 3
1 __ 2 1 __ . 3
2.126 Integral Calculus, 3D Geometry & Vector Booster
( Ú ( p
61. Let
) )
sin n x In = Ú ___________ dx x –p (1 + p ) sinx p
sin n x = __________ dx x –p (1 + p )sinx
( Ú ( Ú ( p
...(i)
)
sin n (– x) = Ú ______________ dx – x –p (1 + p )sin(– x)
p
0
) )
sin n (x) = ____________ dx x –p (1 + p )sin(x)
p x sin n (x) = ____________ dx x –p (1 + p )sin(x)
...(ii)
Adding Eqs (i) and (ii), we get p
(
( p
p
fi
)
)
(
)
sin n (x) = 2 Ú _______ dx sin(x) 0
(
)
sin n (x) In = Ú _______ dx sin(x) 0
( Ú ( p
\
x = 0, y = 0, then A = 0 y = Ae x = 0 f (x) = 0 f (ln 5) = 0
When Thus, fi fi
) )
sin(n + 2)x – sin n x In + 2 – In = Ú _________________ dx sinx 0 p
2cos(n + 1)x × sin x = __________________ dx sinx 0
= Ú 2 cos(n + 1)x dx
(
)
p
sin(n + 1)x = 2 __________ = 0 n+1 0
fi
In + 2 = In
Now,
sinx I0 = 0 and I1 = Ú ____ dx = p sinx 0
Since and
In + 2 = In, so I1 = I2 = I5 = ... = I(2n – 1) = p I0 = I2 = I4 = ... = I2n = p
p
( )
Thus, S I2m + 1 = I3 + I5 + ... + I21 = 10p 10
fi
f ¢(0) – f (0) = ÷ 0 + 1 = 1
fi
f ¢(0) = f (0) + 1
fi
f ¢(0) = 2 + 1 = 3
Let
f –1 = g
y = f (x) ¤ x = g(y)
fi
f ¢(x) = f (x)
_____
when g is the inverse of f, then we shall use 1 g¢(y) = ____ f ¢(x) Putting
y = 2 and x = 0, then
1 1 g¢(2) = ____ = __ f ¢(0) 3
( Ú ( Ú ( Ú ( Ú ( Ú ( Ú ( 1
)
x4 (1 – x)4 64. Let I = Ú ________ dx 0 1 + x2
) )
1
x4 (x4 – 4x3 + 6x2 – 4x + 1) = _______________________ dx 0 x2 + 1
(x8 – 4x7 + 6x6 – 4x5 + x4 ) = _______________________ dx 0 x2 + 1
(x8 + 6x6 + x4 ) = _____________ dx 0 x2 + 1
(x6 (x2 + 1) + x4 (x2 + 1) + 4x6) = __________________________ dx 0 x2 + 1
(x2 (x2 + 1)(x2 + 1) + 4x6 ) = ______________________ dx 0 x2 + 1
4(x6 + 1 – 1) = (x4 + x2 ) + ___________ dx 0 x2 + 1
1
1
)
)
1
1
f (x) = Ú f (t) dt 0
f ¢(x) – e f (x) = Rx4 + 1
x = 0, we get
m = 1
x
–x
Putting
and S I2m = I2 + I4 + ... + I20 = 0 62. Given,
– x
e
10
m = 1
_____
fi
p
0
x
e– x f (x) = 2 + Ú ÷t 4 + 1 dt 0
sin n (x) = Ú _______ dx sin(x) –p
f (0) = Ú f (t)dt = 0
63. Given
(1 + p x)sin n (x) 2 In = Ú _____________ dx (1 + p x)sin(x) –p p
But
0
p
dy fi ___ = y dx dy fi Ú ___ y = Ú dx fi log|y| = x + C fi y = e x + C = Aex
1
) )
Definite Integrals
( Ú ( 1
)
2.127
p/2
8 2 2 = __ p Ú0 ((3 – 4 sin 3q) × (3 – 4 sin q))dq
4 = Ú (x + x ) – _____ + 4(x4 – x2 + 1) dx 2 0 x +1
4 = 5x4 – 3x2 + 4) – _____ 2 dx 0 x +1
8 = __ p Ú [{3 – 2(1 – cos 6q) × 3 – 2(1 – cos2q)}] dq 0
= (x5 – x3 + 4x – 4 tan–1 x)10
= (4 – 4 tan–1 (1)) = (4 – p)
8 = __ p Ú0 [(1 + 2 cos 6q) × (1 + 2 cos 2q)]dq
4
2
1
p/2
)
p/2
p/2
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q + 4 cos 2q cos 2q)dq 0
65. The given limit is
(
x
)
1 t ln(1 + t) lim __ Ú ________ dt x Æ 0 x3 0 t4 + 4
(
p/2
x
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q)dq 0
)
p/2
t ln(1 + t) Ú ________ dt 0 t4 + 4 = lim ___________ x Æ 0 x3 x ln(1 + x) _________ x2 + 4 _________ = lim x Æ 0 3x2
(
)
( (
8 + __ p Ú0 (4 cos 6q cos 2q)dq p/2
8 = __ p Ú0 (1 + 2 cos 2q + 2 cos 6q)dq p/2
16 + ___ p Ú0 (2 cos 6q cos 2q)dq
)
x ln(1 + x) = lim __________ x Æ 0 3x2 (x4 + 4)
ln(1 + x) = lim _________ x Æ 0 3x(x4 + 4)
p/2
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q)dq 0 p/2
)
16 + ___ p Ú0 (cos 8q + cos 4q)dq
( )
( (
( f (x) is an even function)
9x p sin ___ 2 4 = __ p Ú _______ dx x __ 0 sin 2
10
Thus, Ú f (x)cos(p x)dx –10
10
( ) sin 9q sin 3q 8 = __ p Ú ( _____ × _____ sin 3q sinq ) p/2
p/2
2
= 2 ◊ 5 Ú f (x)cos(p x)dx 0
2
)
sin 3 ◊ (3q) sin 3q 8 = __ p Ú ________ × _____ dq sin 3q sinq 0
(
= 2 Ú f (x)cos(p x)dx 0
0
p/2
)
and f is periodic with period 2. Also, since [– x] = 1 – [x] for non Integral values of x, so, f is an even function.
sin 9q 8 = __ p Ú _____ dq sin q 0
(
)
Ï1 – x : 0 £ x < 1 67. We have, f (x) = Ì Óx – 1 : 1 £ x < 2
p
p/2
p __
= 4
p
( (( )))
)
(
8 p 16 = __ p __ + 0 + ___ p (0 + 0) 2
2 __ Ú f (x)dx p – p 4 = __ p Ú f (x)dx, 0
p __
sin6q 2 ___ sin 4q 2 8 16 sin 8q _____ = __ p q + sin2q + _____ + p _____ + 3 0 8 12 0
ln(1 + x) ________ x = lim ________ x Æ 0 3(x4 + 4) 1 = ___ 12 66. The given integral is
= 10 Ú f (x)cos(p x)dx
where
= 10(I + J ),
I = Ú f (x)cos(p x)dx
0
)
(3 sin 3q – 4 sin33q) (3 sinq – 4 sin3q) 8 = __ p Ú ________________ × ______________ dq sin 3q sinq 0
1
0
2.128 Integral Calculus, 3D Geometry & Vector Booster 2
ln3
J = Ú f (x)cos(p x)dx
and
1
1
ln3
0
= Ú (1 – x)cos(p x)dx
1 = __ Ú dt 2 ln2 1 = __ (ln3 – ln2) 2
1
= Ú [1 – (1 – x))cos(p (1 – x)] dx 0
1
= – Ú x cos(p x)dx
2
J = Ú (x – 1)cos(p x)dx
fi
1 1
= Ú t cos(p (t + 1))dt,
0
(Let x – 1 = t)
1
6f (x) = 3f (x) + 3x f ¢(x) – 3x2 3f (x) = 3x f ¢(x) – 3x2 f (x) = x f ¢(x) – x2 dy fi y = x ___ – x2 dx y__ ___ dy fi x = – x dx dy y fi ___ – __x = x dx
= – Ú xcos(p x)dx 0
Thus,
=I I=J 1
= – Ú x cos(p x)dx
(
0
)
1
1 x 1 __ = – __ p sin(p x) 0 + p Ú0 sin(p x)dx 1 = – ___2 (cosp x)10 p 2 ___ = 2 p
Therefore,
which is a linear differential equation. dx ___ 1 So, IF = e – Ú x = e– logx = __ x Hence, the solution is y__ 1 x = Ú __2 dx + c x y 1 fi __x = – __ x + c when x = 1, y = 2, then c = 3 Thus, the equation of the curve is y 1 __x = – __ x + 3 Now, when x = 2, then y = –1 + 3x = –1 + 6 = 5
10
Ú f (x)cos(p x)dx –10
= 10 (I + J) = 20 I 2 = 20 × ___ 2 p 10
p2 Now, ___ × Ú f (x)cos(p x) 10 10 p 2 40 = ___ × ___ 10 p 2 =4 ___
(
÷ln3
5/6
p2 70. Given integral = ___ Ú sec(p x)dx ln 3 7/6
)
x sin(x2) 68. Let I = ___ dx Ú __________________ 2 2 ÷ln2 sin x + sin(ln(6 – x )) ln3
(
x
fi fi fi
0 1
69. Given, 6 Ú f (t)dt = 3x f (x) – x3
1
= – Ú t cos(p t)dt
( ( )) 3 1 I = __ (l n ( __ ) ) 4 2
3 1 = __ ln __ 2 2
0
Also,
)
ln3
0
(
sin(t) + sin(ln – 6) 1 2 I = __ Ú ________________ dt 2 ln2 sin(ln – 6) + sin(t)
1
)
Adding Eqs (i) and (ii), we get
I = Ú f (x)cos(p x)dx
Now,
(
sin(ln – 6) 1 = __ Ú _______________ dt 2 ln2 sin(ln – 6) + sin(t)
5p/6
p2 1 = ___ × __ Ú sec t dt ln 3 p 7p/6
p2 1 = ___ × __ Ú sec x dx ln 3 p 7p/6
Let x2 = t fi x dx = 1/2 dt
)
sin t 1 = __ Ú _______________ dt 2 ln2 sin t + sin(ln6 – t)
...(i)
5p/6
...(ii)
2.129
Definite Integrals 5p/6
1
p = ___ × Ú sec x dx ln3 7p/6
= 0 – 0 – 12[0 – 0] + 12 Ú 2x(1 – x2)5 dx
(1 – x2)6 1 1 = 12 – _______ = 12 0 + __ = 2 6 6 0 3x2 I = _____ x dx – 2 1 + e
( ) ( ) Ú ( ) Ú ( ) Ú ( ) 0
| (
5p/6 p = ___ (log|secx + tanx|) 7p/6 ln3
p 2__ ___ 1 2__ ___ 1 = ___ log – ___ – __ – – ___ + __ ln3 ÷3 ÷ 3 3 ÷3 ÷
74. Let
p 2 = ___ log ___ __ ln3 ÷3
3x2 = ______ dx – x – 2 1 + e
3ex x2 = _____ dx x – 2 1 + e
71. Given,
| | | )
( (
2
)|)
2
2
b
Ú ( f (x) – 3x)dx = a2 – b2 a
3 fi Ú f (x) dx = __ (b2 – a2) – (b2 – a2) 2 a
2
1 fi Ú f (x)dx = __ (b2 – a2) 2 a
fi
p p f __ = __ 6 6
2
–2
2
2
[
(
)]
p+x = Ú x2 + ln _____ cos x dx p – x –p/2
p __ 2 = x2 p – __ 2
[
(
)]
p+x + log _____ cos x dx p – x
= Ú (x2)cos x dx p – __ 2
=2
p __ 2 (x2)cos x 0
Ú
Ú
[ (
p2 = ___ – 4 sq.u. 2
0
( Ú (
)
2
x[x2] 77. I = Ú _________ dx –1 2 + [x + 1]
d2 73. Ú 4x ___ 2 (1 – x2)5 dx 0 dx 3
)
)
2
1
1 d d = 4x3 ___ (1 – x2)5 – Ú 12x2 ___ (1 – x2)5 dx dx dx 0 0
= (4x3 × 5(1 – x2)4 × (– 2x))10
[
f (x) = ax2 + bx
fi 2a + 3b = 6 fi (a, b) = (3, 0) and (0, 2) Thus, the number of polynomials is 2.
1
( Nr = Odd × Even = Odd function)
1
)
(
= 0,
Ú f (x)dx = 1
2sin x)p/2 0
)
p = 2 ___ – 2 4
Now, it is given that,
dx + 0
(
)]
+x dx Ú cos2x ◊ log 1_____ 1 –x –1/2 = _____________________ 1/2 1+x Ú cos2x ◊ log _____ dx 1–x 0
76. Let
= 2 (x sinx + 2x cos x – 2
( ) ( )
1/2
p __ 2 p+x + log p_____ cos x dx – x p – __ 2
2
(
0
= (x3)20 = 8
75. The given integral
p __ 2
0
I = Ú (3x2)dx
fi
= 2 Ú (3x2)dx
72. The given integral
Ú
)
= Ú (3x2)dx
( ) p/2
(
3(ex + 1)x2 2I = Ú _________ dx (1 + ex) – 2
b
f (x) = x
...(ii)
Adding Eqs (i) and (ii), we get
b
fi
...(i)
1
]
– 12 (x2 (1 – x2)5)10 – Ú 2x(1 – x2)5 dx 0
x[x2] = _________ dx – 1 2 + [x + 1] 0
(
)
1
(
)
__
(
÷2
)
x ◊ 0 x ◊ 0 x ◊ 1 = Ú ______ dx + Ú ______ dx + Ú ______ dx + 0 –1 2 + (0) 0 2 + (1) 1 2 + (0) __
÷2
( )
x = Ú __ dx 2 1
2.130 Integral Calculus, 3D Geometry & Vector Booster
( )
__
x2 ÷2 1 1 = __ = 2 – __ = __ 4 1 4 4
Thus, 4I – 1 = 0 78. We have
(
1
)
–1 12 + 9x2 a = Ú ( e9x + 3tan x ) _______ dx 0 1 + x2 Let 9x + 3tan–1 x = t
(
)
)
a = Ú et dt = ( e
3p 9 + ___ 4 –
fi
3p 9 + ___ 4
3p log(a + 1) = 9 + ___ 4 3p fi log|(a + 1)| – ___ = 9 4 79. We have f (x) = 7 tan8x + 7 tan6x – 3 tan2x fi
(
)
= 7 tan6x(tan2x + 1) – 3 tan2x(tan2x + 1)
= (7 tan6x – 3 tan2x)(tan2x + 1) 6
2
2
= (7 tan x – 3 tan x)sec x p/4
(i) Ú f (x)dx 0
Ú
= Ú (7t6 – 3t2)dt
6 1 1 2 1 = 0 – __ – __ = – 4 – ___ = ___ = ___ 24 24 12 6 4
0
7
7t = ___ – 7
fi
2 £ 2 + sin4(p x) £ 3
1 1 1 fi __ £ __________ £ __ 3 2 + sin4(p x) 2 192x3 __________ 192x3 192x3 _____ fi _____ £ £ 3 2 2 + sin4(p x) 192x3 192x3 fi _____ < f ¢ (x) < _____ 3 2 x
)
3 1
3t ___ = 0 3 0
0
=
x
x
x
x
1/2
1/2
192x3 192x3 fi Ú _____ dx £ Ú f ¢ (x)dx £ Ú _____ dx 2 1/2 3 1/2 1/2
)
( (
x
x4 x4 64 __ £ Ú f ¢ (x)dx £ 96 __ 4 1/2 1/2 4 1/2
fi
1 1 16 x – ___ £ Ú f ¢ (x)dx £ 24 x4 – ___ 16 16 1/2
(
(
x
)
4
)
(
1
1
)
1 fi Ú 16 x4 – ___ dx £ Ú f (x)dx 16 1/2 1/2
p __ 4
Ú {Ú
))
Ú
( )
fi
Integrating, we get
p __ 4 x f (x) 0 – f (x)dx dx 0 p __ 4 p __ 7 3 4 x tan x – tan x 0 – (tan7x 0 p __ 4 0 – (tan7x – tan3x)dx 0
( Ú
( )
x
x
– 3 tan x)sec x dx
(ii) Ú x f (x)dx
=
)
0 £ sin4(p x) £ 1
2
p/4
(
1
=
)
t 4 1 __ 4 0
t 6 = 0 – __ – 6
1/2
2
(
fi Ú (64x )dx £ Ú f ¢ (x)dx £ Ú (96x3)dx
=
= 0 – Ú (t2 – 1)t3dt
3
p __ 4 (7 tan6x 0
(
– 1)tan3x sec2x dx Let tan x = t
x
Ú
=0–
81. We have
1 )
a + 1 = e
p __ 4 (tan4x 0
0
3p 9 + ___ 4 0
– 1)tan3x dx
12 + 9x2 fi _______ dx = dt 1 + x2
Thus,
Ú
=0–
1
3 fi 9 + _____ 2 dx = dt 1+x
(
p __ 4 (tan4x 0
1
}
Ú
(
)
1 £ Ú 24 x4 – ___ dx 16 1/2 fi – tan3x)dx
(
)
1
x5 x 1 16 __ – ___ £ Ú f (x)dx 5 16 1/2 1/2
(
)
x5 x1 £ 24 __ – ___ 5 16 1/2 1
fi
2.6 < Ú f (x)dx < 3 × 9 1/2
Definite Integrals p __ 2
81. Ans. (a, c) 4p
Ú e (sin at + cos at)dt 0 L = ____________________ p 6
t
Given
4
83. Let
k p
Let
I(k) = Ú e (sin at + cos at)dt 6
t
4
I ¢(k) = e k p (sin6 (ak p ) + cos6 (ak p )) ◊ p k p
= p e
for a = 2 as well as a = 4
Integrating, we get I (k) = e k p + c
Since
I(0) = 0 , we get c = – 1
fi
I (k) = e
x
t2 Ú _____ 4 dt = 2x – 1 0 t + 1
Let
t2 f (x) = Ú __4 + 1dt – 2x + 1 0 t
fi
x2 f ¢(x) = __ 4 – 2 < 0 " x Œ [0, 1] x
fi
(
)
ex x2 cos x = Ú ________ dx 1 + e x p – __
...(ii)
1 2 t __4 + 0 t
Ú
t2 1 fi Ú __ 4 + 1dt < __ 2 0 t fi f (1) < 0 f (x) = 0 exactly one root in [0, 1]
p – __ 2
p __ 2 = (x2 cos x)dx p – __ 2
Ú
= 2 Ú (x2 cos x)dx 0
p __ 2
1dt
1 __ " t Π[0, 1] 2
1
2I = Ú ((ex + 1)x2 cos x/(1 + ex))dx
p __ 2
x
t2 0 £ _____ 4 < t +1
p __ 2
2
I(4) fi ____ = e 4 p – 1/e p – 1 I(1) 82. Given,
f (0) = 1 and f (t) =
)
x2 cos x = Ú ______ dx – x p 1 + e – __
p __ 2
–1
fi
p __ 2
Adding Eqs (i) and (ii), we get
k p
(
...(i)
2
0
fi
)
2
Ú e t(sin6 at + cos4 at)dt 0
(
x2 cos x I = Ú ______ dx 1 + ex p – __
2.131
fi
I = Ú (x2 cos x)dx 0
p p /2 __ 2 sin x) – (2 x sin x)dx 0 0
Ú
2
= (x
__ p 2 = ___ – 2 [x( – cos x)]2 – Ú ( – cos x)dx 0 4 0
__ p 2 = ___ – 2[ – (0 – 0) + (sin x)]2 0 4
p 2 = ___ – 2 4
[
p
]
p /2
p
(
)
Chapter
3
Area Bounded by the Curves
Concept Booster 3.1 Rules
to
Draw Different Types
of
y = x2
Curves
To find the area of the plane curves, first we need to draw the given curves. So my dear friends, you should remember the basic rules to trace the given curve. Find the domain and range of the given function y = f (x). II: Find the point of intersection on the co-ordinate axes. III: Find the intervals of the monotonocity of the function y = f (x). IV: Find the local maximum and the local minimum values of the function y = f (x). V: Find the concavity and the points of inflection of the function y = f (x). VI: Find out whether the function is periodic or not. VII: Find the symmetry as follows:
Rule I: Rule Rule Rule Rule Rule Rule
1 y = __ x
(a) A curve is symmetrical about x-axis, if the powers of y, which occur in its equation, are all even. y2 = x
(c) A curve is symmetrical about the line y = x, if the curve remains unchanged by interchanging x and y in the equation.
(b) A curve is symmetrical about y-axis, if the powers of x, which occur in its equation, are all even.
(d) A curve is symmetrical about the line y = – x if the curve remains unchanged when x and y are replaced by – x and – y, respectively.
3.2 Integral Calculus, 3D Geometry & Vector Booster
(e) A curve is symmetrical in opposite quadrants, if the equation of the curve remains unchanged when x and y are replaced by – x and – y, respectively.
The vertical asymptote to the given curve is x = 1 The oblique asymptote to the given curve is y = 2x + 3.
( ) ( ) 3 = lim ( _____ + 2 ) = 2 x–1
f (x) y where as m = lim ____ x = lim __ x Æ • x Æ • x
x Æ •
and
c= lim (y – m x) x Æ •
3.2 Area
(
)
3x = lim _____ + 2x – 2x = 3 x Æ • x – 1
of the
Cartesian Curve
Rule VIII: Find the asymptotes if any. Asymptote It is a straight line which touches the curve at infinity. There are three kinds of asymptotes.
(i) Vertical Asymptote: The straight line x = a is a vertical asymptote to the curve y = f (x) if at least one of the values of lim [ f (x)] or lim [ f (x)] tends x Æ a– x Æ a+ to + • or – •.
(ii) Horizontal Asymptote: The straight line y = b is a horizontal asymptote to the curve y = f (x), where lim [ f (x)] = b. lim [ f (x)] = b or x Æ + • x Æ – • (iii) Oblique Asymptote: The staright line y = m x + c is an oblique asymptote to the curve y = f (x), where
( )
If f (x) is a single valued and continuous function of x in the interval [a, b]. Let
f (x) ≥ 0
for every x in [a, b], then the area bounded by the curve t = f (x), the x-axis and the ordinates x = a and x = b, (a < b) is represented by
f (x) m= lim ____ x
and
c= lim (f (x) – m x)
Ú y dx = Ú f(x) dx
x y = _____ . x–1
3.3 area of the region bounded by a single curve and the co-ordinate axes
x Æ •
x Æ •
For examples,
1. Let
The vertical asymptote to the given curve is x = 1. The horizontal asymptote to the curve is
(
)
x y= lim _____ = 1 x Æ • x – 1
1 y = _____ . x2 – 1
2. Let
The vertical asymptotes to the given curve are
x = ± 1
The horizontal asymptote to the curve is 1 y= lim _____ 2 = 0 x Æ • x –1
3. Let
(
( ) )
3x y = _____ + 2x . x–1
b
b
a
a
(i) The area bounded by the curve y = f (x), x-axis and the straight lines x = a and x = b (a < b) is given by
b
b
a
a
A = Ú y dx = Ú f (x) dx
Area Bounded by the Curves
(ii) If f (x) £ 0 for all x in [a, b], the area bounded by the curve y = f (x), x-axis and the straight lines x = a and x = b, (a < b) is
b
b
a
a
d
d
c
c
A = Ú x dy = Ú f (y) dy
(v) When the curve x = f (y) crosses y-axis, i.e. changes its sign at y = c between y = a and y = b, its area is given by
A = Ú (– y) dx = |Ú f (x) dx|
(iii) If the graph of y = f (x) crosses x-axis between x = a and x = b at x = c, d, e, f respectively, its area (which lie below the x-axis is negative but the area is taken positive, so we take their modulus) is
c
b
a
c
A = Ú x dy + Ú (– x) dy
= Ú x dy + Ú (x) dy
| | | |
c
b
a
c
c
3.4 Area b
A = Ú y dx a
c
d
a
c
b
b
e
f
+ Ú f (x) dx + Ú f (x) dx + Ú f (x) dx d
c
| | | |
= Ú f (x) dx + Ú f (x) dx e
d
= Ú f (x) dx – Ú f (x) dx a
b
= Ú f (y) dy + Ú f (y) dy
3.3
a
between
c
two curves
(i) Suppose we are given two curves reprsented by y = f (x) and y = g (x), where f (x) ≥ g (x) in [a, b]. Here the points of intersection of these two curves are given by x = a and x = b is obtained by taking common values of y from the given equation of two curves. Hence the required area is b
A = Ú (f (x) – g (x)) dx a
c
e
b
b
d
e
f
+ Ú f (x) dx – Ú f (x) dx + Ú f (x) dx
(iv) The area enclosed by the curve x = f (y), y-axis and the abscissae y = c and y = d, (c < d) is given by
(ii) If the curves y = f (x) and y = g (x) intersect at x = c in [a, b], the required area is c
b
A = Ú ( f(x) – g (x)) dx + Ú (g (x) – f (x)) dx
a
c
3.4 Integral Calculus, 3D Geometry & Vector Booster
(iii) The area bounded by the curves x = f (y) and x = g (y) and the straight lines y = c and y = d where f (y) ≥ g (y) is given by d
A = Ú (f (y) – g (y)).
3.6 Least value
of a
variable area
Example 1. If the area is bounded by the curve x3 f (x) = __ – x2 + b 3
c
and the straight lines x = 0 and x = 2 and x-axis is minimum, find the value of b. Solution: Given fi
x3 f (x) = __ – x2 + b. 3 f ¢ (x) = x (x – 2)
Clearly f (x) is monotonic in (0, 2). Hence for minimum area, f (x) must cross the x-axis at
3.5 Area
of the
region bounded
by the
several curves
If y = f (x) is a strictly monotonic function in (a, b) such that f ¢ (x) = 0, the area bounded by the ordinates x = a and x = b, y = f (x) and y = f (c), where c Œ (a, b), is minimum a+b when c = _____ . 2
0+2 x = _____ = 1. 2
Thus,
1 f (1) = __ – 1 + b = 0. 3
fi
2 b = __ . 3
Exercises Area bounded by a function which changes sign
(Problems based on Fundamentals)
Area of a linear curve
1. Find the area of the region bounded by the curves y = x2 – 2x + 2, x = – 1 and x = 2. 2. Find the area bounded by the curves y = ln x + tan–1 x and the ordinates = 1 and x = 2. 3. Find the area bounded by y = x |sin x|, x-axis and the ordinates x = 0 and x = 2p. 4. Find the area bounded by y = 1 log__1 x and the x-axis between x = 1 and x = 2.
2
5. Find the area of the region bounded by the curve y = e2x – 3ex + 2 and the x-axis.
6. Find the area bounded by y = x3 and x-axis between the ordinates x = – 1 and x = 1.
7. Find the area bounded by the curve y = sin x and the x-axis, for 0 £ x £ 2p.
8. Find the area bounded by the curve y = cos x and the x-axis, for 0 £ x £ 2p.
9. Find the area bounded by the curve y = x (x – 1) (x – 2) and the x-axis. 10. Find the area bounded by the curve y = (x – 1)(x – 2) (x – 3) and the x-axis. Area of a region between two non-intersecting curves 11. Find the area of the region enclosed by the curves y = x2 and y = 2x – x2.
Area Bounded by the Curves
12. Find the area of the region bounded by the parabolas y2 = x and x2 = y. 13. Find the area of the region bounded by the parabolas 4y2 = 9x and 3x2 = 16y. 14. Find the area of the region bounded by the curves
15. 16. 17. 18. 19. 20. 21. 22.
23.
3x2 y = ___ and the line 3x – 2y + 12 = 0. 4 Find the area bounded by the curve x2 = 4y and the straight line x = 4y – 2. Find the area of the region included between the parabolas y2 = 4ax and x2 = 4ay. Find the area of the region enclosed by the parabola y2 = 4ax and the chord y = mx. Find the area of the region bounded by the curve y = x2 + 2, y = x, x = 0 and x = 3. Find the area bounded by the curve y = 2x – x2 and the straight line y = – x. Find the area bounded by the straight lines x = 0, x = 2 and the curves y = 2x – x2, y = 2x. Find the area bounded by the curves y = 6x – x2 and y = x2 – 2x. In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x? Find the area of the region between the circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.
24. Find the area of the region enclosed between the two circles x2 + y2 = 1 and (x – 1)2 + y2 = 1. 25. Find the area of the smaller region bounded by the 2 x2 y x y ellipse __ 2 + __2 = 1 and the straight line __ a + __ =1. b a b 26. 27. 28.
Find the area of the region {(x, y): x2 + y2 < 1 < x + y}. Find the area of the region {(x, y): y2 < 4x, 4x2 + 4y2 < 9}. Find the area of the region {(x, y): x2 + y2 < 2ax, y2 > ax, x > 0, y > 0}
29. Find the area of the region
{
}
2 x2 y x y (x, y): __ 2 + __ 2 £ 1 £ __ a + __ . b a b 30. Find the area of the region bounded by y = 1 + |x + 1|, x = – 2, x = 3, y = 0. 31. Find the area of the region bounded by y = 1 + |x + 1|, x = – 3, x = 3, y = 0. 32. Find the area enclosed by the curves y = |x – 1| and y = – |x – 1| + 1. 33. Find the area of the region bounded by the curves
3.5
f (x) = log x and g (x) = (log x)2. 34. Find the area enclosed by y2 = x2 – x4. 1 35. Find the area bounded by the curve |y| + __ £ e–|x|. 2 36. Find the area of the region f (x, y) = [(x, y): 0 £ y £ x2 + 1, 0 £ y £ x + 1, 0 £ x £ 2]. 37. Find the area bounded by the curves y = x and y = x3. 38. Find the area bounded by the curves y = x (x – 1)2 the y-axis and the line y = 2. 39. Find the area of the region bounded by y = x2 + 1, y = x, x = 0 and y = 2. 40. Find the area of the region bounded by the curves y = loge x and y = 2x, x = 1/2 and x = 2. 41. Find ther area enclosed by the parabola (y – 2)2 = (x – 1), the tangent to the parabola at (2, 3) and x-axis. 42. Find the area enclosed by the loop of the curve y2 = (x – 1) (x – 2)2. 43. Find the area bounded by the curves y = (x – 1)2, y = (x + 1)2 and y = 1/4. 44. Find the area bounded by the curves y = ex, y = e–x and the straight line x = 1. 45. Find the area bounded by the curve y = log x, y = log |x|, y = |log x| and y = |log |x||. 46. Find the area bounded by the curve |x – 2| + |y + 1| = 1. 47. Find the area of the region bounded by the curve [x] + [y] = 3 in first quadrant, where [,] = GIF. 48. Find the area of the region bounded by the curve |x| + |y| = 1 and |x – 1| + |y| = 1. 49. Find the area bounded by the curve
| |
p y = cos–1 (cos x) and |x – p | + y – __ = 2
p __ . 2
50. Find the area bounded by the curve |y| = – (1 – |x|)2 + 5.
__
51. Find the area bounded by the curves y = ÷ |x| and y = |x|. 52. Find the area of the region bounded by the curves |x + y| + |x – y| £ 2. 53. Find the area bounded by the curve y = x and y = x + sin x, where 0 £ x £ p. 54. Let f (x) be a non-negative continuous function such that the area bounded by the curve y = f (x), p p x-axis and the ordinates x = __ and x = b > __ is 4 4 __ p p b sin b + __ cos b + ÷ 2 b , find f __ . 4 2
(
)
( )
3.6 Integral Calculus, 3D Geometry & Vector Booster Area of a region between two intersecting curves
Least value of a variable area
55. Find the area bounded by the curves y = sin x and p y = cos x, x = 0 and x = __ . 2 56. Find the area enclosed by the curves y = 2 – |2 – x| 3 and y = __ . |x| 57. The line y = m x bisects the area enclosed by the lines 3 x = 0, y = 0, x = __ and the curve y = 1 + 4x – x2. 2 Find m. 1 58. Find the area bounded by the curve |y| + __ £ e–|x|. 2 Area of a region by a horizontal strip 59. Find the area bounded between y = sin–1 x and y-axis between y = 0 and y = p/2. 60. Find the area enclosed by the line y = x – 1 and the parabola y2 = 2x + 6. 61. Find the area enclosed by x = – 2y 2 and x = 1 – 3y2. 62. Find the area enclosed by the curves y = tan–1 x and y = cot–1 x and the y-axis. 63. Find the area bounded by the curve x = – y2 + y + 2 and the y-axis. Area of a region between several graphs
x3 71. If the area bounded by y = __ – x2 + a and the straight 3 lines x = 0 and x = 2 and the x-axis is minimum, find the value of a. 72. Find the value of a, for which the area bounded by the curve y = a2 x2 + ax + 1 and the straight lines y = 0, x = 0 and x = 1 the least. 73. For what value of k, the area enclosed by the curves y = x2 – 3 and y = kx + 2 is the least. Find also the least area. 74. If the area bounded by the curve y = x2 + 2x – 3 and the line y = mx + 1 is least, find the value of m. 75. Find the value of a for which the area bounded by x 1 the curve y = __ + __2 , and the lines y = 0, x = a and 6 x x = 2a is least.
(Mixed Problems) 1. The area bounded by the curve y = 4x – x2 and the x-axis is
30 31 (a) ___ sq.u. (b) ___ sq.u. 7 7
64. Find the area of the plane figure bounded by __ 2 y = ÷x , x Œ [0, 1], y = x , x Œ [1, 2]
32 (c) ___ sq.u. 3
and y = Рx2 + 2x + 4, x Π[0, 2].
65. Find the area of the region bounded by the curves 4|y| = |4 – x2 | and |y| (x2 + 4) = 12. 2
2
66. Find the area bounded by the ellipse x + 2y = 2 and the outside of the parabola y = 1 – x2. 67. Find the area common to the circle x2 + y2 = 4 and the ellipse x2 + 4y2 = 9. 68. Find the area enclosed by the curves y = ln (x + e) 1 and x = ln __ y , and x-axis.
( )
69. Find the area enclosed by the curves x2 + y2 = 4, parabola y = x 2 + x + 1 and the curve x x y = sin2 __ + cos __ and x-axis. 4 4 x2 70. Find the area enclosed by the curves __ + y2 = 1 4 x__2 and – y2 = 1. 2
[ ( )
( ) ]
34 (d) ___ sq.u. 3 ______
2. The area under the curve y = ÷3x + 4 between x = 0 and x = 4, is
56 (a) ___ sq.u. 9
64 (b) ___ sq.u. 9
(c) 8 sq.u.
(d) none of these
3. The area bounded by the curve y = x3, x-axis and two ordinates x = 1 to x = 2 equal to
15 (a) ___ sq.u. 2
15 (b) ___ sq.u. 4
17 (c) ___ sq.u. 2
17 (d) ___ sq.u. 4
4. The measurement of the area bounded by the coordinate axes and the curve y = loge x is (a) 1 (b) 2 (c) 3 (d) • 5. If the area bounded by the curves y2 = 4ax and a2 y = mx is __ , the value of m is 3
Area Bounded by the Curves
(a) 2 (c) 1/2
(b) – 2 (d) none of these
6. The area bounded by the parabola y = 4x2, y-axis and the lines y = 1, y = 4 is (a) 3 sq.u.
7 (b) __ sq.u. 5
7 (c) __ sq.u. (d) none of these 3 7. The area enclosed by the curves y = sin x, y = 0, p x = 0 and __ is 2 (a) p (b) 2p (c) 1 (d) 2 8. The area of the region bounded by the x-axis and the p p curve defined by y = tan x – __ £ x £ __ is 3 3
(
__
(a) log ÷2 (c) 2 log 2
)
__
(b) – log ÷2 (d) 0
9. The ratio of the areas bounded by the curves p y = cos x and y = cos 2 x between x = 0, __ and 3 x-axis is __
(a) ÷2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1 10. The area bounded by y = [x] and the lines x = 1 and x = 1.7 is
17 (a) ___ 10
(b) 1
17 (c) ___ 5
7 (d) ___ 10
11. The area bounded by the x-axis and the curve y = sin x and x = 0, x = p is (a) 1 (b) 2 (c) 3 (d) 4 12. The area bounded by the parabola y2 = 2x and the ordinates x = 1, x = 4 is __
__
4÷2 (a) ____ sq.u. 3
28÷2 (b) _____ sq.u. 3
56 (c) ___ sq.u. 3
(d) none of these 2
x2 y 13. The area of the ellipse __ 2 + __ 2 = 1 is a b 1 (a) p ab sq.u. (b) __ p ab sq.u. 2 1 (c) __ p ab sq.u. (d) none of these 4 14. The area of the smaller segment cut off from the circle x2 + y2 = 9 by x = 1 is
__ 1 (a) __ (9 sec–1 3 – ÷ 8 ) 2
(b) 9sec–1 (3) – ÷ 8
(c) ÷8 – 9 sec–1 3
(d) none of these
__
3.7
__
15. The area of the upper half of the circle whose equation is (x – 1)2 + y2 = 1 is given by 2
______
(a) Ú ÷2x – x2 dx 0
2
______
(c) Ú ÷2x – x2 dx 1
1
______
(b) Ú ÷2x – x2 dx 0
p (d) __ 4 __
16. The area bounded by the curves y = ÷ x , 2y + 3 = x and x-axis in the first quadrant is
(a) 9
27 (b) ___ 4
(c) 36
(d) 18
17. The area of the region {(x, y): x2 + y2 £ 1 £ x + y} is
p 2 (a) ___ 5
p 2 (b) ___ 2
p 2 (c) ___ 3
p 1 (d) __ – __ 4 2
(
)
18. The area bounded by the curves y = |x| – 1 and y = – |x| + 1 is (a) 1 __ (b) 2 (c) 2÷2 (d) 4 19. The area of the figure bounded by y = ex, y = e–x and the straight line x = 1 is
1 (a) e + __ e
1 (b) e – __ e
1 1 (c) e + __ e – 2 (d) e + __ e + 2 20. The area bounded by the curves y = loge x and y = (loge x)2 is
(a) 3 – e
(b) e – 3
1 1 (c) __ (3 – e) (d) __ (e – 3) 2 2 21. The area enclosed by the parabolas y = x2 – 1 and y = 1 – x2 is 1 2 (a) __ (b) __ 3 3 8 4 __ __ (c) (d) 3 3 22. The area bounded by the circle x2 + y2 = 4, line ___ x=÷ 3y and x-axis lying in the first quadrant, is
p (a) __ 2
p (b) __ 4
3.8 Integral Calculus, 3D Geometry & Vector Booster p (c) __ (d) p 3 23. The area formed by triangular-shaped region bounded by the curves y = sin x, y = cos x and x = 0 is
__
(a) ÷2 – 1
(c) ÷2
(b) 1
__
__
(d) 1 + ÷2
24. The area between the curve y = cos x and x-axis when 0 £ x £ 2p is (a) 2 (b) 4 (c) 3 (d) 0 25. The area of the greatest rectangle that can be inscribed 2 x2 y in the ellipse __ 2 + __ 2 = 1 is a b ___ (a) ÷ab (b) a/b (c) 2ab (d) ab 26. The area bounded by the curves y = |x – 2|, x = 1, x = 3 and the x-axis is (a) 4 (b) 2 (c) 3 (d) 1 27. The area bounded by lines y = 2 + x, y = 2 – x and x = 2 is (a) 3 (b) 4 (c) 8 (d) 16 28. The area enclosed by the parabola y2 = 4ax and the straight line y = 2ax, is
a2 (a) __ sq.u. 3
1 (b) ___ 2 sq.u. 3a
1 (c) ___ sq.u. 3a
2 (d) ___ sq.u. 3a
(a) 26a2
(b) 8a2
2
2
26a 104a (c) ____ (d) _____ 3 3 30. The area bounded by the curves y = In x, y = In |x|, y = |In x| and y = |log |x|| is (a) 4 sq.u. (b) 6 sq.u. (c) 10 sq.u. (d) none of these 31. The area bounded by the curves y = |x – 1| and y = 3 – |x|, is (a) 2 sq.u. (b) 3 sq.u. (c) 4 sq.u. (d) 1 sq.u. 32. The area of the region lying inside x2 + (y__ – 1)2 = 1 and outside c2 x2 + y2 = c2, where c = (÷2 – 1) is
__
p (a) (4 – ÷ 2 ) __ + 4
1 ___ __ 2 ÷
__
p (b) (4 + ÷2 ) __ – 4
( )
(a) 2
(b) 1
(c) 4
(d) none of these
34. The area of the region formed by x2 + y2 – 6x – 4y 5 + 12 £ 0, y £ x and x £ __ is 2 __ __ ______ 3 + 1 3 – 1 p ÷ p ÷______ __ __ (a) – (b) + 8 4 8 6 __
– 1 p ÷3 (c) __ – ______ (d) none of these 8 6 35. Let f (x) = maximum [x2, (1 – x)2, 2x (1 – x)] where 0 < x < 1. The area of the region bounded by the curves y = f (x), x-axis, x = 0 and x = 1 is
17 14 (a) ___ (b) ___ 27 27 19 (c) ___ (d) none of these 27 36. The area enclosed by the parabola ay = 3(a2 – x2) and x-axis is (a) 4a2 sq.u. (b) 12a2 sq.u. 3 (c) 4a sq.u. (d) none of these 37. The area bounded by the curve y = k sin x between x = p and x = 2p, is (a) 2k sq.u. (b) 0
29. The area bounded by the curves x = at2, y = 2at and the x-axis in 1 £ t £ 3, is
__ p 1 (c) (4 + ÷ 2 ) __ + ___ __ (d) none of these 4 ÷ 2 33. The area enclosed between the curves y = loge (x + e), 1 x = loge __ y and the x-axis, is
1 ___ __ 2 ÷
k 2 (c) ___ 2
(d) k sq.u.
38. Let f (x) be a non-negative continuous function such that the area bounded by the curve y = f (x), p p x-axis and the ordinates x = __ and x = b > __ is 4 4 __ p p b sin b + __ cos b + ÷2 b , then f __ is 4 2
( )
(
)
( ) p (c) ( __ + ÷ 2 – 1 ) 4
( ) p (d) ( __ – ÷ 2 + 1 ) 4
__ p (a) 1 – __ – ÷ 2 4 __
__ p (b) 1 – __ + ÷ 2 4 __
39. The area in the first quadrant between x2 + y2 = p 2 and y = sin x is
(p 3 – 8) (a) _______ 4 (p3 – 16) (c) ________ 4
p 3 (b) ___ 3 (p 3 – 8) (d) _______ 2
40. For 0 £ x £ p, the area bounded by y = x and y = x + sin x, is
Area Bounded by the Curves
(a) 2 (b) 4 (c) 2p (d) 4p 41. The area bounded by y = x sin x and x-axis between x = 0 and x = 2p, is (a) 0 (b) 2p sq.u. (c) p sq.u. (d) 4p sq.u. 42. The area bounded by the lines y = |x – 2|, |x| = 3 and y = 0 is (a) 13 unit2 (b) 5 unit 2 2 (c) 9 unit (d) 7 unit2 43. The area bounded by the curve x2 = ky, k > 0 and the line y = 3 is 12 unit2. Then k is
__
3 (c) __ (d) none of these 4 44. The area of the portion enclosed by the curve __ __ __ x + ÷ y = ÷a and the axes of reference is ÷
a2 (a) __ 6
e 2 (a) ___ (b) e2 2 (c) 2e2 (d) 1. 50. The area bounded by eln (x + 1), |x| £ 1, is (a) 1 (b) 2 (c) 4 (d) none __ 51. The area common to the region determined by y ≥ ÷x and x2 + y2 < 2 has the value
52.
(b) 3÷3
(a) 3
(b) a2
a2 a2 (c) __ (d) __ 2 4 45. The area bounded by the curve x = cos–1y and the lines |x| = 1 is
53. 54.
(a) sin 1
(b) cos 1
(c) 2 sin 1 (d) 2cos 1 46. The area enclosed between the curves |y| = 1– x2 and x2 + y2 = 1 is
3p – 8 (a) ______ 3 2p – 8 (c) ______ 3
)
1 (c) 2 e + __ e
(d) none
3 56. The area bounded by the curve y = __ and |x| y = 2 – |x – 2| is
(
y = |x – p|, is (a) p 2 (b) 2p 2 (c) p 2/2 (d) none 49. The area bounded by the curves ÏÔ x1/ln x y = Ì ÔÓe
)
p 1 (c) __ – __ (d) none 4 6 The area of the region bounded by ||x| – |y|| £ 1 and x2 + y2 in the xy-plane is (a) p (b) 2p (c) 3p (d) 1. The area bounded by 1 £ |x – 2| + |y + 1| £ 2 is (a) 2 (b) 4 (c) 6 (d) none The area bounded by the curve y = tan x + cot x| – |tan x – cot x| and between the lines x = 0 and p x = __ and the x-axis is 2 __ (a) ln 4 (b) ln__ ÷ 2 (c) 2 ln 2 (d) ÷ 2 ln 2 The area is enclosed by the graph of the curve y = ln2 x – 1 lying in the fourth quadrant is
p – 8 (b) _____ 3
( ) (
(
(b) (2p – 1)
2 (a) __ e
1__ B 1, ___ , C (2, 0) such that min {PA, PB, PC} = 1, 3 ÷ the area bounded by the curve traced by the point P is __ __ 3p p (a) 3÷ 3 – ___ (b) ÷ 3 + __ 2 2 __ __ p 3p __ (c) ÷ 3 – (d) 3÷ 3 + ___ . 2 2 48. The area bounded by the curves y = cos–1 (cos x) and
)
55.
(a) p
47. A point P moves inside a triangle formed by A(0, 0),
(
3.9
: x π1 and y = |x – e| is : x =1
)
(
4 (b) __ e
)
(
)
1 (d) 4 e – __ e .
4 – ln 27 (a) ________ (b) 2 – ln 3 3 (c) 2 + ln 3 (d) none 57. The area enclosed by the curve |x + y – 1| + |2x + y + 1| = 1 is (a) 2 (b) 1 (c) 4 (d) 3/2
58. The area of the region {f ( x, y) : x2 + y2 £ 1 |x| + |y|} is (a) p (b) p – 1 (c) p – 2 (d) p – 3 59. The area of the region of the xy-plane is bounded by |x| + |y| + |x + y| £ 1 is (a) 1/2 (b) 3/4 (c) 1 (d) 4 60. The area of the region bounded by the curves y = [x] and y = {x}, where [,] = GIF and {,} = LIF, is
3.10 Integral Calculus, 3D Geometry & Vector Booster (a) 2 (b) 1 (c) 1/2 (d) 4 61. The smaller area is bounded by the curves |x| + |y| = 1 __ __ and ÷ |x| + ÷ |y| = 1, is (a) 1/3 (b) 2/3 (c) 4/3 (d) 5/3. 62. The area bounded by the curves |x| + |y| = 1 and |x – 1| + |y| = 1 is (a) 1/2 (b) 1/4 (c) 3/2 (d) 3/4 63. The area of the region bounded by the curves x = – 2y2 and x = 1 – 3y2 is (a) 4/3 (b) 3/4 (c) 5/3 (d) 3/5 64. The area enclosed by the parabola (y – 2)2 = x – 1 and the tangent to it at (2, 3) and x-axis is (a) 10 (b) 7 (c) 9 (d) 5 65. The area enclosed by the curves (y – x)2 = x3 and the straight line x = 1 is (a) 5/4 (b) 4/5 (c) 2/3 (d) 3/2
(c) 2 (d) 4 70. A normal to the curve x2 + kx – y + 2 = 0 at the point, whose abscissa ‘1’ is paral let to the line y = x. Then (a) the value of k is – 3 (b) Equation of the normal is x – y = 1 (c) The area bounded by the curve the normal and the x-axis is 7/6 sq u. (d) Equation of tangent is x + y + 1 = 0 71. If the area bounded by the curve y = 3x3 + 2x and the lines x = a and y = 0 is unity, then the value of a is/are __ – 1 2 ___ (a) __ (b) __ 3 3 ÷
(More than one options are correct)
1__ (d) ___ 3 ÷ 72. Let the parabola by y2 = 4x and the solpe of the normal by – 1. Then (a) Equation of the normal is x + y = 3 (b) The points of intersections are (1, 2) and (9, – 6) (c) The area bounded by the curve and its normal is 64/3 sq. units (d) None of these
(Problems for JEE-Advanced)
1. Find the area bounded by the curve a2y2 = x2(a2 – x2) in xy-plane by the method of integration. [Roorkee, 1985]
2. Find the area bounded by the curve y = (x – 1)(x – 3) and x-axis lying between the ordinates x = 0 and x = 3. [Roorkee, 1986] 3. Find the area of the region bounded by the curves y = logex and y = sin 4(px) and x = 0. [Roorkee, 1987]
|x|
)
(b) n = 1 (d) m + n + p = 0 x 2 67. Let the curves y = 4ax, y = ax and y = __ a where 1 £ a £ 2. Then
(a) m = 1 (c) m + n + p = 3
(
)
8 1 (a) Area = __ a5 – __ a 5 (b) Max area = 84 (c) Max area = 75 a (d) Points of intersections are __ , 4 and 4 (4a3, 4a3)
(
{
}
)
[ ]
p 1__ 68. Let f(x) = min tan x, cot x, ___ " x Œ 0, __ . Then 2 3 ÷ the are bounded by the curve y = f (x) and the x-axis is p__ m ____ log __ + sq units, where m, n, p Œ N then n p÷3
(
( )
÷
__
2 (c) – __ 3
– x
66. The area bounded by the curve y = |e – e |, the 1 x-axis and x = 1 is em + __n + p sq unit, then e
(
÷
)
(a) m = 4 (b) n = 4 (c) p = 6 (d) m + n + p = 14 69. The area bounded by the curve y = x – x2 and the line y = m x is a/2 then the value of m is/are (a) 1 (b) – 2
4. Find the area between the curve y = 2x4 – x2, the x-axis and the ordinates of the two minima of the curve. [Roorkee, 1988] 5. Find the area of the portion of the circle x2 + y2 which is exterrior to the parabola y2 = 12x. [Roorkee, 1989] 6. Find the area enclosed between the curves 1 y = loge(x + e), x = loge __ y and the x-axis. [Roorkee, 1990] 7. The line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x – x2. Find the value of m. [Roorkee, 1991]
( )
Area Bounded by the Curves
8. Find the area bounded by the curve y = 2x – x2 and the line y = – x. [Roorkee, 1992] 9. Find the area bounded by the curve y = (x – 1)(x – 2) and the axis of x. [Roorkee, 1993] 10. Find the ratio in which the area bounded by the curves y2 = 12x and x2 = 12y is divided by the line x = 3. [Roorkee, 1994] 11. Find the area given by the curves x + y £ 6. x2 + y2 £ 6y and y2 £ 8x. [Roorkee, 1995] 12. Find the area of the region bounded by the curves x2 + y2 – 6x – 4y + 12 £ 0, y £ x, x £ 5/2. [Roorkee, 1996] 13. Indicate the region bounded by the curves x2 = y, y = x + 2 and x-axis and obtain the area enclosed by them. [Roorkee, 1997] 14. Indicate the region bounded by the curves y = x log x and y = 2x – x2 and obtain the area enclosed by them. [Roorkee, 1998] 15. Find the area of the region lying inside x__2 + (y + 1)2 and outside c2x2 + y2 = c2, where c = ÷ 2 – 1. [Roorkee, 1999] 16. Find the area enclosed by the parabola (y – 2)3 = x – 1, the tangent to the parabola at (2, 3) and the x-axis. [Roorkee, 2000] 17. A point P moves inside a triangle formed by A(0, 0), 1 B 1, ___ __ , C(2, 0) such that {PA, PB, PC} = 1, find 3 ÷ the area bounded by the curve traced by P. 18. Find the area bounded by the curve f(x) = cos–1(cos x), 0 £ x £ 2p with the tangent to the curve f(x) = |cos x| at x = p. 19. Find the area enclosed between the curves |y| = 1 – x2, x2 + y2 = 1. 1 20. Find the area bounded by the curves |y| = e–|x| – __ 2 and |x| + |y| = ln2. 21. Find the area enclosed between the parabolas y2 – 2y + 4x + 5 = 0 and x2 + 2x – y + 2 = 0. 22. Find the area enclosed between the smaller arc of the circle x2 + y2 – 2x + 4y – 11 = 0 and the parabola __ 2 y = – x + 2x + 1 – 2÷3 . 23. Find the smaller of the two areas enclosed between the ellipse 9x2 + 4y2 – 36x + 8y + 4 = 0 and the line 3x + 2y = 10. 24. Let f (x) = x2 + 3x + 2 and g(x) be its inverse. Find the area bounded by g(x), the x-axis and the ordinates x = 1 and x = 4. 25. Find the area bounded by the curve g (x), the x-axis, the ordinates x = – 1 and x = 4, where g (x) is the
( )
13 x3 x2 inverse of the function f(x) = ___ + __ + ____ + 1. 24 8 12x
3.11
26. Find the area bounded by the curves y = xex, y = xe–x and the line x = 1. 27. Find the area of the region bounded by |x + 2y| + |x + 2y| ≥ 8 and xy ≥ 2. 28. Find the area enclosed by |x| + |y| £ 3 and xy ≥ 2. 29. Find the area of the region bounded by |x – y| + |x + y| £ 8 and xy ≥ 2. 30. Find the area of the curve enclosed by |x + y| £ 2 2 + y2 ≥ 2.
(Tougher Problems for JEE-Advanced)
1. Find the area enclosed by 2x2 + 6xy + 5y2 = 1.
2. Find the area enclosed by the curve x2/3 + y2/3 = a2/3.
3. A polynomial function f(x) satisfies the condition f(x + 1) = f(x) + 2x + 1. Find f(x), if f(0) = 1. Find also the equations of the pair of tangents from the origin on the curve y = f(x) and compute the area enclosed by the curve and the pair of tangents. 2 4. Find the area bounded by the curve y = xe–x , the x-axis and the line x = c, where y(x) is maximum. 5. For what value of a, the area bounded by the curve y = a2x2 + ax + 1 and the straight lines x = 0, y = 0 and x = 1 is the least? 6. Find the possible value of a for which the parabola y = x2 + 1 bisects the area of the rectangle with vertices (0, 0), (a, 0), (0, a2 + 1), (a, a2 + 1). 7. Find the area bounded by the curve y = xe–x, xy = 0 and x = c where c is the x-coordinate of the curves inflection point. 8. Find the area of the region bounded by |x| + |y| ≥ 1 and x2 – 2x + 1 £ 1 – y2. 9. Find the area bounded by the curves |x + y| £ 1, |y – x| £ 1 and 3x2 + 3y2 ≥ 1. 10. Find the area bounded by the curves |x – 2| + |y – 2| £ 3, x2 – 4x + y + 3 £ 0 and x2 + y2 + 2x – 9 £ 0. 11. Find the area enclosed by the curves (x2 + y2) £ 4 £ 2(|x| + |y|) 4 – x2 12. Find the area bounded by the curves y = _____ and 4 y = 7 – |x|. 13. Find the area bounded by the curves y = x2 + x – 2, y = 2x for which |x2 + x – 2| + |2x| = |x2 + 3x – 2| is satisfied. 14. Find the area bounded by the curves y2 = 4a (x + a) and y2 = 4b (b – x). 15. Find the area bounded by the parabola y = x2 – 2x + 3, the line tangent to it at the point P(2, – 5) and the y-axis.
3.12 Integral Calculus, 3D Geometry & Vector Booster 16. Find the area enclosed between the circle x2 + y2 – 2x + 4y__– 11 = 0 and the parabola y = – x2 + 2x + (1 – 2÷3 ) . 17. Find the area enclosed by the curves a2x2 + b2y2 = 1 and b2x2 + a2 y2 = 1. 1 18. Find the area bounded by the curves |y| + __ £ e–|x| 2 and max {|x|, |y|} £ 2. 19. Find the area enclosed by the curves y = min {x3, |x – 2|, e3–x}, x-axis, y-axis and x = 4. 20. Find the area of the region bounded by the set S = S3 – S1 – S2, where
S1:{(x, y) : x2 + 2y2 £ 2}
S2:{(x, y) : 2x2 + y2 £ 2} 2
__
4÷2 (a) ____ 3 2 7 3
2
4÷2 (b) ____ 2 2 7 3
__
1. Find the area bounded by the curves x y = sin2x + cos __ , x = – 2 and x = 2 and x-axis. 4 2. Find the area bounded by the curves |x| + |y| = 1. 3. Find the area bounded by the curves |x – 2| + |y – 2| = 1. 4. Find the area bounded by the curve eln(x + 1) ≥ |y|, |x| £ 1. 5. Find the area bounded by the curves y = ln |x| and y = 1 – |x|. 6. Find the area bounded by the curves y = 3 – |x| and y = |x – 1|. 7. Find the area bounded by the curves |x| y = __ x , x π 0 and y = x (x – 1) (x – 3). 8. Find the area bounded by the curves y = ex, x = 0, y = e. 9. Find the area bounded by 1 £ |x – 2| + |y + 1| £ 2. 10. Find the area bounded by the curve |y| = sin (2x), where 0 < x < 2p. 11. Find the area enclosed by the curves
( ) ]
__________
|x + y| + |x – y|, £ 4, |x| £ 1 and y £ ÷x 2 – 2x + 1 . 12. Find the area bounded by the curves 4 £ x2 + y2 £ 2 (|x| + |y|). 13. If the area bounded by the curves 1 1 4 y = __ x , y = ______ , x = 2, x = a is ___ __ , find a. 2x – 1 ÷5
__
4÷2 4÷2 (c) ____ (d) ____ 723 733 2. The area of the region bounded by the curves y = f(x), the x-axis and the lines x = a and x = b, where – • < a < b < – 2, is
Integer type questions
[
assage I P Consider the functions defined implicity y3 – 3y + x = 0 on various interval in the real line. If x Œ (– •, 2) » (2, •), the equation implicity defines a unique real valued differentiable function y = f (x). If x Œ (– 2, 2), the equation implicity define a unique real valued differentiable function y = g (x) satisfying g(0) = 0. On the basis of the above information, answer the following questions. __ __ __ 1. If f (– 10÷2 ) = 2÷2 , then f ¢¢(–10÷2 ) is __
and S3:{(x, y) : x + y £ 2
Comprehensive Link Passages (For JEE-Advanced Examinations)
b
x (a) Ú ___________ dx + bf (b) – af (a) a 3((f (x))2 – 1)
x (b) – Ú ___________ dx + bf (b) – af (a) a 3((f (x))2 – 1)
x (c) Ú ___________ dx – bf (b) + af (a) a 3((f (x))2 – 1)
b
b
b
x (d) – Ú ___________ dx – bf (b) + af (a). a 3((f (x))2 – 1)
1
3. The value of Ú g¢(x) is
(a) 2g (–1) (c) – 2g (1)
–1
(b) 0 (d) 2g (1).
Passage II If x = f (y) and x = g (y) be two functional curves, the area bounded by the curves x = f (y), x = g(y), y = c and y = is given by d
Ú |f(y) – g(y)| dy, c
where d > c. d In case of g(y) always left of f(y), required area = Ú c [ f(y) – g(y)]. On the basis of the above information, answer the following questions. 1. The area bounded by y2 = 2x + 1 and x – y – 1 = 0, will be (a) 16/9 (b) 16/3 (c) 8/3 (d) none.
3.13
Area Bounded by the Curves
2. The area bounded by y = loge x, x-axis and y-axis is given by •
(a) Ú loge y dy
1
0
(b) Ú e–y dy –•
0
(c) Ú e y dy
– •
(d) none
3. The area bounded by y = tan–1 x, y = cot–1 x and y-axis is equal to __
(a) log ÷2
(b) log 4
(c) log 2
(d) none
Passage III Conseider the function 1 Ï : x œI Ô x - [ x] f (x) = Ì , 2 ÔÓ0 : x ŒI where [,] = GIF
and if g(x) = max {x2, f(x), [x]}, where – 2 £ x £ 2 On the basis of the above information, answer the following questions. 1. The area bounded by the curves y = f(x), x-axis and the ordinates x = 1/2 and x = 1 is (a) 1/4 (b) 1/8 (c) 1/2 (d) none. 2. The area bounded by y = g(x), x-axis and the straight line x = 1 is (a) 1/4 (b) 1/2 (c) 1/8 (d) none 3. The area bounded by y = g(x), where – 2 £ x £ 2, is (a) 175/48 (b) 275/48 (c) 175/24 (d) 275/24. Passage IV A normal to the curve x2 + kx – y + 2 = 0 at the point whose abscissa 1 is parallel to the line y = x. On the basis of the above information, answer the following questions. 1. The value of k is (a) –3 (b) 1 (c) 0 (d) 2 2. The equation of the normal is given by (a) y = x + 1 (b) y = x – 1 (c) y = – x + 1 (d) y = – x – 1. 3. The area in the first quadrant bounded by the curves, the normal and the x-axis is (a) 13/6 (b) 7/6 (c) 10/3 (d) 43/6.
Passage V
_____
Consider x 2 – 4, g(x) = |x – 2| and h(x) _____ the functions f (x) = ÷ = ÷x – 2 for x Œ R, a function is defined as F(x) = max or min {f (x), g(x), h(x)}. On the basis of the above information, answer the following questions. 1. The area of F (x) = min {f(x), g(x), h(x)} between the co-odinates axes for x < 0 is (a) 2p (b) p (c) 4p (d) none 2. The area enclosed by F(x) = min{ f(x), g(x), h(x)} and F(x) = max{ f(x), g(x), h(x)} for x Œ [0, 2] is (a) p (b) p + 2 (c) p – 2 (d) p – 1. 3. The area bounded by the curves F(x) = min{f (x), g(x), h(x)}, x = 2, y = 0 and x = 3, is (a) 3/2 (b) 1/2 (c) 1 (d) none
Matrix Match (For JEE-Advanced Examination Only)
1. Match the following Columns: Column I
Column II
(A) The area bounded by y = 3x and (P) y = x2 is
2/3
(B) The area bounded by x = 4 – y2 (Q) and the y-axis is
17/12
(C) The area in squares units of the2 (R) region bounded by the curve x = 4y and the line x = 2 and the x-axis is
32/3
(D) The area bounded by y = x3, (S) y = x2 and x = 1, x = 2 is (T)
32/9 9/2
2. Match the following Columns: Column I
Column II
(A) The area enclosed by the curve (y (P) – sin–1 x)2 = x – x2 is
0
(B) The area of the region represented (Q) by __ 2 £ |x + y| + |x – y| £ 2 is ÷
1
(C) If the area bounded by (R) y = x2 – 3 and the line y = ax + 2 attains its maximum value, the value of a is
p
(D) If k is a positive number and the area (S) of the region bounded by the curves y = x – kx2 and k y = x2 attains its (T) maximum value, the value of k is
6 p/4
3.14 Integral Calculus, 3D Geometry & Vector Booster
3. Match the following Columns: Column I
Column II
(A) The area of the figure (P) bounded by y = tan x, p p y = tan2 x, – __ £ x £ __ 2 2 is
4p
(B) The area bounded by (Q) the curve y = x sin x and the x-axis, x = 0 and x = 2p is
p 1 – __ 4
(C) The area bounded by (R) the curves x = tan–1 __ , x = 0, x = p/4 and ÷y y = 0 is
__
2(p + 1) + ÷ 2
(
)
(
Column II
(A) The area bounded by the (P) curves y = |sin x|, x-axis and the lines |x| = p.
2 s.u.
(B) The area bounded by the (Q) curves y = sin –1 x, y-axis, |y| = p/2
1 s.u.
(C) The area bounded by the (R) x2 curves y = ___ + 2 , 64 [,] = GIF, y = x – 1, x = 0 is
8 s.u.
(D) The area bounded by the (S) curves y = |x – 1| and y = 3 – |x|, is
4 s.u.
[
)
)
)
Questions asked in past IIT-JEE Examinations
4. Match the following Columns: Column I
(
(
p (D) The area bounded by (S) log 2 – 2 + __ e 2 the curves f(x) = max {1 + sin x, 1, 1 – cos x}, between the co-ordinates x = – p and x = p is
2. Assertion (A): The area bounded by the curve y = [x] n(n – 1) and the lines x = 0 and x = n is _______ s.u. 2 Reason (R): The area bounded by the curve y = [x] and the lines x = 0 and x = 10 is 45 s.u. 3. Assertion (A): The area bounded by the curves f(x, y) p 1 = {(x, y) : x2 + y2 £ a2, x + y ≥ a} is __ – __ s.u. 4 2 Reason (R): The area bounded by the curves f(x, y) = {(x, y) : x2 + y2 £ 1, x + y ≥ 1,} p 1 is __ – __ s.u. 4 2 4. Assertion (A): The area bouned by the curves y = min {|x + 2|, |x|, |x – 2|} is 2 s.u. Reason (R): The area bounded by the curves y = 1 – |x| and y = |x| – 1 is 2 s.u. 5. Assertion (A): The area of the region bounded by the curves y = |x – 1| and y = 3 – |x| is 4 s.u. Reason (R): The area of the region bounded by the curves y = ln x, y = ln |x|, y = |ln x| and y = | ln|x| | is 4 s.u.
]
Assertion and Reason (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true. 1. Assertion (A): The area bounded by the curve |x| + |y| = 1 is 1 s.u. Reason (R): The area bounded by the curve |x – 2| + |y – 2| = 1 is also 1 s.u.
1. Find the area bounded by the curve x2 = 4 and the straight line x = 4y – 2. [IIT-JEE, 1981] 2. Find the area enclosed by |x| + |y| = 1. [IIT-JEE, 1981] 3. Find the area bounded by the curve y = f(x), x-axis and the co-ordinates x = 1 and x = b is (b – 1) sin (3b + 4). Then f (x) is (a) (x – 1) cos (3x + 4) (b) sin (3x + 4) (c) sin (3x + 4) (x – 1) cos (3x + 4) (d) none. [IIT-JEE, 1982] 4. Find the area bounded by the x-axis, part of the curve 8 y = 1 + __ 2 and the ordinates at x = 2 and x = 4. x If the co-ordinates at x = a divides the area into two equal parts, find a. [IIT-JEE, 1983]
(
)
5. Find the area of the region bounded by the x-axis and the curves defned by
p p Ï ÔÔ y = tan x : - 3 £ x £ 3 Ì Ô y = cot x : p £ x £ p ÔÓ 6 2
[IIT-JEE, 1984] _____
6. Find the area bounded by the curves y = ÷ 5 – x2 and y = |x – 1| [IIT-JEE, 1985] 7. Find the area bounded by the curves x2 + y2 = 4, __
x2 = – ÷2 y and y = x.
[IIT-JEE, 1986]
Area Bounded by the Curves
8. Find the area bounded by the curves x2 + y2 = 25, 4y = |4 – x2| and x = 0 above the x-axis. [IIT-JEE, 1987]
9. Find the area of the region bounded by the curve p y = tan x, tangent drawn to the curve at x = __ and 4 the x-axis. [IIT-JEE, 1988]
10. Find all maxima and minima of the function y = x (x – 1)2 : 0 £ x £ 2. Also, determine the area bounded by the curve y = x (x – 1)2, the y-axis and the line y = 2. [IIT-JEE, 1989] 11. Find the area of the region bounded by the curves ln x y = e x ln x, y = ___ e x and x = 1. [IIT-JEE, 1990] 12. Sketch the curves and identify the region bounded 1 by y = ln x, y = 2x, x = __ and x = 2. Find the area 2 of the region. [IIT-JEE, 1991] 13. Find the area bounded by the curves y = x2 and 2 y = _____ 2 . [IIT-JEE, 1992] 1+x 14. The area of the region bounded by the curves y = |x – 1| and y = 1 is (a) 1 (b) 2 (c) 1/2 (d) none [IIT-JEE, 1993] 15. In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x? [IIT-JEE, 1994] 16. Consider a square with vertices at (1, 1), (–1, 1) (–1, –1) and (1, –1). Let S be the region consisting of all points inside the square which are near to the origin than to any edge. Sketch the region S and find its area. [IIT-JEE, 1995] 17. The slope of the tangent to the curve y = f(x) at a point (x, y) is (2x + 1) and the curve passes through (1, 2). The area of the region bounded by the curve, the x-axis and the line x = 1 is (a) 5/3 (b) 5/6 (c) 6/5 (d) 6 [IIT-JEE, 1996] 18. Let An be the area bounded by the curve y = tann x, p y = 0, x = 0, x = __ . If n ≥ 2, then (An + An+2) is 4 1 1 (a) ______ (b) __ n (n + 1) 1 (c) ____ (d) none n–1 [IIT-JEE, 1997] 19. Find all possible values of b > 0 so that the area of the region bounded between the parabolas y = x – bx2 x2 and y = __ is maximum. [IIT-JEE, 1997] b
3.15
( )
1__ 20. Let O (0, 0), A (2, 0), B 1, ___ be the vertices of a 3 ÷ triangle. Let R be the region consisting of all those points P inside D OAB, which satisfy d (P, OA) ≥ min {(P, OB), d (P, AB)} where d denotes the distance from the point to the corresponding line. Sketech the region R and find its area. [IIT-JEE, 1997] 2 2 21. Let f (x) = max {x , (1 – x) , 2x (1 – x)} where 0 £ x £ 1. Determine the area of the region bounded by the curves y = f (x), x-axis, x = 0, x = 1. [IIT-JEE, 1997] 22. Let C1 and 2 be the graphs of the functions y = x2, y = 2x, 0 £ x £ 1 respectively. Let C3 be the graph of a function y = f (x), 0 £ x £ 1, f (0) = 0.
For a point P on C1, let the lines through P parallel to the axes meet C2 and C3 at Q and R, respectively. If for every position of P on C1, the area of the shaded regions OPQ and ORP are equal, determine the function f (x). [IIT-JEE, 1998] 23. For what values of m, the area of the region bounded by the curves y = x – x2 and the line y = m x equals 9/2? (a) – 4 (b) – 2 (c) 2 (d) 4 [IIT-JEE, 1999] 24. Let f(x) be a continuous function given by : | x| £ 1 ÏÔ2 x f (x) = Ì 2 . ÔÓ x + ax + b : | x | > 1 Find the area of the region in the third quadrant bounded by the curves x = – 2y2 and y = f (x) lying in the left of the line 8x + 1 = 0. [IIT-JEE, 1999] No questions asked in 2000. 25. Let b π 0 and for j = 0, 1, 2, ..., n, let S1 be the area of the region bounded by the y-axis and the curve ( j + 1)p jp x exy = sin by, __ £ y £ _______ . Show that S0, S1, b b ..., Sn are in GP. Also, find their sum for a = – 1 and b = – 1 and b = p. [IIT-JEE, 2001]
3.16 Integral Calculus, 3D Geometry & Vector Booster 26. The area bounded by the curves y = |x| –1 and y = 1 – |x| is (a) 1 __ (b) 2 (c) 2÷2 (d) 4 [IIT-JEE, 2002] __ 27. The area bounded by the curves y = ÷ x , 2y + 3 = x and x-axis in the first quadrant is (a) 9 (b) 27/4 (c) 36 (d) 18 [IIT-JEE, 2003] 28. Find the curve passing through (2, 0) and having slope
(
)
(x + 1)2 + (y – 3) of tangent at any point P (x, y) as ______________ . (x + 1) Find also the area enclosed by the curve and x-axis in the fourth quadrant [IIT-JEE, 2004] 29. The area enclosed between the curves y = ax2 and x = ay2__ (a > 0) is 1 sq.u., the value of a is (a) 1/÷3 (b) 1/2 (c) 1 (d) 1/3 [IIT-JEE, 2004] 30. The area bounded by the curves y = (x + 1)2 and y = (x – 1)2 and the line y = 1/4 is (a) 4 s.u. (b) 1/6 s.u. (c) 4/3 s.u. (d) 1/3 s.u. [IIT-JEE, 2005] 31. Find the area bounded by the curves x2 = y, x2 = – y, and y2 = 4x – 3 [IIT-JEE, 2005] 32.
33.
È 4 a 2 4 a 1˘ È f (-1) ˘ È3a 2 + 3a ˘ Í ˙ Í ˙ If Í 4b2 4b 1˙ ÍÍ f (1) ˙˙ = Í3b2 + 3b ˙ Í 2 ˙ Í ˙ 4c 1˙˚ ÍÎ f (2) ˙˚ ÍÎ3c 2 + 3c ˙˚ ÍÎ 4c and f(x) is a quadratic function and its maximum value occurs at a point V. A is a point of intersection of y = f (x) with x-axis and point B is such that the chord AB subtends a right angle at V. Find the area enclosed by f (x) and chord AB. [IIT-JEE, 2005] No questions asked in between 2006 to 2007. The area of the region between the curves _______
÷
_______
÷
1 + sin x 1 – sin x y = _______ and y = _______ bounded by the cos x cos x p lines x = 0 and x = __ is __ 4 ÷2 – 1 t _____ (a) Ú _____________ dt 0 (1 + t2) ÷1 – t2
(
( Ú (
__
÷2 –1
) )
dt _____ (b) Ú _____________ 2 0 (1 + t )÷1 – t2 __
+ 1 ÷2
)
4t dt______ (c) _____________ 0 (1 + t2)÷1 – t2
__
÷2 + 1
(
)
t dt______ (d) Ú _____________ 2 0 (1 + t )÷1 – t2
[IIT-JEE, 2008]
Comprehension Consider the functions defined implicitly by the equation y2 – 3y + x = 0 on varoius intervals in the real line. If x Œ (– •, – 2) » (2, •), the equation implicitly defines a unique real-valued differentiable function y = f(x). If x Œ (– 2, 2), the equation uniquely implicity defines a unique real-valued differentiable function y = g(x) satisfying g(0) = 0 __ __ __ 34. If f (– 10÷2 ) = 2÷2 , then f ¢¢(– 10÷2 ) = __
__
4÷2 (a) ____ 73 32
4÷2 (b) – ____ 3 2 7 3
__
__
4÷2 4÷2 (c) ____ (d) – ____ 3 3 73 73 35. The area of the region bounded by the curve y = f (x), the x-axis and the lines x = a and x = b, where – < • < a < b < – 2, is
( Ú ( Ú ( Ú ( b
)
x (a) Ú _________ dx + bf (b) – af (a) a 3(f (x)2 – 1
x (b) – __________ dx + bf (b) – af (a) a 3(f (x)2 – 1)
b
) ) )
b
x (c) __________ dx – bf (b) + af (a) a 3(f (x)2 – 1)
x (d) – __________ dx – bf (b) + af (a) a 3(f (x)2 – 1)
b
1
36. Ú g¢(x) dx = –1
(a) 2 g (–1)
(b) 0
(c) – 2 g(1)
(d) 2 g (1) [IIT-JEE, 2008]
37. Area of the region bounded by y = ex and x = 0 and x = e is e
(b) Ú log (1 + e – y) dy
(a) e – 1 1
(c) e – Ú ex dx 0
1 e
(d) Ú ln y dy 1
[IIT-JEE, 2009]
Consider the polynomial f (x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f (x) and let t = |s|. 38. The real number s lies in the interval 3 1 (a) – __ , 0 (b) –1, __ 4 4
(
)
( )
Area Bounded by the Curves
(
( )
)
3 1 (c) – __ , – __ 4 2
1 (d) 0, __ 4
39. The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = , lies in the interval
( ) 21 (d) ( 0, ___ ) 64
( )
3 (a) __ , 3 4
(c) (9, 10)
21 11 (b) ___ , ___ 64 16
40. The function f ¢(x) is
( (
)
( ) ( )
1 1 (a) inc in – t, – __ and dec. in – __ , t 4 4 1 1 (b) dec. in – t, – __ and inc. in – __ , t 4 4 (c) inc in (– t, t) (d) dec in (– t, t) [IIT-JEE, 2010] 41. Let f : [– 1, 2] Æ [0, •) be a continuous function such that f (x) = (1 – x) for all x Œ [– 1, 2]. Let
)
2
R1 = Ú x f (x) dx and R2 be the area of the region –1
bounded by y = f (x), x = – 1 and x = 2 and the x-axis, Then (a) R1 = 2R2 (b) R1 = 3R2
(c) 2R1 = R2
(d) 3R1 = R2 [IIT-JEE, 2011]
42. Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts R1 (0 £ x £ b) and R2 (b £ x £ 1) such that R1 – R2 1 = __ , Then b is 4 (a) 3/4 (b) 1/2 (c) 1/3 (d) 1/4 [IIT-JEE, 2011] 2 43. Let S be the area of the region enclosed by y = e –x , y = 0, x = 0 and x = 1. Then 1 1 (a) S ≥ __ e (b) S ≥ 1 – __ e 1 1 (c) S ≥ __ 1 + ___ __ e ÷ 4 1 1 1__ __ + ___ (d) S ≥ ___ __ 1 – ___ [IIT-JEE, 2012] e ÷ 2 2 ÷ ÷ 44. The area of the region enclosed by the curves y = sin x + cos x and y = |cos x – sin x| over the p interval 0, __ is 2
(
)
(
)
[ ]
__
__
__
(a) 4(÷__ 2 – 1) (b) 2÷__ 2 ( ÷ __ 2 – 1) (c) 2(÷2 + 1) (d) 2÷2 ( ÷ 2 + 1) [IIT-JEE, 2013] No questions asked in 2014, 2015. 45. Area of the region ______ {(x, y) : y £ ÷|x + 3|, 5y £ x + 9 £ 15} is (a) 1/6 (b) 4/3 (c) 3/2 (d) 5/3 [IIT-JEE, 2016]
Answers Level-II
1. (c) 2. (d) 3. (b) 6. (c) 7. (c) 8. (c) 11. (b) 12. (b) 13. (a) 16. (a) 17. (d) 18. (b) 21. (d) 22. (c) 23. (a) 26. (d) 27. (b) 28. (c) 31. (c) 32. (a) 33. (a) 36. (a) 37. (a) 38. (b) 41. (d) 42. (a) 43. (a) 46. (a) 47. (c) 48. (c) 51. (c) 52. (a) 53. (c) 56. (a) 57. (a) 58. (c) 61. (b) 62. (a) 63. (a) 66. (a, b, d) 67. (a, b, d) 69. (b, d) 70. (a, b, c, d)
4. (d) 5. (a) 9. (d) 10. (d) 14. (b) 15. (a) 19. (c) 20. (a) 24. (b) 25. (c) 29. (d) 30. (a) 34. (c) 35. (a) 39. (a) 40. (a) 44. (a) 45. (c) 49. (b) 50. (c) 54. (a) 55. (b) 59. (b) 60. (b) 64. (c) 65. (d) 68. (a, b, c, d) 72. (b, c) 72. (a, b, c)
6. 4 11. 2
1. 4
2. 2
3. 2
7. 2 12. 8
8. 1 13. a = 8
9. 6
10. 4
Comprehensive Link Passages Passage Passage Passage Passage Passage
I : II : III : IV : V :
1. (b) 1. (b) 1. (b) 1. (a) 1. (b)
2. (a) 2. (c) 2. (a) 2. (b) 2. (c)
3. (d) 3. (c) 3. (b) 3. (b) 3. (b)
Matrix Match
1. 2. 3. 4.
(A)Æ(T), (A)Æ(T), (A)Æ(S), (A)Æ(S),
(B)Æ(R), (B)Æ(S), (B)Æ(P), (B)Æ(P),
(C)Æ(P), (C)Æ(P), (C)Æ(Q), (C)Æ(S),
(D)Æ(Q) (D)Æ(Q) (D)Æ(R) (D)Æ(S)
Assertion and Reason
INTEGER TYPE QUESTIONS
4. 4
5. 3
3.17
1. B
2. A
3. A
4. C
5. B
3.18 Integral Calculus, 3D Geometry & Vector Booster
Hints
and
2
= Ú y dx – 1
– 1
(
)|
=0 7. Hence, the required area
2
(
–1
x3 = __ – x2 + 2x 3 –1 8 1 = __ – 4 + 4 + __ + 1 – 2 3 3
2p
)
2. Hence, the required area 2
= Ú (ln x + tan x) dx
( (
)|
2 1 = (x log x – x tan x – __ log |1 + x2| 2 1 p 1 = (2 log 2 – 1) + 2tan–12 – __ log 5 – __ 2 4 p 1 = 2 (log 2 + tan–1 2) – __ log 5 – __ + 1 s.u. 2 4
x)|21 +
(
–1
)
(
))
3. Hence, the required area
0 p
2p
0
p
= Ú (x sin x)dx – Ú (x sin x)dx
= (– x cos x + sin x)|p0 – (– x cos x + sin x)|2p p = p – (– 2p – p) = 4p. sq.u. 4. Hence, the required area 2
= Ú (log1/2x) dx
1 = – ____ (x log x – x)|21 log2 1 = – ____ (2log 2 – 3) log2 1 = ____ (3 – 2 log2) sq.u. log2 5. Hence, the required area
log2
= Ú (e2x – 3ex + 2)dx 0
(
p
(– cos x)|p0 –
(– cos x)|2p p
= – (cos x)|p0 + (cos x)|2p p
= – (–1 – 1) + (1– (–1)) = 4 sq.u. 8. Hence, the required area 2p
= Ú cos x dx
0
p/2
3p/2
2p
0
p/2
3p/2
= Ú cos x dx – Ú cos x dx + Ú cos x dx (sin x)|p/2 – 0
(sin x)|3p/2 + p/2
)|
log2
e2x = ___ – 3ex + 2x 0 2
(sin x)|2p 3p/2
=
= (1 – 0) – (–1 – 1) + (0 – (–1)) = 4 sq.u.
9. Hence, the required area 2
= Ú {x (x – 1) (x – 2)}dx 0 1
2
= Ú {x(x – 1) (x – 2)}dx – Ú {x (x – 1) (x – 2)}dx 0 1
1
0
=
= Ú (x|sin x|) dx
2p
2p
0 p
= Ú sin x dx – Ú sin x dx
–1
= Ú sin x dx
= 3 + 1 – 2 = 2 sq.u.
1
)
= Ú x3dx
= Ú (x2 – 2x + 2) dx
)
1
2
( (
1 = 4 – 6 + 2 log 2 + 3 – __ 2 1 = __ + 2 log 2 sq.u. 2 6. Hence, the required area
1. Hence, the required area
solutions
1
2
= Ú (x3 – 3x2 + 2x)dx – Ú (x3 – 3x2 + 2x)dx 0
(
) | ( 1
4
1
)|
2
4
x x = __ – x3 + x2 – __ – x3 + x2 0 4 4 1 1 1 __ __ = – 1 + 1 – (4 – 8 + 4) – – 1 + 1 4 4 1 = 2 __ – 1 + 1 4 1 = __ sq.u. 2 10. Hence, the required area
(
) ( )
(
3
(
= Ú {(x – 1) (x – 2) (x – 3)}dx 0
))
Area Bounded by the Curves
3.19
1
= – Ú {(x – 1) (x – 2) (x – 3)}dx 0
2
+ Ú {(x – 1) (x – 2) (x – 3)}dx 1 3
– Ú {(x – 1) (x – 2) (x – 3)}dx
(
2
)|
1
x4 11 = – __ – 2x3 + ___ x2 – 6x 2 0 4
( (
) )
2 x4 11 + __ – 2x3 + ___ x2 – 6x 4 2 1 3 x__4 11 – – 2x3 + ___ x2 – 6x 4 2 2 1 11 = – __ – 2 + ___ – 6 + (4 – 8 + 22 – 12) 4 2 1 11 – __ – 2 + ___ – 6 4 2 99 81 ___ – – 54 + ___ – 18 + (4 – 8 + 22 – 12) 4 2 279 23 = – 2 ___ – 8 + 12 – ____ – 72 4 4 46 279 = ___ – ____ + 100 4 4 233 400 – 233 ____ 167 = 100 – ____ = _________ = sq.u. 4 4 4 11. Given curves are y = x2 and y = 2x – x2
( ( ( (
)
)
)
) (
)
Hence, the required area 1
= Ú (y2 – y1) dx 0 1
__
2 = Ú (÷x – x ) dx 0
(
)|
1
x3 2 = __ x3/2 – __ 3 0 3 2 1 1 = __ – __ = __ sq.u. 3 3 3
(
)
13. Do yourself. 3x2 14. Given curves are y = ___ and 3x – 2y + 12 = 0 4
Hence, the required area 4
= Ú (y2 – y1) dx – 2
Hence, the required area 1
= Ú (y2 – y1)dx 0
1
= Ú (2x – x2 – x2)dx 0
4
= Ú (2x – 2x2)dx 0
( )| 1 2 = ( 1 – __ ) = __ sq.u. 3 3 1 2 = x2 – __ x3 3 0
12. Given curves are y2 = x and x2 = y.
)
3x + 12 ___ 3x3 = Ú _______ – dx 2 4 – 2
3 = __ Ú (2x + 8 – x2) dx 4 – 2
3 x3 4 = __ x2 + 8x – __ 4 3 – 2
4
1
(
(
)
54 8 ___ ) – ( 4 – 16 + __ ) ) ( ( 3 3 ( ) 3 108 – 62 = __ ( ________ ) = ___ 464 = ___ 232 sq.u. 4 3
3 = __ 16 + 32 – 4 3 62 = __ 36 – ___ 4 3
15. Given curves are x2 = 4y and x = 4y – 2
3.20 Integral Calculus, 3D Geometry & Vector Booster
Hence, the required area
Hence, the required area
2
4a
= Ú (y2 – y1) dx
–1 2
0
(
)
x2 __ dx 4
4a
x+2 = Ú _____ – 4 –1
1 = __ Ú (x + 2 – x2) dx 4 –1
(
__ __
= Ú (2÷a ÷x – mx) dx 0
2
= Ú (y2 – y1) dx
4a
)
x3 2 1 x2 = __ __ + 2x – __ 4 2 3 –1 8 1 1 1 = __ 2 + 4 – __ – __ – 2 + __ 4 3 2 3
) ( )) ( ( 9 7 1 10 __ 1 27 = __ ( ___ + ) = __ ( ___ = __ sq.u. 4 3 4 6) 8 6 2
( ( ) ( ) ) ( ( ) ( ) )
2
__ __ x2 2 3/2 __ = 2÷a x – m 3 2 0
__ __ 16a2 2 3/2 ____ = 2÷a (4a) – m 3 2
32 = ___ – 8m a2 sq.u. 3
(
)
18. Given curves are y = x2 + 2, y = x, x = 0 and x = 3.
16. Given curves are y = 4 ax and x = 4 ay.
Hence, the required area
Hence, the required area
4a
3
= Ú (y2 – y1) dx 0
4a
= Ú (y2 – y1) dx 0
(
) ( ) ( ) ( ) ( )
__ __ x2 ___ = Ú 2÷a ÷x – dx 4a 0
x3 4a 2 3/2 4a __ ____ = 2÷a x – 3 12a 0 0
3 x3 = __ + 2x – x2 3 0
27 = ___ + 6 – 9 = 6 sq.u. 3
__
3
0
3
__ __ 64a 2 3/2 ____ = 2÷a (4a) – 3 12a 2
2
2
32a 16a 16a = ____ – ____ = ____ sq.u. 3 3 3
17. Given curves are y2 = 4ax and y = mx.
= Ú (x2 + 2 – 2x) dx
( (
)
)
19. Given curves are
y = 2x – x2 and y = – x.
Area Bounded by the Curves
Hence, the required area 3
= Ú (y2 – y1) dx 0
We have 6x – x2 = x2 – 2x. 2x2 = 8x x = 0, 4 Hence, the required area 4
3
= Ú (2x – x + x) dx 2
4
3
= Ú (3x – x2) dx
(
)
3x2 x3 3 = ___ – __ 2 3 0
(
)
27 = ___ – 9 = 2
= Ú (6x – x2 – x2 + 2x) dx 0 4
0
= Ú (y2 – y1) dx 0
0
= Ú (8x – 2x2) dx 0
9 __ sq.u. 2
20. Given curves are y = 2x – x2, y = 2x and x = 0, x = 2
( ) 128 64 = ( 64 – ____ = ___ sq.u. 3 ) 3 4 2 = 4x2 – __ x3 3 0
22. Let A1 be the area of the curve y = 4x – x2 with x-axis and A2 be the area of the curve y = x2 – x with x-axis 4
A1 = Ú y dx 0
4
= Ú (4x – x2) dx 0
Hence, the required area = Ú (y2 – y1) dx
64 32 = 32 – ___ = ___ 3 3
and
= Ú (2x – x – 2 ) dx 2
x
)
1
)
2x 2 ____ log 2 0
x3 = x – __ – 3
8 4 1 = 4 – __ – ____ + ____ 3 log 2 log 2
3 4 = __ – ____ sq.u. 3 log 2
)
2
( (
= Ú (x – x2) dx 0
)
(
x2 = __ – 2
(
)
x3 1 __ 3 0
)
1 1 = __ – __ = 2 3 Hence, the required ratio,
21. Given curves are 2
A2 = Ú y dx 0
0
( 1
2
(
)
x3 4 = 2x2 – __ 3 0
0
(
2
3.21
2
y = 6x – x and y = x – 2x.
A1 ___ = A2
32 ___ : 3
1 __ = 6
1 __ 6 32 ___ × 3
6 __ = 64 : 1 1
3.22 Integral Calculus, 3D Geometry & Vector Booster 26. Given curves are {(x, y) : x2 + y2 £ 1 £ x + y}.
23. Given curves are
x2 + y2 = 4 and (x – 2)2 + y2 = 4.
Hence, the required area 1
Hence, the required area
(
1
2
__________
1
(
( ) )
(x – 2) __________2 __ x–2 1 4 = 2 _____ ÷4 – (x – 2) + sin–1 ____ 2 2 2 0
(
_____
x + 2 __ ÷ 4 – x2 + 2
)
(
__
( ) )
x 2 4 __ sin–1 __ 2 2 1
( )
)
3 ÷ 1 = 2 – ___ + 2sin–1 – __ – 2sin–1(–1) 2 2
(
__
( ) )
3 ÷ 1 + 2 sin (1) – ___ – 2sin–1 __ 2 2
(
–1
__
__
_____
= Ú (÷1 – x2 – (1 – x)) dx 0
(
_____
( )
3 p 3 p ÷ p ÷ = 2 – ___ – __ + p + __ – ___ – __ 2 3 2 2 3
__ 3p 2p = 2 ___ – ___ – ÷ 3 2 3
__ 5p = ___ – 2÷3 sq.u. 3
( (
( ) ) ( ) )
(
)
)
(
)
24. Do yourself. x2 25. Given curves are __ 2 + a
y2 x __ 2 = 1 and __ a + b
y __ = 1. b
Solving y2 = 4x and 4x2 + 4y2 = 9 we get. 4x2 + 16x – 9 = 0 fi 4x2 – 2x + 18x – 9 = 0 fi 2x (2x – 1) + 9 (2x – 1) = 0 fi (2x + 9) (2x – 1) = 0 1 9 fi x = __ , – __ 2 2 Hence, the required area
Hence, the required area
pab = ____ – 4
(
1 __ ab 2
1 p = __ __ – 1 sq.u. 2 2
)
)
1
x÷1 – x2 __ x x2 1 + sin–1 __ – x + __ = _______ 2 2 2 2 0 1 –1 __ 1 1 __ __ = sin – 1 + 2 2 2 1 1 1 = __ sin–1 __ – __ 2 2 2 1 p = __ __ – 1 sq.u. 2 6 27. Given curves are {x : y2 £ 4x, 4x2 + 4y2 £ 9}.
)
(
= Ú (C – L) dx 0 1
= 2 Ú ÷4 – (x – 2)2 + Ú ÷4 – x2 dx 0
_____
1/2
3/2
0
1/2
= Ú P dx + Ú E dx 1/2
1/2
3/2
______
__ 1 __ = Ú 2÷x dx + Ú ÷9 – 4x2 dx 2 1/2 0 3/2
÷( )
________
3 2 1 2 __ __ = Ú 2÷x dx + Ú – x dx 2 1/2 2 0 __
Area Bounded by the Curves
(
( ÷( ) ) ________
)
1/2 3 2 2 1 x __ = 2 × __ × x2/3 + __ __ – x2 3 2 2 2 0
( )
9 2x 3/2 + __ sin–1 ___ 8 3 1/2
3.23
27 1 = __ (3 + 2 + 21) = ___ sq.u. 2 2 31. Given curves are y = 1 + |x + 1|, x = – 3, x = 3, y = 0.
9 1 4 1 = __ × ____ __ + __ × __ sin–1(1) 3 2÷2 8 2 __ 9 1 ÷2 1 – __ ___ + __ sin–1 __ 2 4 8 3 9p 9 2 1 1 = ____ __ + ___ – ____ __ – ___ sin–1 __ 3 3÷2 32 4÷2 16
(
(
( ) )
( )
( ) )
9p 9 5 1 = _____ __ + ___ – ___ sin–1 __ sq.u. 3 16 12÷2 32 28. Given curves are {x : x2 + y2 £ 2ax, y2 ≥ ax, x ≥ 0, y ≥ 0}
Hence, the required area 1 1 1 = __ × (3 + 1) × 2 + __ × (1 + 2) × 1 + __ × (2 + 5) × 3 2 2 2 28 1 = __ (4 + 3 + 21) = ___ = 14 2 2 32. Given curves are y = |x – 1| and y = – |x – 1| + 1.
Hence, the required area a
= Ú (y2 – y1) dx 0 a
(
__________
__
)
= Ú ÷a 2 – (x – a)2 – a÷x dx
Hence, the required area 1 1 1 __ __ __ a 2a – a = __ sin–1 x_____ – ___ x3/2 = 2 2 × 1 × 2 = 2 sq.u. ÷ a 2 2 3 0 33. Given curves are 2a __ __ a2 –1 = ___ a÷a – sin (–1) f (x) = log x and g (x) = (log x)2 3 2 0
(
__________ x–a 2 _____ a – (x – a)2 +
(
2
(
)
(
)
a
)
)
pa2 2a5/2 ____ = ____ + sq.u. 3 4 29. Do yourself.
Y
30. Given curves are y = 1 + |x + 1|, x = – 2, x = 3, y = 0.
X¢
O
X
Y¢
Hence, the required area e
= Ú (log x – (log x)2)dx 1
Hence, the required area 1 1 1 = __ × (2 + 1) × 1 + __ × (1 + 2) × 1 + __ × (2 + 5) × 3 2 2 2
34.
= (x log x – x – (x (log x)2 – 2 (x log x – x)))e1 = ((3 (x log x – x) – x (log x)2))e1 = (3 – e) sq.u. Given curve is y2 = x2 – x4.
3.24 Integral Calculus, 3D Geometry & Vector Booster Hence, the required area 1
2
= Ú (x2 + 1)dx + Ú (x + 1) dx 0
Hence, the required area 1
= 4 Ú (x – x ) dx 2
4
0
(
)
x5 1 __ 5 0
x3 = 4 __ – 3
8 1 1 = 4 __ – __ = ___ sq.u. 3 5 15
(
1
( ) ( ) 3
1
2 x __ + x 2 1 2
x = __ + x + 3 0
3 1 = __ + 1 + 4 – __ 3 2
(
) ( )
4 5 = __ + __ = 3 sq.u. 3 3 37. Given curves are
y = x and y = x3.
)
35. Given curve is
1 |y| + __ £ e– |x| 2
Hence, the required area 1
= 2 Ú (x – x3) dx 0
(
)
= 4 Ú e dx – x
0
log 2 = 4 ( – e– x )0
= 4 ( 1 – e– log 2 )
)
x2 x4 1 = 2 __ – __ 2 4 0
1 1 2 = 2 __ – __ = 2 × __ = 1 sq.u. 2 4 4
Hence, the required area log 2
(
(
)
38. Given curve is y = x (x – 1)
( )
1 1 = 4 1 – __ = 4 × __ = 2 sq.u. 2 2 36. Given curve is
f (x, y) = [(x, y) : 0 £ y £ x2 + 1,
0 £ y £ x + 1, 0 £ x £ 2] Hence, the required area 2
= Ú (2 – x (x – 1)2) dx 0
2
= Ú (2 – x (x2 – 2x + 1)) dx 0
Area Bounded by the Curves 2
Hence, the required area
0
= Ú (2 – (x3 – 2x2 + x)) dx
(
3
)
x4 2 3 __ x2 2 = 2x – __ + __ x – 4 3 2 0 16 = 4 – 4 + ___ – 2 3 10 ___ = sq.u. 3 39. Given curves are y = x2 + 1, y = x, x = 0 and y = 2.
(
)
= Ú (x2 – x1) dy 0
3
= Ú ((y – 2)2 + 1 – (2y – 4)) dy 0
3
= Ú (y2 – 4y + 4 + 1 – 2y + 4) dy 0
3
= Ú (y2 – 6y + 9)dy 0
(
)
3 y3 = __ – 3y2 + 9y 3 0
= 9 – 27 + 27
= 9 sq.u. 42. Given curve is y2 = (x – 1) (x – 2)2 40. Given curves are y = loge x and y = 2x, x = 1/2 and x = 2.
Hence, the required area 2
_____
= 2 Ú (x – 2)÷x – 1 dx 1
1
Hence, the required area 2
= Ú (2x – log x) dx
1/2
(
)
x
2
2 = ____ – (x log x – x) log 2 1/2
( (
__
( )
)
2 ÷ 4 1 1 1 = ____ – 21 log 2 + 2 – ____ + __ log __ – __ 2 2 log 2 log 2 2 __
)
4 – ÷ 2 __ 5 3 = ______ – log 2 + __ sq.u. 2 2 log 2 41. Given curve is (y – 2)2 = (x – 1) and the equation of the tangent to the curve at (2, 3) is x – 2y + 4 = 0.
= 4 Ú (t2 – 1)t2dt
|
0
1
|
= 4 Ú (t4 – t2)dt 0
| (
)|
t5 t3 1 = 4 __ – __ 5 3 0
1 1 = 4 __ – __ 5 3
| ( ) |
8 = ___ sq.u. 15 43. Given curves are
y = (x – 1)2, y = (x + 1)2 and y = 1/4.
3.25
3.26 Integral Calculus, 3D Geometry & Vector Booster Hence, the required shaded area 0
(
1/2
)
(
)
1 1 = Ú (x + 1)2 – __ dx + Ú (x – 1)2 – __ dx 4 4 –1/2 0
(
) (
46. Given curve is |x – 2| + |y + 1| = 1
)
0 1/2 (x + 1)3 __ (x – 1)3 __ 1 1 = _______ – x + _______ – x 3 4 –1/2 3 4 0
(
)
1 1 1 1 1 1 = __ – ___ + __ – ___ – __ + __ 3 24 8 24 8 3
(
)
1 1 1 1 = __ – 2 ___ + __ + __ 3 24 8 3 2 2 __ 2 1 __ 1 = __ – __ = – __ = sq.u. 3 6 3 3 3
Thus, the length of each side
44. Given curves are
y = ex, y = e– x and the straight line x = 1
________________
__
= ÷(2 – 1)2 + (0 + 1)2 = ÷ 2
Hence, the required area
__
= ( ÷2 )2 sq.u.
47. Given curve is [x] + [y] = 3
Hence, the required area 1
= Ú (ex – e– x ) 0
= (ex + e– x)10
1 = e + __ e – 2 sq.u.
(
)
45. Given curves are y = log x, y = log |x|, y = |log x| and y = |log|x||
If [x] = 0, 1, 2, 3 then [y] = 3, 2, 1, 0. Hence, the required area
= 4 (1 × 1 + 1 × 1 + 1 × 1 + 1 × 1)
=4×4
= 16 sq.u. 48. Given curves are |x| + |y| = 1 and |x + 1| + |y| = 1.
Hence, the required area 0
= 4 Ú x dy
= 4 Ú e ydy
= 4 (e y)0– •
= 4 (e0 – 0) = 4 (1 – 0) = 4 sq.u.
– • 0
– •
Hence, the required area
1 1 = 2 × __ × 1 × __ 2 2 1 = __ sq.u. 2
Area Bounded by the Curves
49. Given curves are
| |
p p y = cos–1 (cos x) and |x – p| + y – __ = __ 2 2
Hence, the required area 1
__
= 2 Ú (÷x – x) dx 0
(
)
x2 1 2 = 2 __ x2/3 – __ 3 2 0
2 1 = 2 __ – __ 3 2
( ) 4 – 3 1 1 = 2 ( _____ = 2 × __ = __ sq.u. 6 ) 6 3
52. Given curve is |x + y| + |x – y| £ 2
Hence, the required area
3.27
p 1 = 2 × __ × p × __ = 2 2
p2 __ sq.u. 2
Ï x £ 1, x ≥ –1 = Ì Ó y £ 1, y ≥ –1
50. Since the given curve is symmetrical about x-axis as well as y-axis.
Hence, the required area = (2)2 = 4 sq.u. 53. Given curves are y = x, y = x + sin x, 0 £ x £ p Hence, the required area
Hence, the required area __
÷5 + 1
= 4 Ú y dx 0
p
__
÷5 + 1
= 4 Ú (5 – (x – 1)2)dx
(
)
__
8 = __ ( 7 + 5÷5 ) sq.u. 3
51. Given curves are
__
+ 1 (x – 1)3 ÷5 = 4 5x – _______ 3 0
__
= Ú (x sin x – x) dx 0 p
0
= Ú (sin x) dx 0
= – (cos x)p0 = – (–1 – 1) = 2 sq.u.
54. Hence, the required area b
y=÷ |x| and y = |x|
(
__ p = Ú f (x)dx = b sinb + __ cosb + ÷ 2 b 4 p __
)
4
Applying, Newton and Leibnitz rule, we get __ p f (b) = b cos b + sin b – __ sin b + ÷ 2 4 __ p p f __ = 0 + 1 – __ + ÷ 2 2 4 __ p p f __ = 1 – __ + ÷2 2 4 55. Given curves are y = sin x and y = cos x p x = 0 and x = __ . 2
( ) ( ) (
)
3.28 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) ( 3 2 = ( 2 7 + __ log 2 – 5 ) + 3log ( ______ 2 2 + 7 )
)
__ 1 3 2 11 = __ + __ log 2 + 3log ______ __ + 2÷7 – ___ 2 2 2 2+÷ 7 __
÷
__
÷
(
)
__ __ 2 = (2÷7 – 5) + 3log ______ __ + 3log ( ÷2 ) 2+÷ 7
(
__
Hence, the required area p __ 4 (cos x 0
Ú
– sin x) dx + Ú (sin x – cos x) dx
fi
p __ 4
p __ cos x)04 –
= (sin x +
(
) ( ( )
)
57. Given curve is y = 4x – x2
p __ 2
__
2÷2 __ = ( 2÷7 – 5 ) + 3log ______ 2+÷ 7
(cos x +
y = 5 – (x – 2)2
p __ 2 sin x)__p 4
)
2 2__ = ___ __ – 1 – 1 – ___ 2 ÷2 ÷ 2 = 2 ___ __ – 1 2 ÷ __
= 2 ( ÷2 – 1 ) sq.u. 56. Given curves are
Now, the area OABO
3 y = 2 – |2 – x| and y = __ . |x|
3/2
= Ú (mx) dx 0
( ) ( )
9m mx2 3/2 = ____ = ___ 2 0 8
Hence, the area OABCDO 3/2
= Ú (1 + 4x – x2) dx 0
Hence, the required area 2
(
__
2 + ÷7
)
(
)
3 3 = Ú__ x – __ x dx + Ú 4 – x – __ x dx 2 ÷3
(
2
) ( 2
2
)
x x = __ – 3log x __ + 4x – __ – 3log x 2 2 ÷3 2
(
3 = 2 – 3log 2 – __ + 2
(
__
39 9m 1 ___ According to the question, ___ = __ ◊ 8 2 8 13 fi m = ___ 6
)
3 __ log 2 2 __
(________ 2 + ÷7 )2
+ 4 ( 2 + ÷7 ) –
__
2 + ÷7
( ) 9 27 3 = ( __ + 2 ◊ __ – ___ 4 24 ) 2 9 39 = ( 6 – __ ) = ___ sq.u 8 8 x3 3/2 = x + 2x2 – __ x 0
__
– 3 log ( 2 + ÷7 )
2
)
– 8 + 2 + 3 log 2
13 Hence, the value of m is ___ . 6 58. Do yourself. 59. Given curve is y = sin–1x and y-axis between y = 0 and y = p/2.
Area Bounded by the Curves
| (
)|
8 1 64 = __ ___ – 48 + __ + 4 – 16 2 3 3
1 = __ (24 – 48 + 4 – 16) 2
1 = __ (28 – 64) 2
36 1 = __ |(28 – 64)| = ___ = 18 sq.u. 2 2
| |
|
|
3.29
61. Given curves are Hence, the required area
p __ 2 (1 0
Ú
=
x = – 2y2 and x = 1 – 3y2.
– sin x) dx p __ 2 cos y)0
= (y +
p = __ – 1 sq.u. 2
(
)
60. Given curves are y2 = 2x + 6 and y = x – 1 On solving, we get,
– 2y2 = 1 – 3y2
y2 = 1
y = ± 1
Hence, the required area 1
= Ú (x2 – x1) dy –1
On solving, we get, (x – 1)2 = 2x + 6
y2 = 2 (y + 1) + 6
1
y – 2y – 8 = 0
(y – 4) (y + 2) = 0
1
= 2 Ú (y2 – 1) dy 0
4
= Ú (x2 – x1) dy 4
= Ú (y2 – 1) dy –1
y = – 2, 4 Hence, the required area
– 2
= Ú ((– 2y2) – (1 – 3y2)) dy –1
2
1
(
)
y2 – 6 = Ú _____ – (y + 1) dy 2 – 2
1 = __ Ú (y2 – 6 – 2y – 2) dy 2 – 2
4
( )
1 y3 = 2 __ – y 3 0
| ( ) |
1 4 = 2 __ – 1 = __ sq.u. 3 3 –1 62. Given curves are y = tan x and y = cot–1x and the y-axis. Hence, the required area
4
1 = __ Ú (y2 – 2y – 8) dy 2 –2
4 3 1 y = __ __ – y2 – 8y 2 3 – 2
(
)
p __ 4 tan y dy 0
Ú
p __ 2
+ Ú cot y dy
=
= (log (sec y))04 + (log (sin y))__p
p __ 4
p __
p __ 2
4
3.30 Integral Calculus, 3D Geometry & Vector Booster
(
( ) )
__ 1 = log ( ÷2 ) + 0 – log ___ __ 2 ÷
28 2 __ 7 = ___ – __ – 3 3 3
= 2log ( ÷2 )
= log (2) sq.u.
28 – 9 = ______ 3
19 = ___ sq.u. 3
__
63. Given curve is x = y2 + y + 2 and the y-axis.
65. Find the area of the region bounded by the curves 4|y| = |4 – x2| and |y| (x2 + 4) = 12. 65. Given curves are 4|y| = |4 – x2| and |y| (x2 + 4) = 12.
Hence, the required area 2
= Ú xdy
–1 2
= Ú (– y2 + y + 2)dy
Hence, the required area
–1
(
)
2 y2 __ + 2y 2 –1
y3 = – __ + 3
8 1 1 = – __ + 2 + 4 + __ – __ – 2 3 3 2
( ) 7 1 7 = ( 4 – __ – __ = __ sq.u. 3 2) 6
[
__
2÷2
__
2 64. Given curves are y = ÷ x , y = x
y = – (x2 – 2x – 4) = 5 – (x – 1)2
( ( ) ( Ú ( ) )
( )
2
4 – x2 12 = 4 Ú _____ 2 dx – Ú _____ dx 4 0 0 x +4 __
2÷2
x2 – 4 + _____ dx 4 2
[ (
( ) )
(
) ( ]
) ] __
2÷2 x ÷2 x3 2 x3 = 4 6 tan __ 2 – x – ___ – ___ – x 2 0 12 0 12 2 –1
__
[
__
__ 4 4 – 2÷2 = 4 6 tan–1 ( ÷2 ) – __ + _______ 3 3
__ __ 4 = __ ( 18 tan–1 ( ÷2 ) – 2÷2 ) sq.u. 3 66. Given curves are x2 + 2y2 = 2 and y = 1 – x2
Hence, the required area 2
1
0
0
2
__
= Ú (– x2 + 2x + 4)dx – Ú ÷ x dx – Ú x2dx
(
1
) ( ) ( )
2 1 x3 x3 2 2 = – __ + x2 + 4x – __ x3/2 – __ 3 3 3 1 0 0
(
) ( ) (
)
18 8 1 2 = – ___ + 12 – __ – __ – __ 3 3 3 3
Hence, the required area 1
(
______
_____
)
= 2 Ú ÷2 – 2y2 – ÷1 – y dy – 1/2
Area Bounded by the Curves
__ 1
1
_____
(
– 1/2
_____
)
__ y 1 1 = 2÷2 __ ÷1 – y2 + __ sin–1(y) 2 2 – 1/2
(
( ( ( ( (
)
1 2 – 2 ___ (1 – y)3/2 – 3 – 1/2
_____
= 2÷2 Ú ÷1 – y2 dy – 2 Ú ÷1 – y dy – 1/2
3.31
__
( ) ) ( )
__ p 1 1 ÷3 1 4 3 3/2 = 2÷2 __ + __ ◊ ___ + __ sin–1 __ – __ __ 4 4 2 2 2 3 2 __
) ( )
__
__
3 p 3 p ÷ 4 3÷__ = 2÷2 __ + ___ + ___ – __ ____ 4 8 12 3 2÷2 __
__
Hence, the required area
) ( ) ) ) __
0
•
1 – e
0
= Ú log (x + e) dx + Ú e– xdx
3 ÷3 p ÷ = 2÷2 __ + ___ – 2 ___ __ 3 8 2 ÷
= ((x + e) log (x + e) – (x + e))01 – e – (e– x)•0
__ __ p 3 ÷ = 2÷2 __ + ___ – ÷6 sq.u. 3 8
= (eloge – e) – (1log1 – (1)) – (e– x)•0
= (1 – (0 – 0))
__
67. Given curves are x2 + y2 = 4 and x2 + 4y2 = 9
= 2 sq.u. 69. Given curves are x2 + y2 = 4, y = x2 + x + 1 and
[ ( )
( ) ]
x x y = sin2 __ + cos __ and y = 0. 4 4
Solving the above equations, we get,
x2 + 4 (4 – x2) = 9
fi
3x2 = 7
fi
7 x = __ 3
÷
Hence, the required area
__
(
___
2
_____
_____
)
1 = 4 Ú __ ÷9 – x2 dx + ___ Ú ÷ 4 – x2 dx 0 2 ÷7/3
[ (
_____
( )
___ ÷7/3
9 x 1 x = 4 __ __ ÷9 – x2 + __ sin–1 __ 2 2 2 3 0 _____
( ÷ ) }
{
68. Given curves are fi
( ÷ )
1 7 1 7 = 4p + 9 sin __ __ – 8 sin–1 __ __ 3 3 2 3
__
( )
1 y = ln (x + e) and x = ln __ y y = ln (x + e), y = e– x, y = 0
2
_____
+ Ú (x2 + x + 1) dx + 2 Ú__ ÷4 – x2 dx –1
( ) ) ]
x x 2 + __ ÷4 – x2 + 2 sin–1 __ ___ 2 4 ÷7/3 __
–1
__
0
Hence, the required area ÷7/3
__
= ( ÷ 3 × 1 ) + ( ÷ 3 – 1 )
(
÷3
)
0 x x = ( 2÷3 – 1 ) + __ + __ + x 3 2 –1 _____ x x 2 2 –1 + 2 __ ÷ 4 – x + 2sin __ __ 2 2 ÷3
__
(
3
2
( ) )
( (
__ 1 = ( 2÷3 – 1 ) + 0 – – __ + 3
{
(
__
))
1 __ – 1 2
)}
3 2p ÷ + 2 (0 + p) – ___ + ___ 2 3 2p ___ – ÷ 3 } { 3 2p 1 = ( ___ + ÷ 3 – __ ) sq.u. 3 6 __ 5 = ( 2÷3 – 1 ) + __ + 6 __
__
3.32 Integral Calculus, 3D Geometry & Vector Booster
(
2
)]
__ 1 1 1 1 + 2 ___ __ + __ log ( ÷2 + 1 )– __ – __ log 3 sq.u. 2 3 4 2 ÷
70. Given curves are 2
x x __ + y2 = 1 and __ – y2 = 1 4 2
71. Let the area be A. 2
Then
2
x __ + 4
) )
(
2 x4 x3 = ___ – __ + ax 12 3 0
4 8 = __ – __ + 2a 3 3
Solving the above equations, we get
(
x3 A = Ú __ – x2 + a dx 3 0
) ( 4 = ( 2a – __ ) 3
dA ___ = 2 > 0 da
2
x __ = 2 2
So, the area is minimum. It will be minimum when A = 0.
fi
3x2 ___ = 2 4
fi
8 x2 = __ 3
fi
Now,
x2 y2 = 1 – __ 4
fi
8 2 1 y2 = 1 – _____ = 1 – __ = __ 4×3 3 3
2 Hence, the value of a is __ . 3 72. Hence, the required area
Thus,
1
1__ y = ___ 3 ÷ Hence, the required area __
1
1/÷3
) )
= 4 Ú (Hyperbola) dy + Ú __ (Ellipse) dy
=
0
1/÷3
__
1
1/÷3 __ _____ 4 ÷ 2 1 + y2 dy 0
Ú
( ÷
)
_____
+ Ú __ 2÷1 + y2 dy 1/÷3
[ (
_____ __ y _____ 1/÷3 1 = 4 ÷2 __ ÷1 + y2 + __ log y + ÷ y 2 + 1 2 2 0
|
(
|)
[ (
1 = 4 ÷2 ____ __ × 2÷3
|
2 ___ __ + 3 ÷
|
1 1 __ log ___ __ + 2 ÷3
(
|)
[ (
a2 a = __ + __ + 1 3 2
fi
dA 2a __ 1 ___ = ___ + da 3 2
fi
d 2A 2 ____2 = __ > 0 3 da
)
2a 1 For maximum or minimum, ___ + __ = 0 3 2
]
fi
|)
2 ___ __ 3 ÷
2a 1 ___ = – __ 3 2
3 a = – __ 4 Hence, the value of a is – 3/4 fi
)]
)
)
1 a2x3 ___ ax2 = ____ + + x 3 2 0
So, the area is minimum.
__ 1 1 1__ ___ 2 __ 1 + 2 ___ __ + __ log ( ÷2 + 1 ) – ____ × __ – log3 2 2÷3 ÷3 4 ÷2 __ 1 1 = 4 ÷2 __ + __ log (3) 3 4
( (
__
_____ 1 y _____ __ 1 + 2 __ ÷1 + y2 + log y + ÷ y 2 + 1 __ 2 1/÷3 2 __
= A = Ú (a2x2 + ax + 1) dx 0
fi
( (
4 2a – __ = 0 3 2 a = __ 3
73. Do yourself. 74. Given curves are y = x2 + 2x – 3 and y = mx + 1
Area Bounded by the Curves
3.33
75. Hence, the required area
( ) ( ) ( ) ( ) 2a
x = A = Ú __ + 6 a
x2 1 2a = ___ – __ 12 x a
Solving, we get x2 + (2 – m) x – 4 = 0 Let a, b be the roots such that a + b = m – 2, ab = – 4
Hence, the required area b
A = Ú (Line – Parabola) dx a
b
= Ú ((mx + 1) – (x2 + 2x – 3)) dx a
(
) ) (
b mx2 x3 = ____ + x – __ – x2 + 3x 2 3 a
( ( ( ( ( ( (
a2 1 1 ___ – ___ + __ 2a 12 a 1 ___ 2a
dA a ___ 1 ___ = __ – da 2 2a2 d 2A 1 1 ____ fi 2 = __ + __ 2 a3 da Now for maximum or minimum, dA ___ = 0 da a 1 __ fi – ___ = 0 2 2a2 a 1 __ fi = ___ 2 2a2 fi a3 = 1 fi a=1 Hence, the value of a is 1, when the area is least.
x2 + 2x – 3 = mx + 1
a2 = __ – 3 a2 = __ + 4
1 __ 2 dx x
))
mb 2 b3 ma2 a3 = ____ + 4b – __ – b 2 – ____ + 4a – ___ – a2 2 3 2 3 (Problems for Jee-Advanced)
)) )) )
(b3 – a3) m = __ (b2 – a2) + 4 (b – a) – ________ – (b2 – a2) 2 3
1. Given curve is a2 y2 = x2 (a2 – x2).
(b3 – a3) m = __ – 1 (b2 – a2) + 4 (b – a) – ________ 2 3
)
( ( ( (
(b2 + ab + a2) m = (b – a) __ – 1 (b + a) + 4 – ____________ 2 3
)
)
(m – 2)2 (m – 2)2 + 4 = (b – a) _______ + 4 – __________ 2 3
a
)
(m – 2)2 ___ 16 = (b – a) _______ + 3 6
(
)
_____________ (m – 2)2 ___ 16 (a + b)2 – 4ab _______ +
= ÷
(
3
6
)
____________ (m – 2)2 ___ 16 (m – 2)2 + 16 _______ +
= ÷
6
Hence, the required area
3
The area is minimum, if m = 2 and the least area is 64 = ___ . 3
= 4 Ú y dx 0 a
0 p/2
(
______
)
x = 4 Ú __ a ÷ a2 – x2 dx = 4 Ú a2 sin2q cosq dq, (Let x = a sinq) 0
1
= 4a2 Ú t2 dt, 0
4a2 = ___ sq.u. 3
(Let t = sinq)
3.34 Integral Calculus, 3D Geometry & Vector Booster
| |
Hence, the required area
2. Given curve is y = (x – 1)(x – 2)(x – 3).
1
1
0
0
= Ú logx dx + Ú sin4 x dx
( 1 – cos2p x = 1 + Ú ( __________ ) dx 2 1
)
1 – cos2p x 2 = |x(logx – 1)|10 + Ú __________ dx 2 0 1
2
0
1
Hence, the required area
1 = 1 + __ Ú (1 – 2cos(2p x) + cos2 (2p x)) dx 4 0
3
= Ú {(x – 1)(x – 2)(x – 3)}dx
1
1
= – Ú {(x –1)(x – 2)(x – 3)}dx 0
2
+ Ú {(x – 1)(x – 2)(x – 3)}dx
1 3
– Ú {(x – 1)(x – 2)(x – 3)}dx
(
2
)|
x 11 = – __ – 2x3 + ___ x2 – 6x 4 2 0
( (
))
(
)
( )
3 11 1 3 = 1 + __ __ – 0 = 1 + __ = ___ 4 2 8 8 4. Given curve is y = 2x4 – x2.
) )
2 x4 11 + __ – 2x3 + ___ x2 – 6x 4 2 1
3 x4 11 – __ – 2x3 + ___ x2 – 6x 4 2 2
(
(
2sin(2p x) ________ sin(4p x) 1 1 3x _________ = 1 + __ ___ – + 4 2 2p 8p 0
1
4
(
1 + cos(4p x) 1 = 1 + __ Ú 1 – 2cos(2p x) + ___________ dx 4 0 2
0
) 1 11 – ( __ – 2 + ___ – 6 ) 4 2 99 81 – ( ___ – 54 + ___ – 18 ) + (4 – 8 + 22 – 12) 4 2 279 23 = – 2( ___ – 8 ) + 12 – ( ____ – 72 ) 4 4 1 11 = – __ – 2 + ___ – 6 + (4 – 8 + 22 – 12) 4 2
46 279 = ___ – ____ + 100 4 4 233 _________ 400 – 233 ____ 167 = 100 – ____ = = sq.u. 4 4 4
3. Given curves are y = loge x, y = sin4 (p x) and x = 0
Hence, the required area
|
1/2
|
= 2 Ú (2x4 – x2) dx 0
| (
)
|
2x5 x3 1/2 = 2 ___ – __ 3 0 5
| (
1 = 2 ___ – 80
)|
1 ___ = 24
7 ____ sq.u. 120
5. Given curves are x2 + y2 = 64 and y2 = 12x.
Area Bounded by the Curves
Hence, the required area
(
4
8
)
= area of a semi-circle – 2 Ú P dx + Ú C dx 0
(
4
8
___ __
4
)
______
= 64p – 2 Ú ÷12 ÷ x dx + Ú ÷64 – x2 dx 0
4
(
(
6.
(
__
( ) )
( ) )
__ 32p 1 = 64p – 24÷3 – 2 ____ – 8÷3 – 32sin–1 __ 2 2 __ 64p = 32p – 8÷3 + ____ 3 __ 16p = ____ – 8÷3 sq.u. 3 Given curves are 1 y = ln(x + e) and x = ln __ y
(
= Ú (mx) dx 0
( ) ( )
9m mx2 3/2 = ____ = ___ 2 0 8
3/2
x x 8 – 2 __ ÷64 – x2 + 32sin–1 __ 2 2 4
3/2
Hence, the area OABCDO
)
__ 4 3 = 64p – 2 2÷3 × __ x3/2 2 0 ______
Now, the area OABO
)
( )
fi y = ln(x + e), y = e– x, y = 0
= Ú (1 + 4x – x2) dx 0
(
)
x3 3/2 = x + 2x2 – __ 3 0
9 27 3 = __ + 2 ◊ __ – ___ 2 4 24
( ) 9 39 = ( 6 – __ ) = ___ sq.u. 8 8
According to the question, 9m ___ = 8 fi
1 39 __ ◊ ___ 2 8
13 m = ___ 6
13 Hence, the value of m is ___ . 6 8. Given curve is
Hence, the required area 0
•
1 – e
0
= Ú log(x + e)dx + Ú e– x dx = ((x + e) log (x + e) – (x + e))01 – e – (e–x)•0 = (eloge – e) – (1log1 – (1)) – (e–x)•0 7.
= (1 – (0 – 1)) = 2 sq.u. Given curve is y = 1 + 4x – x2 fi y = 5 – (x – 2)2
Hence, the required area 3
= Ú (y2 – y1) dx 0
3
= Ú (2x – x2 + x) dx 0
3
= Ú (3x – x2) dx 0
(
)
3x2 x3 3 = ___ – __ 2 3 0
27 27 = ___ – ___ 2 3
( ) 9 27 – 18 = ( _______ = __ sq.u. 2 ) 2
3.35
3.36 Integral Calculus, 3D Geometry & Vector Booster
9. Given curve is
12
(
)
__ __ x2 Also, A2 = Ú 2÷3 ÷ x – ___ dx 12 3
( ( ) )
__ 2 x3 12 = 2÷3 __ x3/2 – ___ 3 36 3
(
) (
)
__ ___ 12 __ __ 27 2 2 = 2÷3 × __ × 12 ÷12 ___ – __ × 2÷3 × 3÷3 ___ 3 3 36 36
(
)
3 = (96 – 48) – 12 – __ 4
| |
3 3 147 = 48 – 12 + __ = 36 + __ = ____ 4 4 4
Hence, the required area 2
|
= Ú {(x – 1)(x – 2)}dx 1
2
|
Thus, A1 : A2 = 45 : 147 = 15 : 49
= Ú (x2 – 3x + 2) dx 1
| (
)|
11. Given curves are x + y £ 6. x2 + y2 £ 6y and y2 £ 8x
2 x3 3x2 = __ – ___ + 2x 3 2 1
| ( | ( | (
8 1 = __ – 6 + 4 – __ + 3 3
)|
3 __ – 2 2
)| 14 + 9 = ______ – 4 ) 6 | 23 1 = ( ___ – 4 ) = __ sq.u. 6 6 7 3 = __ + __ – 4 3 2
Hence, the required area
2
2
10. Given curves are y = 12x and x = 12y and x = 3.
2
______
___
= Ú {( 2÷2x ) – ( 3 – ÷9 – x2 ) } dx 0
3
______
{
}
+ Ú (6 – x) – ( 3 – ÷9 – x2 ) dx 2
(
3
)
__ 2 __ x2 3 2 = 2÷2 ◊ __ ( x3/2 )0 – Ú ( 3 – ÷ 9 – x2 )dx + 6x – __ 3 2 2 0
(
3
)
9 16 = ___ + 18 – __ – 12 + 2 3 2
(
)
__ __ x2 Here, A1 = Ú 2÷3 ÷ x – ___ dx 12 0
( ( ) )
__ 2 x3 3 = 2÷3 __ x3/2 – ___ 3 36 0
__ 2 __ = 2÷3 __ × 3÷3 – 3
3 45 = 12 – __ = ___ sq.u. 4 4
(
)
27 ___ 36
(
_____
( ))
3
x÷9 – x2 __ 9 x – 3x – ________ – sin–1 __ 2 2 3 0 9p 27 16 = ___ – 9 + ___ + ___ – 10 3 4 2
(
)
9p 16 27 = ___ + ___ + ___ – 19 4 3 2
(
)
9p 1 = ___ – __ sq.u. 4 6
Area Bounded by the Curves
12. Given curve is 2
Hence, the required area 2
2
x + y – 6x – 4y + 12 £ 0, y £ x, x £ 5/2
= Ú (x + 2 – x2)dx –1
fi (x – 3)2 + (y – 2)2 £ 1, y £ x, x £ 5/2
(
__________ 1 – (x – 3)2
( (
fi y = 2 ± ÷
__________ 1 – (x – 3)2 as
fi y = 2 – ÷
)
x2 x3 2 = __ + 2x – __ 2 3 –1
fi (y – 2)2 = {1 – (x – 3)2}
)
8 1 1 = 2 + 4 – __ – __ + 2 + __ 3 2 3 1 7 = 8 – __ – __ 2 3
5 x £ __ 2
)
48 – 3 – 14 ___ 31 = __________ = sq.u. 6 6 14. Given curve is y = x log x and y = 2x – x2
Hence, the required area
(
5/2
)
__________
(
)
5 1 1 = __ 2 + __ __ – Ú 2 – ÷1 – (x – 3)2 dx 2 2 2 2 5/2
__________
(
Hence, the required area
)
9 = __ – Ú 2 – ÷1 – (x – 3)2 dx 8 2
( { ( ) ( ( )
1
)
5/2 (x – 3) __________2 __ 9 1 = __ – 2x – _______ 1 – (x – 3) + sin–1 (x – 3) ÷ 8 2 2 2
__
3 ÷ 9 1 1 __ 1 = __ – 2 ◊ __ – __ – × ___ 8 2 2 2 2
)}
1 1 + __ sin–1 – __ – sin–1 (– 1) 2 2
( (
__
(
))
3 1 ÷ 9 p p = __ – 1 + ___ + __ – __ + __ 8 8 2 6 2 __
= Ú ((2x – 2x2) – xlogx) dx 0
( ) ( __ x2 logx – __ x4 ) 2 1 = ( 1 – __ ) + __ 3 4 1 2 = x2 – __ x3 – 3 0
1 = __ + 3 15. Given
2 1
2
0
7 1 __ = ___ sq.u. 4 12 curves are x2 + (y – 1)2 = 1 __
and c2 x2 + y2 = c2 where c = ÷ 2 – 1
)
3 p ÷ 9 = __ – 1 + ___ + __ 8 8 6
(
__
)
3 p 1 ÷ = __ – ___ – __ sq.u. 8 8 6 13. Given curves are x2 = y, y = x + 2 Hence, the required area __
1/÷2
______
______
) = p ◊ (1)2 – 2 Ú ( c ÷1 – x2 – 1 – ÷ 1 – x2 0
__
1/÷2
______
= p ◊ (1) – 2 Ú ( (c + 1) ÷1 – x2 – 1 ) 2
0
3.37
3.38 Integral Calculus, 3D Geometry & Vector Booster
} ) ( { x 1 = p ◊ (1) – 2( 2 { __ ÷1 – x + __ sin x } – x ) 2 2 = p – 2 2 { ____ ( 212 ◊ ___12 + __ 12 ◊ __p4 } – ___ 12 ) ______
__
1/÷2 x 1 = p ◊ (1)2 – 2 (c + 1) __ ÷ 1 – x2 + __ sin–1 x – x 2 2 0
______
__
2
2
÷
__
÷
–1
__
__
÷ ÷
__ 1/÷2
0
__
÷
__
(p – 2)÷2 = p – _________ 4 __ __ 1 = __ { ( 4 – ÷2 )p + 2÷2 } sq.u. 4 16. Given curve is (y – 2)2 = (x – 1) and the equation of the tangent to the curve at (2, 3) is x – 2y + 4 = 0.
Hence, the required area 3
= Ú (x2 – x1) dy 0
3
= Ú ((y – 2)2 + 1 – (2y – 4)) dy 0
3
= Ú (y2 – 4y + 4 + 1 – 2y + 4) dy 0
Hence, the required area
( (
)
1 1__ __ 1 = 3 2 ◊ __ ◊ ___ – p ◊ 1 2 ÷ 3 6
)
p 1 = 3 ___ __ – __ 3 6 ÷
(
__ p = ÷3 – __ sq.u. 2 18. The tangent at x = p to the curve f (x) = |cos x| will be parallel to x-axis and cuts the curve f (x) = cos–1 (cos x) at B and C.
)
Thus, AD = p – 1 Hence, the area of DABC 1 = __ × (2p – 2) × (p – 1) 2 1 = __ × 2(p – 1) × (p – 1) 2 = (p – 1)2 sq.u. 19. Given curves are |y| = 1 – x2, x2 + y2 = 1.
3
= Ú (y2 – 6y +9) dy 0
(
)
3 y3 = __ – 3y2 + 9y 3 0
= 9 – 27 + 27 = 9 sq.u. 17. Here D, E and F are the mid-points of the sides of the trangle ABC respectively.
Hence, the required area
(
1
)
= p – 4 Ú (1 – x2) dx 0
( )
x3 1 = p – 4 x – __ 3 0
( ) 8 = ( p – __ ) sq.u. 3
1 = p – 4 1 – __ 3
Area Bounded by the Curves
20. Given curves are 1 |y| = e–|x| – __ and |x| + |y| = ln 2 2
Hence, the required area
(
ln2
(
22. Given curves are
)
)
1 1 = 4 __ (ln2)2 – Ú e–x – __ dx 2 2 0
(
(
x2 + y2 – 2x + 4y – 11 = 0
and
y = – x2 + 2x + 1 – 2 ÷3
fi
(x – 1)2 + (y + 2)2 = 16
and
y + 2(÷3 – 1) = – (x – 1)2
__
__
( (
ln2
e 1 1 = 4 __ (ln 2)2 – ___ – __ 2 –1 2 0
)
2
= 2 Ú (Parabola – Circle) dx
( )
0
2 = 2 (ln 2) + 2 ln __ e . 21. Given curves are 2
Hence, the required area
) )
–x
3.39
y2 – 2y + 4x + 5 = 0 and x2 + 2x – y + 2 = 0 2
2
fi (y – 1) = – 4(x + 1) and (x + 1) = (y – 1).
2
__
= 2 Ú ((– x2 + 2x + 1 – 2 ÷3 ) 0
____________
(
)) )
– – (x – 1)2 dx – 2 + ÷16
[ (
)
__ 2 x3 = 2 – __ + x2 + x – 2÷3 x 3 0
(
( ) ) ] ÷15 8 1 = 2 ( – __ + 6 – 4 ÷ 3 – ( (– 4) + ____ + 8 sin ( __ ) ) 3 2 4 ÷15 1 + ( – ____ – 8 sin ( __ ) ) ) 2 4 8 = 2 ( 10 – __ – 4 ÷3 ) 3 22 = 2 ( ___ – 4 ÷3 ) sq.u. 3
(x – 1) ___________2 x–1 2 – – 2x + ______ 16 – (x – 1) + 8sin–1 _____ ÷ 2 4 0 ___
__
___
–1
–1
__
Hence, the required area –1
________
Ú __ ( 1 + ÷– 4(x + 1) ) – (1 + (x + 1)2) dx
=
–1 – 3÷4
( (
(
))
–1
(x + 1)3 2 = x + 2 ◊ __ (– (x + 1)3/2 ) – x + _______ 3 __ 3 3 –1 – ÷4
(
3
))
–1
(x + 1) 4 = __ (– (x + 1)3/2) – _______ 3 __ 3 3 –1 – ÷4
( ( (
( ) ) __
( 3 )3
– ÷4 4 = __ – ÷4 ) + _______ 3 3
| (
3 __ 3/2
)|
))
4 4 12 = __ (–2) – __ = ___ = 4 sq.u. 3 3 3
__
23. Given curve is 9x2 + 4y2 – 36x + 8y + 4 = 0 fi
(9x2 – 36x) + (4y2 + 8y) + 4 = 0
fi
9 (x2 – 4x) + 4(y2 + 2y) + 4 = 0
fi
9 (x – 2)2 + 4(y + 1)2 = 36 + 4 – 4 = 36
fi
9 (x – 2)2 + 4(y + 1)2 = 36
(y + 1)2 (x – 2)2 _______ fi _______ + = 1 4 9
3.40 Integral Calculus, 3D Geometry & Vector Booster 1 11 = 6 – __ = ___ sq.u. 2 2 25. The required area will be equal to area enclosed by y = f (x) and the y-axis between the abscissae y = –1 and y = 4 \ f (0) = 1, f (–2) = –1, f (2) = 4 Clearly, f (x) is monotonic in [–2, 2]. Hence, the required area Solving, we get x = 2 and x = 4 Hence, the required area 2
( ÷ ( ) ( ) ) Ú ( ) ÷ ( ) Ú ( ÷ ( ) ) _____________ 2
(x – 2) 10 – 3x = Ú –1 + 3 1 – _______ – _______ dx 4 2 2 ____________ 2
(x – 2) 3x = –1 + 3 1 – _______ – 5 + ___ dx 4 2 2 ____________
4
(x – 2)2 = – 6 + 3 1 – _______ + 4 2
(
3x ___ dx 2
( ÷ ( ) 3x x–2 ( _____ ) ) + ___ ) 2 4
___________
x–2 x – 2 = – 6x + 3 _____ 1 – _____ 4 2 2
1 + __ sin–1 2
(
\
–2
(
)
2
( )
0
(
)
x3 x2 13 + Ú ___ + __ + ___ x + 2 dx 8 12 –2 24
(
)
x4 x3 13x2 2 = 3x – ___ – ___ – ____ 24 0 96 24
(
)
0 13x2 x4 x3 + ___ + ___ + ____ + 2x 24 96 24 –2
(
) (
)
)
)
16 8 52 16 8 52 = 6 – ___ – ___ – ___ – ___ – ___ + ___ – 4 96 24 24 96 24 24
| | |
32 = 10 – ___ – 96
2 4
p = – 24 + 3 __ + 12 + 12 – 3 4 p = 3 __ – 1 sq.u. 4 24. The required area will be equal to area enclosed by y = f (x) and the y-axis between the abscissae y = – 2 and y = 6.
(
0
13 x3 x2 ___ = Ú 3 – ___ – __ – x dx 24 8 12 0
= Ú (E – L) dx
4
0
2
4
4
2
= Ú (4 – f (x)) dx + Ú ( f (x) – (–1)) dx
|
104 ____ 24
|
1 13 = 10 – __ – ___ 3 3 14 = 10 – ___ 3 16 = ___ sq.u. 3 26. Given curves are y = xe x, y = xe–x
)
f (0) = 2, f (–1) = – 2, f (1) = 6
Clearly, f (x) is monotonic in [–1, 1]. Hence, the required area 1
–1
0
0
= Ú (6 – f (x)) dx + Ú ( f (x) – (– 2)) dx 1
–1
0
0
= Ú (4 – x3 – x) dx + Ú (x3 + 3x + 4) dx
) (
(
x2 1 __ + 2 0
x4 = 4x – __ – 4
| ( |
) (
x4 __ + 4
)
0 3x2 ___ + 4x 2 –1
)|
1 1 1 3 = 4 – __ – __ – __ + __ – 4 4 2 4 2 2 = 8 – __ – 2 4
|
Hence, the required area 1
= Ú (xex – xe–x) dx 0
1
= (x (ex + e–x))10 – Ú (ex + e–x) dx 0
Area Bounded by the Curves
(
) (
3.41
)
1 1 __ = e + __ e – e – e
2 = __ e sq.u. 27. Given curves are
|x – 2y| + |x + 2y| £ 8 and xy ≥ 2.
Hence, the required area 4
Hence, the required area 4
(
)
(
)
2 = 2 Ú 4 – __ x dx 1/2
= 2 (4x – 2 log x)21/2
1 = 2 8 – 2log2 – 2 + 2 log __ 2
= 2 (6 – 4 log 2) sq.u.
(
( ) )
30. Given curves are
2 = 2 Ú 2 – __ x dx 1
= 2 (2x – 2 log x)41
= 2 (8 – 2 log 2 – 2)
= 2 (6 – 2 log 2) sq.u.
|x + y| £ 2 and x2 + y2 ≥ 2.
28. Given curves are |x| + |y| ≥ 3 and xy ≥ 2.
Hence, the required area
(
__
)
p ◊ (÷2 )2 1 = 2 __ × 2 × 2 – _______ 2 4
p = 4 – __ sq.u. 2
(
)
Hence, the required area 2
(
)
2 = Ú 3 – x – __ x dx 1
2 x2 = 3x – __ – 2 log x 2 1
1 = 6 – 2 – 2 log 2 – 3 + __ 2
(
)
( 3 = ( __ – 2 log 2 ) sq.u. 2
)
29. Given curves are
|x – y| + |x + y| £ 8 and xy ≥ 2.
(Tougher Problems for JEE-Advanced) 1. Given curve is
2x2 + 6xy + 5y2 = 1
Let x = r cos q, y = r sin q
1 fi r2 = __________________________ 2 2 cos q + 6 sinq cosq + 5 sin2q
Hence, the required area p/2
= Ú r2 dq –p/2
3.42 Integral Calculus, 3D Geometry & Vector Booster p/2
(
)
1 = Ú _________________________ dq 2 2 –p/2 2cos q + 6sinq cosq + 5sin q
sec2q = Ú _________________ dq 2 –p/2 2 + 6 tanq + 5 tan q
p/2
•
(
(
•
(
)
dt 1 = __ Ú __________ 6 5 – • 2 __ 2 t + t + __ 5 5 •
)
dt = Ú __________ 2 , – • 5t + 6t + 2
( (
–1
( ( ) )
= p sq.u.
y – 2 = ± 2 (x – 1)
(Let t = tan q)
y = ± 2x
)
•
– •
Hence, the required area 1
= 4 Ú (a2/3 – x2/3)3/2 dx 0
0
= 4 Ú (a sin3q) (– 3a cos2q sinq) dq p __ 2
(Let x = a cos2q) p __ 2
= 12 a2 Ú sin4q cos2q dq
3 1 1 p = 12a2 × __ ◊ __ ◊ __ ◊ __ 6 4 2 2
( )
Now,
m = ( f ¢(x))x = 0 = 0 y – b = m(x – a)
fi
y – b = 2a (x – a)
Since the point (a, b ) lies on the curve, so
2
1 – 2x2 = 0 1__ fi x = ± ___ 2 ÷ 1 By the sign scheme, it is maximum at x = ___ __ 2 ÷ 1 Thus, c = ___ __ 2 ÷ Hence, the required area
Equation of the tangent at (a, b) is
2
e–x – 2x2 e–x = 0
fi
3. Clearly, f (x) = x2 + 1 f ¢(x) = 2x
)
2 2 dy fi ___ = e–x – 2x2 e–x dx
fi
3pa2 = ____ sq.u. 8
fi
)
dy For maximum or minimum, ___ = 0 dx
0
)
( (
0 a
(
1 x3 x2 = 2 __ – __ + x 3 2 0 1 1 = 2 __ – __ + 1 3 2 4 1 = 2 __ – __ 3 2 5 = 2 × __ 6 5 = __ sq. u. 3 4. Given curve is 2 y = xe–x
= 4 Ú y dx
= 2 Ú (x2 + 1 – x) dx 0
a
a = ± 1, b = 2
Hence, the equation of the tangent is
Hence, the required area
From Eq. (i) and (ii), we get
)
2. Given curve is x2/3 + y2/3 = a2/3.
...(ii)
) ( ) 5t + 3 1 = __ × ( 5 tan ( ______ 1 )) 5
b = a2 + 1
dt = 1/5 Ú _____________ 3 2 1 – • __ t + + __ 5 5
p p = __ – – __ 2 2
c
= Ú x e–x dx 2
0
...(i)
– c2
1 = – __ Ú et dt, (Let – x2 = t) 2 0
Area Bounded by the Curves 2 1 = – __ (et)– c 0 2
2 1 = – __ (e– c – 1) 2
2 1 = __ (1 – e– c ) sq. u. 2 1 The area is maximum, when c = ___ __ 2 ÷ Hence, the maximum area
)
1 – __ e 2
1 = __ 1 – 2 1 1 = __ 1 – ___ __ ÷e 2 5. Let the required area be A.
(
)
0
)
1 a2 x3 ___ ax2 = ____ + + x 3 2 0
a2 a = __ + __ + 1 3 2
)
(
)
dA 2a 1 fi ___ = ___ + __ 3 2 da d 2A fi ____2 = da
fi
a3 = 3a
fi
a2 = 3
A = Ú (a2 x2 + ax + 1) dx
( (
fi
a3 2 __ + a = (a3 + a) 3 2a3 + 6a = 3a3 + 3a
2 __ > 0 3
So, the area is the least. For maximum or minimum,
7. Given curve is
fi x–2=0 fi x=2 So c=2 Hence, the required area is 2
2a fi ___ + 3
1 __ = 0 2 3 fi a = – __ 4 Hence, the value of a is –3/4. 6. Given curve is y = x2 + 1
a
= Ú (x2 + 1) dx 0
= Ú xe– x dx 0
2
= (– xe– x)20 + Ú e– x dx – x
= – (xe
+
0 – x 2 e )0
= (1 – 3e–2) sq.u. 8. Given curves are |x| + |y| ≥ 1 and x2 – 2x + 1 £ 1 – y2 |x| + |y| ≥ 1 and (x – 1)2 + y2 £ 1 fi
Hence, the required area
y = xe– x
dy fi ___ = e– x – xe– x dx d 2y fi ___2 = – e– x – e– x + xe– x dx For point of inflection, d 2y ___2 = 0 dx fi – e– x – e– x + xe– x = 0
dA ___ = 0 da
__
fi a = ÷3 __ Hence, the value of a is ÷ 3 .
1
Thus,
x3 = __ + x 3 0 a3 __ = + a 3 It is given that,
(
( ) ( ) ( )
a
3.43
Ï x + y ≥ 1, x + y £ - 1 and (x – 1)2 + y2 £ 1 Ì Ó x - y ≥ 1, x - y £ - 1
3.44 Integral Calculus, 3D Geometry & Vector Booster Hence, the required area
(
1
1 = 2 Ú ÷1 – (x – 1)2 dx + __ × 1 × 1 2 0 =
(
(x – 1) __________2 2 ______ 1 – (x – 1) +
÷
2
(
___________
| |
|
1
+ 2 Ú ÷10 – (x + 1)2 dx + Ú (1 – x)dx
)
__________
|
___
÷10 – 1
( ) )
1
x–1 1 __ sin–1 _____ + 1 2 2 0
( ) )
(
2
1
)
2
x3 = – __ + 2x2 – 3x 3 1
[
___________ (x + 1) 10 – (x + 1)2 ___________________
÷
( ) ] ( ) ___
÷10 – 1
x + 1 + 5sin–1 _____ ___ 2 ÷10 2
1 1 1 = 2 __ + __ sin–1 __ + 1 2 2 2
+ 2
p = 2 + __ sq.u. 6
x2 2 – x – __ 2 1 8 1 = – __ + 8 – 6 + __ – 2 + 3 3 3
(
)
9. Given curves are |x + y| £ 1, |y – x| £ 1 and 3x2 + 3y2 ≥ 1.
(
(
)
( ) )
3 3 1 + 2 0 + 5sin–1 (1) – __ – 5 sin–1 ____ ___ + __ 2 2 ÷10
( ) ( ) )
3 2 1 = __ + __ + 5p – 3 – 10 sin–1 ____ ___ 3 2 ÷10
(
3 11 = 5p – ___ – 10 sin–1 ____ ___ sq.u. 6 ÷10 11. Given curves are (x2 + y2) £ 4 £ 2(|x| + |y|) Hence, the required area = area of the square – area of the circle p = 2 – __ sq.u. 2 10. Given curves are
(
)
|x – 2| + |y – 2| £ 3, x2 – 4x + y + 3 £ 0
and
x2 + y2 + 2x – 9 £ 0
fi
|x – 2| + |y – 2| £ 3, (x – 2)2 £ – (y + 1)
and
(x + 1)2 + y2 £ 10
fi
(x – 2)2 £ – (y + 1), (x + 1)2 + y2 £ 10
Ï x + y £ 7, x + y ≥ 1 Ì Ó x - y £ 3, x - y ≥ - 3
Hence, the required area = area of a circle-area of a square = 4p – 4 = 4 (p – 1) sq.u. 12. Given curves are |4 – x2| y = _______ and y = 7 – |x| 4 |x2 – 4| fi y = _______ and y = 7 – |x| 4
Hence, the required area 2
= Ú (– x2 + 4x – 3) dx 1
Area Bounded by the Curves
Hence, the required area
[
( ) ( ) ( ) ] 2
)
4
]
4 – x2 x2 – 4 1 = 2 __ (7 + 3) × 4 – Ú ______ dx – Ú ______ dx 2 4 2 0 2
[ (
3 2
x = 2 40 – x – ___ – 12 0
[ ( [
On solving, we get,
____
[ ( ) ( ) ] Ú [ ( ) ] Ú [ ( )] Ú [ ( )] ___
x ___ – x 12 2
) (
____
B (b – a, ÷4ab ) , B ¢(b – a, – ÷4ab )
Hence, the required area
4
2
2÷ab
)]
3.45
y2 = Ú ___ b – ___ – 4b – 2÷ab
y2 ___ – a dy 4a
8 64 8 = 2 20 – 2 – ___ – ___ – 4 – ___ + 2 12 12 12 2 16 2 = 2 18 + __ – ___ + 4 + __ – 2 ] 3 3 3
2 1 y = ___ (a + b) – __ __ a + 4 – 2 ÷ab
= 32 sq.u.
2 1 y = 2(a + b) ◊ 2 ÷ab – __ __ a + 4 0
÷ab y2 y2 1 = 4 (a + b) ÷ab – __ __ a + __ dy 2 0 b
3 1 y = 4 (a + b) ÷ab – __ ___ + 2 3a
]
13. Given curves are y = x2 + x – 2, y = 2x where
___
2 ÷ab
___
x (x + x – 2) ≥ 0
fi
x (x – 1) (x + 1) ≥ 0 2
Solving, y = x + x – 2, y = 2x, we get,
x = – 1, x = 2
___
÷ab
y2 __ dy b
___
2
y2 __ dy b
___
___
( (
)
___
y3 2 ÷ab ___ 3b 0
___
___
)
___ _______ 8ab ÷ab 1 8ab ÷ab = 4 (a + b) ÷ab – __ _______ + 2 3a 3b ___ ___ 4 = 4 (a + b) ÷ab – __ (a + b) ÷ab 3 ___ 8 = __ (a + b) ÷ab sq.u. 3
15. Given curve is y = – x2 – 2x + 3 dy fi ___ = – 2x – 2 dx
( )
dy fi ___ = – 6 dx x = 2
Hence, the required area 0
2
–1
1
= Ú [2x – (x2 + x – 2)] dx + Ú [2x – (x2 + x – 2)]dx
[
] [
The equation of the tangent to the given curve at P (2, – 5) is 6x + y = 7
]
0 2 x2 x3 x2 x3 = __ – __ + 2x + __ – __ + 2x 2 3 2 3 –1 1
[ (
)]
10 13 7 = __ + ___ – ___ 3 6 6 14 7 = ___ = __ sq.u. 3 6 14. Given curves are y2 = 4a(x + a) and y2 = 4b(b – x)
| |
Hence, the required area
3
1 = __ ◊ 2 ◊ 12 – Ú x dy 2 – 5
= 12 – Ú (–1 + ÷ 4 – y ) dy
|
3
_____
– 5
| [
]
|
|
3 2 = 12 – – y – __ (4 – y)3/2 3 – 5
3.46 Integral Calculus, 3D Geometry & Vector Booster
[
]
2 = 12 – 10 – __ 3
Hence, the required area
and
y = – x + 2x + (1 – 2 ÷3 )
fi
(x – 1)2 + (y + 2)2 = 16
and
y = – (x – 1)2 + 2(1 – ÷ 3 )
]
_______
( Ú ( ÷
)
1 Let a = ________ _______ 2 ÷ a + b2
__
2
1 __ b
________
÷1 – a 2 x2 – b 2 x2 ÷ 1 = 4 Ú _________ + Ú ________ dx a b a 0
x2 + y2 – 2x + 4y – 11 = 0
[
a
8 = __ sq.u. 3 16. Given curves are
[
a
( ÷
_______
1 __ b
)
_______
)
]
a b 1 1 = 4 __ Ú __ 2 – x2 dx + __ a __ 2 – x2 dx b 0 a a b
__
(
Solving, we get x = –1, 3
( ) )
a 4 = ___ tan–1 __ sq.u. ab b 18. Given curves are 1 |y| + __ £ e–|x| and {|x|, |y|} £ 2 2
Hence, the required area 3
= Ú (yp – yc) dx –1
Hence, the required area
3
__
= Ú [(– x2 + 2x + 1 – 2 ÷3 ) – (– 2 – –1
___ ÷16 –
(x – 1)2)] dx
ln 2
[
__ (x – 1) ____________2 x3 = – __ + x2 + (3 – 2 ÷3 ) x + ______ ÷16 – (x – 1) 3 2
( ) ] )
16 x–1 3 + ___ sin–1 _____ 4 2 –1 __ 8p 32 = ___ – 4÷3 + ___ sq.u. 3 3 17. Given curves are
(
a2 x2 + b2 y2 = 1 and b2 x2 + a2 y2 = 1 On solving, we get, 1 _______ x = ± ________ =y 2 ÷ a + b2
(
)
1 = 4 Ú e–x – __ dx 2 0 ln 2 x = 4 – e– x – __ 2 0
(
)
( ) ( ) 1 ln2 = 4 ( __ – ___ ) 2 2
ln2 = 4 – e– ln 2 – ___ + 1 2 1 ln2 = 4 – __ – ___ + 1 2 2
= (2 – 2 ln 2) sq.u. 19. Given curves are y = min{x3, |x – 2|, e3 – x}, x-axis, y-axis and x = 4.
Area Bounded by the Curves
On solving, we get,
2
A = (1, 1), B = (3, 1), C = (4, e)
= Ú 1 dx = (x)2–2 = 4 –2
Hence, the required area
1
2
3
4
0
1
2
3
2. Given curve is |x| + |y| = 1
= Ú x3 dx + Ú (2 – x) dx + Ú (x – 2) dx + Ú (e3 – x) dx
( ) (
) (
)
3 x4 1 x2 2 x2 = __ + 2x – __ + __ – 2x + ( – e3 – x )43 4 0 2 1 2 2
( 9 1 = ( __ – __ sq.u. 4 e)
)
1 1 __ 1 1 = __ + __ + + 1 – __ e 4 2 2
Hence, the required area
20. Given curves are
S1:{(x, y): x2 + 2y2 £ 2}
= ar (ABCD)
S2: {(x, y): 2x2 + y2 £ 2}
= 4 (ar D OAB)
and
S3:{(x, y): x2 + y2 £ 2}
1 = 4 × __ × 1 × 1 2
= 2 sq.u.
3. Given curve is |x – 2| + |y – 2| = 1
The area of the first quadrant ___
÷2/3
__ ______
= Ú ( ÷2 ÷ 1 – x2 – x ) dx 0
( (
______
__
( ) )
)
___
Hence, the required area
x x x2 ÷2/3 1 = ÷2 __ ÷1 – x2 + __ sin–1 __ – __ 2 2 2 2 0
1 2 1 1 = ___ __ sin–1 __ + __ – __ 3 3 3 2 ÷
= 2 sq.u. 4. Given curve are
1 2 = ___ __ sin–1 __ 3 2 ÷ Hence, the required area
eln (x + 1) ≥ |y|, |x| £ 1
x + 1 ≥ |y|, |x| £ 1
|
( ÷ ) ( ÷ ) __
(
2p = 8 ___ – 8
(
|
__
1 = 4 × __ × 1 × 1 2
( ÷ ) ) ( ÷ ) ) __
1 2 ___ __ sin–1 __ 3 2 ÷ __
__ 2 = 2p – 4 ÷2 sin–1 __ sq.u. 3
Integer Type Questions
[
( ) ]
x x Œ [–2, 2], y = sin2x + cos __ = 1 4 Hence, the required area
1. For
Hence, the required area
3.47
3.48 Integral Calculus, 3D Geometry & Vector Booster
1 = 2 × __ × 2 × 2 2
= sq.u.
5. Given curves are
y = ln |x| and y = 1 – |x|.
Hence, the required area 1
= Ú (e – e x) dx 0
Hence, the required area
[
]
0
1 = 2 __ × 1 × 1 + Ú e x dx 2 – •
(
)
y = 3 – |x| and y = |x – 1|
Here,
AB = ÷1 2 + 12 =÷ 2
and
BC = ÷2 2 + 22 =÷ 8 = 2 ÷2
__
_______
__
= AB × BC
= ÷2 × 2 ÷2 = 4 sq.u.
__
= (e – e + 1)
= 1.
9. Given curves are
1 £ |x – 2| + |y + 1| £ 2
Putting
X = x – 2, Y = y + 1
Thus, the equation redices to 1 £ |X| + |Y| £ 2
Hence, the required area __
__
7. Given curves are
|x| y = __ x , x π 0 and y = x (x – 1)(x – 3)
Ï1 : x > 0 y= Ì and y = x (x – 1) (x – 3) Ó-1: x < 0
8. Given curves are y = e x, x = 0, y = e
Hence, the required area
= (e – e x )10
3 1 = 2 __ + 1 = 2 × __ = 3 sq.u. 2 2 6. Given curves are
_______
(
3 = 4 × __ = 6 sq.u. 2 10. Given curve is |y| = sin (2x), where 0 < x < 2p
)
1 1 = 4 __ × 2 × 2 – 2 × __ × 1 × 1 2 2
Area Bounded by the Curves
Hence, the required area p/2
= 4 Ú (sin (2x)) dx 0
(
)
cos(2x) p/2 = 4 – _______ 2 0
= – 2 (–1 – 1)
= 4 sq.u. 11. Given curve is
Hence, the required area,
( ) ( ( ) ) ( ) ( ( ) ( ) ) ( ) ( ) ( ) ( ) a
|x + y| + |x – y| £ 4, |x| £ 1, y ≥ |x – 1|
(
)
1 1 4 Ú __ x – ______ dx = log ___ __ 2x – 1 ÷5 2 a
fi fi fi fi
Hence, the required area
1 = __ × 2 × 2 2
= 2 sq.u.
4 £ x2 + y2 £ 2 (|x| + |y|)
(
64 a2 log ______ = log ___ 2a – 1 15
a2 ______ 2a – 1
| |
64 = ___ 15
1. Given curves are x2 = 4y and x = 4y – 2
Hence, the required area
Hence, the required area __
16 a2 4 log ______ – log __ = log ___ 2a – 1 3 5
fi a=8 Hence the value of a is 8.
12. Given curves are
x2 1 4 __ log ______ = log ___ __ 2 2x – 1 2 ÷5
)
2
= Ú (y2 – y1) dx
p (÷2 ) 2 = 4 ______ – (p – 2) 2
= 4 (p – (p – 2))
x + 2 = Ú _____ – 4 –1
= 8 sq.u.
1 = __ Ú (x + 2 – x2) dx 4 –1
x3 2 1 x2 = __ __ + 2x – __ 4 2 3 –1
13. Given curves are
1 1 ______ y = __ , x = 2, x = a, where a > 2 x , y = 2x – 1
–1 2
(
)
x2 __ dx 4
2
(
)
3.49
3.50 Integral Calculus, 3D Geometry & Vector Booster
) ( )) ( ( 9 7 1 10 __ 1 27 = __ ( ___ + ) = __ ( ___ = __ sq.u. 4 3 4 6) 8 6
8 1 1 1 = __ 2 + 4 – __ – __ – 2 + __ 4 3 2 3
2. The required area 1 = 4 × __ × 1 × 1 = 2 sq.u. 2
b
3. Given Ú f (x) dx = (b – 1) sin (3b + 4) 1
Differentiating both sides w.r.t. b, we get f (b) = sin (3b + 4) + 3 (b – 1) cos (3b + 4)
Hence, the required area
fi f (x) = sin (3x + 4) + 3 (x – 1) cos (3x + 4)
| | 0
4. Given curve is
= Ú tan x dx +
= (log secx)0 __p + (log sec)04
p – __ 3
|
It is given that
(
4
)
(
)
8 8 Ú 1 + __ 2 dx = Ú 1 + __ 2 dx a 1 x x
Ú
p __ 2
+ Ú cot x dx p __ 4
p __
|
– 3
p __
+ (log (sin x))p2
a
p __ 4 tan x dx 0
( )
( )
1 1 = log (2) + log ___ __ – log ___ __ 2 2 ÷ ÷
= log (2) sq.u. 6. Given curves are
( ) ( ) 8 8 fi ( a – __ a – 2 + 4 ) = ( 4 – 2 – a + __ a )
8 a 8 4 fi x – __ x 2 = x – __ x a
16 fi 2a – ___ a = 0 fi 2a2 = 16
Hence, the required area
fi a2 = 8
2
__
fi a = 2÷2
p p Ï ÔÔ y = tan x : - 3 £ x £ 3 Ì Ô y = cot x : p £ x £ p ÔÓ 6 2
2
_____
= Ú ÷5 – x2 dx – Ú |x – 1|dx –1
__
Hence, the value of a is 2÷2 . 5. Given curves are
__ 4
–1
( ) 3 5p = ( __ + ___ ) – 2 2 4 5p 1 = ( ___ – __ ) sq.u. 4 2 3 5p 5 = __ + ___ – __ 2 4 2
7. Given curves are
Area Bounded by the Curves __
Hence, the required area
(
__
0
( )
÷2
)
(
)
(
)
__
÷2 _____
x2 = Ú __ x dx + Ú – ___ dx – Ú __ ÷5 – x2 dx R2 0 – ÷2 – ÷2
( ) ( )
5 = – __ + (2 + 2p) 3 1 = 2p + __ sq.u. 3
Hence, the equation of the tangent is p y – 1 = 2 x – __ 4 p __ y – 1 = 2x – 2 p y = 2x + 1 – __ 2
x2 + y2 = 4, x2 = – ÷2 y and y = x
Hence, the required area p __ 4
x2 + y2 = 25, 4y = |4 – x2| and x = 0
( ( __ p4 – __ 12 ) ) × 1 p 1 1 p (log (sec x)) – __ ( __ – __ – __ 2 4 ( 4 2 ))
1 p Ú tan x dx – __ __ – 2 4 0
8. Given curves are
p __ 4 0
__ 1 log ( ÷2 ) – __ 4
1 1 __ log (2) – __ 2 4 10. Given curve is y = x (x – 1)2
Hence, the required area
( ( ) ( ) ( Ú ( ) Ú ( ) 2
4
5 ______
) )
p ◊ 52 x2 x2 = ____ – Ú 1 – __ dx + Ú __ – 1 dx + Ú ÷25 – x2 dx 4 4 0 2 4 4 2
4
5 _______
25p x2 x2 = ____ – 1 – __ dx + __ – 1 dx + Ú ÷25 – x2 d x 4 4 0 2 4 4
(
( ) )
25p 25p ___ 25 4 8 ____ 4 = ____ – __ + __ + – sin–1 __ + 6 4 3 3 4 2 5
Hence, the required area
25 4 = 2 + ___ sin–1 __ sq.u. 2 5
(
( ) )
9. Given curve is y = tan x dy fi ___ = sec2x dx
= Ú (2 – x (x2 – 2x + 1))dx 0
2
( )
dy p fi ___ x = __p = sec2 __ = 2 4 dx 4 p When x = __ , then y = 1 4
0
2
( )
2
= Ú (2 – x (x – 1)2)dx
= Ú (2 – (x3 – 2x2 + x))dx 0
(
)
x4 2 3 __ x2 2 = 2x – __ + __ x – 4 3 2 0
3.51
3.52 Integral Calculus, 3D Geometry & Vector Booster
(
)
16 = 4 – 4 + ___ – 3 3
10 = ___ sq.u. 3 11. Given curves are ln x y = ex ln x, y = ___ ex and x = 1
1
( 1
)
( )
4 – ÷2 __ 5 3 = ______ – log 2 + __ sq.u. 2 2 log 2
1
1 x2 x2 1 1 (log x) = __ e ______ + __ log x – __ 2 2 4 1/e 1/e 1
1 1 1 1 = – ___ – e ___ 2 – __ + ___ 2e 4 4e2 2e
e 1 1 1 = – ___ – ___ + __ – ___ 2e 2e 4 4e
) (
{
)
( ( 1
log x 1 = __ e Ú ____ x dx – e Ú x log x dx 1/e 1/e 2 1/e
__
2 13. Given curves are y = x2 and y = ______ 2 1 + x
(
( ) )
Hence, the required area
log x = Ú ____ ex – ex log x dx 1/e
) (
Hence, the required area
( (
__
2 ÷ 1 1 ____ – 2log 2 – 2 – __ log __ 2 2 log 2
4 = ____ – log 2
) ( ) }
)
2 = 2 Ú _____ 2 – x2 0 1 + x
)
x3 1 = 2 2tan–1x – __ 3 0 p 1 = 2 2 ◊ __ – __ 4 3 2 = p – __ sq.u. 3 14. Given curves are y = |x – 1| and y = 1
(
(
)
)
| |
e2 – 5 = _____ 4e 12. Given curves are
1 y = ln x, y = 2x, x = __ and x = 2 2 Hence, the required area 1 = __ × 2 × 1 = 1 sq.u. 2 15. Given curves are y = 4x – x2 and y = x2 –
Hence, the required area 2
= Ú (2x – log x) dx 1/2
( )
2x 2 = ____ – (x log x – x)21/2 log 2 1/2
Area Bounded by the Curves 5/2
Now,
A1 = Ú ((4x – x2) – (x2 – x)) dx
3.53
x2 = 1 + 2y, x2 = 1 – 2y
0
5/2
= Ú (5x – 2x2) dx
0
(
)
5x2 2x3 5/2 = ___ – ___ 2 3 0
125 ____ 250 125 = ____ – = ____ sq.u. 8 24 24
(
)
1
and
A2 = Ú (0 – (x2 – x)) dx 0
(
)
x2 x3 1 = __ – __ 2 3 0
1 = __ – 2
1 __ = 3
1 __ sq.u. 6
Now, Area of the above x-axis = A1 – A2
125 = ____ – 24
1 __ = 6
121 ____ 24
Hence, the required ratio
= (A1 – A2) : A2
121 1 = ____ : __ = 121 : 4 24 6
16.
Equation of the sides of a square are x = 1, x = –1, y = 1 and y = –1 Let (x, y) be any point inside the region S. Thus, according to the given conditions ______
÷x 2 + y2 < |1 + x|, |1 – x|, |1 + y|, |1 – y| fi
(x2 + y2) < (1 + x)2, (1 – x)2, (1 + y)2, (1 – y)2
y < 1 – 2x, y2 < 1 + 2x
and
2
2
x < 1 – 2y, x < 1 + 2y
The region S is the region lying inside the four parabolas y2 = 1 – 2x, y2 = 1 + 2x
The parabola y2 = 1 – 2x, x2 = 1 – 2y intersect the line y __= x and the point of intersection is (a, a), where a=÷ 2 – 1. Let A = Area of the region OPQO = Area of OPR + area RPQR 1/2
_______ 1 – 2x) dx = __ a2 + Ú ÷(1 2 a
1/2 1 2 1 × __ × (1 – 2x)3/2 = __ a2 – __ 2 3 2 a
1 1 (1 – 2a)3/2 = __ a2 + __ 2 3
(
)
a2 a3 = __ + __ 2 3 2 a = __ (3 + 2a) 6 __ __ 1 ) ( 3 + 2÷2 – 2 ) = __ ( 3 – 2÷2 6 __ 1 2 – 5 ) sq.u. = __ ( 4÷ 6 Hence, the required area
= Area of the shaded region
= 4(2A) = 8A
__ 8 = __ ( 4÷2 – 5 ) 6
__ 4 = __ ( 4÷2 – 5 ) sq.u. 3 17. It is given that, dy ___ = (2x + 1) dx
dy = (2x + 1)dx Integrating, we get
y = x2 + x + c
which is passing through the point (1, 2), so 2=1+1+c fi c=0
3.54 Integral Calculus, 3D Geometry & Vector Booster Hence, the equation of the curve is
y = x2 + x
Hence, the required area = A t
x2 = Ú x – __ dx t 0
x2 x3 = __ – __ 2 3t 01
t2 t2 = __ – __ 2 3
0
(
)
x3 x2 1 = __ + __ 3 2 0 1 = __ + 3
1 __ = 2
5 __ sq.u. 6
18. Given curves are
p y = tannx, y = 0, x = 0, x = __ . 4 p __ 4
Now,
An = Ú (tan x) dx n
0
Thus,
p __ 4
p __ 4
0
0
= Ú (tannx)dx + Ú (tann – 2x)dx p __ 4
= Ú (tann – 2x) (tan2x + 1)dx 0
p __ 4
= Ú (tann – 2x)sec2x dx 0
1
= Ú tn – 2dt, t = tan x 0
( )
t n – 1 1 = _____ n – 10
1 = _____ n–1
| |
19. Given curves are
(
)
( (
) )
x2 y = x – bx2 and y = __ b
t
3t2 – 2t2 = ________ 6 t2 = __ 6 2 b 1 = __ ______ 2 6 b +1 Taking logarith of both sides, we get
(
An + An – 2
)
t
= Ú (x2 + x) dx
) (
x2 b = Ú x – bx2 – __ dx, Let t = ______ 2 b 0 b +1
Hence, the required area 1
(
)
log A = 2log b – 2log (b2 + 1) – log 6
4b 1 dA 2 _______ fi __ ___ = __ – A db b (b2 + 1)
(
)
2 (b2 – 2b + 1) dA fi ___ = A _____________ db b (b2 + 1) dA For maximum or minimum, ___ = 0 gives db b = 1 Since b > 0, so b = 1 Thus A is maximum, when b = 1 20. Let the co-ordinates of P be (x, y) Equation of the line OA : y __= 0 Equation of the line OB : ÷__3 y = x Equation of the line AB : ÷3 y = 2 – x d (P, OA) = Distance of P from the line OA = y d (P, OB) = Distance of P from the line OB |R3y – x| = ________ 2 d (P, AB) = Distance of P from the line AB
|R3y – x – 2| = ___________ 2
Area Bounded by the Curves
It is given that, d (P, OA) £ min {d (P, OB), d (P,OA)}
3.55
From the graph, it is clear that
1 Ï 2 : 0£ x£ Ô(1 – x ) 3 Ô 1 Ô f (x) = Ì2 x(1 – x ) : £ x £ 1 3 Ô 1 Ô 2 : £ x £1 Ôx 3 Ó
Hence, the required area
{
__
}
|R3y – x| |___________ y – x + 2| ÷3 y £ min _______ , 2 2 |R3y – x| |R3y – x + 2| y £ ________ and y £ ___________ 2 2
|R3y – x| , then Case I: When y £ ________ 2 __ y ) ( x – ÷ 3 y £ ________ 2
( 2 + ÷3 ) y £ x
fi
y £ ( 2 – ÷3 ) x
fi
y £ (tan15°) x
__
__
|÷3 y – x + 2| , then Case II: When y £ ____________ 2 __ 2y £ 2 – x – ÷ 3 y __
fi
( 2 + ÷3 ) y £ 2 – x
fi
y £ ( 2 – ÷3 ) (2 – x)
fi
y £ – (tan (15°)) (x – 2)
__
2/3
1
0
1/3
2/3
(
) (
) ( )
(1 – x)3 1/3 2x3 2/3 = – _______ + x2 – ___ + 3 3 1/3 0
1 2 3 __ 1 = – __ __ + 3 3 3
__
fi
1/3
= Ú (1 – x)2dx + Ú 2x (1 – x)dx + Ú x2dx
x3 1 __ 3 2/3
( ( ) ) 2 2 2 1 2 1 + ( __ ) – __ ( __ – ( __ ) + __ ( __ 3 3 3) 3 3 3) 1 1 __ 2 + ( __ – __ 3 3(3) ) 2
3
2
3
3
17 = ___ sq.u. 27
22. Let C1 and C2 be the graphs of dunctions y = x2, y = 2x, 0 £ x £ 1 respectively. Let C3 be the graph of a function
y = f (x), 0 £ x £ 1, f (0) = 0.
Hence, the area of the shaded region
1 = __ × base × height 2
1 = __ × 2 × (1 ◊ tan15°) 2
= tan 15°
= ( 2 – ÷3 ) sq.u.
__
21. Given curve is
f (x) = max{x2, (1 – x)2, 2x (1 – x)}
Let the co-ordinates P be (x, x2), where 0 £ x £ 1. x
Area (ORPO)
= Ú t2 0
=
0
( ) ( ) ( ) ( ) x
Area (ORPO)
x
– Ú f (t)dt
_ ÷t dt
Ú 0
x2
t2 – Ú __ dt 0 2
x t2 x 2 = __ t3/2 – __ 3 4 0 0
2x3 x4 = ___ – __ 3 4
2
3.56 Integral Calculus, 3D Geometry & Vector Booster According to the question,
(
x
)
x3 2x3 x4 __ – Ú f (t)dt = ___ – __ 3 0 3 4 Differentiating both sides w.r.t. x, we get
x2 – f (x) = 2x2 – x3 3
(
)|
0
x2 x3 = (1 – m) __ – __ 2 3 1 – m
1 = – __ (1 – m)3 6
It is also given that
2
fi f (x) = x – x Hence, the required curve is f (x) = x3 – x2, where 0 £ x £ 1. 23. Case-I: When m < 0
9 1 – __ (1 – m)3 = __ 2 6
fi
(1 – m)3 = – 27
fi
1 – m = – 3
fi m = 4. 24. We have, Ï x 2 + ax + b : x > –1 Ô : –1 £ x £ 1 f (x) = Ì2 x Ô 2 Ó x + ax + b : x > 1 As f is continuous on R, f is continuous at –1 and 1,
Area of the region 1 – m
= Ú ((x – x2) – mx) dx 0
(
)|
1 – m
x2 x3 = (1 – m) __ – __ 2 3 0
1 1 = __ (1 – m)3 – __ (1 – m)3 2 3
1 = __ (1 – m) 6 Also, it is given that,
9 1 __ (1 – m)3 = __ 2 6 fi fi fi
(1 – m)3 = 27 (1 – m) = 3 m = – 2.
Case II: When m > 0
Area of the region
lim f (x) = lim f (x) = f (–1) – + and lim + f (x) = lim f (x) = f (1) –
= Ú (x – x2 – mx) dx 1 – m
x Æ 1
x Æ 1
Thus,
1 – a + b = – 2 and 1 + a + b = 2
fi
a – b = 3 and a + b = 1
fi
a = 2 and b = –1.
Hence,
Ï x 2 + 2 x – 1 : x > –1 Ô f (x) = Ì2 x : –1 £ x £ 1 Ô 2 Óx + 2x – 1 : x > 1
Let us find the point of intersection of
x = – 2y2 and y = f (x).
These two curves meet at (– 2, –1)
The required area
0
x Æ – 1
x Æ – 1
– 1/8
( ÷
___
)
– x = Ú ___ – f (x) dx 2 – 2
Area Bounded by the Curves – 1/8
( ÷ )
– 1/8
( ÷ )
– 1
___
Y
– x = Ú ___ dx – Ú f (x) dx 2 – 2 – 2 –1/8
___
P(9, 3)
M(0, 3)
– x = Ú ___ dx – Ú (x2 + 2x – 1) dx 2 – 2 – 2
– Ú f (x) dx
X¢
X
O
– 1/8
Y¢
– 1
) (
(
– 1/8 – 2 = ____ __ (– x)3/2 – 3÷2 – 2
Hence, the required area 3
( ( )
) (
)
2__ __ 1 3/2 1 = – ____ – 23/2 – – __ + 1 + 1 8 3 3÷2
(
) (
)
8 1 + – __ + 4 + 2 – ___ – 1 3 64 __
2 __ ÷ = ___ ( ÷2 – 2– 9/2 ) + 3
881 = ____ 192
5 __ + 3
= Ú (2y + 3 – y2) dy 0
– (– x2)– 1/8 – 1
)
– 1 x3 __ + x2 – x 3 – 2
63 ___ 64
25. Let b π 0 and for j = 0, 1, 2, ..., n, let S1 be the area of the region bounded by the y-axis and the curve jp (j + 1)p xe xy, __ £ y £ _______ . Show that S0, S1, ..., Sn are b b in G.P. Also, find their sum for a = –1 and b = p [IIT-JEE, 2001] 26. The given curves are y = |x| – 1 and y = 1 – |x|
(
)
y3 3 = y2 + 3y – __ 3 0 =9+9–9 = 9 sq.u.
dy (x + 1)2 + (y – 3) 28. Given ___ = _______________ (x + 1) dx dy y 3 fi ___ = (x + 1) + ______ – ______ (x + 1) (x + 1) dx dy y 3 fi ___ – ______ = (x + 1) – ______ (x + 1) dx (x + 1) which is a linear differential equation. Thus,
dx – Ú____ x + 1 =
IF = e
1 e– log(x + 1) = ______ (x + 1)
Therefore, the solution is y 3 _____ = Ú 1 – _______ dx x+1 (x + 1)2
(
fi
)
y 3 _____ = x + _____ + c x+1 x+1
Put x = 2 and y = 0, then c = – 3 Hence, the equation of the curve is y 3 _____ = x + _____ – 3 x+1 x+1 Hence, the required area
= ar (Quad ABCD)
= 4ar (DOAB)
1 = 4 × __ × 1 × 1 2
fi
y = x2 + x + 3 – 3x – 3 = x2 – 2x.
= 2 sq.u. 27. The given curves are
_
y = ÷x , 2y + 3 = x and y = 0 Hence, the required area
3.57
3.58 Integral Calculus, 3D Geometry & Vector Booster Hence, the required shaded area
2
= Ú (0 – (x2 – 2x))dx
0
0
(
)
x3 2 = x – __ 3 0
8 = 4 – __ = 3
2
( )
4 __ sq.u. 3
29. The given curves are
y = ax2 and x = ay2, where a > 0
(
)
1/2
(
)
1 1 = Ú (x + 1)2 – __ dx + Ú (x – 1)2 – __ dx 4 4 –1/2 0
0 (x + 1)3 __ 1 1 1/2 = _______ – x + (x – 1)3/3 – __ x 3 4 –1/2 4 0
1 1 1 1 1 1 = __ – ___ + __ – ___ – __ + __ 3 24 8 24 8 3
1 1 1 1 = __ – 2 ___ + __ + __ 3 24 8 3
2 2 __ 2 1 __ 1 = __ – __ = – __ = sq.u. 3 6 3 3 3
(
) (
(
)
)
(
)
31. The given curves are
x2 = y, x2 = – y and y2 = 4x – 3
Hence, the required area 1/a
__
(÷
)
x = Ú __ a – ax2 dx 0
1/a 1 2 1 = ___ __ ◊ __ x3/2 – __ ax3 ÷a 3 3 0
1 2 1 1 = ___ __ ◊ __ a3/2 – __ a ◊ __3 ÷a 3 3 a
(
)
(
)
1 = ___ 2 3a It is also given that, 1 ___ 2 = 1 3a 1 fi a2 = __ 3
÷
__
1 a = ± __ 3 1 fi a = ___ __ , since a > 0. 3 ÷ 30. The given curves are fi
y = (x + 1)2, y = (x – 1)2 and y = 1/4
Hence, the required area
(
1
1
0
3/4
______
)
= 2 Ú x2dx – Ú ÷4x – 3 dx
( ( ) (
))
3/2 1 x3 1 2 (4x – 3) = 2 __ – __ ◊ _________ 4 3 0 3 3/4
1 1 = 2 __ – __ – 0 3 6
1 1 = 2 ◊ __ = __ 6 3
( ( ) ( ) )
32.
The given system of equations 4x2f (–1) + 4x f (1) + f (2) = 3x2 + 3x is satisfied for 3 distinct real numbers a, b and c. Comparing the co-efficients of x2, x and constant terms, we get 4f (–1) = 3, 4f (1) = 3, f (2) = 0 Let
f (x) = ax2 + bx + c
Given
3 f (–1) = __ 4
fi
3 a – b + c = __ 4
3 f (1) = __ 4
fi
3 a + b + c = __ 4
b=0 f (2) = 0 4a + 2b + c = 0 c = – 4a we get
Thus, And fi fi Solving,
3.59
Area Bounded by the Curves
| |
|
130 = 80 + 8 – 3 – ____ 3 130 = 85 – ____ 3 125 = ____ 3
|
_______
÷
1 + sin x 33. We have C1 : y = _______ cos x
÷ ÷
______________
1 a = – __ , b = 0, c = 1 4
Thus,
1 f (1) = – __ x2 + 1 4
(
p 1 – cos __ + x 2 = _____________ p __ sin + x 2
(
)
)
______________________
(
p x 2sin2 __ + __ 4 2 = _____________________ p x p x 2 sin __ + __ cos __ + __ 4 2 4 2
(
) ) (
)
÷ ( ) p x 1 – sin x Similarly, C : y = ÷ _______ = tan ( __ cos x ÷ 4 – __2 ) _________
p __ x = tan __ + 4 2
_______
__________
2
Hence, the area of the shaded region
(
)
Let
1 – 0 1 m1 = m (AV) = _____ = – __ 0–2 2 t2 1 – __ –1 t 4 m2 = m (BV) = ________ = – __ t–0 4
m1 × m2 = –1
fi
t 1 – __ – __ = –1 2 4
p/4
( (
__
÷2 – 1
( ) ( )
__
( ( ) ) Ú ( ) ( )
____________ 2
____________ 2
(
)
)
)
(
) (
( ) )
4t dt_____ x = Ú ____________ , Let tan __ = t 2 2 2 0 (1 + t )÷1 – t
34. Given curve is
2
2
x2 3 = 4 – __ – __ x dx 4 2 – 8
x3 = 4x – ___ – 12
8 128 = 8 – ___ – 3 – – 32 + ____ – 48 12 3
3x2 2 ___ 4 – 8
) (
__________
2t (2dt) _____ = Ú ____________ 0 (1 + t2)÷1 – t2 ÷2 – 1
3 x2 = Ú 1 – __ – __ (x – 2) dx 4 2 – 8
( (
0
fi t = – 8 Therefore, the co-ordinates of B = (– 8, –15). Hence, the required area
0
p – AVB = __ , so 2
__________
p __ p __ x x = Ú tan __ + – tan __ – dx 4 2 4 2 0 p/4
( ÷ (
) ÷ ( ) ) x x ( 1 + tan ( __ ) ) – ( 1 – tan ( __ ) ) 2 2 = Ú ________________________ dx x ( 1 – tan ( __ ) ) 2 ÷ x 2tan ( __ ) 2 = Ú _____________ dx x ( 1 – tan ( __ ) ) 2 ÷ p/4
t2 Let the co-ordinates of B be B t, l – __ 4
))
y3 – 3y + x = 0
fi
3y2y ¢ – 3y ¢ + 1 = 0
fi
(3y2 – 3)y ¢ + 1 = 0
fi
1 1 y ¢ = – ________ = – ________ 2 2 (3y – 3) 3(y – 1)
...(i)
fi
2yy ¢ y ≤ = – ________ 2 2 3(y – 1)
...(ii)
3.60 Integral Calculus, 3D Geometry & Vector Booster From Eq. (i), we get
Also, the shaded area
From Eq. (ii), we get
1
__ 1 1 y ¢ (– 10 ÷2 ) = _______ = – ___ 3(1 – 8) 21 __
0
__ 4÷2 _______
2 (2÷2 ) (–1) y ≤(– 10÷2 ) = ___________ = – 2 3(1 – 8)2(21) (3) (7)3 __
35. Hence, the required area
1
1
= Ú e dx – Ú exdx
0
0
1
= e – Ú e xdx
0
b
38. Given f (x) = 1 + 2x + 3x2 + 4x3 f ¢(x) = 2 + 6x + 12x2 > 0 for all x in R Thus, f (x) is an increasing function on R. So, f (x) can have atmost one root. It is clear that f (x) cannot have a poistive real root. We have
= Ú f (x)dx
= (x f (x))ba – Ú (x f ¢(x))dx
x 1 = (bf (b) – af (a)) – __ Ú _____ 2 dx 3 a 1–y
x 1 = (bf (b) – af (a)) + __ _____ dx 3 a y2 – 1
x 1 = (bf (b) – af (a)) + __ _________ dx 3 a ( f (x))2 – 1
36. For
x Π(Р2, 2),
(g (x))3 – 3g (x) + x = 0
a
b
a
( ) Ú ( ) Ú ( ) b
b
3
(g (– x)) – 3g (– x) – x = 0
( ) 3 1 __ 1 1 and f ( Р__ ) = 1 Р1 + __ Р__ = > 0 2 4 2 4 3 1 Hence, f (x) has a root in ( Р__ , Р__ ). 4 2 3 1 39. Let a Π( Р__ , Р__ ) and t = |x| = |a| 4 2
b
= Ú (e – ex ) dx
...(i) ...(ii)
Adding Eqs. (i) and (ii), we get fi {(g (x))3 + (g (x))3} – 3{g (x) + g (– x)} = 0
fi {g (x) + g (– x)}{(g (x))2 + (g (– x))2 – g(x)g(– x)) – 3} = 0
fi {g (x) + g (– x)} = 0
As fi
1
3 3 27 ___ 27 1 f – __ = 1 – __ + ___ – = – __ < 0 4 2 16 16 2
3 1 – __ < a < – __ 4 2 3 1 – __ < t < – __ 4 2 t
1/2
Now, Ú g ¢(x)dx
fi
–1
= (g (x))1–1 = g (1) – g (–1) = g (1) + g (1) = 2g (1) 37. We have
3/4
Ú f (x)dx < Ú f (x)dx < Ú f (x) dx 0
0
2
3
0
x4)1/2 0
fi
(x + x + x +
< Ú f (x)dx < (x + x2 + x3 + x4)3/4 0
t
0
(
)
t
fi
1 1 __ 1 1 __ + __ + + ___ < Ú f (x)dx 2 4 8 16 0
9 3 81 27 < __ + ___ + ___ + ____ 4 16 64 256
fi
15 525 ___ < Ú f (x)dx < ____ 16 0 256
fi
3 __ < 4
(
)
t
Hence, the required area e
= Ú x dy 1 e
= Ú ln y dy
1 e
= Ú ln (1 + e – y) dy 0
t
15 525 ___ < Ú f (x)dx < ____ < 3 16 0 256
40. We have,
f (x) = 1 + 2x + 3x2 + 4x3
fi f ¢(x) = 2 + 6x + 12x2
Area Bounded by the Curves
fi f ≤(x) = 6 + 24x By the sign scheme, we get that
(
fi
(
fi 1 1 f ¢(x) decreases in – •, – __ and increases in – __ , • 4 4 fi 1 In particular f ¢(x) decreases in – t, – __ and increases 4 fi 1 in – __ , t . 4 43.
(
)
(
)
)
)
3.61
2 1 __ (1 – b)3 = ___ 3 12 1 (1 – b)3 = __ 8 1 (1 – b) = __ 2 1 b = __ 2
2
41. We have R1 = Ú x f (x) dx
– 1 2
= Ú (1 – x)f (1 – x) dx –1 2
= Ú (1 – x)f (x) dx –1
2
2
–1
–1
= Ú f (x)dx – Ú x f (x) dx
= Ú f (x)dx – R1 –1
2
fi
1 S ≥ __ e (Since area of a rectangle OCDS = 1/e),
2
2R1 = Ú f (x)dx = R2
2
Since
e– x ≥ e– x, " x Œ [0, 1]
fi
1 S ≥ Ú e– xdx = 1 – __ e 0
1
–1
42. The given curves are y = (1 – x)2, y = 0 and x = 0
(
)
Area of a rectangle OAPQ + Area of a rectangle QBRS > S 1__ 1 –__1 ___ 1 fi S £ ___ (1) + _____ __ e ÷ 2 ÷2 ÷
( ) ( )
(
)
1 1 1 Since __ 1 + ___ __ £ 1 – __ e ÷e 4 Thus, option (c) is incorrect. 44. The given curves are y = sin x + cos x and y = |cos x – sin x|
b
We have, R1 = Ú (1 – x)2dx 0
(
)
(1 – x)3 b = – _______ = 3 0
1 __ (1 – (1 – b)3) 3
1
Also,
R2 = Ú (1 – x2)dx b
Now, fi fi
(
)
(1 – x)3 1 __ 1 = – _______ = ((1 – b)3) 3 3 b 1 R1 – R2 = __ 4 1 1 1 __ (1 – (1 – b)3) – __ (1 – b)3 = __ 3 3 4 2 1 1 1 __ (1 – b)3 = __ – __ = ___ 3 3 4 12
Hence, the required area p/2
= Ú (sin x + cos x) dx 0
(
p/4
p/2
0
p/4
)
– Ú (cos x – sin x)dx + Ú (sin x – cos x)dx
3.62 Integral Calculus, 3D Geometry & Vector Booster
= – (cos x)p/2 + (sin x)p/2 0 0
__
(
– (sin x)p/4 + 0
(cos x)p/4 – 0
)
(sin x)p/2 p/4
(cos x)p/2 p/4
(
Area = { ( x, y) Œ R2 : y ≥ ÷|x| , 5y £ x + 6 £ 15 }
)
1 1 = – (0 – 1) + (1 – 0) – ___ __ + ___ __ – 1 2 ÷2 ÷
(
) (
)
1__ 1 – 0 – ___ – 1 – ___ __ 2 2 ÷ ÷
45. Shifting the origin to (– 3, 0), we get the
(
)
__ 1 1 = 2 – ÷2 – 1 + ___ __ – 1 + ___ __ 2 2 ÷ ÷ __
= 2 – ( 2÷2 – 2 )
= 4 – 2÷2
= 2÷2 ( ÷2 – 1 ). Sq.u.
__
__
__
Hence, the required area = Region (OPK) + Region (QLKR) + Region (OLQ) – Triangle (PQR)
8 5 3 1 5 = __ + 1 + __ – __ = 4 – __ = __ sq.u. 3 3 2 2 2
Chapter
4
Differential Equation
Concept Booster 1. Introduction A differential equation is a mathematical equation for an unknown function of one or more variables that relates the values of the function itself and its derivatives of various orders. It plays a prominent role in engineering, physics, economics, biology, and other disciplines. Differential equations arise in many areas of science and technology, specifically whenever a deterministic relation involving some continuously varying quantities (modelled by functions) and their rates of change in space and/or time (expressed as derivatives) is known or postulated. This is illustrated in classical mechanics, where the motion of a body is described by its position and the velocity as the time value varies. Newton’s laws allow us (given the position, velocity, acceleration and various forces acting on the body) to express these variables dynamically as a differential equation for the unknown position of the body as a function of time. In some cases, this differential equation (called an equation of motion) may be solved explicitly. An example of modelling a real world problem using differential equations is the determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The ball’s acceleration towards the ground is the acceleration due to gravity minus the acceleration due to air resistance. The gravity is considered constant, and the air resistance may be modelled as proportional to the ball’s velocity. It means that the ball’s acceleration, which is a derivative of its velocity, depends on the velocity (and the velocity depends on time). Finding the velocity as a function of time involves solving a differential equation. Differential equations are mathematically, studied from different perspectives, mostly concerned with their
solutions—the set of functions that satisfy the equation. Only the simplest differential equations admit solutions given by explicit formulae; however, some properties of solutions of a given differential equation may be determined without finding their exact form.
2. Definition An equation that involves the dependent variables, independent variables and the derivatives of dependent variables is called a differential equation, i.e.
{
{
}
}
dy dx x, y, ___ = 0 or x, y, ___ = 0 dx dy For examples, dy 1. ___ = sin x dx dy3 d 2y dy 2. ___3 + 4 ___2 + 5 ___ + 10 y = 0 dx dx dx ________
÷ ( ) ÷ ________
( )
dy 2 d2y 3. 1 + ___ = 2 + ___ 2 dx dx dw du dv ___ 4. ___ + ___ + = u + v + w dx dx dx
each are differential equations.
3. Ordinary Differential Equation A differential equation that involves the derivative with respect to a single independent variable is called an ordinary differential equation. It can also be expressed as a function of variables x and y, and the derivatives of y w.r.t. x, i.e.
(
)
dy f x, y, ___ = 0. dx
Thus, dy 1. ___ + 10 y = 0 dx
4.2 Integral Calculus, 3D Geometry & Vector Booster
dy 2. ___ + 2y = sin x dx
Example 6. The order of the differential equation whose
d2y dy 3. ___2 + 2 ___ + 5y = 0 dx dx
y = (c1 + c2) cos (x + c3) – c4 ex + c5,
where c1, c2, c3, c4 and c5 are arbitrary constants
d ny 4. ___n + ny = 0 dx
is 3.
are ordinary differential equations.
4. Partial differential equation A differential equation which contains two or more independent variables and partial derivatives w.r.t. them is called a partial differential equation. For examples,
∂z ∂ z 1. ___ + ___ = 0 ∂ x ∂ y
∂2 z ____ ∂2 z 2. ____2 + 2 = x ∂ x ∂ y
5. Order
of a
6. Degree
genral solution is given by
of a
differential equation
The index power of the highest order derivative of a differential equation is called the degree of a differential equation, and is free from any radical sign.
Example 1. The degree of the differential equation
( )
( )
d3y 2102 d 4y dy 2013 ___ 3 + ___4 + ___ + 5005y = 0 dx dx dx
is 1.
Example 2. The degree of the differential equation
÷(
) ÷(
_________
differential equation
The order of a differential equation is the highest order derivative of a differential equation. Example 1. The order of the differential equation
__________
)
5 d2y d2y 1 + ___ 2 = 10 + ___ dx dx2
4
is 5.
Example 3. The degree of the differential equation
dy dy ___2 + 2 ___ + 5y = 0 dx dx is 2.
dy ___ + dx
Example 2. The order of the differential equation
Example 4. The degree of the differential equation
d 2y 100 ___2 + dx
edx = x + 1
2
( )
d 4y ___ 4 + dx
dy ___ + 50y = 0 dx
is 4.
Example 3. The order of the differential equation
( )
dy sin ___ = x dx
is 1.
Example 4. The order of the differential equation of the
curve
y = (a + b) e3x + (c + d) e–2x
a, b, c, d, ΠR
is 2. Example 5. The order of the differential equation of the curve
y = (a + b) cos ((c + d) x + e),
where
a, b, c, d, e, ΠR
is 3.
dx ___ = y dy
is 2. dy ___
is not defined.
Example 5. The degree of the differential equation
(
)
dy log 1 + ___ = x dx is not defined.
Example 6. The order and the degree of the differential
equation
(
)
( )
dy 2/3 d3y 1 + 3 ___ = 4 ___ 3 dx dx
are 3 and 3, respesctively
Example 7. The order and the degree of the differential
equation
÷
_______
÷( ( ) )
__________
d2y 4 dy 5 1 + ___ 2 = y + ___ dx dx 5
are 2 and 4, respectively.
Differential Equation
7. Linear
and
non-linear differential equation
A differential equation in which the dependent variables and all its derivatives present occur in the first degree only and no products of dependent variables and/or derivatives occur is known as linear differential equation. A differential equation which is not linear is called a nonlinear differential equation. For examples,
dy 1. ___ + 5y = x is linear. dx
d2y dy 2. ___2 + 5 ___ + 6y = 0 is linear. dx dx
d 2y 3. ___2 + dx
d2013y dy 4. _____ + 5 ___ + 10y = 0 is linear. 2013 dx dx
8. Formation
first order
(ii) (iii) (iv) (v) (vi) (vii) (viii)
fi
dy = f (x) dx
Integrating, we get
Ú dy = Ú f (x) dx
dy ___ = f (x) ◊ f (y) dx fi
dy ____ = f (x) dx f (y)
Integrating, we get
and
first degree
All differential equations of the first order cannot be always solvable. However they can be solved by suitable methods if they belong to any of the following standard forms. (i)
dy ___ = f (x) dx
Let the differential equation be
A differential equation of the first order and the first degree can be written as the form of dy ___ = f (x, y) or M dx ± N dy = 0, where M and N are dx functions of x and y or constants.
9.2 Variable Separable Form
Let the equation of the curve be f (x, y, c1, c2 ..., cn) = 0, ...(i) where c1, c2, ..., cn are n arbitrary constants apart from x and y. To obtain the differential equation of the curve (i), we have to differentiate (i) up to n times w.r.t. x and eliminate those arbitrary constants from (n + 1) equations, which yield the required differential equation. of
9.1 differential equation of the form
Thus, y = j (x) + c is the required solution.
differential equation
9. Differential equation
(ix) Reducible to exact differential equation. (x) Clairauts differential equation.
fi
dy y ___ + 10x = 0 is not linear. dx
of a
4.3
dy A differential equation is of the form ___ = f (x). dx Variable separable form. Reducible to variable separable form. Homegeneous differential equation. Reducible to homegeneous differential equation. Linear differential equation. Bernoulli’s differential equation. Exact differential equation.
dy
= f (x) dx Ú ____ f (y) Ú
fi j (y) = y (x) + c, which is the required solution.
9.3 Reducible to variable separable form Let a differential equation is of the form dy ___ = f (ax + by + c) dx Let
(ax + by + c = v)
(
)
dy 1 ___ dv fi ___ = __ – a dx b dx The Eq. (i) reduces to
(
)
1 dv __ ___ – a = f (v) b dx dv fi ________ = dx b f (v) + a
Integrating, we get
dv fi Ú ________ = dx bf (v) + a Ú fi j (v) = x + c fi
j (ax + by + c) = x + c,
which is the required solution.
...(i)
4.4 Integral Calculus, 3D Geometry & Vector Booster
9.4 Homogeneous differential equation Homogeneous equation If the degree of each term throughout the equation is the same, it is called a homogeneous equation. For examples, 2
2
2
x3 + xy2 + x2y + y3 = 0 are the homogeneous equation of 3rd degree. A homogeneous differential equation is of the form f (x, y) dy ___ = ______ ...(i) dx g (x, y) where the degree of f (x, y) is the same as the degree of g (x, y). In this case, divide the numerator and the denominator by the highest power of x. Thus the given differential equation reduces to dy y ___ = f __ x ...(ii) dx
( )
dy dv ___ = v + x ___ ◊ dx dx y __ Let x = v fi y = vx. fi
dv
j (v) = log |x| + c
fi
y j __ x = log |x| + C
a1a + b1b + c1 = 0 and a2a + b2 b + c2 = 0.
Solving, we get
b1c2 – b2c1 a2c1 – a1c2 a = __________ and b = __________ . a1b2 – a2b1 a1b2 – a2b1 Then the Eq. (ii) reduces to ...(iii)
X V dY ___ = V + ___ dX dX
Then the Eq. (iii) reduces to dV a1 + b1V V + X ___ = ________ . dX a2 + b2V
dx
fi
Let us choose a and b in such a way that
fi
Integrating, we get = ___ Ú _______ f (v) – v Ú x
...(ii)
which is a homogeneous equation of the first degree. Y Let __ = V X
dv dx _______ = ___ x f (v) – v
(a1X + b1Y) + (a1a + b1b + c1) = ___________________________ (a2X + b2Y) + (a2a + b2b + c2)
dY a1X + b1Y ___ = _________ dX a2X + b2Y
The Eq. (ii) reduces to dv v + x ___ = f (v) dx
dx = dX, dy = dY
dY a1(X + a) + b1(Y + b) + c1 ___ = ________________________ dX a2(X + a) + b2 (Y + b) + c2
2. x3 + x2y + y3 = 0,
fi
fi
dy dY Thus, ___ = ___ . dx dX Now, Eq. (i) can be reduced to
2
1. x + y = 0, x + xy + y = 0 are the homogeneous equations of 2nd degree.
The Eq. (i) can be reduced into homogeneous equation form by means of suitable substitutions. x = X + a, y = Y + b, where a, b are constants.
Integrating, we get j (V) = log |X| + k
( )
fi
Y j __ = log (X) + c X
which is the required solution.
fi
y–b j x_____ – a = log (x – a) + c
9.5 Reducible to homogeneous differential equation
which is the required solution of the given differential equation.
( )
Let a differential equation is of the form dy a1x + b1y + c1 ___ = _____________ , dx a2x + b2y + c2 a1 where __ a2 π
b1 __ , b2
( )
9.6 Linear differential equation ...(i)
A differential equation is said to be linear differential eqation of the first order, when the dependent variable and its derivatives appear only in the first degree. Let a differential equation is of the form
Differential Equation
dy ___ + Py = Q ...(i) dx where P and Q are either functions of x or constants.
Ú pdx
Here IF = e = e . Multiplying both sides of Eq. (i) by IF, we get dy e px ___ + P e px y = Q ◊ e px dx d fi ___ (y e px) = Q ◊ epx dx Integrating, we get y ◊ e px = Ú Q ◊ epx dx + c
fi
y ◊ (I.F) = Ú Q ◊ (I.F) dx + c
fi
y ◊ (I.F) = j (x) + c
fi
y ◊ e px = j (x) + c
fi
y = (j (x) + c) e– px
Note: Sometimes it is convenient to express the differential equation in the form dx ___ + Px = Q dy where P and Q are either functions of y or constants. IF = e Ú Pdy = e Py
...(i)
where P and Q are either functions of x alone or constants.
dv Thus, ___ + P (1 – n) v = Q (1 – n) dx which is a linear differential equation. IF = e Ú P (1 – n) dx = e (1 – n) Ú pdx Thus the required solution is
v (IF) = Ú Q ◊ (IF) dx + c
(
)
fi y (1 – n) ◊ e (1 – n) Ú pdx = Ú Q e(1 – n) Úpdx dx + c
Its solution is given by u (x, y) = c, " c Œ R. For examples, 1. ex ◊ cos y dy + ex ◊ sin y dx = 0 is an exact differential equation. Since, it is derived by d (ex sin y) = 0 fi
ex sin y = c
which is the required solution. 2.
x dy + y dx = 0 is an exact differential equation. fi d (x y) = 0 fi xy = c which is the required solution.
Dear friends you should remember the follwoing exact differentials. These help us to find the integrating factors.
9.7 Bernouli’s differential equation
Dividing both sides of Eq. (i) by yn, we get dy y–n ___ + P y1–n = Q, dx dy dv 1 ___ Let y1– n = v fi y–n ___ = ______ . dx (1 – n) dx
du = Mdx + Ndy = 0
General form of variable seperation (By Inspection)
Thus the solution is x (IF) = Ú Q (I.F) dx + c
A differential equation is of the form dy ___ = Py = Qyn dx
9.8 Exact differential equation
which is the required solution of the given differential equation.
Then
is the required solution of the given differential equation.
A differential equation is said to be an exact differential equation, if it is formed by the equation, an exact differential to zero. Thus a differential equation is of the form M (x, y)dx + N (x, y) dy = 0 said to be exact, if the expression M (x, y) dx + N (x, y) dy is the exact differential of some function u (x, y) i.e.
px
4.5
1. d (x ± y) = dx ± dy
2. d (x y) = x dy + y dx
y dx – x dy x 3. d __ y = _________ y2
( ) y x dy – y dx 4. d ( __ x ) = _________ x 2
dx + dy 5. d [log (x + y)] = _______ x+y
x dy + y dx 6. d [log (xy)] = __________ xy
y dx – x dy x 7. d log __ y = _________ xy
y x dy – y dx 8. d log __ x = _________ xy
x dx + y dy 1 9. d __ log ( x2 + y2 ) = __________ 2 2 x + y2
( ( ) )
[ ( ) ]
[
]
4.6 Integral Calculus, 3D Geometry & Vector Booster
[ ( ) ]
x+y x dy – y dx 1 10. d __ log _____ = _________ 2 x – y 2 x – y2
( ) y dx – x dy x 12. d [ tan ( __ y ) ] = _________ x +y
x dy + y dx 1 _________ 11. d – __ xy = x2y2
fi
)
Æ (x2 + y2) x = r cos q, y = r sin q,
( )
y x2 + y2 = r2, q = tan–1 __ x x dx + y dy = r dr, x dy – y dx = r2 dq
16. when the expression is in the form of: Æ (x2 – y2) Let
x = r sec q, y = r tan q,
where
y x2 – y2 = r2, q = sin–1 __ x
fi
x dx – y dy = r dr, x dy – y dx = r2 sec q dq
( )
10. Orthogonal Tragectories If a curve intersects a family of curves at right angles, then previous one is the orthogonal trajectory to the later one. The differential equation of the orthogonal trajectories of dy the curves f x, y, ___ = 0 is the family of the curves, whose dx
(
)
(
pn + f1 pn–1 + f2 pn–2 + ...
15. when the expression is in the form of
fi
dy + fn–1 ___ + fn = 0 dx
______ x dx + y dy 14. d ÷x 2 + y2 = _________ ______ . ÷x 2 + y2
where
( )
dy + fn–1 (x, y) ___ + fn (x, y) = 0 ...(i) dx
( ) ( ) ( ) ( )
2
[ ( ) ]
Let
( )
dy n dy n – 1 dy n–2 fi ___ + f1 ___ + f2 ___ + ... dx dx dx
y x dy – y dx 13. tan–1 __ x = _________ 2 x + y2
(
( )
–1
2
( )
dy n dy n – 1 dy n – 2 ___ + f1 (x, y) ___ + f2 (x, y) ___ + ... dx dx dx
)
dx differential equation is f x, y, Р___ = 0. dy ule to find out the orthogonal trajectory: R (i) Let the equation of the family of curves is f (x, y, c) = 0, where c ΠR.
+ fn – 1 p + fn = 0 ...(i)
dy p = ___ dx and f1, f2, ..., fn are functions of x and y. The Eq. (i) can aslo be written as f (x, y, p) = 0 where
...(ii)
For examples, 1. The equation dy 5 dy y ___ + 2013 x ___ – y = 0 dx dx is a differential equation of 5th degree. 2. The equation dy 2014 dy ___ + x ___ – 2015y = 0 dx dx is a differential equation of 2014th degree. The differential equation f (x, y, p) = 0 can be solvable (i) for p (ii) for x (iii) for y (iv) for the first degree in x and y. Now, we shall discuss the varoius methods to solve the above types of equations.
( )
( )
( )
( )
(i) Equations solvable for p
Consider a differential equation of the form (p – f1 (x, y)) (p – f2 (x, y)) ... (p – fn (x, y)) = 0,
(ii) Form a differential equation of the given curve. dy dx (iii) Replace ___ by – ___ . dx dy dx (iv) Solve the equation f x, y, – ___ = 0, dy
which is solvable for p.
We will get the required orthogonal trajectories.
Each of these equations is of it first order and of first degree. Let the solutions of these equations are given by j (x, y, c1) = 0,
(
)
11. First Order Higher Degree Differential Equation The most general form of a differential equation of the first order and of higher degrees (say nth degree) is
fi [p – f1 (x, y)] = 0,
(p – f2 (x, y)) = 0
(p – fn (x, y)) = 0
j 2(x, y, c2) = 0
4.7
Differential Equation
j n (x, y, cn) = 0,
where c1, c2,
dy where P = ___ , dx
cn are arbitrary constants. Hence, the general solution is
y = Px + f (P),
is called the Clairaut’s differential equation.
j1(x, y, c1) ◊ j2(x, y, c2) ... qn (x, y, cn) = 0
Rule to find the Clairaut’s differential equation
which contains n arbitrary constants, whereas an equation of the first order and of first degree should contain only one arbitrary constant. So, we can take, without loss of generality, c1 = c2 = cn = c. Hence, the genral solution of Eq. (i) is
j 1(x, y, c) ◊ j 2 (x, y, c)
j n (x, y, c) ◊ j2 (x, y, c) = 0.
When the differential equation f (x, y, p) = 0 is solvable for x, it can be expressed in the form of x = f (y, p) ...(i) Differentiating w.r.t. y, we get dp dx ___ = f y, p, ___ dy dy
( ) ( )
dp 1 ___ fi __ ...(ii) p = f y, p, dy which is a differential equation of two variables y and p and hence it can be solved. Let us consider its solution be j (y, p, c) = 0 ...(iii) Eliminating p between (i) and (iii) gives the required solution. (iii) Equations solvable for y If the given equation is solvable for y, it can be expressed in the form y = f (x, p) ...(i) Differentiaing w.r.t. x, we get dy dp ___ = p = j x, p, ___ = 0 dx dx which is a differential equation in two variables x and p and hence its solution is possible. Let us consider its solution be
)
y (x, p, c) = 0
...(ii)
Eliminating p between (i) and (ii) we get the required solution of Eq. (i).
(iv) Clairaut’s Differential Equation
A differential equation of the first order in the form
y = Px + f (P)
...(i)
Differentiating w.r.t. x, we get dy dP dP ___ = P + x ___ + f ¢(P) ___ dx dx dx fi fi
(ii) Equations solvable for x
(
1. Given
fi
dP dP P = P + x ___ + f ¢(P) ___ dx dx dP (x + f ¢(P)) ___ = 0 dx dP ___ = 0 or (x + f ¢(P)) = 0 dx
fi
P = c
...(ii)
or,
(x + f ¢(P)) = 0
...(iii)
Eliminating P between Eq. (i) and (ii), we get,
y = cx + f (c)
where is the general solution of Eq. (i). Eliminating P between (i) and (iii), we get an equation, which has no constant. This is the singular solution of (i).
12. Higher Order Differential Equation
(i) Let a differential equation is of the form
d 2y ___2 = f (x) dx d dy ___ fi ___ = f (x) dx dx
( )
Integrating, we get dy ___ = f (x) + c1 dx Integrating again, we get
y = Ú f (x) dx + c1 x + c2
fi y = j (x) + c1 x + c2 which is the required solution. (ii) Let a differential equation is of the form
d 2y ___2 = f (y) dx
...(i)
dy Let ___ = p dx fi
d 2y ___ 2 = dx
dp ___ = dx
dp dp dy ___ ◊ ___ = p ◊ ___ dy dy dx
4.8 Integral Calculus, 3D Geometry & Vector Booster From Eq. (i), we get, dp p ___ = f (y) dy fi p dp = f (y) dy
Let the curve be y = f (x) and the point be P (x, y). dy Then tan (y) = ___ . dx (i) Length of the tangent (PT ) y siny = ___ PT fi PT = y cosecy
Integrating, we get
p2 __ = Ú f (y) dy + c1 2
fi
p2 = 2 Ú f (y) dy + 2c1
= j (y) + 2c1 fi
__________
__________ dy ___ = ± ÷ j (y) + 2c1 dx __________ fi dy = ± ÷ j (y) + 2c1 dx Integrating, we get
fi
y = y (x) + c2
which is the required solution. (iii) Let a differential equation is of the form
( )
dx 2 = y 1 + ___ dy
÷ ( )
( )
_________
y = Ú ± ÷j (y) + 2c1 dx + c2
d2y dy j ___ 2 , ___ = 0 dx dx
________
(ii) Length of the sub-tangent (TM): y tan y = ___ TM fi TM = y coty dx = y ◊ ___ dy (iii) Length of the normal (PN ): y cos y = ___ PN fi PN = y secy
fi
= y ÷1 + cot2y
p = ± ÷ j (y) + 2c1
_________
________
...(i)
( )
dy d2y dp d dy Let ___ = p fi ___ 2 = ___ ___ = ___ dx dx dx dx dx The Eq. (i) reduces to dp j ___ , p = 0 ...(ii) dx which is an equation of the first order. Then solve Eq. (ii) and get the required solution. Also, Eq. (i) can be written in the form dp j p ___ , p = 0 ...(iii) dy which is also an equation of the first order. Solve it and get the required solution.
( )
= y ÷1 + tan2y
dy 2 = y 1 + ___ dx
÷ ( )
________
(iv) Length of the sub-normal (MN): y tany = ____ MN
( )
dy fi MN = y ___ . dx (ii) Length of Intercepts of the
tangent by the axes:
( )
13. Applications
of
Differential Equation
Geometrical Application (i) Lengths of tangent, sub-tangent, normal and subnormal to the curve at a point
(i) The equation of the tangent to a curve at (x1, y1) is dy y – y1 = ___ (x – x1) dx at (x1, y1) (ii) The length of the x-intercept is
( )
( ) ( )
dx x1 – y1 ___ dy at (x1, y1) (iii) The length of the y-intercept is dx = x – y ___ dy at (x1, y1)
Differential Equation
4.9
Exercises (Problems based on Fundamentals) Find the order and the degree of each of the following differential equations: 2
d y 1. ___2 + 4y = 0 dx
dy 2. ___ + dx
d2y 3. ___2 + dx
( )
2
19. y = aex + be– x, where a and b are parameters.
20. y = sin (bx + c), b and c being parameters. 21. y = ae2x + be3x, where a and b are parameters.
dx ___ = 2 dy
( ) ÷ ( ) ÷ ÷ ( ) ÷ ( ) dy 2 ___ + xy = 0 dx
_____
_________
d2y dy 2 3 ___ 4. 1 + ___ = c ◊ 2 dx dx
dy dy 2 5. x + ___ = 1 + ___ dx dx
dy dy 2 6. y = x ___ + a 1 + ___ dx dx
d 4y d3y 7. ___4 + sin ___ 3 = 0 dx dx
d2y 2 8. ___ 2 + dx
9. edx = (x + 1)
_________
_________
( )
( ) ( )
( )
dy 6 d2y ___ = x sin ___ 2 dx dx
dy ___
( )
17. y = m x, where m is a parameter. 18. y = A cos x + B sin x, where A and B are parameters.
( )
dy dy 10. sin ___ + cos ___ = x dx dx Formation of a differential equation 11. Find the differential equation of all non-vertical lines in a plane. 12. Find the differential equation of the family of parabolas having vertex at the origin and the axis along positive x-axis. 13. Find the differential equation of the family of all circles, whose centre lies on x-axis and touches the y-axis at the origin. 14. Find the differential equation of all parabolas whose axes are parallel to the x-axis and have latus rectum 4a. 15. Find the differential equation corresponding to the family of curves y = c (x – c)2, where c is an arbitrary constant. 16. Find the differential equation of the system of y2 x2 ellipses __ 2 + __2 = 1, where a and b are arbitrary a b constants.
22. Find the differential equation of all circles touching the (a) x-axis at the origin. (b) y-axis at the origin. 23. Obtain the differential equation of all circles of the radius r. 24. Find the differential equation of all the circles in the first quadrant which touch the co-ordinate axes. 25. Find the differential equation of all the circles which pass through the origin and the centre lies on x-axis. 26. Find the differential equation of all non-vertical lines in a plane. 27. Find the differential equation of all the parabolas with the latus rectum 4a and whose axes are parallel to the x-axis. 28. Find the differential equation of all the ellipses having foci on the x-axis and the centre at the origin. 29. Find the differential equation of the family of parabolas having vertex at the origin and the axis along positive y-axis. 30. Find the differential equation of the family of curves __ y2 = 2c (x + ÷c ). 31. Find the order and the degree of the differential equation of all parabolas whose axis of symmetry is parallel to x-axis. Differential equation of first order and first degres dy 32. Solve: ___ = dx
x2 – 1 _____ . x2 + 1
dy 1 33. Solve: ___ = _____ . dx ex + 1 dy cos 2 x – cos x 34. Solve: ___ = ____________ . 1 – cos x dx dy 1 35. Solve: ___ = ________ . dx x (x4 + 1) ____
dy ÷sin x 36. Solve: ___ = ________ . dx sin x cos x dy x2 37. Solve: ___ = ________ . dx (3 + 2x)2
4.10 Integral Calculus, 3D Geometry & Vector Booster dy 38. Solve: ___ = dx dy 39. Solve: ___ = dx dy 40. Solve: ___ = dx
÷
_______ __ 1 – ÷x ________ . 1 + ÷x _____
61. Solve: (x2 – yx2) dy + (y2 + xy2) dx = 0. Reducible to variable separable Form
____
(÷ tan x + ÷cot x ) .
dy 62. Solve: ___ = (x + y + 1)2. dx
1 ____________ 3 . sin x + cos3 x
dy 63. Solve: ___ = sin (x + y) + cos (x + y). dx
dy 1 41. Solve: ___ = _______________ ____ 4 . _____ dx (÷sin x + ÷ cos x )
64. Solve: (x + y) (dx – dy) = (dx + dy).
Variable separable Form
dy 65. Solve: tan y ___ = sin (x + y) + sin (x – y). dx dy 66. Solve: ___ – x tan y ( y – x) = 1. dx
dy 42. Solve: ___ = 1 + x + y + xy. dx dy 43. Solve: ___ = ex–y + x2 ◊ e–y. dx x
2
x
44. Solve: 3 e tan y dx + (1 – e ) sec y dy = 0. ________________ dy 45. Solve: ÷ 1 + x2 + y2 + x2y2 + xy ___ = 0. dx
)
dy dy 47. Solve: x ___ = a y2 + ___ . dx dx dy 48. Solve: xy2 ___ = 1 – x2 + y2 – x2y2. dx dy 49. Solve: (x + 1) ___ = 2xy. dx 50. Solve: sec2x tan y dx + sec2y tan x dy = 0. 51. Solve: (1 + e2x) dy + (1 + y2) ex dx = 0. dy 52. Solve: ___ = ex+y + x2ey. dx
_____
54. Solve: ÷ 1 + x2 dy + ÷ 1 + y2 dx = 0. ________________ 1 + x2 + y2 + x2y2 +
55. Solve: ÷
xy = 0.
dy 56. Solve: xy ___ + 1 + x + y + xy = 0. dx 57. Solve: (1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0. _____
_____
58. Solve: x ÷1 – y2 dx + y ÷1 – x2 dy = 0.
(
) (
)
dy dy 59. Solve: y – x ___ = a y2 + ___ . dx dx dy 60. Solve: xy ___ = dx
dy 69. Solve: ___ = sec (x + y). dx dy 70. Solve: sin–1 ___ = x + y. dx dy 71. Solve: ___ = cos (x + y + 1). dx dy 72. Solve: (x2 + 2 xy + y2 + 1) ___ = 2 (x + y). dx dy 73. Solve: ___ = sin (10x + 6y). dx y dy ____________ 2 (x2 + y2) – 1 74. Solve: __ x ___ + 2 = 0. dx x + y2 + 1 Homogeneous differential equation 75. Solve: (x3 + y3) dy – x2 y dx = 0.
_____ _____ dy 53. Solve: y ÷1 + x2 + x÷1 + x2 ___ = 0. dx _____
68. Solve: (x + y) (dx – dy) = dx + dy.
( )
dy sin y + cos y 46. Solve: ___ = ___________ . dx x (2log x + 1)
(
dy (x – y) + 3 67. Solve: ___ = ___________ . dx 2 (x – y) + 5
1 + y2 _____2 (1 + x + x2). 1+x
______
76. Solve: x dy – y dx = ÷ x 2 + y2 dx. 77. Solve: (x2 – y2) dx = 2xy dy. 78. Solve: (1 + 2ex/y) dx + 2e x/y (1 – x/y) dy = 0.
(
) (
)
dy dy 79. Solve: x ___ + 1 = y 1 – ___ . dx dx dy 80. Solve: x ___ = x + y. dx dy y – x 81. Solve: ___ = _____ . dx y + x dy 82. Solve: 2 xy ___ = x2 + y2. dx 83. Solve: (y2 – 2xy) dx = (x2 – 2xy) dy.
Differential Equation
84. Solve: x2y dx – (x3 + y3) dy = 0. 85. Solve: (x3 – 3xy2) dx = (y3 – 3x2y) dy.
÷
______ 2
dy y y 86. Solve: ___ = __ – __ 2 – 1 . dx x x dy y y 87. Solve: ___ = __ x + sin __ x . dx dy y 88. Solve: x ___ = y – x tan __ x . dx
( ) ( ) x 89. Solve: ( 1 + e ) dx + e ( 1 – __ y ) dy = 0. x __ y
x __ y
Reducible to homogeneous differential equation dy 90. Solve: ___ = dx
x + 2y + 3 __________ . 2x + 3y + 4
dy (x + y)2 91. Solve: ___ = ____________ . (x + 2) (y – 2) dx 93. Solve: (x + 2y + 3) dx = (2x + 3y + 4) dy. dy x + 2y – 5 94. Solve: ___ = _________ . dx 2x + y – 4 dy 4x + 6y – 5 95. Solve: ___ = __________ . dx 6x + 9y + 7 dy 96. Solve: ___ = dx
6x – 2y – 7 __________ . 3x – y + 4
Linear differential equation 97. Solve: y dx – x dy + ln x dx = 0. dy 98. Solve: x2 ___ – 3xy = 4x4 + 2x2. dx dy 99. Solve: ___ = y tan x – sin x. dx dy 2 100. Solve: x log x ___ + y = __ x log x. dx x 101. Solve: x dx = __ 2 – y dy. y
( )
dy y 107. Solve: ___ + __ = x3. dx x dy 108. Solve: ___ = dx dy 109. Solve: ___ – dx
y tan x – 2 sin x. y __ x = 2x2.
dy 2 110. Solve: x log x ___ + y = __ x log x. dx
111. Solve: y dx – (x + 2y2) dy = 0. 112. Solve: y dx + (x – y3) dy = 0. dy 113. Solve: x ___ – ay = x + 1. dx dy 114. Solve: (x + 1) ___ – ny = ex (x + 1)n + 1. dx 115. Solve: (1 + y2) dx = (tan–1 y – x) dx. dy 116. Solve: (x + 3y + 2) ___ = 1. dx 117. Solve: (1 + y2) dx = (xy + y3 + y) dy. Bernoulli’s differential equation dy y y 118. Solve: ___ + __ x log x = __ 2 (log y)2. dx x dy 119. Solve: ___ + xy = x2 y6. dx dy sin2 y 120. Solve: ___ + _____ x = x3 cos2 y. dx dy 121. Solve: ___ – x3 y2 + xy = 0. dx dy 122. Solve: (1 – x2) ___ + xy = xy2. dx dy 123. Solve: x ___ + y = x3 y6. dx –1 dy 124. Solve: (1 + x2) ___ + y = e tan x. dx
dy 102. Solve: (x + 3y + 2) ___ = 1. dx
dy 125. Solve: ___ (x2 y3 + xy) = 1. dx
dy 103. Solve: ___ + 2y = e3x. dx
126. Solve: (y log x – 1) y dx = x dy
dy 104. Solve: x ___ = x + y. dx
dy 105. Solve: x ___ + y = xex. dx
dy 106. Solve: x ___ + y = x log x. dx
dy 127. Solve: ___ = x2 y3 – xy dx dy xy __ 128. Solve: ___ + _____ 2 = x÷y . dx 1 – x dy y y 129. Solve: ___ + __ x ◊ log y = __ 2 (log y)2 dx x
4.11
4.12 Integral Calculus, 3D Geometry & Vector Booster dy 1 130. Solve: ___ + __ ◊ sin 2y = x3 ◊ cos 2y. dx x
2
131. Solve: (xy2 – e 1/x ) dx – xy2 dy = 0. dy 132. Solve: ___ + x (x + y) = x3 (x + y)3 – 1. dx Exact differential equation 133. Solve: x dx + y dy = x dy – y dx. 134. Solve: x dy – y dx = x4dx.
dy 158. Solve: ___ = dx
= (2xy + sin x ◊ cosec2 y) dy
(
)
2x – 3y + 1 __________ . 3x – 2y – 2
159. Solve: x (dy + dx) = y (dx – dy).
(
137. Solve: x dy + y dx + xy2 dx – x2y dy = 0.
dy y – x ___ dx 1 1 162. Solve: ________ = __ 2 + __ 2 . dy x y y + x ___ dx
138. Solve: (4x – 3y) dx + (2y – 3x) dy = 0.
163. Solve: x2 y dy – xy2 dx = x4 dy + x3 y dx.
135. Solve: x dy + y dx = sin y dy. 136. Solve: x dy + y dx + y2(x dy – y dx) = 0.
(
)
164. Solve: dx + x (y dx + x dy) = e– ey dx.
1 139. Solve: sin y + y sin x + __ x dx.
(
)
1 + x cos y – cos x + __ y dy = 0.
x dx + y dy _________ y dx – x dy 140. Solve: _________ _______ = . 2 2 x2 ÷x + y
(
)
xy 160. Solve: x (tan–1 x – log x) dy + y dx = _____ 2 dx. 1+x 161. Solve: x (dy – x dx) + y dx = 0.
(
x – __
165. Solve: y (y2 dx + (x2 dy – xy dx)) = e y dy. 166. Solve: y dx – x dy + xy2 dx = 0.
(
)
(
)
sin 2 x sin2 x 141. Solve: _____ y + x dx + y – _____ 2 dy = 0. y dy x + y ___ y4 dx 142. Solve: _______ = x2 + 2y2 + __2 . dy x y – x ___ dx dy a2 x ___ – y dy dx 143. Solve: x + y ___ = ___________ 2 . dx x + y2
168. Solve: (x2 + y2 + a2) y dy = (x2 + y2 – a2) x dx.
144. Solve: x dx + y dy = (x2 + y2) y dy.
172. Solve: x dx + y dy = m (x dy – y dx).
145. Solve: (x + y) dx + (x – y) dy = 0.
dy x + y ___ x sin2 (x2 + y2) dx ____________ 173. Solve: ________ = . dy y3 ___ y – x dx dy a2 x ___ – y dy dx 174. Solve: x + y ___ = ___________ 2 . dx x + y2
)
(
)
146. Solve: y dx + x (x – 1) dy = 0. 147. Solve: y dx + x (1 – xy) dy = 0. 148. Solve: (x + y) (dx – dy) = (dx + dy). 149. Solve: dx + dy = x dy + y dx. 150. Solve: x dy – y dx = (x2 + y2) dx.
(
)
1 1 151. Solve: __ x + __ y (x dy + y dx) = dx + dy. ______
152. Solve: (x dy + y dx) ÷x 2 + y2 = x2 y dx + xy2 dy. 153. Solve: (x – x3) dy = y (dx + x2 dy). 154. Solve: (x + 2y) dy + y dx = 0.
169. Solve: (1 + xy) y dx + x (1 – xy) dy = 0.
(
)
x dy y 170. Solve: ______ 2 = ______ 2 2 – 1 dx. 2 x +y x +y dy x + y ___ y4 dx 171. Solve: _______ = x2 + 2y2 + __ 2 . dy x y – x ___ dx
(
)
÷ ÷
___________
x dx + y dy a2 – x2 – y2 175. Solve: _________ = ___________ 2 . x dy – y dx x + y2 __________
x dx – y dy 1 + x2 – y2 176. Solve: _________ = __________ . x dy – y dx x2 – y2
( )
155. Solve: (x + sin y) dy + y dx = 0.
x 177. Solve: sin __ y (y dx – x dy) = xy2 (x dy + y dx).
156. Solve: ey dx + (xe y – 2y) dy = 0.
178. Solve: x dy + y dx + y2 (x dy – y dx) = 0.
157. Solve: 2y dy + (cos x ◊ cot y – y2) dx.
)
dy dy 167. Solve: 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ . dx dx
179. Solve: (4x3 + ex sin y) dx + ex cos y dy = 0.
Differential Equation
Orthogonal Trajectories 180. Find the orthogonal trajectories of the family of the straight lines which are passing through the origin. 181. Determine the 45∞ trajectories of the family of concentric circles x2 + y2 = a2. 182. Find the orthogonal trajectories of the family of the curves ax2 + y2 = 1. 183. Find the orthogonal trajectories of the circles x2 + y2 – ay = 0, where a is a parameter. 184. Find the orthogonal trajectories of the family of parabolas y2 = 4ax, where a is a parameter. 185. Find the orthogonal trajectories of the family of rectangular hyperbola xy = c2. 186. Find the orthogonal trajectories of the family of curves x2 – y2 = c2. First Order Higher Degree Differential Equation. Equation Solvable for p 187. Solve: (x + y + p) (2x + p) = 0. 188. Solve: p2 – p (ex + e–x) + 1 = 0. 189. Solve: p2 + 2 py cot x = y2. 190. Solve: p2 – px – xy – y2 = 0. 2
191. Solve: p y + (x – y) p – x = 0. 192. Solve: (p2 – 1) xy = (x2 – y2) p. 193. Solve: x y p2 – (x2 + y2) p + xy = 0. Equation solvable for x 194. Solve: p x – y p2 = a p. 195. Solve: y = 2 p x + y2 p3. 196. Solve: y2 log y = x y p + p2. 197. Solve: p2 y + 2p x = y. 198. Solve: p3 – 4 x y p + 8p2 = 0. 199. Solve: y p = 2p2 x + y2 p4. 200. Solve: x p2 – y p – p + 1 = 0. Equations solvable for y 201. Solve: y = (1 + p) x + a p2. 202. Solve: y = y p2 + 2px. 203. y = 2p x + tan–1 (xp2). 204. x2 p4 + 2x p = y. 205. y + p x = x4 p2. 206. y = p sin + cos p. Clairaut Differential Equation 207. Solve: (y + 1) P – xP 2 + 2 = 0. 208. Solve: P3x – P2y – 1 = 0.
209. Solve: (y + 1) P – xP2 + 2 = 0. 210. Solve: sin y ◊ cos Px – cos y ◊ sin Px – P = 0. 211. Solve: (x – a) P2 + (x – y) P – y = 0. ______
212. Solve: y = px + a ÷1 + p2 . a 213. Solve: y = P (x – b) + __ . P Higher Order Differential Equation d2y Differential equation is the form ___ 2 = f (x) dx d 2y ___ 214. Solve: 2 = x + sin x. dx d 2y 215. Solve: ___ 2 = e2x + ex + 2014. dx d 2y 216. Solve: ___2 = sin2 x. dx d 2y 217. Solve: ___ 2 = cos3 x. dx d2y 1 218. Solve: ___ 2 = _________ 2 . dx sin x cos2 x d2y 219. Solve: ___ 2 = sin4 x + cos4 x. dx d2y 220. Solve: ___ 2 = xex. dx
d2y Differential equation of the form ___ 2 = f (y) dx d2y 221. Solve: ___ 2 + y = 0. dx d2y 1 222. Solve: ___ 2 = __ 3 . dx y d2y 1 223. Solve: ___ 2 = ____ __ . 4÷y dx d2y 224. Solve: a2 ___ – y = 0. dx2 d2y 225. Solve: ___ 2 = e2y. dx d2y 226. Solve: 2 ___2 = 3y2, dx
( )
dy y (– 2) = –1, ___ = – 1, dx x = – 2
4.13
4.14 Integral Calculus, 3D Geometry & Vector Booster
( )
d2y 227. Solve: y3 ___ 2 = –1, dx
242. A curve passes through (2, 0) and the slope of the
( )
dy y (1) = –1, ___ = 0. dx x = – 1
( )
d 2y ___ dy Differential equation is the form j ___ 2 , = 0 dx dx 2 d y dy 228. Solve: x ___2 + ___ + x = 0. dx dx d 2y 1 ___ dy 229. Solve: ___2 = __ x + x. dx dx d2y 230. Solve: y ___2 + dx
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dy 2 ___ = 1. dx
d 2y dy 2 dy 231. Solve: y ___2 – ___ = y2 ___ . dx dx dx d2y dy 2 232. Solve: (x + a) ___2 + x ___ = dx dx d 2y 1 ___ dy 2 233. Solve: x ___2 – __ = 4 dx dx
dy ___ . dx
dy ___ . dx
d2y dy 2 234. Solve: y ___2 = 1 – ___ . dx dx d2y dy 235. Solve: ___ 2 = 2y ___ . dx dx
d2y dy 2 dy 236. Solve: y ___2 – ___ = y2 ___ . dx dx dx d2y 1 ___ dy a 237. Solve: ___ 2 + __ x – __ 2 = 0. dx dx x
d2y dy dy 2 238. Solve: y ___2 – y ___ ln y = ___ . dx dx dx Geometrical Applications
239. The slope of the curve at any point is the reciprocal of twice the ordinate at the point. The curve passes through the point (4, 3). Find the equation of the curve. 240. Find the equation of the curve whose slope at any point is y + 2 x and which passes through the origin. 241. Find the equation of the curve which passes through the origin and the tangent to which at every point (x, y) has slope equal to 4
x + 2xy – 1 ___________ . x2 + 1
(x + 1)2 + y – 3 tangent at P (x, y) is ______________ . (x + 1) Find the equation of the curve. 243. A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve. 244. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve. 245. If the length of the tangent at any point on the curve y = f (x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve. 246. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve, given that it passes through (– 2, 1). 247. The ordinate and the normal at any point P on the curve meet the x-axis at points A and B respestively. Find the equation of the family of the curves satisfying the condition AB is the AM of abscissa and the ordinate of P. 248. Find the nature of the curve for which the length of the normal at any point P is equal to the radius vector.
(Mixed Problems) 1. The order and the degree of the differential equation
÷
___
dy dy ___ – 4 ___ – 7y = 0 dx dx are
(a) 1 and 1/2 (b) 2 and 1 (c) 1 and 1 (d) 1 and 2 2. The order of the differential equation whose general solution is given by
y = c1e2x + c 2 + c3 ex + c4 sin (x + c5) 3.
is (a) 5 (b) 4 (c) 3 (d) 2 The differential equation of all straight lines passing through the origin is
÷
____
dy (a) y = x ___ dx
dy (b) ___ = y + x dx
Differential Equation
dy y (c) ___ = __ dx x
(d) none of these
4. A differential equation associated by the promitive y = a + be5x + ce–7x is
(a) y3 + 2y2 + y1 = 0
(b) 4y3 + 5y2 – 20y1 = 0
(c) y3 + 2y2 – 35y1 = 0
(d) none of these 5. The differential equation of the family of curves y = a cos (x + b) is
d2y (a) ___2 – y = 0 dx
d2y (b) ___2 + y = 0 dx
d2y (c) ___2 + 2y = 0 (d) none of these dx 6. The differential equation, whose general solution is y = A sin x + B cos x, is
d2y (a) ___2 + y = 0 dx
d 2y (b) ___2 – y = 0 dx
dy (c) ___ + y = 0 dx
(d) none of these
7. The differential equation of all straight lines passing through the point (1, – 1) is
10. The solution of the differential equation dy 1 + x2 ___ + _____ = 0 x dx is 1 (a) y = – __ tan–1 x + c 2 x2 (b) y + ln x + __ + c = 0 2 1 (c) y = __ tan–1x + c 2 x2 (d) y – ln x – __ = c 2 11. The solution of the differential equation
x cos y dy = (x ex ln x + ex) dx is
1 (a) sin y = __ x ex + c
(b) sin y + ex ln x + c = 0
(c) sin y = ex ln x + c
(d) none of these
12. The solution of the differential equation x (e2y – 1) dy + (x2 – 1) ey dx = 0 is
x2 (a) ey + e–y = ln x – __ + c 2 x2 (b) ey – e–y = ln x – __ + c 2
dy (a) y = (x + 1) ___ + 1 dx
dy (b) y = (x + 1) ___ – 1 dx
dy (c) y = (x – 1) ___ + 1 dx
dy (d) y = (x – 1) ___ – 1 dx
8. The differential equation of the family of parabolas with focus at the origin and the x-axis as axis is
( ) ( ) ( ) ( )
dy dy (a) y ___ + 4x ___ = 4y dx dx 2
x2 (c) ey + e–y = ln x + __ + c 2 (d) none of these
13. Solution of the equation (1 – x2) dy + xy dx = xy2dx is
dy 2 dy (b) – y ___ = 2x ___ – y dx dx 2 dy dy (c) y ___ + y = 2xy ___ dx dx
(a) (y – 1)2 (1 – x2) = 0
(b) (y – 1)2 (1 – x2) = c2y2
(c) (y – 1)2 (1 + x2) = c2y2
(d) none of these
dy dy (d) y ___ + 2xy ___ + y = 0 dx dx
14.
The general solution of y dx + (1 + x2) tan–1 x dy is (a) y tan–1 x = c (c) y + tan–1x = c
2
9. The differential equation of all lines in the xy-plane is
dy (a) ___ – x = 0 dx
d2y (b) ___2 – dx
d 2y (c) ___2 = 0 dx
d2y (d) ___2 + x = 0 dx
dy x ___ = 0 dx
4.15
the differential equation = 0, (b) x tan–1y = c (d) x + tan–1 y = c
dy 15. For solving ___ = (4x + y + 1), suitable substitution dx is
(a) y = vx
(b) y = 4x + v
4.16 Integral Calculus, 3D Geometry & Vector Booster
(c) y = 4x
(d) y + 4x + 1 = v
16. The solution of the differential equation dy xy ___ = ______ dx x2 + y2 is 2
2
(a) ay2 = e x /y
(b) ay = ex/y
(c) y = ex2 + ey2 + c
(d) y = ex2 + y2 + c
17. (x2 + y2) dy = x y dx, if y (x0) = e, y (1) = 1, the value of x0 = ______ __ 1 (a) ÷3 e (b) e2 – __ 2 ______ ______
÷
÷
÷
e2 – 1 e2 + 1 (c) ______ (d) ______ 2 2 18. The solution of (1 + xy) y dx + (1 – xy) x dy = 0 is x x 1 1 (a) __y + __ xy = k (b) log __ y = __ xy + k
x (c) __y = exy + k
( ) x (d) log ( __ y ) = xy + k
19. The solution of the equation (x + log y) dy + y dx = 0 is
(a) xy + y log y = c
(b) xy + y log y – y = c (c) xy + log y – x = c (d) none of these
20. The solution of (x – y3) dx + 3xy2dy = 0 is
x (a) log x + __ 3 = k y y (b) log x + __ 3 = k x x (c) log x – __ 3 = k y (d) log xy – y3 = k
21. The solution of the differential equation dy 3x2 sin2 x ___ + _____ 3 y = _____ dx 1 + x 1 + x3 is
1 (a) y (1 + x3) = x + __ sin 2 x + c 2 1 (b) y (1 + x3) = cx + __ sin 2 x + c 2 1 3 (c) y (1 + x ) = cx – __ sin 2 x + c 2 x 1 (d) y (1 + x3) = __ – __ sin 2 x + c 2 2
22. The solution of the equation
dy x ___ + 3y = x dx is
x4 (a) x3y + __ + c = 0 4
x4 (b) x3y = __ + c 4
x4 (c) x3y + __ + c (d) none of these 4 23. The solution of the differential equation dy ___ + 2y cot x = 3x3 cosec2 x dx
is
(a) y sin2 x = x3 + c
(b) y sin x = c
2
(d) y sin x2 = c
(c) y cos x = c
24. The integrating factor of the differential equation x dy – y dx = xy2 dx is
1 (a) __2 x
1 (b) __2 y
1 (c) __ xy
1 (d) ____ x2y2
25. The equation of the curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx – xy dy = 0, is
(a) x2 + y2 = 1
(b) x2 – y2 = 1
(c) 2x2 + y2 = 2
(d) none of these
( )
7 26. The equation of a curve passing through 2, __ and 2 1 having gradient 1 – __2 at (x, y) is x (a) y = x2 + x + 1 (b) xy = x2 + x + 1 (c) xy = x + 1 (d) none of these 27. The differential equation of the family of circles passing through the fixed points (a, 0) and (– a, 0) is
( )
(a) y1(y2 – x2) + 2xy + a2 = 0
(b) y1y2 + xy + a2x2 = 0
(c) y1(y2 – x2 + a2) + 2xy = 0
(d) none of these
y+1 28. The number of solutions of y = _____ , y (1) = 2 x–1 is
(a) none (c) two
(b) one (d) infinite
Differential Equation
29. The solution of
(a) order 1
(b) order 2
dy ___ = 1 + x + y2 + xy2, y (0) = 0 dx is
(c) degree 3
(d) degree 4
2
(a) y =
(
2
)
x x + __ – 1 e 2
2
(
(b) y = 1 +
(c) y = tan (c + x + x2)
(
2
(b) y2 = ax + b
y = (c1 + c2) cos (x + c3) – c4 ex + c 5 where c1, c2, c3, c4, c5 are arbitrary constants, is (a) 5 (b) 4 (c) 3 (d) 2 A solution of the differential equation
( )
dy 2 dy ___ – x ___ + y = 0 dx dx is
(a) y = 2 (c) y = 2x – 4
(a) – 1/2 (c) e – 1/2
(b) y = 2x (d) y = 2x2 – 4
(b) e + 1/2 (d) 1/2
( )
dy 2 + sin x ___ 34. If y = y (x) and ________ = – cos x y (0) = 1, y + 1 dx
( )
p then y __ equals 2 1 (a) __ 3 1 (c) – __ 3 35. 36.
_____
39.
_____
)
(
_____
_____
)
is (a) 2 (b) 3 (c) 4 (d) none The degree of the differential equation
( )
d3y 2/3 d2y dy ___ 3 – 3 ___2 + 5 ___ + 4 = 0 dx dx dx is
(a) 1 (c) 3
2 (b) __ 3 (d) 1
If x dy = y (dx + y dy), y (1) and y (x) > 0, then y (– 3) = (a) 3 (b) 2 (c) 1 (d) 0 The differential equation representing the family __ of curves y2 = 2c (x + ÷c ), where c is a positive parameter, is of
(b) 2 (d) none.
(More than one options are correct) 40. If m and n be the order and the degree of the differential equation
( )
( ) ( ) ( )
d2y 3 ___ d2y d3y dx2 ___ 2 + 4 ______ + ___ 3 = x2 – 1, 3 dx dx dy ___ 3 dx 5
dy 33. If y (t) is a solution of (1 + t) ___ – ty = 1 and dt y (0) = – 1 then y (1) is equal to
(b) (x + 1) + ex
(a) x + 2
(
(c) y = log x (d) y = ex + C 31. The order of the differential equation whose general solution is given by 32.
÷1 + x2 + ÷ 1 + y2 = A x ÷1 + y2 – y÷1 + x2
)
x (d) y = tan x + __ 2 2 d y 30. If ___2 = 0, then dx (a) y = ax + b
37. If f ¢(x) + f (x) = x, when f (0) = 2, then f (x) is (c) (x – 1) + 3e–x (d) none. 38. The degree of the differential equation
)
x2 x + __ 2 c e
4.17
then
(a) m = 3 and n = 5 (c) m = 3 and n = 3
(b) m = 3 and n = 1 (d) m = 3 and n = 2
41. If y = e4x + 2e–x satisfies the relation d 2y dy ___2 + A ___ + By = 0, then dx dx
(a) A = – 3
(b) B = – 4
(c) A – B = 1
(d) A + B = – 7
dy x2 + y2 + 1 42. The solution of ___ = _________ satisfying 2xy dx y (1) = 1 is given by
(a) a system of hyperbola (b) a system of circles (c) y2 = x (1 + x) – 1 (d) (x – 2) + (y – 3)2 = 5.
4.18 Integral Calculus, 3D Geometry & Vector Booster 43. The function f (x) satisfying the relation 2
2
({f (x)} + 4 ◊ f (x) ◊ f ¢(x) + {f ¢(x)} ) = 0 is given by __
__
) x (a) ke (2 + ÷3
) x (b) ke (2 – ÷3
) x (c) ke (4 – ÷3
) x (d) ke(4 + ÷3
__
(a) m = 3 (b) n = 2 (c) m + n = 5 (d) m – n = 1 45. The value of k such that the family of parabolas y = cx2 + k is the orthogonal trajectory of the family of ellipses x2 + 2y2 – 2y = c is (a) 1/2 (b) 1/4 (c) 2 (d) 4. 46. A differential function f satisfies the relation x+y f ______ = f(x) f(y) " x, y Œ R –{– 1}, 1 + xy
)
where f(0) π 0 and f ¢(0) = 1. Then dy y (a) f(0) = 1 (b) ___ = _____ 2 dx 1 – x 1–x (c) f(x) = _____ (d) None of these 1+x d 47. Let ___ (x2y) = x – 1, where x π 0 and x = 1, y = 0. dx Then 1 1 ___ 1 (a) The solution of the curve is y = __ – __ + 2 x 2x2 (b) The set of values of x, where x is increasing is (– •, 0) (1, •) (c) The set of values of x where x is decreasing is (0, 1) (d) None of these. dy dy dy 2 48. Let y = x ___ + ___ – ___ . then dx dx dx
( ) ( ) ( )
(a) The genral solution is y = cx + c – c2 (b) The particular solution is y = 2x – 4 (c) The singular solution is 4y = (x + 1)2 (d) None of these dy 1 – 3x – 3y 49. Let ___ = _________ . dx 1 + x + y If the solution of the above differential equation is mx + y + n loge|1 – x – y| + C then (a) m = 2 (b) n = 3 (c) m + n = 5 (d) n – m = 1 dy ax + b 50. Let ___ = ______ . If the solution of the given differdx cy + d ential equation is a parabola, then
(a) a (b) c (c) a (d) c
= = = =
2, 0, 0, 0,
b d b a
= = = =
0 4 2, c = 4 2, b = 2, d = 4
__
44. If the order and degree of the differential equation y = c1 sin–1x + c2 cos–1x + c3 tan–1x + c4 cot–1x where c1, c2, c3 and c4 are arbitrary constants, are m and n respectively, then
(
(Problems for JEE-Advanced)
1. Find the equation of the curve which passes through the origin and the tangent to which at every point
x4 + 2xy – 1 (x, y) has slope equal to ___________ . 1 + x2 [Roorkee-JEE, 2001]
2. Solve the differential equation y y y cos __ x (x dy – y dx) + x sin __ x (x dy + y dx) = 0
( )
( )
p when y (1) = __ . [Roorkee-JEE, 1997] 2 3. Solve the differential equation dy cos2 x ___ – (tan 2 x) y = cos4 x dx
( )
__
3÷3 p p p where – __ < x < __ and y __ = ____ 4 4 8 6 [Roorkee-JEE, 1996]
d 4. If y + ___ (xy) = x (sin x + log x), find y (x). dx [Roorkee-JEE, 1995] 5. Solve the differential equation
x (1 – x2) dy + (2x2y – y – 5x3) dx = 0 [Roorkee-JEE,1994] 6. A tangent to the curve y = f (x) cuts the line y = x at a point which is at a distance of one unit from y-axis. Find the equation of the curve. ______
7. Solve: x dy – y dx = ÷x 2 + y2 dx. y y 8. Solve: xe y/x – y sin __ x dx + x sin __ x dy = 0. 9. Solve: (x + 2y) (dx – dy) = dx + dy.
(
( ) )
( )
10. Solve: (1 + y2) dx = (tan– 1 y – x) dy. dy 2xy 11. Solve: ___ = __________ . dx x2 – 2y – 1 12. Solve: x dy – {y + xy3(1 + log x)} dx = 0. 13. Solve: sec2x ◊ tan y dx + sec2y ◊ tan x dy = 0. ________ dy 14. Solve: ÷ x + y + 1 ◊ ___ = x + y – 1. dx
(
)
x 15. Solve: (1 + ex/y) dy + 1 – __ y ex/ydy = 0.
Differential Equation
16. Solve: (4x – y + 3) dy + (2x – 3y – 1) dx = 0. dy 17. Solve: (1 + y2) + (x – etan–1y) ___ = 0. dx 18. Solve: y dx – x (1 + xy) dy = 0. 19. Prove that the equation of a curve whose slope at – (x + y) (x, y) is _______ and which passes through the point x 20.
21. 22. 23. 24. 25.
26.
27.
(2, 1), is x2 + 2xy = 8. If the square of the intercept cut by any tangent on the y-axis is equal to the product of the coordinate of the point of contact, find the equation of such curves. Find the curves for which the length of the normal is equal to the radius vector. dy 2 dy Solve: ___ – (ex + e– x) ___ + 1 = 0. dx dx 2 dy dy Solve: ___ + 2x ___ = 3x2. dx dx 2 dy dy Solve: xy ___ – 1 = (x2 – y2) ___ . dx dx Find the equation of the curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve, given that it passing through (– 2, 1). Find the equation of the curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.
( ) ( ) { ( ) }
28. Find the equation of the curve which touches the line y = 1 and passes through the point (0, 1) and satisfies
( ) 2
dy the differential equation y3 ___ 2 – y = 1. dx 29. The ordinate and the normal at any point P on the curve meet the x-axis at points A and B respectively. Find the equation of the family of curves satisfying the condition AB = arithmetic mean of abscissa and ordinate of P. 30. A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. PQ is of the constant length k. Find the coordinates of Q. Also find the differential equation of the given curve. 31. Find the equation of the curve, the slope of whose 2y tangent at any point (x, y) is ___ x , for all x, y > 0 and which pass through the point (1, 1).
4.19
32. Let C be a curve such that the normal at any point P on it meets x-axis and y-axis at A and B, respectively. If PB : PA = 1 : 2 (internally) and the curve passes through the point (0, 4), prove that the curve is a ___
hyperbola and it passes through the point (÷10 , 6). 33. A normal is drawn at any point P (x, y) of a curve, it meets the x-axis and the y-axis in points A and B 1 1 respectively, such that ___ + ___ = 1, where O is OB OA the origin. Find the equation of such a curve passing through (5, 4).
(Tougher Problems for JEE-Advanced)
1. Find the order of the differential equation whose general solution is given by y = C1cos (2x + C2) – (C3 + C4)3x + C 5 + C6sin (x – C7) 2. Form the differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis. 3. The slope of the tangent to a curve at a point P (x, y) 2y is ___ x , x, y > 0 and which passes through the point (1, 1), find the equation of the curve. 4. A conic C passes through the point (2, 4) and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangentcy. Find the equation of the conic. 5. Solve the differential equation x2 + y2 dy x2 + y2 ______ 2y ___ = e x + ______ x – 2x . dx 6. Find the integral curve of the differential equation dy x (1 – x lny) ___ + y = 0, dx 1 which passes through the the point e, __ e . 7. Solve the differential equation dy dy 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ dx dx d 8. If y + ___ (x y) = x (sin x + log x), find y as a function dx of x. 9. Solve the differential equation (x dy + y dx) sin (xy) + (x2y dx + x y2dx) cos (x y) = 0.
(
)
(
)
( )
(
)
(
10. Solve the differential equation dy 2y2cos x + y sin 2x + 2cos x ◊ sin2x ___ = ____________________________ . dx sin2x 11. Solve the differential equation
)
4.20 Integral Calculus, 3D Geometry & Vector Booster
( )
( )
y y y cos __ x (x dy – yd x) + x sin __ x (x dy + y dx) = 0, p where y (1) = __ . 2 12. Solve the differential equation
÷
______________________________
dy x4y2 – x6 + 2x4y – x6y2 – 2x6y + x4 ___ = ______________________________ dx y2 – x2y2 + x3y2 – x5y2 x
log t 13. If f (x) = Ú ________ 2 dt, x ≥ 1, prove that 1 1 + t + t
( )
1 f (x) = f __ x . 14. Solve the differential equation dy x2 + y2 + 3x + 3y + 2xy + 1 ___ = _________________________ . dx x2 + y2 – 3x – 3y + 2xy + 2 15. A differentiable function f satisfies the relation f (x + y) + f (x) ◊ f (y) = f (xy + 1). Given f (0) = –1, f ¢(0) = f ¢(1) =1, find the function f. 16. A differentiable function f satisfies the relation f (xy) = x f (y) + yf (x) for every x, y in R. If f ¢(1) = 1, the find the function f. y y x 1 17. Solve: __ y sin __ y – __ 2 cos __ x + 1 dx x
(
( ) ) ( ) y x x 1 1 + __ x cos ( __ x ) – __ sin ( __ y ) + __ dy = 0. ( y y )
2
2
dy (1 + y2) 18. Solve: ___ = _________ . dx xy (1 + x2) y dy ________________ 1 19. Solve: __ x + ___ = 4 . dx sin (xy) + cos2 (xy) 20. Solve: (x2 + y2 + 1) dy + 2xy dx = 0.
(
)
2 2 2 dy 1 x + y 21. Solve: y ___ + x = __ ______ x . 2 dx y + sin x cos2(xy) x dy 22. Solve: _______________ dx + _______ 2 = 0. 2 cos (xy) cos (xy)
23. Find the general solution of the linear equation of dy the first order ___ + p (x) y = q (x), if one particular dx solution y1(x) is known. 24. Find the curve such that the square of the intercept cut by any tangent off the y-axis is equal to the product of the co-ordinates of the point of tangency. 25. If f (x) is a function such that x
x
0
0
x Ú (1 – t) f (t) dt = Ú t f (t) dt, f (1) = 1, find f (x).
26. Find the equation of the curve passing through (3, 4) and satisfying the differential equation dy 2 dy y ___ + (x – y) ___ – x = 0. dx dx y j (y/x) 27. Solve: y¢ = __ x + ______ . j¢(y/x) 28. A differentiable function f satisfies the relation x+y f ______ = f (x) ◊ f (y) for all x, y in R – {–1}, where 1 + xy
( )
(
)
f (0) π 0 and f ¢(0) = 1. Find f (x). A differentiable function f satisfies the relation f (xy) = xf (y) + yf (x) for all x, y in R+. If f (1) = 1, find f (x). Find the curve for which the portion of the tangent included between the coordinate axes is bisected at the point of contact. 31. Find the nature of the curve for which the length of the normal at a point P is equal to the radius vector of the point P. 32. Find the equation of the curve passing through (2, 2) such that the slope of the tangent at any point to the curve is reciprocal of the ordinate of the point. 33. A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is dy y2 – 2xy ___ – x2 = 0 dx and hence find the equation of the curve. 34. Find the orthogonal trajectories to the curve y2 = a (x + a). 35. Find the equation of the curve in which the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact. 29. 30.
Integer type questions
1. Find the degree of the differential equation
( ) ( ) ( )
d3y 2/3 d2y dy ___ 3 – 3 ___ 2 + 5 ___ + 4y = 0. dx dx dx 2. If the order of the differential equation of the curve y = (c1 + c2) sin (x + c3) – c4ex + c 5 is M, find the value of (M – 1). 3. If y = e4x + 2e–x satisfies the differential equation d3y dy ___3 + A ___ + By = 0, the value of A – B + 4. dx dx dy a x + b 4. The solution of ___ = ______ , where a < 0 and dx b y + k b ≥ 0, represents a parabola, find the value of a + b + 4.
Differential Equation
( )
dy 5. The solution of the differential equation ln ___ = dx 3x + 4y, y (0) = 0 is
Ae3x + Be– 4y + C = 0, where A + B + C + 5 is... 6. The polynomial f (x) satisfies the equation f (x + 1) = x2 + 4x. The area enclosed by y = f (x – 1) and the __ 16÷2 2 _____ curve x + y = 0 is , where k is ... k 7. If the number of straight lines which satisfy the dy dy 2 differential equation ___ + x ___ = y is m, the value dx dx of m + 2 is...
( )
8. If f (x) + f ¢(x) = x, where f (0) = 2, then f (x) is
(x – 1) + ke– x, where k + 2 is ... 9. The solution of the differential equation dz z z 1 1 ___ + __ ln z = __ 2 (ln z)2 is ____ = ____ 2 + C, dx x x ln z x mz
(
)
Comprehensive Link Passages
(b) –7 (d) –1
1. The degree of the differential equation
( )
( )
( )
( )
Passage I The differential equation corresponding to
y = c1em 1x + c2em 2x + c3em 3x,
where c1, c2 and c3 are arbitrary constants and m1, m2 and m3 are the roots of m3 – 7m + 6 = 0 is
( ) ( ) ( )
where a, b, g and d are arbitrary constants. On the basis of the above information, answer the following questions. 1. The order of the differential equation is (a) 1 (b) 2 (c) 3 (d) 4 2. The degree of the differential equation is (a) 1 (b) 2 (c) 3 (d) 4 3. The value of a and b, respectively, are
÷ ( ) ÷ ( ) _________
_________
4 d2y d3y 1 + ___ 2 = 1 + ___ 3 is dx dx
d 3y d2y dy a ___ 3 + f ___ 2 + g ___ + d = 0, dx dx dx
dy 3 ! xy ___ + ... to • is dx (a) 0 (b) 1 (c) 2 (d) not defined 2. The dgree of the differential equation
3
In these questions, a passage (paragraph) has been given followed by questions based on each of the passage. You have to answer the questions based on the passage given.
(b) –7 (d) –1
dy dy 2 dy 3 1 1 1 x = 1 + xy ___ + __ ! xy ___ + __ ! xy ___ + __ 2 3 4 dx dx dx
10. The solution of the differential equation e xy2 – __ 3 dx – x2 y dy = 0, x e 1 given y = 0 when x = 1, is y2 = __ __ – x2 , where k x4 k + 5 is ...
)
(b) 1, 0 (d) 0, –1
Passage II The order of the differential equation is the highest order derivative appearing in the equation and the degree of the differential equation is the index power of the highest order derivative and is free from any type of radical sign. On the basis of the above information, answer the following questions.
where m + 4 is
(
(a) 0, 1 (c) –1, 0 4. The value of g is (a) 6 (c) 2 5. The value of d is (a) 6 (c) 2
4.21
(a) 3 (b) 1 (c) 2 (d) not defined 3. The order and the degree of the differential equation of the curve y = mx + m2 + m3 + m4 is (a) 1, 3 (b) 1, 4 (c) 2, 3 (d) 2, 4 4. The order and the degree of the differential equations dy in ___ = x + 1 is dx (a) 1, 4 (b) 2, 4 (c) 1, not defined (d) 2, not defined
( )
Passage III Differential equations are solved by reducing them to get the exact differential of an expression x and y, i.e. they are reduced to the form d [ f (x, y)] = 0 For example, x dx + y dy _________ x dy – y dx _______ _________ = x2 ÷ x 2 + y2 + 2y dy _________ x dy – y dx 1 2x dx fi __ × ___________ ______ = 2 x2 ÷x 2 + y2
4.22 Integral Calculus, 3D Geometry & Vector Booster d (x2 + y2) y ______ fi _________ = – d __ x 2 2 ÷x + y _______ y fi d (÷x 2 + y2 ) = – d __x Thus the required solution is _______ y ÷x 2 + y2 + __ x = c. On the basis of the above information, answer the following questions. 1. The general solution of (2x3 – xy2) dx + (2y3 – x2y) dy = 0, is
( ) ( )
2
2 2
4
(a) x + x y – y = c
2
2 2
4
(b) x – x y + y = c
(c) x2 – x2y2 – y4 = c (d) x2 + x2y2 + y4 = c 2. The general solution of the differential equation x dy y ______ 2 + 1– ______ 2 2 dx = 0, 2 x +y x +y
(
is
)
( ) y ( __ x ) = c
( )
(1 – x)2 ______ (a) e 2
(1 – x)2 ______ (b) e 2 –
(c) ln (1 + x) – 1
(d) 1 + x.
Passage V A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. PQ is of constant length k. On the basis of the above information, answer the following questions. 1. The co-odinates of Q are
y (a) x + tan–1 __ x = c
x (b) x + tan–1 __ y = c
(c) x – tan–1
(d) none.
3. The general solution of the differential equation ey dy + (xey – 2y) dy = 0, is
(a) xey – y2 = c y
(c) ye + x = c
(b) xe y – x2 = c y
2
(d) xe – 1 = cy .
the function and get required result. On the basis of the above information, answer the following questions. 1. The equation of the curve to the point (1, 0) which satisfies the differential equation (1 + y2) dx – xy dy = 0, is
(a) x2 + y2 = 1
(b) x2 – y2 = 1
(c) x2 + y2 = 2
(d) x2 – y2 = 2
–1
–1
dx (d) x – y ___ , 0 . dy
2. The differential equation of the curve is
( ) ( )
( ) ( )
dy 2 dy 2 (a) y ___ = y2 – k2 (b) y ___ = k2 – y2 dx dx 2 dy dy 2 (c) ___ = y2 – k2 (d) ___ = k2 – y2 dx dx 3. The cartesian equation of the curve, if it passes through the point (0, k) is (a) x2 + y2 = k2
(b) x2 + y2 = 2k2 y (c) log y + ÷y 2 – k2 = 0 (d) sin–1 __ = x + c. k
(
______
( )
)
Given below are matching type questions, with two columns (each having some items) each. Each item of Column I has to be matched with the items of Column II, by encircling the correct match(es). Note: An item of Column I can be matched with more than one items of Column II. All the items of Column II have to be matched.
1. Observe the following Columns Column I
(
–1
(a) tan y + sin x = c (b) tan x + sin y = c
(c) tan–1 y ◊ sin–1 x = c
) )
dx (b) x + y ___ , 0 dy
Column II 3
(B) The degree of the differential (Q) Not dy defined equation log 1 + ___ = x is dx
1 + y2 _____ _______ = 0, is ÷ 1 + x2 –1
( (
) )
(A) The degree of the differential (P) equation of the curve __ y2 = 2c (x + ÷c ) is
2. The solution of the differential equation
dy ___ + dx
( (
dy (a) x + y ___ , 0 dx dy (c) x – y ___ , 0 dx
Matrix Match (For JEE-Advanced Examination only)
Passage IV
dy The differential equation ___ = f (x) ◊ g (y) can be solved by dx dy seperating the variable ____ = f (x) dx and then integrating g (y)
1
(d) tan–1 y ◊ – sin–1 x = c.
dy 3. If ___ = 1 + x + y + xy and y (–1) = 0, then y is dx
)
(C) The order (L) and the degree (R) (M) of the differential equation of the curve y = a sin (bx + c) where a, b and c are arbitrary constants, the value of L + M is
4
Differential Equation
(D) The order (L) and the degree (S) (M) of the family of parabolas having vertex at origin and axis along positive y-axis, then L + M + 3 is
5
(B)
(A) The integrating factor of the (P) – x2
e
(B) The integrating factor of the (Q) differential equation dy ___ + x (x + y) = x3 (x + y)3 – dx 1 is
3
1 __ 4 x 1 – __4 y
3. Observe the following Columns: Column II
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) R is true and A is false. (d) R is false and A is true. 1. Assertion (A): The order of all the non-vertical lines in a plane is 3. Reason (R): The order of all the non-vertical lines in a plane passing through the origin is 1. 2. Assertion (A): The order of the differential equation dy sin ___ = x is 1. dx Reason (R): The degree of the differential equation
( )
(B) x2 dy + (xy + y2) dx = 0, (Q) cos (y/x) = y = 1 when x = 1 log |cx| x dy y y ________ (C) ___ – __x + cosec __ x = 0, (R) y = 2 1 – log |x| dx (x π 0, x π e) y = 0, when x = 1
( )
dy (S) log (x2 + y2) + 2xy + y2 – 2x2 ___ = 0, dx 2 tan–1 (y/x) y = 2 when x = 1 p = __ + log 2 . 2
(
)
4. Observe the following Columns: Column I
Column II
(A) If m and n are the order and (P) the degree of the equation d 4y dy 4 ___4 = y + ___ , then (m dx dx + n) is
( )
1/3
Assertion and Reason
(A) (x + y) dy + (x – y) dy = (P) y + 2x = 3x2y 0, y = 1, when x = 1
(S)
(D) The integrating factor of the (S) differential equation 3 (xy2 – e1/x ) dx – x2 y dy = 0 is
(D)
(D) The solution of dy y ___ – dx ________ _______ = 2x2 is 2 ÷x + y2
Codes
3
Column I
(R) sin–1 (y/x) = x2 + c
3
x y dx – (x + y ) dy = 0 is
(C) The solution of dy y x ___ = y + x tan __x is dx
e –x
(C) The integrating factor of the (R) differential equation 2
5
( )
Column II
differential equation dy ___ = x3 y2 – xy is dx
(Q)
p y = 1 at x = 0, then y __ 2 is
2. Observe the following Columns: Column I
dy 2 + sin x ___ If ________ = – cos x, i + y dx
4.23
sin (y/x) = cx
dy ___
edx = (x + 1) is 2. 3. Assertion (A): The integrating factor of the differential dy equation ___ + xy = 10 is ex. dx
Reason (R): The integrating factor of the differential dy equation ___ + Py = Q is e Ú Pdx. dx 4. Assertion (A): The orthogonal trajectories of the curve xy = c is x2 – y2 = c. Reason (R): The solution of he differential equation x d y + y d x = 0 is x y = c 5. Assertion (A): The solution of the differential
2
3
3
equation x y dx = (x + y ) dy is y =
x3 ___ 3 3y ce .
Reason (R): The integrating factor of the differential 1 equation M dx + N dy = 0 is ________ . Mx + Ny
4.24 Integral Calculus, 3D Geometry & Vector Booster 6. Assertion (A): The order of the differential equation of the curve y = c1ex + c2e2x + c3ex + 5 is 2. Reason (R): The order of a differential equation is the highest order derivative of a differential equation.
Questions asked in Previous Years’ JEE-Advanced Examinations
1. A normal is drawn at a point P (x, y) of a curve, it meets the x-axis at Q. If PQ is of constant length k,
prove that the differential equation describing such ______ dy curve is y ___ = ÷k 2 – y2 . dx Find the equation of such a curve passing through (0, k) [IIT, 1994] Let y = f (x) be a curve passing through (1, 1) such that the triangle formed by the co-ordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves. [IIT, 1995] Determine the equation of the curve passing through the origin in the form y = f (x) which satisfies the dy differential equation ___ = sin (10x + 6y) dx [IIT, 1996] A curve y = f (x) passes through the point P(1, 1). The normal to the curve at P is a (y – 1) + (x – 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also, obtain the area bounded by the y-axis, the curve and the normal to the curve at P. [IIT, 1996] A spherical rain drop evaporates at a rate proportional to its surface area at any instant t. The differential equation giving the rate of change of the radius of the rain drop is ... [IIT, 1997]
2.
3.
4.
5.
6. Let u (x) and v (x) satisfy the differential equations
dv du ___ + p (x)u = f (x) and ___ + p (x)v = g (x), where dx dx p(x), f (x) and g (x) are continuous functions. If u (x1)
> v (x1) for some x1 and f (x) > g (x) for all x > x1, prove that any point (x, y) where x > x1 does not satisfy the equation y = u (x) and y = v (x). [IIT, 1997] 7. A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, then P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve. [IIT, 1998] 8. The order of the differential equation whose general solution is given by y = (c1 + c2) cos (x + c3) – c4ex + c 5,
where c1, c2, c3, c4, and c5 are arbitrary constants is ... (a) 5 (b) 4 (c) 3 (d) 2 [IIT, 1998] 9. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve. [IIT, 1999] 10. A solution of the differential equation dy 2 dy ___ – x ___ + y = 0 dx dx is (a) y = 2 (b) y = 2x (c) y = 2x – 4 (d) y = 2x2 – 4 [IIT, 1999] 11. The differential equation representing the family __ of curves y2 = 2c(x + ÷c ), where c is a positive parameter is of (a) order = 1 (b) order = 2 (c) degree = 3 (d) degree = 4. [IIT, 1999] 2 2 12. If x + y = 1, then (a) yy ≤ – 2(y ¢)2 + 1 = 0 (b) yy ≤ + (y ¢)2 + 1 = 0 (c) yy ≤ + (y ¢)2 – 1 = 0 (d) yy ≤ + 2(y ¢)2 + 1 = 0. [IIT, 2000] dy 13. If y (t) is a solution of (1 + t) ___ – t y = 1 and dt y (0) = –1, then y (1) is (a) –1/2 (b) e + 1/2 (c) e – 1/2 (d) 1/2 [IIT, 2003] 14. A right circular cone with radius R and height H contains a liquid which evaporates at a rate propeortional to its surface in contact with air (proportonality constant (k > 0). Find the after what time which the cone is empty. [IIT, 2003]
( ) ( )
( )
dy 2 + sin x ___ 15. If y = y (x) and ________ = – cos x and y + 1 dx
( )
p y (0) = 1, then y __ equals 2
(a) 1/3 (c) –1/3
(b) 2/3 (d) 1. [IIT, 2004]
Differential Equation
16. A curve C passes through (2, 0) and the slope at (x, (x + 1)2 + y – 3 y) as _____________ . Find the equation of the curve (x + 1)
(d) the fixed radius 1 and variable centres along the y-axis. [IIT, 2007]
and the area enclosed by the curve and the x-axis in the fourth quadrant. [IIT, 2004] 17. If y = y (x) and it follows the relation x cos y + y cos x = p, the value of y ≤ (0) is
24. Let f (x) be a differentiable on (0, •) such that
(a) 1 (c) p
(b) –1 (d) – p [IIT, 2005]
18. If the length of the tangent at any point on the curve y = f (x) intercepted between the point and the x-axis is of length 1, find the equation of the curve. [IIT, 2005] 19. The solution of the primitive integral equation 2
2
(x + y ) dy = xy dx is y = y (x). If y (1) = 1 and y (x0) = e, then the value of x0 is ________
(a) ÷2 (e – 1)
(c) ÷3 e
2
__
________ 2 (e2 + 1) ______
(b) ÷
÷
e2 + 1 (d) _____ 2
[IIT, 2005] 20. For the primitive integral equation y dx + y2dy = x dy, x ŒR, y > 0, y = y (x), y (1) = 1, then y (– 3) is
(a) 3 (c) 1
(b) 2 (d) 5
[IIT, 2005] 21. A curve y = f (x) passes through (1, 1) and the tangent at P (x, y) cuts the x-axis and y-axis at A and B, respectively such that BP : AP = 3 : 1, then (a) the equation of the curve is x y¢ – 3y = 0. (b) the normal at (1, 1) is x + 3y = 4. (c) the curve passes through (2, 1/8). (d) the equation of the curve is xy¢ + 3y = 0. [IIT, 2006] dy 2 22. If y satisfies the differential equation ___ = __ x + y, dx
y (1) = 1, then (x + y + 2)2 e – y equals ... ______ [IIT, 2006] dy 1 – y2 23. The differential equation ___ = _____ determines a y dx
÷
family of circles with (a) the variable radii and fixed centre at (0, 1) (b) the variable radii and fixed centre at (0, –1) (c) the fixed radius 1 and variable centre along the x-axis.
4.25
(
)
t2f (x) – x2f (t) f (1) = 1 and lim ____________ = 1 for each x > t Æ x t – x 0, then f (x) is
1 (a) ___ + 3 x 1 (c) – __ x +
2x2 ___ 3 2 __ 2 x
1 (b) – ___ + 3x 1 (d) __ x
4x2 ___ 3
[IIT, 2007] 25. Let a solution y = y (x) of the differential equation ______
______
2 x ÷x 2 – 1 dy – y ÷y 2 – 1 dx = 0 satisfies y (2) = ___ __ . 3 ÷ p –1 Assertion (A): y = (x) = sec sec x – __ 6
(
)
__
÷
_____
1 2÷3 1 Reason (R): y (x) is given by __ y = ____ x – 1 – __ 2 x
26.
27.
28.
29.
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explannation of A. (c) A is true and R is false. (d) A is false and R is true. [IIT, 2008] If y ¢ = y + 1 and y (0) = 1, the values of y (ln2) is ... [IIT, 2009] Find the domain of the definition of non-zero solution of the differential equation (x – 3)2 y ¢ + y = 0 [IIT, 2009] Let f be a real-valued differentiable function on R (the set of real numbers) such that f (1) = 1. If the y-intercept of the tangent at any point P (x, y) on the curve y = f (x) is equal to the cube of the abscissa of P, the value of f (– 3) is ... [IIT, 2010] Let y ¢(x) + y (x)g ¢(x) = g (x)g ¢(x), y(0) = 0, x ŒR,
df (x) where f ¢(x) denotes _____ and g (x) is a given dx non-constant differentiable function on R with g (0) = g (2) = 0. The value of y (2) is ... [IIT, 2011] 30. Let f : [1, •) Æ [2, •) be a differentiable function such that f (1) = 2.
4.26 Integral Calculus, 3D Geometry & Vector Booster __
x
If 6 Ú f (t) dt = 3x f (x) – x – 5 for all x ≥ 1, the value 3
0
of f (2) is ... [IIT, 2011] 31. If y (x) satisfies the differential equation y¢ – y tan x = 2x sec x and y (0) = 0, then
( ) p p (b) y¢ ( __ ) = ___ 4 18 p p (c) y ( __ ) = ___ 3 9 p 4p 2p (d) y¢ ( __ ) = ___ + ____ 3 3 3÷3 p p 2 (a) y __ = ____ __ 4 8÷2 2
2
2
__
[IIT, 2012] p 32. A curve passes through the point 1, __ . Let the 6 y y slope of the curve at each point (x, y) be __x + sec __ x ,
( )
( )
x > 0. The eaquation of the curve is y 1 (a) sin __ x = log x + __ 2 y (b) cosec __x = log x + 2 2y (c) sec ___ x = log x + 2 2y 1 (d) cos ___ x = log x + __ 2 [IIT-JEE, 2013]
( ) ( )
( ) ( )
33. The function y = f (x) is the solution of the differential dy xy x4______ + 2x equation ___ + _____ = _______ in (–1, 1) satisfying 2 dx x – 1 ÷ 1 – x2 __
3 ÷ ___ 2
f (0) = 0, then Ú __ f (x) = dx is
3 p ÷ (b) __ – ___ 3 4
p (c) __ – 6
p (d) __ – 6
__
3 ÷ ___ 4
Py ≤ + Qy ¢ + 1 = 0, where P, Q are functions of x, y and y dy dy Here y ¢ = ___ , y ≤ = ___ , which of the following dx dx
(
)
statements (is/are) true?
(a) P (b) P (c) P (d) P
= = + –
y+x y–x Q = 1 – x + y + y + (y ¢)2 Q = x + y – y ¢ – (y ¢)2
[IIT-JEE, 2015] 36. A solution curve of the differential equation dy (x2 + xy + 4x + 2y + 4) ___ – y2 = 0, x > 0 dx
passes through the point (1, 3). The solution curve (a) intersects y = x + 2 exactly at one point (b) intersects y = x + 2 exactly at two points (c) intersects y = (x + 2)2 (d) does not intersect y = (x + 3)2 [IIT-JEE, 2016]
Answers 198. c (4x – c)2 = 64 y
190. (y – cex) (y + x – ce– x) = 0
199. y2 = 2cx + c3
191. (x – y + c) (x2 + y2 + c) = 0
200. y = cx – 1 + 1/c
192. (xy – c) (x2 – y2 – c) = 0
–1 203. y = 2 ÷cx + tan (c)
193. (y – cx) (y2 – x2 – c) = 0
2 204. y = 2 ÷cx + c c 205. y = c2 – __ x 206. x = sin + c
196. ln y = cx + c2. 197. y2 = 2cx + c2
__
3 ÷ ___ 2 [IIT-JEE, 2014] 34. Let y (x) be a solution of the differential equation (1 + ex) y ¢ + yex = 1. If y (0) = 2, which of the following statements (is/are) true? (a) y (– 4) = 0 (b) y (2) = 0 (c) y(x) has a critical point in (–1, 0) (d) y(x) has no critical point in (–1, 0). [IIT-JEE, 2015] 35. Consider the family of all circles whose centres lie on the straight line y = x. If this family of circles is represented by the differential equation
3 ÷ – ___ 2
LEVEL I
__
3 p ÷ (a) __ – ___ 3 2
__ __
Differential Equation
24. (x y – c) (x2 – y2 – c) = 0
Level II
1. (d) 2. (b) 3. (c) 4. (c) 5. (b) 6. (a) 7. (d) 8. (b) 9. (c) 10. (b) 11. (c) 12. (a) 13. (b) 14. (a) 15. (d) 16. (a) 17. (a) 18. (b) 19. (b) 20. (b) 21. (d) 22. (b) 23. (a) 24. (b) 25. (d) 26. (b) 27. (c) 28. (a) 29. (d) 30. (a) 31. (a) 32. (b) 33. (c) 34. (d) 35. (b) 36. (c) 37. (c) 38. (a) 39. (b) 40. (a, c) 41. (a, b, c, d) 42. (a, c) 43. (a, b) 44. (a, b, c, d) 45. (b) 46. (a, b, c, d) 47. (a, b, c) 48. (a, b, c) 49. (a, b, c, d) 50. (b, c, d)
|
Level III
______
|)
_____ 1 + ÷1 – y2 ______ 1. ÷1 – y2 ln __________ = ± x + c 2 1–÷ 1 – y 3 3 1 4 –1 –1 3. __ tan __ tan 4x + tan __ – __ – 3 4 5 5 4 4. __ sq.u. 3 5. x2 + y2 = 2x
7. y + ÷ x 2 + y2 = cx2
[ (
]
–1
5x ___ 3
}
tan y
10. xe
]
–1
–1
tan y
+ c = (tan y –1) e
x2 1 __ 11. __ y + 2 ln y – y = c. 2
3
(
)
x 2x __ 2 12. – __2 = ___ + ln x + c 3 3 y 13. tan x . tan y = c ________ ________ 2 14. x + c = 2 ÷x + y + 1 + __ ln (÷x + y + 1 – 1) 3 _____ 8 – __ ln ÷x + y + 1 + 2 3 15. x + y ◊ ex/y = c
16. (x + y + z)5 = c (y – 2x – 1)2 –1
17. xetan y = c (y – 2x –1)2 1 18. ln y = __ xy + c ___
20. x = ce ± 2 ÷y/x 2
2
______
28. y2 + ÷y 4 – 1 = e± 2x 29. x2 – y2 = kx dy 30. Q x + y ___ , 0 , x2 + y2 = 2k2 dx
(
)
31. y = x2 32. y2 = 2x2 + 16 33. x2 + y2 – 8 (x + y) + 16 = 0
LEVEL IV
1. 5
( )
d 2y 2. 2a ___ 2 + dx
3. y = x2 4. xy = 8.
5. cx = e – e–
6. ((ln y + 1) + ey) x = 1
y 7. log |(x2 – y2)| + tan–1 __ x = C
8. y ◊ x2 = – x2cos x + 2x sin x + 2cos x
dy 2 ___ = 0 dx
( )
x2 + y2 _____ x
( ( ) )
x3 x3 + __ log x – __ + C 3 9
9. (xy) sin (xy) = C
(
)
4y + sin x 2 ___ 10. ____ tan–1 _________ ___ = log |sin x| + C ÷15 ÷15 sin x
( )
y p 11. y sin __x = ___ 2x 2 12. y – log |y + 1| = __ R1 + x3 + C 3 3 14. y – __ log |(x + y)2 + 1| = x + C 2 15. y = x – 1 16. y = x log |x|
2
21. x ± y = a x
2
27. y = Cex /2 – 1
35. y = cx3, x = cy3
y y 1 8. log x = c + __ e– y/x sin __x + cos __x 2 1 4 __ 9. x + c = x + 2y + __ ln (3x + 6y – 1) 3 3
[
26. (y + 3) = (x + 4)2
34. (x – 1)2 + (y – 1)2 = 25
______
{
25. x2 – y2 + 4 = 0
– x
22. (y – e + c) (y + e
+ c) = 0
23. (2y – x2 – c) (2y + 3x2 – c) = 0
( )
( )
y x 1 17. – cos __ y + sin __ x + x – __ y = c 18. (1 + y2) (1 + y2) = c2 x2
4.27
4.28 Integral Calculus, 3D Geometry & Vector Booster
(
)
35. x2 + y2 = cx
sin (4xy) 1 1 19. __ (xy – sin (2xy)) + __ xy + _______ 4 8 4
(
36. x2 + y2 = 25, y = x + 1
)
sin2 (xy) 1 + __ xy + _______ = log |x| + c 2 2
Integer Type
y3 20. __ + y + x2y = c 3
1 _______ 1 21. __ 2 2 = c x – (x +y)
23. y ◊ ep(x)x = Ú (ep(x)x ◊ q (x)) dx + c __
y 24. 2 __x + log |x| = c 1 – x ____
( )
31. x2 ± y2 = a2 32. y2 = 2x 33. x + y = cx
(
)
34. ÷x 2 + y2 = 2x + c
Hints
P-I : P-II : P-III : P-IV : P-V :
1. (c) 1. (b) 1. (b) 1. (b) 1. (a)
1. Clearly, the order is 2 and the degree is 1. 2. Clearly, the order is 1 and the degree is 3. 3. Clearly, the order is 2 and the degree is 1. 4. We have,
÷ ( ) ÷ ( ( ) ) ( ) ( ( ) ) ( ) _________
_____
d2y dy 2 3 ___ 1 + ___ = c ◊ 2 dx dx fi fi
2 1/2
2
1/3
dy dy 1 + ___ = c ___2 dx dx dy 2 3 1 + ___ = c dx
d2y 2 ___ dx
Thus, order is 2 and degree is also 2.
2. (a) 2. (a) 2. (a) 2. (a) 2. (b)
1. (A)Æ(P); 2. (A)Æ(P); 3. (A)Æ(S), 4. (A)Æ(Q),
and
1. (d) 6. (a)
3. (b) 3. (b) 3. (a) 3. (b) 3. (a)
(B)Æ(Q); (B)Æ(P); (B)Æ(P), (B)Æ(S),
2. (c)
5. (5) 10. (8)
4. (b) 4. (c)
5. (a)
(C)Æ(R); (C)Æ(S); (C)Æ(Q), (C)Æ(P),
(D)Æ(S) (D)Æ(R) (D)Æ(R) (D)Æ(R)
5. We have,
( ) ) ( ( ) )
3. (d)
4. (b)
5. (a)
solutions fi
4. (4) 9. (6)
Assertion and Reason
2
_______
3. (3) 8. (5)
Matrix Match
25. f (x) = x3e( x ) y 27. j __ x = cx 30. xy = c
2
2. (2) 7. (4)
Comprehensive Link Passages
22. y ◊ ep(x)x = Ú (ep(x)x ◊ q (x)) dx + c
÷
1. (2) 6. (3)
dy x2 + 2x ___ = 1 dx
So, the order is 1 and the degree is also 1.
6. We have,
÷ ( ) ) ( ( ) ) _________
dy dy 2 y = x ___ + a 1 + ___ dx dx
fi
dy 2 dy 2 y – x ___ = a2 1 + ___ dx dx
(
So, the order is 1 and the also degree is 1.
7. We have,
( ) ( ) ( ) ( )
d 4y d3y ___4 + sin ___ 3 = 0 dx dx
dy dy 2 x + ___ = R1 + ___ dx dx
d 4y ___4 + dx
fi
dy 2 dy 2 x + ___ = 1 + ___ dx dx
So, the order is 4 and the degree is not defined.
(
d3y ___ 3 – dx
3
3
dy 1 d3 5 1 ___ 3 __ ! + ___ 3 __ ! + ... 3 5 dx dx
Differential Equation
8. We have,
( ) ( ) 2
2
( ) 2
dy dy dy ___ 2 + ___ = x sin ___ 2 dx dx dx So, the order is 2 but the degree is not defined.
2
9. We have, dy ___
edx = (x + 1)
( )
dy x + y ___ = a. dx Thus, the required differential equation is fi
( )
(
)
dy 2 dy 2 y ___ + y2 = x + y ___ dx dx dy 2xy ___ = y2 – x2. dx 14. The equation of all parabolas whose axes parallel to x-axis is (y – k)2 = 4a (x – h) Here, h and k are two arbitrary constants. Differentiating both sides, we get dy (y – k) ___ = 2a ...(i) dx Differentiating again w.r.t. x, we get fi
( )
( )
dy 1 dy 2 __ 1 dy 3 1 + ___ + __ ! ___ + ! ___ + ... 2 dx 3 dx dx
So, the order is 1 and the degree is not defined. 10. We have,
( ) ( ( ) ( (
4.29
( ) ( ) ( ) ) ) ( ) ( ) )
dy dy sin ___ + cos ___ dx dx
dy 1 dy 3 __ 1 dy 5 ! ___ + ! ___ – ... = x ___ – __ 3 dx dx 5 dx
1 dy 2 __ 1 dy 4 __ 1 dy 6 + ! ___ – ! ___ + ... = x + 1 – __ ! ___ 2 dx 4 dx 6 dx So, the order is 1 and degree is not defined. 11. Let the equation of the family of non-vertical lines in a plane is y = mx + c, where m and c are arbitrary constants. dy d2y 2 = 0. Now ___ = m and ___ dx dx From the above three equations, we should eliminate m. Thus, the required differential equation is d2y ___2 = 0. dx 12. Let the equation of the parabola be y2 = 4ax. where a is an arbitrary constant. Differentiating w.r.t. x, we get dy 2y ___ = 4a dx y dy __ fi ___ = a 2 dx Thus, the required differential equation is dy y2 = 2xy ___ . dx 13. The equation of the circle be (x – a)2 + y2 = a2, where a is an arbitrary constant. fi x2 + y2 – 2ax = 0 dy fi 2x + 2y ___ – 2a = 0 dx
( )
d2y dy 2 (y – k) ___2 + ___ = 0 dx dx
...(ii)
Eliminating k from Eq. (i) and (ii), we get
( ) ( )
d2y dy 3 2a ___ 2 + ___ = 0 dx dx which is the required differential equation.
15. The given curve is y = c(x – c)2 Differentiating w.r.t. x, we get dy ___ = 2c (x – c) dx
...(i)
...(ii)
( )
1 dy 2 From Eq. (i) and (ii), we get, c = ___ ___ 4y dx Put the value of c in Eq. (ii), we get
( ) ( ( ) ) ( ( ) ) ( )
dy 1 dy 2 1 dy 2 ___ = 2 ◊ ___ ___ x – ___ ___ 4y dx 4y dx dx fi
dy 1 dy 2 2y = ___ x – ___ ___ 4y dx dx
fi
dy dy 3 8y2 = 4xy ___ – ___ dx dx
which is the required differential equation. 16. The given curve is 2 x2 y __2 + __ 2 = 1 a b Differentiating w.r.t. x, we get
2x 2 yy ≤ ___2 + _____ 2 = 0 a b fi
yy ¢ x __2 + ___ 2 = 0 a b
...(i)
4.30 Integral Calculus, 3D Geometry & Vector Booster Again differentiating w.r.t. x, we get 1 __2 + a fi
yy ≤ ____ 2 + b
(y ¢)2 ____ 2 = 0 b
x yy ≤ _____ x (y ¢)2 1 __2 + _____ 2 + 2 = 0 a b b 2
fi
yy ¢ – ___ + b2
x yy ≤ _____ x (y ¢) _____ 2 + 2 = 0 b b
fi
d2y dy 2 dy xy ___ 2 + x ___ – y ___ = 0 dx dx dx
( ) ( )
which is the required differential equation. 17. Given curve is y = mx dy fi ___ = m dx y = __ x which is the required differential equation. 18. Given curve is y = A cos x + B sin x
fi fi
dy ___ = – A sin x + B cos x dx d2y ___ 2 = – A cos x – B sin x = – y dx
Hence, the required differential equation is d 2y ___2 + y = 0 dx 19. We have, y = aex + be– x fi
dy ___ = aex – be– x dx 2
dy ___2 = aex + be– x = y dx Hence, the required differential equation is fi
d2y ___2 – y = 0 dx 20. We have, y = sin (bx + c) fi
y¢ = b cos (bx + c)
fi
y≤ = – b2sin (bx + c)
fi
y ≤ (1 – y2) + y (y ¢)2 = 0
( )
d2y dy 2 (1 – y2) ___ 2 + y ___ = 0 dx dx which is the required differential equation. fi
21. We have,
y = ae2x + be3x
fi
dy ___ = 2ae2x + 3be3x dx
fi
d 2y ___2 = 4ae2x + 9be3x dx
= 2 (2ae2x + 3be3x) + 3be3x
d2y dy ___ 2 = 2 ___ + 3be3x dx dx
dy dy = 2 ___ + 3 ___ – 2y dx dx
fi
dy 2 d2y 1 + ___ + (y – a) ___ 2 = 0 dx dx
fi
d2y dy ___ 2 = 5 ___ – 6y dx dx
fi
d 2y dy ___ 2 – 5 ___ + 6y = 0 dx dx
(
( )
( )
22. Let the equation of the circle be
x2 + (y – a)2 = a2
fi
x2 + y2 – 2ay = 0
fi
dy dy 2x + 2y ___ – 2a ___ = 0 dx dx
dy dy x + y – ___ = a ___ dx dx dy x + y ___ x + y y dx _______1 ________ fi a = = y dy 1 ___ dx Hence, the required differential equation is x + yy1 fi x2 + y2 = 2 _______ y y 1 dy ___ fi (x2 + y2 – 2y2) = 2xy dx fi
(
dy 2xy ___ = _______ dx (x2 – y2)
= – b2y
fi
2 y ¢ = – _______ ______ y ÷1 – y2
(b) Do yourself.
( )
)
)
Differential Equation
23. Let the equation of the circle be 2
2
2
(x – h) + (y – k) = r
where h and k are parameters but not r. dy fi 2 (x – h) + 2(y – a) ___ = 0 dx dy fi (x – h) + (y – k) ___ = 0 dx 2 dy d 2y fi 1 + ___ + (y – k) ___2 = 0 dx dx
( )
24. Let the equation of the circle be
(x – a)2 + (y – a)2 = a2
fi
x2 + y2 – 2a (x + y) + a2 = 0
fi
2x + 2y – 2a (1 + y1) = 0
fi
x + y – a (1 + y1) = 0 x+y fi a = _____ 1 + y1 Hence, the required differential equation is x+y x+y 2 (x + y) + _____ = 0 x2 + y2 – 2 _____ 1 + y1 1 + y1
( )
( )
fi (1 + y1)2 (x2 + y2) – 2 (x + y)2(1 + y1) + (x + y)2 = 0
25. Do yourself 26. Let the line be y = mx + c dy ___ fi = m dx
fi fi fi fi
fi
27.
fi
dy (y – k) ___ = 2a dx
( ) ( )
fi
d2y (y – k) ___2 + dx
fi
d2y 2a ___ 2 + dx
y d2y __ 2 ___ = 0 b dx2
( )
y ___ dy 1 dy 2 – ___ + __ 2 ___ + 2 dx bx b dx
y d2y __ 2 ___ = 0 b dx2
y dy – __x ___ + dx
( )
which is the required differential equation. 30. Given curve is
d2y ___ 2 = 0 dx which is the required differential equation. Let the equation of the parabola be (y – k)2 = 4a (x – h), where h and k are parameters. dy fi 2 (y – k) ___ = 4a dx
fi
( )
dy 2 d2y ___ + y ___ 2 = 0 dx dx which is the required differential equation. 29. Let the equation of the parabola be (x – h)2 = 4a (y – k) where h and k are parameters dy fi 2 (x – h) = 4a ___ dx dy fi (x – h) = 2a ___ dx 2 dy fi 1 = 2a ___2 dx 2 d y 1 ___ fi 2 = ___ 2a dx fi
fi
dy 2x 2y ___ ___2 + ___ 2 = 0 a b dx y dy x __ 2 + __ 2 ___ = 0 a b dx 1 1 dy 2 __ 2 + __ 2 ___ + a b dx
dy 2 ___ = 0 dx
fi
__
y2 = 2c (x + ÷c ) dy 2y ___ = 2c dx dy y ___ = c dx d 2y y ___2 + dx
( )
dy 2 ___ = 0 dx
which is the required differential equation. 31. Do yourself. 32. Given differential equation is x2 – 1 dy = ______ 2 dx. x +1 Integrating, we get
x2 – 1
dx Ú dy = Ú x______ 2 +1
which is the required differential equation 28. Let the equation of the ellipse be
fi
1–2 y = Ú _____ dx x2 + 1
y x2 __2 + __ 2 = 1 a b
dy 3 ___ = 0 dx
( )
= x – 2tan–1x + c
which is the required solution.
4.31
4.32 Integral Calculus, 3D Geometry & Vector Booster 33. Given differential equation is
1 dy = _____ dx ex + 1
which is the required solution.
Integrating, we get
36. The given differential equation is
1 x dx Ú dy = e_____ +1
fi
ex y = Ú ((ex + 1) – __ x + 1) dx e
fi
(
____
)
1 – ex y = Ú _____x dx 1+e = x – log |e + 1| + c,
(
)
cos 2x – cos x dx dy = ____________ 1 – cos x
Integrating, we get cos 2x – cos x
dx Ú dy = Ú ____________ 1 – cos x
fi
(2cos x + 1) (1 – cos x) y = – Ú ___________________ dx (1 – cos x)
= – Ú (2cos x + 1) dx
= – (2 sin x + x) + c
which is the required solution of the given differential equation. 35. The given differential equation is dy 1 4 ___ = __ (x + 1) dx x dx fi dy = ________ 4 x (x + 1) Integrating, we get
fi
____
fi
dx ________ Ú dy = Ú x (x 4 + 1) 3
x dx y = Ú ________ 4 4 x (x + 1)
dt 1 = __ Ú _______ 4 t (t + 1)
____
÷sin x dx Ú dy = Ú ________ sinx cos x
fi
sin x ÷ y = Ú ________ dx sin x cos x
2 sec x ____ = Ú _____ dx tan x ÷
____
(Dividing the numerator and the denuminator by cos2x) ____
fi y = 2 ÷tan x + C which is the general solution. 37. The given differential equation is dy ___ = x2 (3x + 2x)2 dx x2 fi dy = ________ dx (3 + 2x)2 Integrating, we get ________ dx Ú dy = Ú (3 + 2x)2
fi
x2dx y = Ú ________ (3 + 2x)2
Let
3 + 2x = t
1 1 t – 3 2 __ = __ Ú ____ ◊ 2 dt 2 2 t
1 t–3 2 = __ Ú _____ dt t 8
1 1 __ 1 = __ Ú __ – + 1 dt t t 4
t 1 = __ log ____ + c 4 t+1
| |
)
x2
(
sin x ÷ dy = ________ dx sin x cos x
Integrating, we get
34. The given differential equation is dy cos 2x – cos x ___ = ____________ 1 – cos x dx
dy ÷sin x ___ = ________ dx dx sin x cos x
x
which is the required solution.
fi
| |
x4 1 = __ log _____ 4 + c 4 x +1
( )
( ) 3 1 = __ Ú ( 1 – __ ) dt t 8 2
(
)
6 9 1 = __ Ú 1 – __ + __ 2 dt t 8 t 1 1 = __ 1 – 6log t – __ + c t 8
(
(
)
)
1 1 = __ (3 + 2x) – 6 log (3 + 2x) – _______ + c 8 (3 + 2x)
Differential Equation
which is the required differential equation. 38. The given differential equation is
Let
sin x – cos x = t
÷ 1 – x dy = ______ ÷1 + x dx
fi
(cos x + sin x) dx = dt
Also,
1 – sin 2x = t 2
fi
sin 2x = 1 – t2 __ dt _____ =÷ 2 Ú ______ ÷ 1 – t2
Ú dy = Ú ÷
= ÷2 sin–1(t) + c
= ÷2 sin–1 (sin x – cos x) + c
______ __
dy 1 – ÷x ___ = ______ __ 1+÷ x dx fi
______ __ ÷ __ ÷
Integrating, we get
______ __ 1 – ÷x ______ dx __ 1+÷ x
÷
fi Let
x = cos22q
fi
dx = – 2sin 4q dq
fi
Integrating, we get
÷
_________
fi
1– cos 2q y = Ú _________ × – 2sin 4q dq 1 + cos 2q = Ú tanq × – 2 sin 4q dq
= – 2 Ú cos 2q (1 – cos 2q ) dq
= – 2 Ú (2cos 2q – 1– cos 4q ) dq
sin 4q = – 2 sin 2q – q – _____ + c 4
= (– 2 sin 2q + 2q + sin 2q cos 2q ) + c
= – 2 ÷1 – x + cos–1 ÷x + ÷ x (1 – x) + c
(
)
_____
__
__
which is the required solution of the given differential equation. 39. The given differential equation is ____ ____ dy ___ = (÷tan x + ÷ cot x ) dx fi
____
____ ____ (÷tan x + ÷ cot x ) dx
Ú
fi
y=
sin x + cos x ________ = Ú __________ dx sin x cos x ÷
__ sinx + cos x _________ = ÷2 Ú __________ dx 2sin x cos x ÷
__ sin x + cos x =÷ 2 ___________ _____ dx sin 2x ÷
(
dx = Ú ____________ 3 sin x + cos3x
dx = Ú ________________________ (sin x + cos x) (1 – sinx cosx)
sin x + cos x 1 2 = __ Ú ___________ + _____________ dx 3 (sin x + cos x) (1 – sin x cos x)
dx 2 = __ Ú ___________ 3 (sin x + cos x)
____
+ ÷ cot x ) dx Ú dy = Ú (÷tan x
)
( Ú (
dx dy = ____________ 3 sin x + cos3x
dx Ú dy = Ú ____________ 3 sin x + cos3x
Integrating, we get
__
Integrating, we get
____ ____ (÷tan x + ÷ cot x ) dx
dy =
__
which is the required differential equation. 40. The given differential equation is dy 1 ___ = ___________ dx sin3x + cos3x
______ __
1–÷ x y = Ú ______ dx __ 1+÷ x
4.33
(
)
1 sin x + cos x + __ Ú _____________ dx 3 (1 – sin x cos x) dx 2 = ____ __ Ú _________ p 3÷2 sin x + __ 4
(
)
2 sin x + cos x + __ Ú __________ dx 3 (2 – sin 2x)
| |
x p 2__ = ____ log __ + __ 2 8 3÷2 2 + __ tan–1(sin x – cos x) + c 3
)
which is the required solution of the given differential equation. 41. The given differential equation is
)
dy 1 ___ = ______________ ____ 4 ____ dx (÷sin x + ÷ sin x )
4.34 Integral Calculus, 3D Geometry & Vector Booster fi
dx dy = _______________ 4 ____ _____ (÷sin x + ÷ cos x )
Integrating, we get fi
dx
____ 4 Ú dy = Ú _______________ _____ (÷sin x + ÷cos x )
sec x dx = Ú ___________ ____ (÷tan x + 1)4
2t dt = ______ , (Let tan x = t2) (t + 1)4
(t + 1) – 1 = 2 Ú _________ dt (t + 1)4
1 1 = 2 Ú ______ – ______ dt (t + 1)3 (t + 1)4
1 1 = 2 _________ 2 + ________ + c – 2(t + 1) 3 (t + 1)3
1 1 = 2 _____________ + ____________ + c ____ ____ – 2(÷tan x + 1)2 3(÷tan x + 1)3
)
fi
Ú e dy = Ú (e + x ) dx y
x
_____________ dy ÷(1 + x2)(1 + x2) + xy ___ = 0 dx
fi
_______ _______ dy ÷(1 + x2) + y2) = – xy ___ ÷ (1 dx
)
÷(1 + x2) y _______ dx = – ________ dx ________ x + y2) ÷ (1 Integrating, we get _______
+ x2) y ÷ (1 _______ dx = – Ú _________ dy Ú ________ x ÷(1 + y2) fi
x
44. The given differential equation is
______ ÷1 + x2 +
|
|
_____
______
1 + x2 – 1 ÷ _____ 1 __ + ÷1 + y2 =c log __________ 2 2 ÷1 + x + 1
which is the required solution of the given differential equation. 46. The given differential equation is dy sin y + cos y ___ = ___________ dx x (2log x + 1) fi
dy dx __________ = ___________ sin y + cos y x (2 log x + 1)
Integrating, we get dy dx __________ = ___________ Ú sin y + cos y Ú x (2log x + 1) fi fi
2
x3 fi e = e + __ + c, 3 which is the required solution. y
fi
_______
which is the required solution of the given differential equation. 42. Given differential equation is dy ______ = (1 + x) dx Ú (1 + y) Ú
dy = Ú (ex + x2) dx Ú ___ e– y
sec2y 3ex _______ _____ d x = dy Ú (1 – ex) Ú tan y
_______________ dy ÷1 + x2 + y2 + x2y2 + xy ___ = 0 dx
)
dy fi ___ = (ex + x2) dx e– y Integrating, we get
sec2y 3ex _______ _____ d x = dy tan y (1 – ex)
fi – 3 log |1 – e x| = log |tan y| + c which is the required solution of the given differential equation. 45. The given differential equation is
Integrating, we get dy ______ = (1 + x) dx. (1 + y) x2 fi log |1 + y| = x + __ + c, 2 which is the required solution. 43. Given differential equation is dy ___ = e– y(ex + x2) dx
fi
2
( (
3ex tan y dx + (1 – ex) sec2y dy = 0
On Integrating, we get
dx ____ y = Ú _______________ 4 _____ (÷sin x + ÷ cos x )
(
dy dx 1 ___ __ Ú _________ = Ú ___________ p x (2log x + 1) __ 2 sin y + ÷ 4 y p 1 1 ___ __ log tan __ + __ = __ log |2 log x + 1| 2 8 2 ÷2
(
|
(
)
)|
which is the required solution of the given differential equation. 47. The given differential equation is
(
)
dy dy x ___ = a y2 + ___ dx dx
Differential Equation
dy (x – a) ___ = ay2 dx a fi y – 2dy = ______ dx (x – a) Integrating, we get 1 fi – __ y = a log |x – a| + C
Integrating, we get
fi
)
fi
y2 1 – x2 _______ 2 dy = _____ dx x (1 + y )
fi
1 1 1 – _____ dy = __ x – x dx 1 + y2
(
)
(
)
Integrating, we get
x2 y – tan–1y = log |x| – __ + c 2
which is the required solution of the given differential equation. 49. Given differential equation is dy (x + 1) ___ = 2xy dx dy 2x ___ fi y = _____ dx x+1 dy 1 ___ fi y = 2 1 – _____ dx x+1 Integrating, we get
( ) ( )
tan–1(y) + tan–1(e x) = tan–1 c
fi
y + ex tan–1 _______ = tan–1c 1 – exy
(
log |y| = 2x – 2log |x + 1| + c
( )
dy ___ = e x + y + x2ey dx fi
dy ___ = (ex + x2)ey dx
fi
e– ydy = (ex + x2) dx
Integrating, we get x3 – e– y = ex + __ + c 3
which is the required differential equation. 53. Given differential equation is ______ ______ dy ___ y ÷1 + x2 + x ÷1 + x2 = 0 dx dy y + x ___ = 0 dx dy dx ___ x + ___ y = 0
fi fi
Integrating, we get
log |x| + log |y| = logc
fi xy = c which is the required solution. 54. Given differential equation is _____
_____
÷ 1 + x2 dy + ÷ 1 + y2 dx = 0
which is the required solution. 50. Given differential equation is
dy dx ______ _______ + _______ ______ = 0 2 ÷1 + y ÷1 + x2
sec2x tan y dx + sec2y tan x dy = 0
fi
fi
sec2y sec2x _____ dx + _____ dy = 0 tan x tan y
Integrating, we get
|
log |tan x| + log |tan y| = log c
fi
tan x tan y = c
which is the required solution. 51. Given differential equation is 2
_____
|
|
_____
|
log y + ÷y 2 + 1 + log x + ÷x 2 + 1 = logc
Integrating, we get
2x
)
y + ex ______ = c 1 – exy which is the required solution. 52. Given differential equation is fi
which is the required solution of the given differential equation. 48. The given differential equation is dy xy2 ___ = 1 – x2 + y2 – x2y2 dx dy fi xy2 ___ = (1 – x2) (1 + y2) dx
(
x
(1 + e ) dy + (1 + y ) e dx = 0
fi
dy ex _____ 2 + _______ dx = 0 1+y 1 + ex2x
(
_____
)
_____
) = c fi y + ÷ y 2 + 1 ( x + ÷x 2 + 1 which is the required solution. 55. Given differential equation is ________________ dy ÷1 + x2 + y2 + x2y2 + xy ___ = 0 dx
fi
______________ dy ÷(1 + x2) (1 + y2) + xy ___ = 0 dx
4.35
4.36 Integral Calculus, 3D Geometry & Vector Booster fi
______________ dy xy ___ = – ÷(1 + x2) (1 + y2) dx
Integrating, we get
______
fi
ydy ÷1 + x _______ ______ = – _______ dx x 2 ÷ 1 + y 2
|
Intergrating, wet get _____ 1 + y2 =
÷
_____ ÷x 2 + 1 +
fi
_____
|
x 2 + 1 – 1 ÷ _____ 1 __ log __________ + c 2 ÷x 2 + 1 + 1
which is the required solution. 56. Given differential equation is dy xy ___ + 1 + x + y + xy = 0 dx dy xy ___ + (1 + x)(1 + y) = 0 dx y 1+x fi _____ dy + _____ dx = 0 x 1+y
log |x + a| + logc = log |y| – log |1 – ay| y c (x + a) = ______ 1 – ay
which is the required solution. 60. We have,
dx xy ___ = dy
y dy (1 + x + x2) fi _____ 2 = ___________ dx 1+y (1 + x2) x
(1 + x + x2) = __________ d x (1 + x2) x
dx dx = ___ x + _____ 2 1+x
fi
( ) ( ) ( ) ( )
1 1 fi 1 – _____ dy + 1 + __ x dx = 0 1+y Integrating, we get y – log |1 + y| + x + log |x| = c fi
| |
x (x + y) – log _____ = c 1+y
57. Given differential equation is
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
) ( )
(
1+y 1+x fi _____ dx + _____ dy = 0 2 1+x 1 + y2 Integrating, we get 1 tan–1 x + tan–1 y + __ log |(x2 + 1) (y2 + 1)| = c 2 58. Given differential equation is _____
_____
x ÷1 – y2 dx + y ÷1 – x2 dy = 0 y x _____ fi ______ dx + ______ _____ dy = 0 2 ÷1 – y2 ÷ 1 – x Integrating, we get
______
_____
÷ 1 – x2 + ÷1 – y2 = c which is the required solution. 59. Given differential equation is dy dy y – x ___ = a y2 + ___ dx dx dy fi y(1 – ) = (x + a) ___ dx dy dx fi _____ = ________ x + a y (1 – ay)
(
) (
(
)
)
dx a 1 fi _____ = __ y + _____ x+a 1 – ay
1 + y2 _____2 (1 + x + x2) 1+x
1 fi __ log |1 + y2| = log |x| + tan–1 x + c 2 which is the required solution. 61. We have,
(x2 – yx2) dy + (y2 + xy2) dx = 0
fi
x2 (1 – y) dy + y2 (1 + x) dx = 0
(1 – y) (1 + x) fi ______ dy + ______ dx =0 2 y x2 Integrating, we get 1 1 __ – __ y – log |y| – x + log |x| = c
| |
(
)
x 1 1 log __ y = c + __ x + __ y which is the required solution. 62. The given differential equation is fi
Let
dy ___ = (x + y + 1)2 dx x+y+1=v
dy fi ___ = dx dv ___ – dx
dv ___ – 1 dx 1 = v2
dv fi _____ = dx 2 v +1 Integrating, we get
dv = Ú dx Ú _____ 2 v +1
fi fi
tan–1 (v) = x + c tan–1 (x + y + 1) = x + c
...(i)
Differential Equation
is the required solution. 63. The given differential equation is dy ___ = sin (x + y) + cos (x + y) dx
...(i)
dy dv Put x + y = v fi 1 + ___ = ___ dx dx dy dv fi ___ = ___ – 1 dx dx
dv ___ – 1 = sin v + cos v dx
Integrating, we get
(v + log |v|) = 2x + c
fi
(x + y + log |x + y|) = 2x + c
fi
(y + log |x + y|) = x + c
which is the required solution of the given differential equation. 65. The given differential equation is dy tan y ___ = sin (x + y) + sin (x – y) dx
tan y fi ____ dy = 2 sin x dx cos y
dv = x + c Ú ______________ 1 + sin v + cos v
dv fi Ú ________________________________ = x + c 2 1 – tan (v/2) 2tan(v/2) ___________ ___________ 1 + + 1 + tan2 (v/2) 1 + tan2 (v/2)
) (
fi
dy sin (x + y) + sin (x – y) fi ___ = ____________________ tan y dx 2 sin x cos y = _________ tan y
dv fi ______________ = dx 1 + sin v + cos v
(
Ú ( 1 + __ 1v ) dv = 2x + c
dv fi ___ = 1 + sin v + cos v dx
)
sin y fi _______ = 2 sin x dx cos2 y dy
sec2 (v/2)dv fi Ú ____________ = x + c 2(1 + tan(v/2))
Integrating, we get sin y = Ú 2 sin x dx Ú _______ cos2y dy
fi log |1 + tan(v/2)| = x + c
1 fi ____ = c – 2 cos x cos y
fi log |1 + tan(v/2)| = x + c
|
( ) |
x+y fi log 1 + tan _____ = x + c 2 which is the required solution of the given differential equation. 64. The given differential equation is
(x + y) (dx – dy) = (dx + dy)
fi
(x + y – 1) dx = (x + y + 1) dy
dy fi ___ = dx Let
(x + y – 1) __________ (x + y + 1)
dy dv x + y = v fi ___ = ___ – 1 dx dx
dv v–1 fi ___ – 1 = _____ v+1 dx dv v – 1 2v fi ___ = _____ + 1 = _____ v+1 dx v + 1
(
)
v+1 fi _____ dv = 2 dx v Integrating, we get
4.37
which is the required solution of the given differential equation. 66. The given differential equation is dy ___ – x tan (y – x) = 1 dx dy fi ___ = 1 + x tan (y – x) dx dy dv Let y – x = v fi ___ = ___ + 1 dx dx
...(i)
dv fi ___ + 1 = x tan v + 1 dx dv fi ___ = x tan v + 1 dx fi
cot v dv = x dx
Intergrating, we get
Ú cot v dv = x + c
x2 log |sin v| = __ + c 2 x2 fi log |sin(x – y)| = __ + c 2 which is the required solution of the given differential equation. fi
4.38 Integral Calculus, 3D Geometry & Vector Booster fi
67. We have, dy (x – y) + 3 ___ = ___________ dx 2(x – y) + 5
...(i)
Let x – y = v
dy dv fi 1 – ___ = ___ dx dx
dy dv fi ___ = 1 – ___ dx dx
(
)
– cosec (x + y) + cot (x + y) = 2x + y + c
which is the required solution. 70. We have,
( )
dy sin–1 ___ = x + y dx dy fi ___ = sin(x + y) dx dv fi ___ – 1 = sin v, (Let v = x + y) dx
dv v+3 fi 1 – ___ = ______ dx 2v + 5
dv fi ___ = 1 + sin v dx
dv v+3 fi ___ = 1 – ______ dx 2v + 5
dv fi _________ = dx (1 + sin v)
( (
2v + 5 – v – 3 ______ v+2 = _____________ = 2v + 5 2v + 5
)
2v + 5 fi ______ dv = dx v + 2 1 fi 2 + _____ dv = dx v+2
)
Integrating, we get (2v + log |v + 2|) = x + c fi
(2(x – y) + log |x – y + 2|) = x + c
fi
(x – 2y + log |x – y + 2|) = c
which is the required solution. 68. We have, (x + y) (dx – dy) = dx + dy
(1 – sin v)dv fi __________ = dx cos2v fi
(sec2v – sec v tan v) dv = dx
Integrating, we get
tan v – sec v = x + c
fi
tan (x + y) – sec (x + y) = x + c
which is the required solution. 71. We have dy ___ = cos (x + y + 1) dx dv fi ___ – 1 = cos v, (Let x + y + 1 = v) dx
d(x + y) fi _______ = d (x – y) (x + y) Integrating, we get log |x + y| = (x – y) + c
dv fi ________ = dx 1 + cos v
dy 69. We have ___ = sec (x + y) dx
fi
dv fi ___ – 1 = sec v (Let v = x + y) dx
= 1 + sec v
dv fi _______ = dx 1 + sec v cos v dv fi ________ = dx 1 + cos v cos v(1 – cos v) dv fi ________________ = dx sin2v fi [cosec v cot v – (cosec v – 1)] dv = dx Integrating, we get
– cosec v + cot v – v = x + c
1 – cos v fi _______ = dx sin2v dv (cosec2v – cosec v . cot v) dv = dx
Integrating, we get cosec v – cot v = x + c fi
cosec (x + y + 1) – cot (x + y + 1)
72. We have, dy 2(x + y) ___ = __________ dx 1 + (x + y)2 dv 2v fi ___ – 1 = _____ 2 ,v=x+y dx 1+v dv 2v fi ___ = 1 + _____ dx 1 + v2
2v = 1 + _____ = 1 + v2
(1 + v)2 _______2 1+v
Differential Equation
( (
)
)
2 2 fi 1 – _____ + _______ dv = dx 1 + v (1 + v)2 Integrating, we get 2 v – 2 log |1 + v| – ______ = x + c (1 + v) 2 fi y – 2 log |1 + x + y| – _________ = c (1 + x + y) which is the required solution. 73. We have, dy ___ = sin (10x + 6y) dx
)
1 dv fi __ ___ – 10 = sin v 6 dx dv fi ___ = 6 sin v + 10 dx dv fi __________ = dx 6 sin v + 10 dv fi __________________ = dx 2 tan(v/2) 6 ___________ + 10 1 + tan2(v/2)
(
)
5 tan (5x + 3y) + 3 5 fi __ tan–1 _______________ = 5x + c 4 4
2 fi 1 – _______ dv = dx (1 + v)2
(
)
5 tan(v/2) + 3 5 fi __ tan–1 ___________ = 5x + c 4 4
1 + v2 fi _______ dv = dx (1 + v)2
( (
)
which is the required solution. 74. We have, y dy 2(x2 + y2) – 1 __x ___ + ____________ = 0 dx x2 + y2 + 1 2v – 1 1 dv fi __ ___ – 1 + ______ = 0 2 dx v+1 2v – 1 1 dv fi __ ___ = 1 – ______ 2 dx v+1 2–v = _____ v+1
( ) v+1 fi ( _____ dv = – 2dx v – 1) v – 2 + 3 fi ( ________ dv = – 2dx v–2 ) 3 fi 1 + ( _____ dv = – 2dx v – 2) v + 1 fi _____ dv = 2dx 2–v
Integrating, we get
(x2 + y2) + log |(x2 + y2) – 2| = c – 2x
sec2(v/2)dv fi __________________________ = dx 10 (1 + tan2(v/2)) + 12 tan (v/2)
which is the required solution.
2 dt fi _____________ = dx, t = tan(v/2) 10(1 + t2) + 12t
dy x2y ___ = ______ 3 3 . dx x + y
dt fi ___________ = dx 5(1 + t2) + 6t
y __ x dy , fi ___ = _______ y 3 dx 1 + __ 3
dt fi __________ = dx 5t2 + 6t + 5 dt fi __________ = 5dx 6 2 __ t + t + 1 5 dt fi _____________ 2 = 5dx 3 2 4 __ t + + __ 5 5
( ) ( )
Integrating, we get
(
)
t + (3/5) 5 __ tan–1 _______ = 5x + c 4 4/5
(
)
4.39
5 5t + 3 fi __ tan–1 _____ = 5x + c 4 4
75. The given differential equation can be written as
( )
( )
y Let __x = v fi y = v x
dy dy fi ___ = v + x ___ . dx dx
fi fi
dv v v + x ___ = _____ dx 1 + v3 dv v x ___ = _____ – v dx 1 + v3
( )
fi
v3 + 1 dx _____ dv = – ___ x v4
fi
dx 1 __ v + v– 4 dv = – ___ x
(
)
4.40 Integral Calculus, 3D Geometry & Vector Booster
(
dy 1 __x fi ___ = __ – dx 2 y
Integrating, we get
1 log |v| – ___ 3 = c – log |x| 3v
fi
y x3 log __ x – ___ 3 = c – log |x| 3y
| |
1 log |v| – ___ 3 = c, 3v which is the required fi
y Let __x = v fi y = vx
dy dv fi ___ = v + x ___ dx dx
fi
x dy – y dx = ÷x 2 + y2 dx
fi
______ dy x ___ – y = ÷x 2 + y2 dx
______
x 2 + y2 dy y ÷_______ x = x fi ___ – __ dx
fi
_______
÷ ( )
y y 2 __ x = 1 + __ x
...(i)
y Let __x fi y = vx
dy dy fi ___ = v + x ___ dx dx _____
fi
dy v + x ___ – v = ÷ 1 + v2 dx
fi
_____ dy x ___ = ÷ 1 + v2 dx
(
)
dv 1 __ 1 v + x ___ = __ – v dx 2 v dv 3v 1 – 3v2 1 x ___ = ___ – ___ = ______ 2 2v dx 2v
2v dx fi _______ dv = ___ x 1 – 3v2 Integrating, we get
______
dy fi ___ – dx
fi
solution of the given differential equation. 76. The given differential equation is
)
y __ x
fi
2v dx 2 dv = Ú ___ Ú 1______ x – 3v 1 – __ log |1 – 3v2| = log c + log |x| 3 y 2 1 – __ log 1 – 3 __ x = log c + log|x| 3
|
( ) |
which is the required solution of the given differential equation. 78. The given differential equation is
(1 + 2e x/y) dx + 2e x/y (1 – x/y) dy = 0
dy fi ___ = dx
(1 + 2ex/y) __________ x 2e x/y __ y – 1
( ) x 2e ( __ y – 1 ) ___________ x/y
dv dx _____ fi _______ = ___ x 2 ÷ 1 + v
dy fi ___ = dx (1 + 2ex/y)
Integrating, we get
fi
dv dx _____ = Ú ___ Ú _______ x 2
|
_____
|
fi ( v + ÷1 + v ) = cx 2
( ÷ ( ) ) ______
)
dv v + y ___ = dy
fi
dv y ___ = dy
2vev – 2ev – v – 2vev = __________________ 1 + 2ev
2ev + v = ______v 1 + 2e
fi y + ÷x 2 + y2 = cx2 which is the required solution of the given differential equation. 77. The given differential equation is
(x2 – y2) dx = 2xy dy
fi
______ dy x ___ – y = ÷ x 2 + y2 dx
...(i)
2ev (v – 1) _________ (1 + 2ev)
fi
_______
y y 2 fi __ x + 1 + __ x = cx
(
dv dx fi ___ = v + y ___ dy dy
log v + ÷ 1 + v2 = log|x| + log c _____
x Let __y = v fi x = vy
÷1 + v
...(i)
2ev (v – 1) _________ –v (1 + 2ev)
dy 1 + 2ev fi _______v dv = – ___ y v + 2e
Differential Equation
fi fi
log |v + 2ev| + log|y| = log c
81. We have,
v
fi
y (v + 2e ) = c x y __ y + 2ex/y = c
fi
(x + 2yex/y) = c
(
dy ___ = dx
)
which is the required solution of the given differential equation. 79. The given differential equation can be written as dy ___ = dx
y–x _____ y+x y (Let __x = v fi y = vx)
dy dv fi ___ = v + x ___ dx dx
fi
dv v + x ___ = dx
v–1 _____ v+1
v–1 = _____ – v v + 1
v – 1 – v2 – v = _____________ v+1
1 + v2 = _____ v+1
fi
dv v – 1 x ___ = _____ – v v+1 dx v–1 = _____ – v v+1
1 fi __ log |v2 + 1| + tan–1 (v) = c – log |x| 2 y 1 fi __ log |x2 + y2| + tan–1 __x = c 2 which is the required solution of the given differential equation.
( )
80. We have,
dy x + y y fi ___ = _____ x = 1 + __ x dx dv fi v + x ___ = 1 + v, v = dx dv fi x ___ = 1 dx dx fi dv = ___ x Integrating, we get
y __ x
log |v – 1| + log |x| = log c
y fi __x = log |x| + c which is the required solution.
1+v dx fi ______2 dv = – ___ x 1+v Integrating, we get 1 tan–1 (v) + __ log |1 + v2| = c – log |x| 2 y 1 tan–1 __ x + __ log |x2 + y2| = c 2 which is the required solution. 82. We have, dy 2xy ___ = x2 + y2 dx
( ) ( )
y 2 2 + __x dy x______ + y2 1_______ ___ fi = = x 2xy dx 2 __ y fi
dv 1 + v2 v + x ___ = _____ 2v dx
fi
dv 1 + v2 1 – v2 x ___ = _____ – v = _____ 2v 2v dx
2v dv dx fi _____ = – ___ x v2 – 1 Integrating, we get
dy x ___ = x + y dx
2
1 + v2 = – ______ 1+v
v+1 dx fi _____ dv = – ___ x v2 + 1
dv v – 1 v + x ___ = _____ dx v + 1
( )
(y/x) – 1 _______ (y/x) + 1
fi
...(i)
y–x _____ = y+x
log |v2 – 1| + log |x| = log c
fi
(v2 – 1) x = c
fi (y2 – x2) = cx which is the required solution. 83. We have, dy ___ = dx
y2 – 2xy _______ x2 – 2xy
( )
( ) ( )
y 2 y __ x – 2 __ x dy fi ___ = __________ y dx 1 – 2 __ x fi
dv v2 – 2v v + x ___ = _______ 1 – 2v dx
4.41
4.42 Integral Calculus, 3D Geometry & Vector Booster fi
dv 3v2 – 3v x ___ = _______ 1 – 2v dx
1 – 2v fi ________ dv = 3v(v – 1)
dx ___ x
( (
)
dx 1 1 2 fi __ _______ – ______ dv = ___ x 3 v(v – 1) (v – 1)
)
dx 1 1 1 2 fi __ _____ – __ – ______ dv = ___ x 3 v – 1 v (v – 1) fi
(
)
dx 1 1 1 – __ _____ + __ dv = ___ x 3 v–1 v
Integrating, we get fi
1 – __ log |v(v – 1)| = log |cx| 3 y y 1 – __ log __ x __ x – 1 = log |cx| 3
| (
fi fi
x2y dx – (x3 + y3) dy = 0
1 – 3v2 – v4 + 3v2 = _______________ v3 – 3v
1 – v4 = _______ v3 – 3v
( ) ( )
Integrating, we get
| |
v2 – 1 3 1 – __ log |1 – v4 | + __ log _____ = log |x| + log c 4 4 v2 + 1
(
)
y dv v v + x ___ = _____ , Let v = __ x dx 1 + v3 dv v x ___ = _____ – v dx 1 + v3 v – v – v4 = _________ 1 + v3 4
v = – _____ 1 + v3
( )
x3 – ___3 + log |y| = c 3y which is the required solution 85. Given differential equation is fi
(x3 – 3xy2) dx = (y3 – 3x2y) dy
dy x – 3xy fi ___ = ________ dx y3 – 3x2y
dv 1 – 3v2 x ___ = ______ – v dx v3 – 3v
v3 3v dx fi _____ 4 dv – _____ 4 dv = ___ x 1–v 1–v
( )
2
fi
Integrating, we get
y __ x = _______ y 3 1 + __ x
3
)
v3 3v dx fi _____ 4 dv – _____ 4 dv = ___ x 1–v 1–v
1 + v3 dx fi _____ dv = – ___ x 4 v Integrating, we get 1 – ___ 3 + log |v| = c – log |x| 3v
dv v + x ___ = dx
( ) ( ) ( )
)|
dy x2y fi ___ = ______ dx x3 + y3
(
y 1 – 3v2 _______ 3 , Let v = __ x v – 3v
fi
v3 – 3v dx fi _______ dv = ___ x 1 – v4
which is the required solution. 84. Given differential equation is
( ) ( ) ( )
y 2 1 – 3 __ x __________ = y 3 y __ – 3 __ x 3
| | ( ) ( )
y 2 __ x – 1 y 4 3 1 __ __ __ _______ fi – log 1 – x + log y 2 = log |xc| 4 4 __ x + 1
|
( ) |
which is the required solution. 86. The given differential equation is
÷
_____
dy ___ = dx
y __ x –
y2 __ 2 – 1 x
fi
_____ y dv v + x ___ = v – ÷v 2 – 1 Let __x = u dx
fi
dv x ___ = – ÷v 2 – 1 dx
(
_____
dv dx _____ fi ______ = – ___ x 2 ÷ v – 1 Integrating, we get
_____
| _____ | fi log | v + ÷v – 1 | = log | __ cx | _____ ) = ( __ cx ) fi ( v + ÷ v – 1 _______ y y fi ( __ x + ÷ ( __ x ) – 1 ) = ( __ cx )
log v + ÷v 2 – 1 = log c – log c 2
2
2
)
Differential Equation ______
fi y + ÷ y 2 – x2 = c
which is the required solution. 87. The given differential equation is
vev – ev – v – vev = ________________ 1 + ev
– ev – v = _______v 1+e
(
dy fi ___ = dx fi fi
)
( )
y y __ x + sin __ x
(
)
y dv v + x ___ = v + sin (v) Let __x = u dx dv x ___ = sin (v) dx
dv dx fi _____ = ___ x sin (v)
fi
|
)
dv 1 + ev fi _____v dv = – ___ y v+e Integrating, we get
log |v + ev| = log c – log|y|
(
) ( )
x c fi __y + ex/y = __ y fi (x + y ex/y ) = c which is the required solution. 90. The given differential equation is
Integrating, we get fi
(
( ) |
v log tan __ = log c + log |x| 2 y tan ___ = cx 2x
( )
dy x + 2y + 3 ___ = __________ . 2x + 3y + 4 dx Let
which is the required solution. 88. The given differential equation is dy y x ___ = y – x tan __ x dx
x = X + a and y = Y + b
dY (X + a) + 2(Y + b) + 3 Thus, ___ = _____________________ dX 2(X + a) + 3(Y + b) + 4
( ) dy y y fi ___ = ( __ x ) – tan ( __ x ) dx
(X + 2Y) + (a + 2b + 3) dY ___ = _______________________ dX (2X + 3Y) + (2a + 3b + 4)
(
)
y dv v + x ___ = v – tan (v) Let __x = u dx dv fi x ___ = – tan (v) dx dv dx fi _____ = – ___ x tan(v)
Let us choose a and b in such a way that
Integrating, we get
dY X + 2Y ___ = ________ 2X + 3Y dX which is a homogeneous equation.
fi
fi fi
| |
( ) | p v c log tan ( __ + __ = log | __ x | 4 2 )|
p v log tan __ + __ = log c – log|x| 4 2
(
)
( )
(
x x __ __ x ( 1 + ey ) dx + e y 1 – __y dy = 0
(
)
x __ 1 ey
)
x __ y – dx ___ ________ fi = dy ( 1 + e__ xy ) fi
v dv (v – 1) e v + x ___ = ________ , v = v dy 1+e
fi
v dv (v – 1) e x ___ = ________ – v v dy 1+e
Solving, we get a = 1, b = – 2. Equation (i) reduces to
Equation (ii) reduces to dV 1 + 2V V + X ___ = _______ dX 2 + 3V 2 + 3V dX fi _______2 dV = ___ X 1 – 3V Integrating, we get
x __ y
...(i)
a + 2b + 3 = 0 and 2a + 3b + 4 = 0
Y Put __ = V X dV dY fi ___ = V + X ___ . dX dX
y p c tan __ + ___ = __ x 4 2x which is the required solution. 89. The given differential equation is fi
4.43
2 + 3V dX dV = Ú ___ Ú _______ 2 X 1 – 3V
|
__
|
1 + ÷3 V 1__ 1 __ fi ___ log ________ – __ log |(1 – 3V 2)| 2 3 1 – ÷3 V ÷
= log C + log |X|
...(ii)
4.44 Integral Calculus, 3D Geometry & Vector Booster
| | __
fi
X + ÷3 Y 1 1 ___ __ __ log _______ – __ log |(X2 – 3Y2)| = log c 2 3 X – ÷3 Y ÷
fi
(x – 1) + ÷ 3 (y + 2) 1 ___ __ __ log _________________ 3 (x – 1) – ÷3 ( y + 2) ÷
|
__
|
1 – __ log|(x2 – 3y2 – 2x – 12y – 11)| = log c 2 which is the required solution of the given differential equation. 91. The given differential equation is
dy (x + y)2 ___ = ____________ (x + 2)(y – 2) dx
Let
X = x + 2 and Y = y + 2
...(i)
Equation (i) reduces to 2 (X + Y)2 dY (X – 2 + Y + 2) ___ = ______________ = _______ ...(ii) XY XY dX Let
dV dY Y = V X fi ___ = V + X ___ dX dX
Equation (ii) reduces to 2 dV (1 + V) V + X ___ = _______ V dX
fi
dy y – x ___ + ln x = 0 dx
fi
dy x ___ – y = ln x dx
dy y ___ ln x fi ___ – __ = x dx x
...(i)
which is a linear differential equation. dx ___ 1 IF = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating, we get
y ◊ (IF) = Ú Q . (IF) dx + c
ln x 1 ___ y ◊ __ x = Ú x2 dx + c log x __ y 1 fi __x = – ____ x + x + c fi
(
fi
)
y = – (log x + 1) + c
which is the required solution of the given differential equation. 98. The differential equation is
fi
2 dV (1 + V) X ___ = _______ – V V dX
fi
dV (1 + 2V) X ___ = _______ V dX
dy y fi ___ – 3 __x = 4x2 + 2 dx
Integrating, we get
)
dX 1 1 fi __ Ú 1 – _______ = Ú ___ X 2 (2V + 1) 1 fi __ (V – log (2V + 1)) = log |X| + c 2
(
|
|)
1 Y Y fi __ __ – log 2 __ + 1 = log |X| + c X 2 X
( ( ) | ( ) | )
y–2 1 y–2 fi __ _____ – log 2 _____ + 1 2 x+2 x+2
= log |x + 2| + c
which is the required solution of the given differential equation. 97. The given differential equation is
dx ___ 1 IF = e – 3 Ú x = e – 3 log x = __ 3 x
Multiplying both sides of Eq. (i) by IF and integrating, we get
VdV dX _______ = ___ Ú (2V + 1) Ú X
(
...(i)
which is a linear differential equation.
VdV dX fi _______ = ___ X (2V + 1)
dy x2 ___ – 3xy = 4x4 + 2x2 dx
y dx – x dy + ln xdx = 0
y ◊ (IF) = Ú Q.(IF) dx + c
fi
4x2 + 2 1 y ◊ __3 = Ú ______ dx + c x x3
(
(
)
) )
y 4 2 fi __3 = Ú __ x + __ 3 dx + c x x
(
1 = 4 log x – __ 2 + c x
which is the required solution of the given differential equation. 99. The given differential equation is dy ___ = y tan x – sin x dx dy fi ___ + (– tan x) y = – sin x dx
...(i)
Differential Equation
which is a linear differential equation.
Multiplying both sides of Eq. (i) by IF and integrating, we get
IF = e– Ú tan xdx = elog(cos x) = cos x
Multiplying both sides of Eq. (i) by I.F and integrating, we get
x ◊ (IF) = Ú Q.(IF) dy + c x ◊ e– y = Ú (3y + 2) e– y dy + c
y ◊ (I.F.) = Ú Q.(I.F) dx + c
fi
fi
y (cos x) = – Ú (sin x cos x) dx + c
fi
1 = – __ Ú sin 2 x dx + c 2
cos 2 x = ______ + c 4 which is the required solution of the given differential equation. 100. The given differential equation is
dy y 2 ___ + ______ = __ dx x log x x2
...(i)
which is a linear differential equation.
dx Ú ______
IF = e x log x = elog(log x) = log x
Multiplying both sides of Eq. (i) by IF and integrating, we get y (IF) = Ú Q (IF) dx + c 2 fi y ◊ log x = Ú __2 log x dx + c x 2 = – __ ...(i) x (1 + log x) + c which is the required solution of the given differential equation. 101. The given differential equation is x x dx = __ 2 – y dy y
( )
fi
dx x ___ + y = xy2 dy
dy y fi ___ + __ = y2 dx x
...(i)
which is a linear differential equation.
IF = e– Ú dy = e– y
x = – 3 (y + 2) + cey
which is the required solution of the given differential equation. 103. The given differential equation is dy ___ + 2y = e3x ...(i) dx which is a linear differential equation. IF = e Ú 2 dx = e2x Multiplying both sides of Eq. (i) by IF and integrating, we get
y ◊ (IF) = Ú Q.(IF) dx + C
fi
y (ex) = Ú (e2x ◊ ex) dx + c
fi
= Ú (e3x) dx + c
( ) ( )
e3x = ___ + c 3
e3x y = ___ + ce– x 3
which is the required solution. 104. The given differential equation is dy x ___ = x + y dx dy y fi ___ = 1 + __ x dx
dy y fi ___ – __ = 1 dx x
...(i)
which is a linear differential equation. dx ___ 1 IF = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating we get
dy (x + 3y + 2) ___ = 1 dx
dx fi ___ – x = 3y + 2 dy
= 3 (– ye– y + 2e– y) + c
Thus,
which is a linear differential equation. 102. The given differential equation is
4.45
...(i)
y ◊ (IF) = Ú Q ◊ (IF) dx + c
y dx fi __x = Ú ___ x + c y fi __x = log |x| + c which is the required solution.
4.46 Integral Calculus, 3D Geometry & Vector Booster
105. The given differential equation is
dy x ___ + y = xe x dx
dy fi ___ + dx
which is the required solution. 108. The given differential equation is dy ___ = y tan x – 2 sin x dx
y __ x = ex
dy fi ___ + (– tan x) y = – 2 sin x dx
which is a linear differential equation.
Thus,
which is a linear differential equation
dx
___ IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú Q.(IF) dx + c
fi
y ◊ x = Ú xex dx + c
fi (xy) = ex (x – 1) + c which is the required solution. 106. The given differential equation is dy x ___ + y = x log x dx dy fi ___ + dx
y __ x = log x
...(i)
which is a linear differential equation. Thus,
dx
___ IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú x log x dx + c
fi
y ◊ x = Ú x log x dx + c
– IF = e Ú tan x dx = e– log(sec x) = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
y ◊ cos x = Ú (– sin x ◊ tan ◊ x) dx + c
1 – cos2x = – Ú _________ cos x dx + c
= – Ú (sec x – cos x) dx + c
(
)
= – log |sec x + tan x| + sin x + c
which is the required solution. 109. The given differential equation is dy y ___ – __ = 2x2 dx x
...(i)
which is a linear differential equation. dx ___ 1 I.F. = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating we get
( )
Thus,
Thus,
Thus,
1 = Ú x log x dx – __ Ú x dx + c 2
x2 x2 = __ log x – __ + c 2 4 which is the required solution. 107. The given differential equation is dy y ___ + __ = x3 dx x which is a linear differential equation.
...(i)
y ◊ (I.F) = Ú Q ◊ (IF) dx + c
1 y ◊ __ x = 2 Ú x dx + c y fi __x = x2 + c which is the required solution. 110. The given differential equation is dy 2 x log x ___ + y = __ x log x dx fi
...(i)
dx ___
IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
dy y 2 fi ___ + ______ = __ dx x log x x2
...(i)
which is a linear differential equation. dx – Ú ______ x log x =
y ◊ (IF) = Ú Q.(I.F.) dx + c
Thus,
fi
xy = Ú x4 dx + c
Multiplying both sides of Eq. (i) by IF and integrating we get
x5 = __ + c 5
IF = e
elog (log x) = x
y ◊ (IF) = Ú Q ◊ (IF) dx + c
Differential Equation
fi
log x y ◊ log x = 2 Ú ____ dx +c x2
which is the required solution. 113. The given differential equation is
= 2 Ú (te– t) dt + c, t = log x
= – 2 (te– t – e– t) + c
log x – 1 = – 2 _______ + c x
dy a 1 __ fi ___ + – __ x y = x + x dx
(
y dx – (x + 2y2) dy = 0
fi
dx y ___ + (x – y2) = 0 dy
fi
dx y ___ + x = y2 dy
dx x fi ___ + __ = y dy y
Thus,
Thus,
IF =
elog y = y
Multiplying both sides of Eq. (i) by IF and integrating we get x ◊ (I.F) = Ú y2dy + c
fi
x ◊ y = Ú y2dy + c
y = __ + c 3 which is the required solution. 112. The given differential equation is fi
dx y ___ + (x – y3) = 0 dy dx y ___ + x = y3 dy
fi
dx fi ___ + dy
x __ y = y2
...(i)
Multiplying both sides of Eq. (i) by IF and integrating we get
fi
x ◊ y = Ú y3 dy + c
y4 = __ + c 4
)
...(i)
dx
– n _____ I.F. = e Ú x + 1 = e– n log (x + 1)
(
)
1 log _______ (x + 1)n =
= e
1 _______ (x + 1)n
y ◊ (IF) = Ú Q ◊ (IF) dx + c
y fi _______ = Ú ex dx + c (x + 1)n
dy
x ◊ (IF) = Ú Q ◊ (IF) dy + c
(
Multiplying both sides of Eq. (i) by IF and integrating we get
___ IF = e Ú y = elog y = y
)
dy n fi ___ – _____ y = ex (x + 1) n x+1 dx
which is a linear differential equation Thus,
(
x– a + 1 x– a = _____ – a + ___ – a + c 1
which is the required solution. 114. The given differential equation is dy (x + 1) ___ – ny = ex (x + 1) n + 1 dx
Thus,
y dx + (x – y3) dy = 0
which is a linear differential equation
3
y ◊ (IF) = Ú Q.(IF) dx + c
y fi __a = Ú (x– a + x– a –1) dx + c x ...(i)
dx – a ___ 1 ) IF = e – a Ú x = e– a log x = e log(x = __ a x
Multiplying both sides of Eq. (i) by IF and integrating we get
which is a linear differential equation. dy ___ e Ú y =
...(i)
which is a linear differential equation
which is the required solution. 111. The given differential equation is
dx x ___ – ay = x + 1 dy
( )
)
4.47
= ex + c
which is the required solution. 115. The given differential equation is
(1 + y2) dx = (tan–1 y – x) dy
–1 dx (tan y – x) fi ___ = __________ dy 1 + y2
dx x fi ___ + _____ = dy 1 + y2
tan–1 y ______ 1 +y2
which is a linear differential equation
...(i)
4.48 Integral Calculus, 3D Geometry & Vector Booster dy
Thus,
2 Ú _____ –1 IF = e 1 + y = e tan y
Multiplying both sides of Eq. (i) by IF and integrating we get
x ◊ (IF) = Ú Q ◊ (IF) dy + c
fi
–1 tan–1y tan–1 y x ◊ etan y = Ú _____ e dy 1 + y2
fi
x ◊ etan y = Ú tet dt + c, t = tan–1y
fi
x ◊ etan y = et (t – 1) + c
fi
x ◊ etan y = e tan y (tan–1y – 1) + c
–1
DIviding both the sides by y (log y)2, we get dy 1 1 __ 1 1 _______ 2 ◊ ___ + ____ ◊ x = __ 2 dx log y y(log y) x
–1
which is the required solution. 116. The given differential equation is dx (x + 3y + 2) ___ = 1 dy dx fi ___ = (x + 3y + 2) dy dx fi ___ – x = (3y + 2) dy which is a linear differential equation
...(i)
Thus, IF = e Ú dy = e– y Multiplying both sides of Eq. (i) by IF and integrating we get –
x ◊ (IF) = Ú Q.(IF) dy + c
fi
x ◊ e– y = Ú {3(ye– y) + 2(e– y)} dy + c – y
– y
= 3 (– ye
= – 3 (y + 1) e– y – 2e– y + c
fi
– e ) – 2e
+c
ydy – Ú _____ y2 +1 =
1 – __ log |y2 + 1|
we get
e 2
dv 1 __ 1 – ___ + v ◊ __ x = x2 dx dv v 1 fi ___ – __ = – __2 dx x x which is a linear differential equation. fi
1 –1 __ 1 IF = e – Ú x dx = e– log x = e log x = __ x . Hence the solution is
Thus,
v ◊ (IF) = Ú Q.(IF) dx + c
fi
1 1 1 v ◊ __ x = Ú – __2 ◊ __ dx + c x x
...(i)
1 = _______ _____ 2 ÷ y + 1 Multiplying both sides of Eq. (i) by IF and integrating IF = e
dy dv 1 fi _______ 2 ◊ ___ = – ___ dx dx y(log y)
which is a linear differential equation Thus,
( )
( )
x 1 dy fi __6 ___ + __ 5 = x3 dx y y
(1 + y2) dx = (xy + y3 + y) dy xy dx fi ___ = _____ 2 + y dy y + 1
)
1 Let ____ = v log y
119. The given differential equation is dy ___ + xy = x3y6 dx
which is the required solution. 117. The given differential equation is
(
1 1 fi ______ – ___ = c x log y 2x2
x = – 3(y + 1) – 2 + cey
y dx fi ___ + – _____ x = y 2 dy y +1
_____
which is the required solution. 118. The given differential equation is dy y y ___ + __x log x = __ 2 (log y)2 dx x
–1
– y
( ) ( ÷ )
ydy x _____ fi _______ _____ = Ú _______ + c ÷ y 2 + 1 ÷y 2 + 1 x fi _______ _____ = ÷ y 2 + 1 + c 2 y + 1
( )
–1
x ◊ (IF) = Ú Q ◊ (IF) dy + c
...(i)
1 1 dy 1 dv Let __5 = v fi __ 6 ___ = – __ ___ dx 5 dx y y
1 dv – __ ___ + vx = x3 5 dx dv fi ___ – 5vx = 5x3 dx which is a linear differential equation. fi
5x2 – ___ 2
IF = e– 5 Ú xdx = e
...(ii)
Differential Equation
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + C
fi
v . e
5x2 – ___ 2 = 2
5x – ___
5x2 – ___ 2 dx
– 5 Ú x3 ◊ e
(
)
+C
2
5x – ___ 5x2 e 2 __ 2 ___ 2 + C fi ____ = + 1 e 5 2 y5
which is the required solution of the given differential equation. 120. The given differential equation is dy sin 2 y ___ + _____ x = x3cos2y dx fi
dy 2 tan y sec2y ___ + ______ x = x3 dx
...(i)
Let tan y = v
dy dv fi sec2y ___ = ___ dx dx
dv 2v fi ___ + ___ x = x3 dx
...(ii)
which is a linear differential equation
dx ___
IF = e 2 Ú x = e2 log x = x2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
v ◊ x2 = Ú x5 dx + c
fi
which is a linear differential equation. IF = e– 2 Ú xdx = e – x Multiplying both sides of Eq. (ii) by IF and integrating, we get 2
v ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
v ◊ e– x = Ú x3 e– x dx + c
fi
2 1 v ◊ e– x = – __ (x2 + 1) + c 2
6
x x2 tan y = __ + c 6
which is the required solution of the given differential equation. 121. The given differential equation is dy ___ – x3y3 + xy = 0 dx dy fi ___ + xy = x3y3 dx 1 dy fi __3 ___ + xy– 2 = x3 ...(i) y dx
2
2
2
2 e– x 1 fi ____ = – __ (x2 + 1) e– x + c 2 2 y
which is the required solution of the given differential equation. 122. The given differential equation is
4.49
dy (1 – x2) ___ + xy = xy2 dx
dy xy xy2 fi ___ + _____ 2 = _____ dx 1 – x 1 – x2 x 1 dy x fi __2 ___ + __ (1 – x2) = _____ y dx y 1 – x2
dy dv 1 1 ___ __ ___ Let __ y = v fi y2 dx = – dx
fi
...(i)
dv vx x ___ – _____ = _____ dx 1 – x2 1 – x2
...(ii)
which is a linear differential equation.
x – Ú ______ dx 1 – x2 2 =
IF = e
1 __ log |1 – x2|
e 2
_____
= ÷1 – x2
Multiplying both sides of Eq. (ii) by IF and integrating, we get, fi
v ◊ (IF) = Ú Q ◊ (IF) dx + C _____
_____
x ÷1 – x2 v ÷1 – x = – Ú ________ dx 1 – x2 2
x _____ = – Ú ______ dx ÷ 1 – x2
= ÷1 – x2 +C
_____
_____ 1 dy 1 dv Let y –2 = v fi __ 3 ___ = – __ ___ = ÷1 – x2 + C 2 dx y dx which is the required solution of the given differential dv 1 ___ 3 __ equation. fi – + vx = x 2 dx 123. The given differential equation is dy dv fi ___ – 2vx = – 2x3 ...(ii) x ___ + y = x3y6. dx dx
4.50 Integral Calculus, 3D Geometry & Vector Booster dy fi ___ + dx
y __ x = x2y6
–1
fi
– 5 1 dy y fi __6 ___ + ___ x = x2 y dx
put
...(i)
y– 5 = v
dy – 5 ___ fi ___ = 6 dx y
1 dv v – __ ___ + __ = x2 5 dx x
1 dx fi __2 ___ – x dy ...(ii)
which is a linear differential equation dx – 5Ú ___ x
IF = e
= e– 5 log x
1 = __ 5 x
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + c
...(i)
which is a linear differential equation dx –1 Ú _____ I.F = e x2 + 1 = e tan x
Multiplying both sides of Eq. (i) by IF and integrating, we get
( e ) –1 y ◊ etan x = Ú _______ dx + c (x2 + 1) tan x 2 –1
= Ú e2t dt + c, t = tan–1 x
e2t = Ú ___ + c 2
fi
1 Let v = – __ x
dv fi ___ = dy
dv fi ___ + vy = y3 dx which is a linear differential equation
y2 __
v ◊ e2 = Ú y3 e2 dy + c
fi
y y2 __ e 2 __ 2 dy + c – ___ = 2 y e Ú x 2
( )
...(ii)
( )
2
y2
__ e 2 2 fi – ___ = (y – 2) e 2 dy + c x which is the required solution.
126. The given differential equation is (y log – 1) y dx = x dy
y2 log x dx – y dx = x dy
dy x ___ + y = y2 log x dx
dy ___ + dx
y y2 log x __ x = ______ x
log x 1 dy 1 __2 ___ + __ xy = ____ x y dx
x log |x2 + c1| + 2a tan–1 ___ + y = c2 C1
1 dx __ 2 ___ x dy
y2 __
y2 __
2
Thus,
y2 __
which is the required solution. 124. The given differential equation is
...(i)
y2
5 ___ 2 + c 2x
–1 dy (1 + x ) ___ + y = etan x dx –1 dy y etan x fi ___ + _______ 2 = _______ 2 dx (x + 1) (x + 1)
y __ x = y3
__ Thus, IF = e Ú y dy = e2 Multiplying both sides of Eq. (ii) by IF and integrating, we get
dx 1 fi ____ 5 = – 5 Ú ___3 + c x (xy) 1 fi ____ = (xy)5
which is the required solution of the given differential equation. 125. The given differential equation is
dx fi ___ = x2y3 + xy dy dx fi ___ – xy = x2y3 dy
dv 5v fi ___ – ___ x = – 5x2 dx
Thus,
dy ___ (x2y3 + xy) = 1 dx
dv ___ dx
1 dy 1 dv fi __6 ___ = – __ ___ dx 5 dx y fi
–1 e2tan x y ◊ etan x = ______ + c 2
dv – ___ + dx
(
)
v log x 1 __ x = ____ x , Let v = __ y
log x dv v ___ – __ x = – ____ x dx
...(i)
Differential Equation
which is a linear differential equation 1 __ 1 I.F. = e – Ú x dx = e– log x = __ x Multing both sides of Eq. (i) IF and integrating, we get
Thus,
log x 1 ____ v ◊ __ x = – Ú x2 dx v fi __x = – Ú te– t dt, (Let t = log x)
)
log x +1 1 ________ fi __ + c xy = x
1 Let __2 = v y 2 dy dv fi – __3 ___ = ___ y dx dx
fi
1 dv – __ ___ + vx = x2 2 dx
dv fi ___ – 2vx = – 2x3 dx which is a linear differential equation Thus,
...(i) __
Let ÷y = v
dy dv 1 ___ fi ____ __ = ___ 2 ÷y dx dx
dy dv 1__ ___ fi ___ = 2 ___ ÷y dx dx
fi
dv vx 2 ___ + _____ = x dx 1 – x2
(
)
dv x x fi ___ + ________ v = __ 2 dx 2(1 – x2)
which is the required solution of the given differential equation. 127. The given differential equation is dy ___ = x3y3 = xy dx dy fi ___ + xy = x3y3 dx x 1 dy fi __3 ___ + __ = x3y3 ...(i) y dx y2
__
x ÷y _____ = x 1 – x2
= (t + 1) e– t + c
(
dy 1__ ___ fi ___ + ÷y dx
...(ii)
IF = e– 2 Ú x dx = e – x
4.51
...(ii)
which is a linear differential equation. Thus,
xdx 2 Ú ________
1 – __ log |1 – x2|
IF = e 2(1 – x ) = e 4
1 = _______ 4 _____ 1 – x2 ÷
Multiplying both sides of Eq. (ii) by IF and integrating, we get
1 _____ v ◊ _______ = 4 ÷1 – x2
x dx 1 __ _____ Ú _______ + c 2 4÷1 – x2
__
÷y (1 – x2)3/4 _____ fi 4_______ = ________ +c 3 ÷ 1 – x2 which is the required solution. 129. The given differential equation is dy y y ___ + __ ◊ log y = __ 2 (log y)2 dx x x dy 1 1 1 ___ + _______ fi ________ = __ y (log y)2 dx x (log y) x2
...(i)
2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v (IF) = Ú Q (IF) dx + c
fi
v ◊ e– x = Ú 2x3 e – x dx + c
= (1 + x2) e– x + c
fi
v = (1 + x2) + cex
2
1 Let ______ =v (log y) fi
dy 1 ___ – ________ 2 = y (log y) dx
dv ___ dx
2
2
2
2 1 fi __2 = (1 + x2) + cex y which is the required solution of the given differential equation. 128. The given differential equation is
dy xy __ ___ + _____ = x ÷y . dx 1 – x2
dy dv 1 ___ fi _______ 2 = – ___ dx y(log y) dx fi
dv v __ 1 – ___ + __ = dx x x2
dv v 1 fi ___ – __ = – __2 dx x x which is a linear differential equation. Thus,
dx ___ 1 IF = e – Ú x = e– log x = __ x
...(ii)
4.52 Integral Calculus, 3D Geometry & Vector Booster Multiplying both sides of Eq. (ii) by IF and integrating, we get dx v __x = – Ú ___ 3 + c x
3
e1/x 1 dv v fi __ ___ – __ x = – ____ 2 dx x2 3
dv 2v 2e1/x fi ___ – ___ x = – _____ dx x2
1 = ___ 2 + c 2x
...(ii)
which is a linear differential equation dx ___ 1 IF = e – 2 Ú x = e– 2 log x = __ 2 x
1 1 fi _______ = ___ + c x (log y) 2x2
Thus,
which is the required solution. 130. The given differential equation is
Multiplying both sides of Eq. (ii) by IF and integrating, we get
dy 1 ___ + __ ◊ sin2y = x3 ◊ cos2y dx x
e1/x v __2 = – 2 Ú ____ 4 dx + c x x
3
fi
dy 2 tan y x = x3 sec2y ___ + ______ dx
...(i)
Let tan y = v dy dv fi sec2y ___ = ___ dx dx
dv 2v fi ___ + ___ x = x3 dx
...(ii)
which is a linear differential equation Thus,
IF = e 2 Ú x = e2 log x = x2 dx ___
Multiplying both sides of Eq. (ii) by IF and integrating, we get fi
v ◊ x2 = Ú x5 dx + c
x6 (tan y) x2 = __ + c 6 which is the required solution of the given differential equation. 131. The given differential equation is fi
3
(xy2 – e 1/x ) dx – x2 ydy = 0
fi
3 dy x2y ___ = xy2 – e 1/x dx
dy y ___ = dx
y e1/x __ x – ____ 2 x
y2 – __ x =
e1/x ____ 2 x
dy ___ + x(x + y) = x3 (x + y)3 – 1 dx
( (
) )
dy + dx x 1 _______ 3 _______ + _______ = x3 dx (x + y) (x + y)2 d(x + ) x 1 _______ 3 _______ + _______ = x3 ...(i) dx (x + y) (x + y)2 1 put _______ = v (x + y)2 d(x + ) ___ dv 2 fi – _______ 3 _______ = dx dx (x + y)
d(x + ) 1 1 dv fi _______ 3 _______ = – __ ___ 2 dx dx (x + y)
fi
1 dv – __ ___ + vx = x3 2 dx
dv fi ___ – 2x v = – 2x3 dx which is a linear differential equation Thus,
3
fi
which is the required solution. 132. The given differential equation is
x6 = __ + c 6
2
( )
y 2 2 1/x3 fi __x = __ e + c 3
IF = e– 2Ú x dx = e – x 2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
3
...(i) Let y2 = v dy 1 ___ dv fi y ___ = __ dx 2 dx
fi
v ◊ e– x = – 2 Ú x3 e– x dx + c 2
2
2
= (x2 + 1) e– x + c 2
v = (x2 + 1) + cex
Differential Equation 2 1 fi _______ = (x2 + 1) + cex 2 (x + 1)
fi
( )
( )
1 __x d – __ xy = – d y
which is the required solution of the given differential equation. 133. The given differential equation is
Integrating, we get
1 fi __ xy =
x dx + y dy = x dy – y dx
1 fi __ d (x2 + y2) = x dy – y dx 2 fi
d (x2 + y2) = 2(x dy – y dx)
d (x2 + y2) ____________ 2(x dy – y dx) fi _________ = 2 2 (x + y ) (x2 + y2)
( ) ( )
y d __ x = 2 _______ 2 y 1 + __ x
Integrating, we get
( )
y log |x2 + y2| = 2 tan–1 __ x + c
which is the required solution of the given differential equation. 134. The given differential equation is
x dy – y dx = x4 dx
fi
y d __ x = x2 dx
( )
fi
4.53
1 __y – __ xy = – x – c y __ x + c
which is the required solution of the given differential equation. 137. The given differential equation is
x dy + y dx + xy2 dx – x2y dy = 0.
fi
d (xy) = xy (x dx – y dx)
d (xy) (x dy – y dx) = __________ fi _____ xy 2 2 xy fi
( ) (
dy 1 ___ d – __ xy = y –
)
dx ___ x
Integrating, we get
( ) 1 __y fi __ xy + log ( x ) + c = 0 fi
1 __y – __ xy = log x + c
which is the required solution of the given differential equation. 138. The given differential equation is
(4x – 3y) dx + (2y – 3x) dy = 0
y x3 fi __x = __ + c 3
fi
4x dx + 2y dy – 3 (x dy + y dx) = 0
fi
4x dx + 2y dy – 3d (xy) = 0
which is the required solution of the given differential equation. 135. The given differential equation is
Integrating, we get
x dy + y dx = sin y dy
fi
d(xy) = sin y dy
which is the required solution of the given differential equation. 139. The given differential equation is
Integrating, we get
Integrating, we get fi
xy = c – cos y
which is the required solution of the given differential equation. 136. The given differential equation is
x dy + y dx + y2 (x dy – y dx) = 0.
fi
d(xy) = y2 (y dx – x dy)
d(xy) ____________ y2 (y dx – x dy) fi _____ = x2y2 x2y2
fi
2x2 + y2 – 3xy = c
(
)
1 sin y + y sin x + __ x dx
(
)
1 + x cos y – cos x + __ y dy = 0 fi (sin y dx + x cos y dy) – (cos x dy – y sin x dx)
(
dx + ___ x +
)
dy ___ y = 0
(
dx fi d (x sin y) – d (y cos x) + ___ x + Integrating, we get
)
dy ___ y = 0
4.54 Integral Calculus, 3D Geometry & Vector Booster fi x sin y – y cos x + log (xy) = c
Integrating, we get
which is the required solution of the given differential equation. 140. The given differential equation is
which is the required solution of the given differential equation. 143. The given differential equation is
d(x2 + y2) y ______ fi _________ = 2d __ x 2 2 ÷x + y
( )
______ y d ÷x 2 + y2 = 2d __ x
(
( )
)
Integrating, we get ______ y fi ÷x 2 + y2 = 2 __ x + c
(
( )
)
which is the required solution of the given differential equation. 141. The given differential equation is
(
)
(
2
)
sin 2 x sin x _____ y + x dx + y – _____ 2 dy = 0 y
(
)
fi
sin 2 x sin2x (x dy + y dy) + _____ y dx – _____ 2 dy = 0 y
fi
sin2x d (xy) + d _____ = 0 y
( )
which is the required solution of the given differential equation. 142. The given differential equation is
2 2
x dx + y dy _____________ x + 2x y + y fi _________ = y dx – x dy x2 2 2
(x + y ) = ________ x2
x dy + y dx _________ y dx – x dy fi _________ = . 2 2 2 (x + y ) x2 2
2
( )
fi
and x dx + y dy = r dr, x dy – y dx = r2 dq
fi
y x2 + y2 = r2, tan q = __ x
a2 ◊ r2 dq r dr = _______ 2 . r
Integrating, we get
Ú r dr = Ú a2 ◊ dq
d(x + y ) y fi _________ = – 2d __ x , 2 2 2 (x + y )
(x2 + y2) y fi _______ = a2 tan–1 __ x + c 2 fi
4
...(i)
Let x = r cos q and y = r sin q,
(x2
( ) y + y ) = 2 a tan ( __ x ) + k 2
2
–1
which is the required solution. 144. The given differential equation is
dy x + y ___ y4 dx _______ = x2 + 2y2 + __ 2 . dy x y – x ___ dx
2
)
r2 fi __ = a2 q + c 2
sin2x xy + _____ y = c
4
(
dy a2 x ___ – y dy dx x + y ___ = __________ 2 dx x + y2
Integrating, we get fi
y 1 – _______ 2 = c – __ x 2 (x + y )
y 1 fi __x – _______ 2 2 = c (x + y )
x dx + y dy _________ y dx – x dy ______ __________ = x2 ÷ x 2 + y2
fi
x dx + y dy = (x2 + y2) y dy
x dx + y dy fi _________ = y dy x2 + y2 1 fi __ d (log(x2 + y2)) = y dy 2 Integrating, we get y2 c 1 __ (log (x2 + y2)) = __ + __ 2 2 2 fi
log(x2 + y2) = y2 + c
which is the required solution. 145. The given differential equation is
Differential Equation
(x + y) dx + (x – y) dy = 0
fi
fi
(x dx – y dy) + (x dy – y dx) = 0
Integrating, we get
(x dx – y dy) + d (xy) = 0
d (x + y) = d (xy) (x + y) = xy + c
Integrating, we get
which is the required solution. 150. The given differential equation is
2 x2 y __ – __ + xy = c 2 2
which is the required solution. 146. The given differential equation is
y dx + x (x – 1) dy = 0
fi
y dx – xdy = – x2 dy
fi
xdy – ydx = x2 dy
xdy – ydx fi _________ = dy x2 y fi d __ x = dy
( )
Integrating, we get
( )
xdy – ydx = (x2 + y2) dx
xdy – ydx fi _________ = dx (x2 + y2) ydx – xdy fi _________ = – dx (x2 + y2) ydx – xdy ________ x2 _________ fi = – dx y 2 1 + __ x
( ( ) ) x d ( __ y ) fi _________ = – dx y ( 1 + ( __ x ) ) 2
y __ x = y + c
Integrating, we get
which is the required solution. 147. The given differential equation is
which is the required solution. 151. The given differential equation is
ydx – x (1 – xy) dy = 0
ydx – xdy fi _________ = ydy x2 fi
( )
y d __ x = ydy
Integrating, we get
( )
2
y y __ x = __ + c 2 which is the required solution. 148. The given differential equation is
(x + y) (dx – dy) = (dx + dy)
fi
(x + y) d (x – y) = d (x + y)
fi
d(x + y) d (x – y) = _______ (x + y)
Integrating, we get
(x – y) = log |(x + y)|
which is the required solution. 149. The given differential equation is dx + dy = xdy + ydx
( )
y tan–1 __ x = c – x
( ) x+y fi ( _____ ) (xdy + ydx) = d(x + y) xy
1 1 __ x + __ y (xdy + ydx) = dx + dy
( )
x+y fi _____ d (xy) = d (x + y) xy d (xy) fi _____ = xy
d (x + y) _______ (x + y)
Integrating, we get
log |xy| = log |x + y| + log c
fi
(xy) = c(x + y)
which is the required solution. 152. The given differential equation is ______
(xdy + ydx) ÷x 2 + y2 = x2ydx + xy2dy
xdy + ydx xdx + ydy ______ fi _________ = _________ xy ÷ x 2 + y2
4.55
4.56 Integral Calculus, 3D Geometry & Vector Booster d(xy) __ d(x2 + y2) 1 _________ ______ fi _____ = xy 2 x 2 + y2
÷
(
)
d ÷ x + y ______ = __________ 2 2 ÷x + y
Integrating, we get
______ 2
2
______
log |xy| = ÷ x 2 + y2 + c
which is the required solution. 153. The given differential equation is
(x – y3) dy = y (dx + x2dy)
fi
xdy – y3dy = ydx + x2ydy
xdy – ydx fi ________ = y2dy + x2dy y xdy – ydx fi _________ = y dy (x2 + y2)
( ) ( ( ) )
x d __ y _________ fi = y dy y 2 1 + __ x Integrating, we get
y y2 tan–1 __ x = __ + c 2
( )
which is the required solution. 154. The given differential equation is
(x + 2y) dy + ydx = 0
fi
(eydx + xe ydy) = 2y dy
fi
d (xey) = 2y dy
Integrating, we get
y2 (xey) = __ + c 2
which is the required solution. 157. The given differential equation is 2ydy + (cos x ◊ cot y – y2) dx = (2xy + sin x ◊ cosec2y) dy
fi
2ydy – d (xy2)
– (sin x cosec2y dy – cos x cot y dx) = 0
fi
2ydy – d (xy2) + d (sin x cot dy) = 0
Integrating, we get
y2 – xy2 + (sin x cot dy) = c
which is the required solution. 158. The given differential equation is dy ___ = dx
2x – 3y + 1 __________ 3 – 2y – 2
fi
3xdy – 2(y + 1) dy = 2xdx – 3ydx + dx
fi
3 (xdy + ydx) – 2(y + 1) dy = (2x + 1) dx
fi
3d (xy) – 2(y + 1) dy = (2x + 1) dx
Integrating, we get
( ) ( )
y2 x2 3(xy) – 2 __ + y = __ + x + c 2 2
fi
(xdy + ydx) + 2y dy = 0
fi
d (xy) + 2y dy = 0
which is the required solution. 159. The given differential equation is
Integrating, we get
(xy) + y2 = c
which is the required solution. 155. The given differential equation is
(x + sin y) dy + ydx = 0
fi
(xdy + ydx) + sin y dy = 0
fi
d (xy) + sin y dy = 0
Integrating, we get
(xy) – cos y = c
which is the required solution. 156. The given differential equation is
eydx + (xe y – 2y) dy = 0
x (dy + dx) = y (dx – dy)
xdy – ydx xdx + ydy fi _________ = _________ 2 2 2 (x + y ) (x + y2) xdy – ydx _________ 2 y d(x2 + y2) 1 _________ fi _________ = __ 2 2 x 2 1 + __ y 2 (x + y )
( ( ) ) x d ( __ y ) 1 d(x + y ) _______ fi = __ ________ x__ 1 + ( y ) 2 (x + y ) 2
Integrating, we get
2
2
2
2
Differential Equation
( )
ydx – xdy ________ y2 d(xy) _________ fi = _____ 2 x2 __ 1 + y (xy)
x 1 tan–1 __ y = __ log |(x2 + y2)| + c 2
which is the required solution. 160. The given differential equation is
(
)
xy x (log x – tan–1 x) dy + ydx = _____ 2 dx 1+x
fi
dy xy x (log x – tan–1 x) ___ + y = _____ 2 dx 1+x
fi
dy xy x(log x – tan x) ___ = _____ – y dx 1 + x2
(
fi
dy x x(log x – tan–1 x) ___ = y _____ 2 – 1 dx 1+x
fi
dy x 1 (log x – tan–1 x) ___ = y _____ 2 – __ x dx 1+x
d fi ___ {y(log x – tan–1 x)} = 0 dx Integrating, we get
y (tan x – log x) = C
x (dy – x dx) + ydx = 0
fi
xdy + ydx = x2dx 2
d (xy) = x dx
Integrating, we get
)
)
2
x xy = __ + c 3
which is the required solution. 162. The given differential equation is dy y – x ___ dx 1 fi ________ = __ 2 + dy x y + x ___ dx
1 __ 2 y
ydx – xdy _______ x2 + y2 fi _________ = ydx + xdy (xy)2 ydx – xdy ydx + xdy fi _________ = _________ 2 2 (x + y ) (xy)2
2
( )
x 1 tan–1 __ y + __ xy = c
which is the required solution. 163. The given differential equation is
x3 xdy – x2y2dx = x4dy + x3ydx
x2y(xdy – ydx) fi ____________ = (xdy + ydx) x3
(
)
fi
xdy – ydx xy ________ = d (xy) x2
fi
y d(xy) d – __x = _____ (xy)
–1
which is the required solution. 161. The given differential equation is
fi
2
Integrating, we get
–1
(
( ( ) ) x d ( __ y ) d(xy) _______ fi = _____ x__ 1 + ( y ) (xy)
( )
Integrating, we get
y log |xy| – __x = c
which is the required solution. 164. The given differential equation is
dx + x (ydx + xdy) = e– xy dx
fi
dx + x d (xy) = e– xy dx
fi
(1 – e– xy)dx + x d(xy) = 0
d(xy) dx ________ fi ___ = 0 x + (1 – e– xy ) Integrating, we get
log |x| + log |e xy – 1| = log c
fi
x (exy – 1) = c
which is the required solution. 165. The given differential equation is x – __
y (y2 dx + (x2dy – xydx)) = e y dy
fi
y3dx + xy(xdy – ydx) = e y dy
fi
x xdy – ydx – __y y3dx + xy3 ________ = e dy y2
x – __
(
)
4.57
4.58 Integral Calculus, 3D Geometry & Vector Booster fi
x – __
Integrating, we get
y3 dx + xy3 d (– xy) = e y dy
(xy)2 ____ + 2
166. The given differential equation is
ydx – xdy + xy2dx = 0
fi
xdy – ydx = xy2dx
( )
x d __ y = xdx
Integrating, we get
( )
(1 + xy) ydx + x (1 – xy) dy = 0
fi
ydx + xdy + xy (ydx – xdy) = 0
fi
d (xy) + xy (ydx – xdy) = 0
d(xy) fi _____ = (xdy – ydx) (xy)
which is the required solution. 167. The given differential equation is
( (
d(xy) fi _____ + (ydx – xdy) = 0 (xy)
x x2 __ y = __ + c 2
a2y2 __ x4 a2x2 ____ = – ____ + c 2 4 2
which is the required solution. 169. The given differential equation is
xdy – ydx fi _________ = xdx x2 fi
y4 __ + 2
(
)
)
dy dy 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ dx dx
) (
)
(
)
d(xy) xdy – ydx dy dx = ___ y – ___ x fi _____2 = ________ xy (xy) Integrating, we get
| |
1 __y – __ xy = log x + c
dy dy 2 x – y ___ y – x ___ dx dx fi _________ = ________ 2 2 2 (x – y ) (x + y2)
2(xdx – ydy) (ydx – xdy) fi ___________ = __________ 2 2 2 (x – y ) (x + y2)
xdy y 2 = ______ 2 2 – 1 dx ______ 2 x +y x +y
(
)
which is the required solution. 170. The given differential equation is
(
)
ydx – xdy ________ 2 2 y2 d(x – y ) _________ __________ fi 2 = x2 (x – y2) 1 + __ y
fi
xdy = ydx – (x2 + y2) dx
fi
xdy – ydx = – (x2 + y2) dx
xdy – ydx ________ x2 ________ fi = – dx y2 1 + __ x
( ( ) ) x d ( __ y ) ________ = x ( 1 + ( __ y ) ) 2
Integrating, we get
( )
x log |x2 – y2| = tan–1 __ y + c
(x2 + y2 + a2)y dy = (x2 – y2 – a2)x dx
fi x2ydy + y2xdx + (y3 + a2y) dy = (x3 – a2x) dx fi xy (xdy + ydx) + (y3 + a2y)dy = (x3 – a2x) dx 3
2
3
( ( ) ) x d ( __ y ) fi ________ = – dx y __ ( 1 + ( x ) ) 2
which is the required solution. 168. The given differential equation is
xdy – ydx = – dx fi ________ (x2 + y2)
2
fi xyd (xy) + (y + a y) dy = (x – a x) dx
Integrating, we get
( )
y tan–1 __ x = c – x
which is the required solution. 171. The given differential equation is
Differential Equation
dy x + y ___ y4 dx ________ = x2 + 2y2 + __ 2 dy x y – x ___ dx 2
2 2
xdx + ydy (x + y ) fi _________ = ________ ydx – xdy x2 2 2 ydx – xdy 1 d(x + y ) _________ fi __ ________ = 2 2 2 2 (x + y ) x2 2
2
Integrating, we get y 1 __x – ________ 2 2 = c 2(x + y )
x dx + y dy = m (x dy – y dx)
xdx + ydy ____________ m (xdy – ydx) fi ________ = 2 2 (x + y ) (x2 + y2)
( ) ( ( ) )
y m d __ x d (x2 + y2) 1 _________ __ _________ fi 2 = y 2 2 (x + y2) 1 + __ x Integrating, we get y 1 tan– 1 __ x + _________ 2 2 = c 2 (x + y ) which is the required solution. 173. The given differential equation is
( )
dy x + y ___ x sin2 (x2 + y2) dx ________ = ____________ dy y3 ___ y – x dx xdx + ydy ____________ x sin2(x2 + y2) fi _________ = ydx – xdy y3 xdx + ydy x (ydx + xdy) fi __________ = ____________ 2 2 2 sin (x + y ) y3 2 2 y x 1 d (x + y ) fi __ ___________ + __ d __ x 2 sin2(x2 + y2) y
( )
Integrating, we get
( )
1 x2 __ __ + 2 y
a2(x dy – y dx) x dx + ydy = ____________ (x2 + y2)
fi
(
)
x dy – y dx a2 _________ y2 _____________ = x 2 1 + __ y
( ( ) ) x a d ( __ y ) = ________ x ( 1 + ( __ y ) ) x a d ( __ y ) + y ) = ________ x ( 1 + ( __ y ) ) 2
1 fi __ d (x2 2
2
2
which is the required solution. 172. The given differential equation is
)
dy a2 x ___ – y dy dx x + y ___ = ___________ 2 dx x + y2
( )
y 1 d (x + y ) fi __ _________ + d __ x = 0 2 (x2 + y2)2
(
4.59
1 __ cot (x2 + y2) = c 2
which is the required solution. 174. The given differential equation is
2
2
Integrating, we get
( )
x 1 fi __ (x2 + y2) = tan– 1 __ y + c 2 which is the required solution. 175. The given differential equation is
÷
___________
x dx + y dy a2 – x2 – y2 __________ = __________ x dy – y dx x2 + y2
Let x = r cosq, y = r sinq
y fi x2 + y2 = r2, __ x = tanq
...(i)
x dy – y dx x dx + y dy = r dr, _________ = sec2 q dq x2
÷ ÷
_____
r dr 1 – r2 fi ____ = _____ 2 r dq r2 _____
dr 1 – r2 fi ___ = _____ rdq r2
dr fi ___ = ÷ 1 – r2 dq
dr _____ fi ______ = dq ÷ 1 – r2
_____
Integrating, we get
sin– 1(r) = q + c
fi
______ y sin– 1 ÷x 2 + y2 = tan– 1 __ x + c
(
)
( )
which is the required solution. 176. The given differential equation is
4.60 Integral Calculus, 3D Geometry & Vector Booster
÷
__________
x dx – y dy 1 + x2 – y2 _________ = __________ x dy – y dx x2 + y2
...(i)
Let x = r secq , y = r tanq
fi
x2 – y2 = r2
fi
x dx – y dy = r dr, x dy – y dx = r2 secq dq ______
÷
x4 + exsin y = c
which is the required solution. 180. Let the equation be y = m x, where m is an arbitrary constant Differentiating w.r.t. x, we get,
...(i)
r dr 1 + r2 ______ fi ________ = r r2sec q dq
dy ___ = m dx
dr ______ fi _______ = secq dq ÷ 1 + r2
Eliminating m, between Eqs (i) and (ii), we get dy ___ = dx
Integrating, we get
_____
log |r + ÷ r 2 + 1 | = log |secq + tanq | + log c
_____ fi r + ÷r 2 + 1 = c(secq + tanq) ______ ___________ x+y ______ fi x2 – y2 + (x2 – y2) + 1 = c _______ x2 – y2
(
)
( ÷
÷
)
( ÷ )
...(ii)
y __ x
...(iii)
whcih is the differential equation of a family of lines. dy dx Now replacing ___ by – ___ in Eq. (iii), we get, dx dy
dx x – ___ = __ dy y
which is the required solution. 177. The given differential equation is x sin __ y ( y dx – x dy) = xy3(x dy + y dx)
dy dx ___ fi ___ y = – x
fi
dx Ú ___y = – Ú ___ x
fi
fi
c log |y| = log |c| – log |x| = log __ x xy = c
( ) x y dx – x dy sin ( __ y ) _________ ( y ) = x y d (x y) x x sin ( __ y ) d ( __ y ) = x y d (x y) 2
Integrating, we get (xy)2 x – cos __ y = ____ + c 2 which is the required solution. 178. The given differential equation is
( )
x dy – y dx + y2 = y2(x dy + y dx) = 0 2
fi
y d (xy) = y dx – x dy
fi
y dx – x dy d (x y) = _________ y2
( )
x = d __y
Integrating, we get
(
)
x fi xy – __ y = c
which is the required solution. 179. The given differential equation is
(4x3 + exsin y) dx + excos y dy = 0
(4x3) dx + (exsin y dx + excos y dy) = 0
fi
fi (4x3) dx + d (exsin y ) = 0 Integrating, we get
Integrating, we get
fi
dy
| |
which is required orthogonal trajectories. 181. The given family of corves are
x2 + y2 = a2
Differentiaint w.r.t. x, we get
dy 2x + 2y ___ = 0 dx
dy x + y ___ = 0 dx dy ___ – tan 45° dy dx Replacing ___ by ____________ = dx dy ___ 1 + tan 45° dx dy ___ – 1 dx ______ in Eq. (i), we get dy ___ 1 + dx fi
( )
dy ___ – 1 dx x + y ______ = 0 dy ___ 1 + dx
...(i)
Differential Equation
) ( )
(
y2 __ + log |y| = c 2
dy dy x 1 + ___ + y ___ – 1 = 0 dx dx
x2 fi __ – 2
fi
dy (x + y) ___ = (y – x) dx
fi (x2 – y2) + 2 log |y| = c which is the required orthogonal trajectories.
dy fi ___ = dx
y–x _____ y+x
...(ii)
183. Given curve is
which is a homeogeneous differential equation. fi dy dv Put y = v x fi ___ = v + x ___ dx dx dv _____ v–1 ___ fi v + x = dx v + 1 fi dv x ___ = dx
v–1 v – 1 – v2 – v _____ – v = _____________ v+1 v+1
fi
v+1 dx fi _____ dv = – ___ x v2 + 1
( )
1 __ log |v2 + 1| + tan– 1v = c – log x 2 y fi log |x2 + y2| + tan– 1 __ x = c which is the required trajectories of the given family of curves. 182. The given curve is
( )
x2 + y2 = 1
fi
dy 2a x + 2y ___ = 0 dx
fi
dy a x + y ___ = 0 dx
...(i)
...(ii)
fi fi fi
fi
dy xy ___ = 1 – y2 dx
dy 1 – y2 fi ___ = _____ xy dx dy dx Replace ___ by – ___ , we get dx dy fi fi fi
2 dx 1 – y – ___ = _____ xy dy 1 x dx = y – __ y dy
(
)
dy x dx – y dy + ___ y = 0
Integrating, we get
dy 2x + 2y ___ dx a = __________ dy ___ dx
( )
dy 2x + 2y ___ dx (x + y ) – _________ y = 0 dy ___ dx 2
2
(
)
dy dy (x2 + y2) ___ – 2x + 2y ___ y = 0 dx dx dy (x2 + y2 – 2y2) ___ = 2xy dx dy (x2 – y2) ___ = 2xy dx
dy 2xy fi ___ = _______ 2 2 dx (x – y ) Replace dy/dx by – dx/dy, we get
Solving Eqs (i) and (ii), we get dy xy ___ + y2 = 1 dx
dy dy 2x + 2y ___ – a ___ = 0 dx dx
Eliminating a between (i) and (ii), we get
Integrating, we get
x2 + y2 – ay = 0
2xy dx – ___ = _______ 2 2 dy (x – y )
dy y2 – x2 fi ___ = ______ 2xy dx y2 __ x – 1 _______ = y 2 __ x
( )
( )
fi
y dv v2 – 1 v + x ___ = ______ , v = __ x 2v dx
fi
dv v2 – 1 x ___ = _____ – v 2v dx
v2 – 1 = – ______ 2v
2v dv dx fi _______ = – ___ x 2 (v + 1) Integrating, we get
log |v2 + 1| + log |x| = log c
4.61
4.62 Integral Calculus, 3D Geometry & Vector Booster fi
x (v2 + 1) = c 2
Integrating, we get log |x| + log |y| = log (c2)
2
fi x + y – c x = 0 which is the required orthogonal trajectories. 184. The given curve is fi fi
y2 = 4ax dy 2y ___ = 4a dx y dy a = __ ___ 2 dx
Eliminating a from Eqs (i) and (ii), we get y dy y2 = 4x __ ___ 2 dx dy fi y = 2x ___ dx
( )
Replacing dy/dx by – dx/dy, we get dy y = 2x × – ___ dx fi y dy + 2x dx = 0 Integrating, we get
...(i)
...(ii)
fi xy = c2 which is the required orthogonal trajectories. 187. The given equation is (x + y + p)(2x + p) = 0 fi
(x + y + p) = 0
...(i)
and
(2x + p) = 0
...(ii)
From Eq. (i), we get, dy ___ + y + x = 0 dx dy fi ___ + y = – x dx which is a linear differential equation.
fi
fi
y . ex = – Ú x ex dx + c1 = – ex(x – 1) + c1 y = – (x – 1) + c1e– x
y2 __ + x2 = c 2 which is the required orthogonal trajectory. 185. Given curve is xy = c2 dy fi x ___ + y = 0 dx dy y fi ___ = – __x dx
Also, from Eq. (ii), we get,
Replacing dy/dx by – dx/dy, we get y dx – ___ = – __x dy fi x dx – y dy = 0
p2 – p (ex + e– x) + 1 = 0
fi fi
1 p2 – p ex + ___ – x + 1 = 0 e p2 ex– p (e2x + 1) + ex = 0
fi
p2 ex – pe2x – p + ex = 0
fi
pex (p – ex) – 1(p – ex) = 0
fi
(p – ex) (pex – 1) = 0
fi
(p – ex) = 0
fi 2x dx – 2y dy = 0 Integrating, we get x2 – y2 = a2 which is the required orthogonal trajectories. 186. The given curve is x2 – y2 = c2. dy fi 2x – 2y ___ = 0 dx dy x fi ___ = __ dx y Replacing dy/dx by – dx/ , we get dx x – ___ = __ dy y dx dy ___ fi x + ___ y = 0
...(iii)
dy ___ + 2x = 0 dx fi
dy = – 2x dx
y = c2 – x2
fi
...(iv)
Hence, the required solution is
(y + (x – 1) – ce– x) (y + x2 – c) = 0.
188. The given equation is
and
(
x
( pe – 1) = 0
...(i)
)
...(ii) ...(iii)
From Eq. (i), we get, dy ___ = ex dx fi dy = ex dx Integrating, we get fi
y = ex + c1
From Eq. (iii), we get,
...(iv)
Differential Equation
fi fi When
dy ___ ◊ ex – 1 = 0 dx fi ex dy = dx fi dy = e– xdx Integrating, we get fi y = c2 – e– x Hence, the general solution of Eq. (i) is (x – e– x – c) (y – e– x – c) = 0. 189. The given equation is fi
p2 + 2py cot x = y2 2
2
2
2
...(i)
2
( p + y cot x) = y (1 + cot x) 2
( p + y cot x) = y cosec x
fi
( p + y cot x) = ± y cosec x
p = y (cosec x – cot x)
...(ii)
p = – y (cosec x – cot x)
...(iii)
and
From Eq. (ii), we get, dy ___ = y (cosec x – cot x) dx dy fi ___ y = (cosec x – cot x) dx
(
)
dy 1_______ – cos x fi ___ dx y = sin x dy x __ fi ___ y = tan 2 dx Integrating, we get x log |y| = 2 log |sec __ | + 2log (c1) 2 x fi y = c21 sec2 __ 2 From Eq. (iii), we get
( )
( ) ( ( ) )
dy ___ = – y (cosec x + cot x) dx dy fi ___ y = – (cosec x + cot x) dx
( )
( ( ) )
x log y = 2 logc2 – 2 log sin __ 2 x fi y sin2 __ = c2 2 Hence, the required solution is x x y – c sec2 __ y sin2 __ – c = 0. 2 2
fi
( ( ) )
( ( ( ) ) ) ( ( ) )
190. The given differential equation is fi
When
log |y| = x + c1 ( p + x + y) = 0,
dy fi ___ + (x + y) = 0 dx dy fi ___ + y = – x dx which is a linear differential equation Thus,
IF = eÚdx = ex
Thus, the solution is
y . e x = – Ú x ex dx + c
fi
y . e x = – (x – 1)ex + c2
Hence, the required solution is
(log |y| – x c1) ( y . ex + (x – 1) ex – c2) = 0
191. The given differential equation is p2y + (x – y) p – x = 0 fi p2y + px – py – x = 0 fi py( p – 1) + x( p – 1) = 0 fi ( py + x)( p – 1) = 0 fi ( py + x) = 0, ( p – 1) = 0 When
( py + x) = 0
y dy fi ____ + x = 0 dx fi y dy + x dx = 0 Integrating, we get x2 fi __ + 2
Integrating, we get dy x __ fi ___ y = – cot 2 dx
( p – y) ( p + x + y) = 0 ( p – y) = 0, ( p + x + y) = 0 (p – y) = 0
dy fi ___ = y dx dy fi ___ y = dx Integrating, we get
fi fi
p2 + px – xy – y2 = 0 ( p2 – y2) + x( p – y) = 0
4.63
fi When
y2 c2 __ = __ 2 2
x2 + y2 = c2 (p – 1) = 0
p=1 dy fi ___ = 1 dx fi dy = dx Integrating, we get y = x + c1 Hence, the solution is (x2 + y2 – c2) (y – x – c1) = 0
4.64 Integral Calculus, 3D Geometry & Vector Booster 192. The given differential equation is ( p2 – 1) xy = (x2 – y2) p
fi
p2xy – xy – px2 + py2 = 0
fi
p2xy – px2 + py2 – xy = 0
fi
px ( py – x) + y ( py – x) = 0
fi
( px + y) ( py – x) = 0
fi When
( px + y) = 0, ( py – x) = 0 ( px + y) = 0
(
dy fi ___ x dx dx fi ___ x +
)
+ y = 0 dy ___ y = 0
log x + log y = log c
fi
x y = c
When
(py – x) = 0
(
Integrating, we get
fi
x2 – y2 = c 21
193. The given differential equation is x y p2 – (x2 + y2) p + x y = 0 2
2
2
fi
x y p – x p – y p + x y = 0
fi
px( py – x) – y( py – x) = 0
fi
( px – y)( py – x) = 0
fi
( px – y) = 0, ( py – x) = 0
When
( px – y) = 0 dy fi ___ x – y = 0 dx dy dx ___ fi ___ x – y = 0 Integrating, we get log x – log y = log c fi x = c y
(
)
( )
log |1 – p2| = log c – 2 log |y| c fi log |1 – p2| = log __ 2 y c fi (1 – p2) = __ 2 y c 2 fi p = 1 – __2 ...(ii) y Eliminating p between Eqs (i) and (ii), we get _____ c x = a ± y 1 – __ 2 y
| |
(
Hence, the required solution is (xy – c) (x2 – y2 – c21) = 0
)
( )
xdx – ydy = 0 2 c 21 x2 y __ – __ = __ 2 2 2
...(i)
– 2dy – 2p fi _____ dp = _____ y 2 1–p Integrating, we get
)
(xy – c) (x2 – y2 – a2) = 0 194. The given equation can be written as x = y p + a Differentiaing w.r.t. y, we get dp dx ___ = p + y ___ dy dy dp 1 ___ fi __ p = p + y dy dp 1 fi __ p – p = y ___ dy dy p fi _____ 2 dp = ___ y 1–p
dy fi y ___ – x = 0 dx fi
x2 – y2 = a2 Hence, the required solution is
(
Integrating, we get
Integrating, we get
)
÷
which is the required solution of Eq. (i) 195. The given equation can be written as 1 y x = __ __ – y2p2 2 p
(
)
...(i)
Differentiaing w.r.t. y, we get
dx ___ = dy
( ( (
)
y dp dp 1 __ 1 __ – __ ___ – 2yp2 – 2py2 ___ 2 p p2 dy dy
)
y ___ dp dp 1 __ 1 __ 1 __ 2 2 ___ fi __ p = 2 p – p2 dy – 2yp – 2py dy
fi
dp dp 2 p = p – y ___ – 2yp4 – 2p3y2 ___ dy dy
fi
( py – x) = 0 dy fi y ___ – x = 0 dx
dp dp p + 2 yp4 = – y ___ – 2p3y2 ___ dy dy
fi
dp p (1 + 2yp3) = – y ___ (1 + 2p3y) dy
fi
fi
dp (1 + 2yp3) p + y ___ = 0 dy
When
(
)
xdx – ydy = 0
(
(
)
)
)
Differential Equation
(
)
dp fi p + y ___ = 0 dy
fi
4.65
Differentiating, w.r.t. y, we get
(
)
dp dp 2p 1 – p2 – 2py ___ – 2 (y – p2y) ___ dy dy dx ___ = ______________________________ 2 dy 4p
d ( py) = 0
Integrating, we get p y = c y fi p = __ c
...(ii)
(
)
dp dp p 1 – p2 – 2py ___ – (y – p2y) ___ dy dy 1 __ p = _____________________________ 2 2p
fi
Eliminating p between Eq. (i) and (ii), we get
c c3 y = 2 __y . x + __ 3 . y2 y
fi
dp dp dp 2p = p – p3 – 2p2y ___ – y ___ + p2y ___ dy dy dy
y2 = 2c x + c3
fi
dp dp p = – p3 – p2y ___ – y ___ dy dy
fi
which is the required solution. 196. The given differential equation is
y2 log y = x yp + p2
fi
y2 log y – p2 x = ___________ y p
...(i)
fi
dp fi ___ p +
Differentiaing w.r.t. y, we get
(
dp p ( p2 + 1) = – ( p2 + 1) y ___ dy dp p = – y ___ dy
fi
(
)
(
)
dy ___ p = 0
fi p y = c ...(ii) dp dp yp 2 y logy + y – 2p ___ – (y2 log y – p2) p + y ___ Eliminating p between Eqs (i) and (ii), we get dy dy dy ___ = ___________________________________________ 2 dx y2 + 2c x = c2 (yp) which is the required solution. dp 2 ___ fi y p = yp 2y log y + y – 2p 198. c (4x – c)2 = 64y dy
)
(
)
199. The given differential equation is dp – (y2 log y – p2) p + y ___ dy y p = 2p2x + y2p4
fi
dp y ___ ( y2log y – p2) = p ( y2 log y – p2) dy
fi
dp y ___ = p dy
dy fi ___ p =
y – y2p3 x = ________ 2p
fi
...(i)
Differentiating, w.r.t. y, we get
(
)
dp dp p 1 – 2yp3 – 3y2p2 ___ – (y – y2p3) ___ dy dy dx 1 ________________________________ fi ___ = __ 2 dy 2 p
dp ___ p
Integrating, we get
dp dp fi p + 2yp4 + 2y2p3 ___ + y ___ = 0 dy dy
Eliminating p between Eqs (i) and (ii), we get log y = c x + c2
dp fi p (1 + 2yp3) + y ___ (1 + 2 yp3) = 0 dy
which is the required solution of (i) 197. The given differential equation is p2y + 2px = y
dp fi p + y ___ (1 + 2yp3) = 0 dy
dp fi p + y ___ = 0 dy
dy dp ___ fi ___ p + y = 0
fi
log p = log y + log c
fi
p = c y
fi
2p x = y – p2y
fi
y – p2y x = _______ 2p
...(ii)
...(i)
( (
) )
4.66 Integral Calculus, 3D Geometry & Vector Booster
fi
p y = c
...(ii)
Eliminating p between Eqs (i) and (ii), we get
Hence, the required solution of Eq. (ii) is
which is the required solution. 200. The given differential equation is 2
xp – yp – p + 1 = 0
fi
yp + p – 1 x = _________ p2
...(i)
(
x.e p = – 2a Ú pe p dp = – 2ae p (p – 1) + c
fi
x = – 2a (p – 1) + ce– p
...(iii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i). 202. The given equation is y = yp2 + 2px
Differentiating, w.r.t. y, we get
IF = e Údp = e p
y2 = 2c x + c3
which is a linear differential equation.
2p x y = _____ 1 – p2 Differentiaing w.r.t. x, we get
fi
)
dp dp dp p2 p + y ___ – ___ – 2 (yp + p – 1) p ___ dy dy dy dx ___ = _________________________________ 4 dy p
2 dp dx ________ = ___ x p ( p2 – 1)
dp dp fi p3 = p3 + p2y ___ – p2 ___ dy dy
(
1 1 fi _____ + _____ – p–1 p+1
)
dx 2 ___ __ p dp = x
dp dp dp – 2p2y ___ – 2p2 ___ + 2p ___ dy dy Integrating, we get dy dp dp dp log | p – 1| + log | p + 1| – log | p2| = log |x| + log c fi p2y ___ – p2 ___ – 2p2y ___ dy dy dy p2 – 1 dp dp fi log _____ = log |c x| 2 ___ ___ – 2p + 2p = 0 p2 dy dy p2 – 1 dp fi ______ = cx fi ___ = 0 dy p2
| |
fi dp = 0
fi p = c
...(ii)
Eliminating p between Eqs (i) and (ii) we get
p2 – 1 = p2c x
fi
p2 (1 – c x) = 1
fi
1 p2 = _______ (1 – c x)
(
)
÷
_______
2
201. The given equation is y = (1 + p) x + ap
...(i)
Differentiaing w.r.t. x, we get
fi
fi
dp (x + 2 ap) ___ = – 1 dx
dx fi ___ = – (x + 2ap) dp
dx fi ___ + x = – 2ap dp
1 p = _______ (1 – c x)
fi
...(ii)
Eliminating p between Eqs (i) and (ii), we get ______
2x ÷1 – c x + c x y = 0.
which is the required solution.
dy dp dp fi ___ = (1 + p) ◊ 1 + x ___ + 2ap ___ dx dx dx dp dp p = (1 + p) ◊ 1 + x ___ + 2ap ___ dx dx
fi
yc = c2x – c + 1
which is the required solution.
203. The given differential equation is y = 2 p x + tan– 1(xp2) Differentiating w.r.t. x, we get
...(ii)
dy dp ___ = 2p + 2x ___ + dx dx
(
...(i)
dp p2 + 2px ___ dx __________ 1 + (xp2)2
)
dp dp fi – p – 2x ___ (1 + p4x2) = p2 + 2px ___ dx dx
dp dp dp fi – p – 2x ___ – p5x2 – 2x3p4 ___ = p2 + 2px ___ dx dx dx
Differential Equation
dp dp dp fi – p – p5x2 – p2 = + 2x ___ + 2x ___ + 2x3p4 ___ dx dx dx dp fi – p (1 + p4x2 – p) = + 2x (p + 1 + x2p4) ___ dx dp fi – p = 2x ___ dx
2dp ___ dx fi ____ p + x = 0
fi 2 log p + log x = log c
fi p2x = c
dy dp dp ___ + p + x ___ = 2px4 ___ + 4x3p2 dx dx dx
fi
dp dp p + p + x ___ = 2px4 ___ + 4x3p2 dx dx
fi
fi
fi
dp x (1 – 2px3) ___ = – 2p (1 – 2x3p) dx
fi
dp x ___ = – 2p dx
dp ____ 2dx fi ___ = 0 p + x
...(ii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i). __
( ÷ )
c y = 2 __ x x + tan– 1(c)
fi
– 1 y = 2÷c x + tan (c)
___
fi
2 4
x p + 2xp = y 4 2
y = p x + 2px
dy dp dp fi ___ = 2x p4 + 4x2p3 ___ + 2p + 2x ___ dx dx
fi
fi
fi fi fi
dp (x – 2px4) ___ = 4x3p2 – 2p dx dp x (1 – 2px3) ___ = 2p (2x3p – 1) dx
Integrating, we get
which is the required solution. 204. The given differential equation is
...(i)
dp dp p = 2x p4 + 4x2p3 ___ + 2p + 2x ___ dx dp – 2x p4 – p = (4x2p3 + 2x) ___
log p + 2log x = log c
fi
p x2 = c
dp 2x (2xp3 + 1) ___ = – p (2xp3 + 1) dx dp 2x ___ = – p dx dp dx ___ 2 ___ p = – x
...(ii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i) c y + __ x + c2 which is the required solution. 206. The given differential equation is
y = p sin p + cos p
...(i)
dy dp dp dp fi ___ = sin p ___ + p cos p ___ – sin p ___ dx dx dx dx dp fi p = p cos p ___ dx
fi
dp cos p ___ = 1 dx
fi
cos p dp = dx
Integrating, we get
Integrating, we get
2 log |p| = log c – log |x|
fi
c 2 log |p| = log __x
Eliminating p from Eqs (i) and (ii), we get
| |
___
2 y = 2÷c x + c
y + px = x4p2
Differentiating w.r.t. x, we get
...(ii)
y = (x + c) sin– 1(x + c) + ÷ 1 – (x + c)2
which is the required solution. 207. The given differential equation can be written as
which is the required solution. 205. The given differential equation is
sin p = x + c
__________
x p2 = __ c ...(ii) Eliminating p between Eqs (i) and (ii), we get
4.67
2 y = Px – __ – 1 P
...(i)
Differentiating w.r.t. x, we get ...(i)
dP dP ___ = P + x ___ + dx dx
2 dP ___ 2 ___ P dx
4.68 Integral Calculus, 3D Geometry & Vector Booster fi
dP 2 dP P = P + x ___ + __ ___ dx p2 dx
(
)
211. The given differential equation is
2 dP fi x + ___ 2 ___ = 0 P dx
(
)
dP 2 fi ___ = 0 or x + ___ 2 = 0 dx P fi
P = c
...(ii)
or
2 x = – __2 p
...(iii)
(x – a) P2 + (x – y) P – y = 0
fi
P2x + Px – aP2 = ( P + 1) y
fi
( P + 1) y = P (P + 1) x – aP2
aP2 y = Px – _______ ( P + 1)
fi
which is a Clairaut differential equation. Hence, the solution is ac2 y = cx – ______ (c + 1)
Eliminating p between Eqs (i) and (ii), we get
2 y = cx – __ c – 1 which is the genral solution of Eq. (i).
212. The given differential equation is
Eliminating p between Eqs (i) and (iii), we get
which is a Clairaut differential equation. Hence, the solution is _____
(y + 1)2 + 8x = 0 which is the singular solution of Eq. (i). 208. The given differential equation is 3
2
Px–Py–1=0
fi
P2y = P3x – 1
1 y = P x – ___ 2 P
fi
( y + 1) P – x P2 + 2 = 0
fi
( y + 1) P = p2x – 2
fi
2 ( y + 1) = Px – __ p
fi
2 y = P x – __ + 1 P
(
)
which is a Clairaut differential equation. Hence, the solution is 2 y = c x – __ c + 1 210. The given differential equation is
)
sin y cos P x – cos y sin Px – P = 0
fi
sin (y – Px) = P
fi
(y – Px) = sin– 1 (P)
fi
y = Px + a ÷1 + P2
y = cx + a ÷1 + c2
213. The given differential equation is a y = p (x – b) + __ p a fi y = px + __ p – bp
(
which is a Clairaut differential equation. Hence, the solution is 1 y = c x – __ 2 c 209. The given differential equation is
(
______
– 1
y = Px + sin (P)
which is a Clairaut differential equation. Hence, the solution is y = c x + sin– 1(c)
)
which is a Clairaut differential equation. Hence, the solution is a y = cx + __ c – bc 214. The given differential equation is
(
)
d 2y ____2 = x + sin x
( )
d dy fi ___ ___ = x + sin x dx dx Integrating, we get dy x2 ___ = __ – cos x + c1 2 dx Again integrating, we get x3 y = __ – sin x + c1x + c2 6 which is the required solution. 215. The given differential equation is fi
d2y ___ 2 = e2x + ex + 2014 dx
d dy fi ___ ___ = e2x + ex + 2014 dx dx
( )
Integrating, we get
dy ___ = dx
e2x ___ + ex + 2014 + c1 2
Differential Equation
Again integrating, we get 2x
e y = ___ + ex + 1007x2 + c1x + c2 4 which is the required solution. 216. The given differential equation is
d 2y ___2 = sin2x dx d dy fi ___ ___ = sin2x dx dx
( )
Integrating, we get
( ) Ú ( ) ( dy ___ = dx
1 __ 2 sin 2x dx + c1 2
dy 2sin2x 1 fi ___ = __ x – ______ + c1 2 2 dx
)
Again Integrating, we get
(
)
cos2x 1 x2 _____ y = __ __ + + c1x + c2 2 2 4
which is the required solution. 217. The given differential equation is d 2y ___2 = cos3x dx d dy fi ___ ___ = cos3x dx dx
( )
( ) ( )
dy ___ = Ú cos3x dx dx dy fi ___ = Ú cos3x dx dx
(
)
)
x2 1 cos3x y = __ + __ x + _____ + c1x + c2 2 3 3
(
)
which is the required solution. 218. The given differential equation is d 2y 1 ___2 = _________ 2 2 dx sin x cos x
( ) ( )
d dy sin2x + cos2x fi ___ ___ = ____________ dx dx sin2x cos2x
d dy fi ___ ___ = sec2x + cosec2x dx dx
Integrating, we get
y = log (cos x) – log (sin x) + c1x + c2
fi
y = log (cot x) + c1x + c2
which is the required solution. 219. The given differential equation is d2y ___2 = sin4x + cos4x dx = 1 – 2 sin2x cos2x 1 = 1 – __ sin2x 2 1 = 1 – __ (1 – cos4x) 4 3 __ 1 __ = + cos4x 4 4 3 1 d dy fi ___ ___ = __ + __ cos4x 4 4 dx dx
( )
Integrating, we get dy 3x 1 ___ = ___ + ___ sin4x + c1 4 dx 16 Again Integrating, we get
( )
3x2 cos (4x) y = ___ + _______ + c1x + c2 8 64
d 2y ___2 = x ex dx
sin3x y = Ú x – _____ dx + c1x + c2 3
fi
fi
which is the required solution. 220. The given differential equation is
sin3x = x – _____ + c1 3 Again Integrating, we get
(
Again Integrating, we get
Integrating, we get
( )
dy ___ = tan x – cot x + c1 dx
( )
d dy fi ___ ___ = x ex dy dx Integrating, we get
( )
dy fi ___ = (x – 1)ex + c1 dx Again integrating, we get fi
y = (x – 2)ex + c1x + c2
which is the required solution. 221. The given differential equation is d 2y ___2 + y = 0 dx dp fi p ___ + y = 0 dy
fi
dp p ___ = – y dy
fi
2p dp = – 2y dy
On integrating, we get
4.69
4.70 Integral Calculus, 3D Geometry & Vector Booster p2 = c 21 – y2
fi
dy 2 fi ___ = c 21 – y2 dx
_______ dy fi ___ = ± ÷( c21 – y2) dx
( )
( )
fi
dy 4p dp = ___ __ y ÷
1 ____ __ 4÷y
__
2p2 = 2÷y + 2c1
p2 = ÷y + c1
which is the required solution. 222. The given differential equation is 1 __ 3 y
dp p ___ = dy
Integrating, we get c1 2 sin– 1 __ y = c ± x
dy ___2 = dx
fi
Integrating, we get
dy _______ fi ________ = ± dx ÷( c21 – y2)
2
fi
__
( ) ( )
dy 2 __ fi ___ = ÷y + c1 dx _______ dy __ fi ___ = + ÷ ÷ y + c1 dx dy ___ fi ________ = ± dx __ ÷÷ y + c1
Again integrating, we get dy ___ c2 ± x = Ú ________ dy dp __ __ 1 ___ ___ fi p = y , Let p = + c1 ÷ ÷ y dy dx 2t dt dy _____ = Ú ______ , where y = t2 fi p dp = – ___3 ÷t + c1 y 2{(t + c1) – c1}dt Integrating, we get _____ = Ú _______________ ÷t + c1 2 c p 1 1 _____ __ = ___ 2 + __ = 2 Ú (÷t + c1 – c1(t + c1)– 1/2) dt 2 2 2y fi
(
c1y2 + 1 p2 = _______ y2
( )
dy 2 c1y2 + 1 fi ___ = _______ dx y2
÷
________ 2
c1y + 1 dy fi ___ = ± _______ dx y2
y ____ fi _____ + 1dy = ± dx ÷ c 1y2
)
_______
which is the required solution. 223. The given differential equation is d2y ___2 = dx
1 ____ __ 4÷y
( ) ( )
d dy fi ___ ___ = dx dx
dy d dy ___ fi ___ ___ = dy dx dx
1 ____ __ 4÷y 1 ____ __ 4÷y
__
÷
1
3/2
which is the required solution. 224. The given differential equation is d2y a2 ___ – y = 0 dx2
Again integrating, we get ÷c 1y2 + 1 ________ = c2 ± x c1
[ 1 = 4 [ __ ( y + c ) 3
d2y y fi ___2 = __ 2 dx a y dp fi p ___ = __ dy a2
y dy p dp = ____ 2 a Integrating, we get
fi
p2a2 = y2 + c 21
fi
y2 + c21 p2 = _______ 2 a
] – c ÷ y + c ]+ c
_____ 1 = 4 __ (t + c1)3/2 – c1÷t + c1 + c2 3
( )
y2 + c21 dy 2 _______ ___ fi = 2 dx a
1
_______ __ ÷ 1 2
Differential Equation
( ) ÷
______
y2 + c21 dy fi ___ = ± ______ dx a2
dy dx ______ fi _______ = ± ___ 2 2 2 a ÷y + c 1
|
_______
|
fi
2p dp = 3y2 dy
p2 = y3 + c1
( ) ( )
______ dy fi ___ = ± ÷y 3 + c1 dx
When
\
( )
dy x = – 2, y = – 1 and ___ = – 1 dx x = – 2 c1 = 2
( )
_____ dy Thus, ___ = ± ÷y 3 + 2 dx
dy _____ fi _______ = ± dx ÷ y 3 + 2
c1 p2 e2y __ __ = ___ + 2 2 2
227. The given differential equation is
p2 = e2y c1
( ) ( )
dy 2 fi ___ = e2y + c1 dx _______ dy fi ___ = ± ÷e 2y + c1 dx
( )
d2y y3 ___ 2 = – 1 dx
d2y 1 fi ___2 = – __3 dx y
dy _______ fi ________ = ± dx 2y ÷e + c1
Again integrating, we get
dy 2 fi ___ = y3 + c1 dx
d2y ___2 = e2y dx dp fi p ___ = e2y dy fi p dp = e2y dy Integrating, we get
dp 2p ___ = 3y2 dy
x log y + ÷y 2 + c21 = c2 ± __ 2 a which is the required solution. 225. The given differential equation is
fi
fi
Integrating, we get
Again integrating, we get
fi
dp 1 2p ___ = – __3 dy y
dy 2p dp = – ___3 y Integrating, we get 1 p2 = – ___ 2 + c1 2y
dy _______ c2 ± x = Ú ________ ÷e 2y + c1
fi
e– ydy _________ = Ú __________ ÷1 + c1e– 2y
dt _______ = – Ú ________ , t = e– y 2 ÷1 + c1t
dy 2 1 fi ___ = – ___ 2 + c1 dx 2y
dt 1__ __________ = – ___ Ú _________ c 1 2 ÷ 1__ 2 ___ t + ÷c 1
dy 2 2c1y2 – 1 fi ___ = _________ dx 2y2
÷ ( ) | ÷ | ÷
_______
|
When
1__ 1 = – ___ log t + t2 + ___ __ ÷c 1 ÷c 1
1__ 1 = – ___ log e– y + e– 2y + ___ __ ÷c 1 ÷c 1
_________
226. The given differential equation is
d 2y 2 ___2 = 3y2 dx
( ) ( )
\
|
( )
dy x = – 1, y = – 1, ___ = 0 dx x = –1 1 c1 = __ 2
( )
÷
______
dy y2 – 1 \ ___ = ± ______ 2 dx 2y
y dy dx _____ fi _______ = ± ___ 2 ÷ y – 1 2
4.71
4.72 Integral Calculus, 3D Geometry & Vector Booster
_____
x fi ÷y 2 – 1 = c2 ± __ 2
Again integrating, we get
which is the required solution. 228. The given differential equation is
d2y x ___2 + dx
dy ___ + x = 0 dx
dy d2y Let ___ = p fi ___ 2 = dx dx
fi
dp x ___ + p + x = 0 dx x dp + p dx + x dx = 0
fi
d (xp) + xdx = 0
fi
Integrating, we get
)
x2 y = – __ + c1 log |x| + c2 4 which is the required solution. 229. The given differential equation is
2
d y 1 dy x ___ + x ___2 = __ dx dx dp p + x fi ___ = __ dx x
fi
x dp = p dx + x2dx
x dp – p dx fi __________ = dx x2 p fi d __ x = dx Integrating, we get p __ x2 __ = + c1 x 2
( )
3
x p = __ + c1x 2 dy x3 fi ___ = __ + c1x 2 dx fi
fi
dp y p ___ + p2 = 1 dy
(
dy 2 ___ = 1 dx
dp y p ___ = 1 – p2 dy dy p fi _____ 2 dp = ___ y 1–p fi
1 – __ log |1 – p2| = log y + log c1 2
1 fi _____ = (c1y)2 1 – p2
fi
1 1 – p2 = ____ 2 2 c1 y
fi
1 p2 – 1 = ____ 2 2 c1 y
dy 1 fi ___ = 1 – ____ 2 2 dx c 1 y
c1y _______ fi ________ dy = dx 2 2 ÷c 1 y – 1
÷
_______
Again integrating, we get _______
fi
d2y y ___2 + dx
fi
( )
Integrating, we get
x x p = – __ + c1 2
(
dp ___ dx
)
c1x2 ____ + c2 2
which is the required solution. 230. The given differential equation is
2
dy x2 x ___ = – __ + c1 2 dx dy x c1 x fi ___ = – __ + __ 2 dx x c1 x dx fi dy = – __ + __ 2 Again Integrating, we get
...(i)
(
x4 y = __ + 6
÷c 21 y2 – 1 _________ = x + c2 c21 which is the required solution. 231. The given differential equation is
fi
fi
dp y ___ – p = y2 dy
fi
y dp – p dy = y2dy
)
x3 dy = __ + c1x dx 2
( )
( )
d2y dy 2 dy y ___2 – ___ = y2 ___ dx dx dx dp py ___ – p2 = y2p dy
ydp – pdy fi _________ = dy y2 p fi d __ y = dy
( )
Differential Equation
Integrating, we get p __ y = y + c1 fi p = y2 + c1y
dy fi ___ = y2 + c1y dx
dy fi _______ = dx 2 y + c1y
dy fi ________ = dx y(y + c1)
(
Integrating, we get
( )
x log |x2 + c1| + 2a tan– 1 __ c + y = c2 1
which is the reqired solution. 233. Given differential equation is
)
1 1 fi __ y – _____ dy = dx y + c1
Again integrating, we get y log _____ = x + c2 y + c1
| |
which is the required solution. 232. Given differential equation is
( ) ( )
( ) ( )
d2y 1 ___ dy dy x ___2 – __ 2 = ___ 4 dx dx dx
fi
dp 1 x ___ – __ p2 = p dx 4
fi
dp p2 p x ___ = __ + p = p __ + 1 4 4 dx
4 dp dx fi _______ = ___ x p(p + 4)
dx 1 1 fi __ p – _____ dp = ___ x p+4
(
(
)
Integrating, we get
| |
d 2y dy 2 dy (x + a) ___2 + x ___ = ___ dx dx dx
fi
dp (x + a) ___ + xp2 = p dx
p fi _____ = c1x p+4
p log _____ = log |x| + log c1 p+4
p dp xp2 fi ___ + ______ = _____ dx (x + a) x + a
p + 4 ___ 1 fi _____ = c x p 1
dp p xp2 fi ___ – _____ = – ______ (x + a) dx x + a
fi
x 1 dp ________ 1 fi __2 ___ – = – ______ (x + a) p dx p/(x + a)
4 ______ 1 fi __ p = c1x – 1
– c1x 4 1______ fi __ p = c1x
c1x p fi __ = ______ 4 1 – c1x
fi
Put
1 – __ p = v
dv v x fi ___ + ______ = ______ dx (x + a) (x + a) which is a linear differential equation. \
dx _____ IF = eÚ x + a = elog |x + a| = (x + a)
fi
v (x + a) = Ú x dx
fi
x2 v(x + a) = _____ 2+c
fi
x2 1 __ – __ p (x + a) = 2 + c
2 (x + a) fi ________ dx + dy = 0 x2 + C1 2x 2a fi ______ dx + ______ 2 dx + dy = 0 x2 + C1 x + C1
)
4 ___ 1 1 + __ p = c1x
4c1x dy = ______ dx 1 – c1x
Again integrating, we get 1 y = – 4x + __ c log |c1x – 1| + c2 1 which is the required solution. 234. Given differential equation is fi
( )
d2y dy 2 ___ y ____ = 1 – dx dx2 dp py ___ = 1 – p2 dx
dy 2p dp fi _______ = – 2 ___ y 2 (p – 1)
4.73
4.74 Integral Calculus, 3D Geometry & Vector Booster Integrating, we get
fi
fi
fi
2
log |p – 1| = log C1 – log |y |
| |
C1 log |p – 1| = log ___ 2 y C 1 (p2 – 1) = ___ 2 y C1 p2 = ___ 2 + 1 y 2
C1 + y2 = ______ y2
÷ ÷
_______
C1 + y2 p = ± _______ 2 y
fi
C1 + y2 dy fi ___ = ± ______ dx y2
y dy ______ fi ________ = ± dx 2 ÷y + C1
fi ÷y 2 + C1 = C2 ± x
_______
______
which is the required solution. 235. Given differential equation is
( )
d 2y dy ___2 = 2y ___ dx dx
2
fi
dp y p ___ – p2 = y2p dy
fi
dp y ___ – p = y2 dy
dp fi ___ – dy
p __ y = y
...(i)
which is a linear differential equation. \
dy
– ___ 1 IF = e Ú y = e– log y = __ y
Multiplying both sides of Eq. (i) by IF and integrating, we get p __ y = Ú dy + c1 p __ y = y + c1 p = y2 + c1y dy ___ = y2 + c1y dx dy ________ = dx y( y + c1) Integrating, we get
| |
y 1 _____ __ | = x + c2 c1 log | y + c1 which is the required solution. 237. Given differential equation is
( ) ( )
d2y 1 ___ dy a ___2 + __ x – __ 2 = 0 dx dx x
dp fi ___ = 2y p dx dp fi p ___ = 2y p dy
dp fi ___ = 2y dy
fi
fi
fi
dy d2y x2 ___ + x ___ = a 2 dx dx dp x2 ___ + xp = a dx a dx x dp + p dx = ____ x
fi
a dx d (xp) = ____ x
fi
dp = 2y dy
Integrating, we get p = y2 + C1
dy fi ___ = y2 + C1 dx
Integrating, we get xp = a log |x| + c1
dy fi ________ = dx 2 (y + C1)
fi
dy x ___ = a log |x| + c1 dx
y 1 – 1 ____ ___ fi ____ = x + C2 tan ___ ÷C 1 ÷C 1
fi
c1 a log |x| ___ dy = ______ x + x dx
( )
which is the required solution. 236. Given differential equation is fi
d2y y ___2 – dx
( )
( )
dy 2 dy ___ = y2 ___ dx dx
(
)
Again integrating, we get
(log |x|)2 y = a _______ + C1 log |x| + C2 2
which is the required solution.
Differential Equation
238. Given differential equation is
( )
2
dy dy y ___2 – y ___ ln y = dx dx dp fi y ___ – yp ln y = p2 dx lny 1 1 dp ___ fi __2 ___ + – p = __ y p dx
fi
240. Given
( )
dy ___ = y + 2x dx dy fi ___ + (– y) = 2x dx
dy ___ dx 2
which is a linear differential equation.
fi
dp fi ___ – dy
p __ y = ln y
...(i)
dy
___ 1 IF = e – Ú y = e– log|y = __ y
Multiplying both sides of Eq. (i) by IF and integrating, we get log y p ____ __ dy + c1 y = Ú y
(log y)2 p ______ fi __ = + c1 y 2 dy fi ______________ = dx 2 (log y) + 2c1 y ____________ 2
(
Integrating, we get
÷
(Let log y = t)
( )
( )
log y 2 fi ___ tan– 1 _____ ____ = x + c2 c1 ÷2c 1
which is the required solution. 239. The slope of the curve = The slope of the tangent
dy = tany = ___ = dx
fi
2y dy = dx
y2 = x + c
fi
fi
y ◊ e– x = – 2e– x (x + 1) + c
241. Given
y ◊ e– x = – 2e– x (x + 1)
dy x4 + 2x y – 1 ___ = __________ dx x2 + 1 2xy = x2 – 1 + _____ 2 x +1 dy 2x fi ___ – _____ y = x2 – 1 dx x2 + 1
t 2 – 1 _____ ___ fi ____ = x + c2 tan ____ 1 ÷2c 1 ÷ 2c ___
y ◊ e– x = 2 Ú xe– x dx + c
which is a linear differential equation.
)
2 dt fi ________ = dx, (t2 + 2c1)
which is passing through (0, 0) then c = 0. Hence, the required equation of the curve is
which is a linear differential equation. \
IF = e – Údx = e– x
Hence, the solution is
dp y p ___ – y p ln y = p2 dy
dp y ___ – y ln y = p dy dp p fi ___ – ln y = __ y dy
4.75
1 ___ 2y
which is passing through (4, 3), so c = 5 Hence the equation of the curve is y2 = x + 5.
2x – Ú _____ dx x2 + 1 =
IF = e
2
e – log |x
1 = _____ 2 x +1
+ 1|
Hence, the solution is 1 y ◊ _____ = x – 2 tan– 1x + c 2 x +1 which passes through (0, 0), then c = 0. Hence, the equation of the curve is
y = (x2 + 1) (x – 2 tan– 1x)
242. Given dy (x + 1)2 + y – 3 ___ = _____________ (x + 1) dx
Let
dy fi ___ = dx
fi
...(i)
x + 1 = X, y – 3 = Y dY ___ dX
dY Y ___ = X + __ X dX
( )
dY 1 fi ___ + – __ Y = X X dX which is a linear differential equation. \
1 __ 1 IF = e – Ú x dX = e– log X = __ X
4.76 Integral Calculus, 3D Geometry & Vector Booster Hence, the solution is
( )
1 1 Y ◊ __ = Ú X ◊ __ dX + c = X + c X X
Y fi __ = X + c X y–3 fi _____ = (x + 1) + c x+1 which is passing through (2, 0) then c = – 4 Hence, the required equation of the curve is y–3 _____ = (x + 1) – 4 x+1 fi
y – 3 = (x + 1)2 – 4(x + 1)
fi
y = x2 – 2x
243. The equation of the tangent at any point (x, y) on the curve is dy Y – y = ___ (X – x) dx dy It meets x-axis at A x – y ___ , 0 dx
(
(
)
)
dy and y-axis at B 0, y – x ___ dx \ Mid-point of AB
( (
) (
))
dy 1 dy 1 = __ x – y ___ , __ y – x ___ 2 dx 2 dx
It is given that,
( (
) )
dy 1 and __ x – y ___ = x, 2 dx dy 1 __ y – x ___ = y 2 dx dy Thus, x ___ = – y dx dy dx ___ fi ___ y = – x dy ___ dx fi ___ y + x = 0 fi log y + log x = log c fi
log (y x) log c
fi x y = c which is passing through (1, 1), then c = 1. Hence, the equation of the curve is
x y = 1.
244. The equation of the normal at the point (x, y) is dx Y – y = – ___ (X – x) ...(i) dy The distance of perpendicular from the origin to the normal (i) is
|
|
dy y + x ___ dx ________ = _________ dy 2 ___ 1 + dx
÷ ( )
Also the distance between P and x-axis is |y|.
|
|
dy y + x ___ dx ________ Thus, _________ = |y| dy 2 ___ 1 + dx
÷ ( ) ( ) ( ) ( ( ) ( )
( ( ) )
fi
dy dy 2 dx 2 y2 + ___ x2 + 2xy ___ = y2 1 + ___ dy dx dx
fi
dx 2 dx (x2 – y2) ___ + 2xy ___ = 0 dy dy
)
dx dx fi ___ (x2 – y2) ___ + 2xy = 0 dy dy y2 – x2 dx dx fi ___ = 0, ___ = ______ 2xy dy dy dx Now, ___ = 0 gives x = k dy which is passing through (1, 1), so k = 1. Thus, the equation of the curve is x = 1 y2 – x2 dx Also, ___ = 0 = ______ 2xy dy which is a homogeneous differential equation. dy dv Let y = v x fi ___ = v + x ___ dx dx fi
dv v2 – 1 v + x ___ = _____ 2v dx
fi
dv v2 – 1 v2 + 1 x ___ = _____ – v = – _____ 2v 2v dx
2v dx fi ___2 + 1dv = – ___ x v fi log |v2 + 1| = log c – log x c fi |v2 + 1| = __ x fi
x2 + y2 = c x
which also passes through (1, 1), so c = 2. Hence, the equation of the curve is x2 + y2 = 2x. 245. The Equation of the tangent at any point (x, y) is dy Y – y = ___ (X – x) dx dy It meets the x-axis at x – y ___ , 0 dx
(
Given
AP = 1
fi
AP 2 = 1
)
Differential Equation
( ) ( )
dx 2 fi – y ___ + y2 = 1 dy
2v dx fi _________ dv = ___ x 2 1 + v – 2v
2 dx 2 1 – y fi ___ = _____ dy y2
2v dx fi _________ dv = – ___ x 2v2 – v – 1
dy y _____ fi ___ = ± ______ dx ÷1 – y2 ______
÷ 1 – y2 fi _______ dy = ± dx y
|
Integrating, we get
|
______ y _____ log __________ + ÷ 1 – y2 = c ± x 1 + ÷1 – y2
which is the required equation of the curve. 246. The slope of the line segment joining the points (x, y) and (– 4, – 3) is y+3 _____ x+4
( )
dy y+3 Thus, ___ = 2 _____ x+4 dx
(4v – 1)dv dv 2dx fi ___________ + ___________ = – ____ x (2v2 – v – 1) (2v2 – v – 1)
|
|
2v – 2 1 fi log |2v2 – v – 1| + __ log ______ = c – log x 3 2v – 1
|
| | |
2y2 fi log ___ 2 – x
2y – 2x 1 fi log |2y2 – xy – x2| + __ log _______ – log x = c 3 2y – x
log |y + 3| = log c + 2 log (x + 4)
fi
( y + 3) = c (x + 4)2
which is passing through (– 2, 1), so c = 1 Hence, the required equation of the curve is 2
( y + 3) = (x + 4)
247. The equation of the normal at P is dx Y – y = ___ (X – x) dy Thus, the points A and B are (x, 0) and
(
(4v – 1) + 1 2dx fi ___________ dv = – ____ x 2v2 – v – 1
y 2y – 2x 1 __ x – 1 + __ log ______ = c – log x 3 2y – x
| |
which is the required equation of the curve. 248. The equation of the normal at P is
dy 2 ◊ dx fi _____ = _____ y+3 x+4 Integrating, we get
4.77
)
dy x + y ___ , 0 dx Given condition is
dy x+y x + y ___ – x = _____ 2 dx
fi
dv 1 + v 1 + v – 2v2 x ___ = _____ – v = __________ 2v 2v dx
dx Y – y = – ___ (X – x) dy
(
)
dy It meets the x-axis at A x + y ___ , 0 dx Given condition is
OP2 = OA2
( )
dy 2 x2 + y2 = y2 ___ + y2 dx dy x fi ___ = ± __y dx fi
Taking positive sign, dy x ___ = __ dx y fi
y dy = x dx
x2 fi __ – 2
y2 c2 __ = __ 2 2
fi x2 – y2 = c2, which represents a rectangular hyperbola. dy x + y Taking negative sign, ___ = _____ 2y dx dy x ___ = – __y which is a homogeneous equation. dx dy dv y dy = – x dx Put y = vx fi ___ = v + x ___ fi dx dx y2 __ x2 __ a2 __ dv 1_____ +v fi + = ___ fi v + x = 2 2 2 2v dx fi
x2 + y2 = a2.
which represents a circle.
4.78 Integral Calculus, 3D Geometry & Vector Booster
(Problems for JEE-Advanced)
when
1. It is given that
2
e – log (x
1 = _______ 2 (x + 1)
+ 1)
Hence, the sloution is
which is a linear differential equation. 4
1 x – 1 y ◊ _______ = Ú ________ dx + c 2 (x + 1) (x2 + 1)2
dy cos2 x ___ – (tan2x) y = cos4 x dx dy (tan2x) fi ___ – ______ y = cos2 x dx cos2 x
which is a linear differential equation. – Ú_____ 2x dx 1 + x2 =
p p x = 1, y = __ then c = __ 2 2
( )
dy 2x x4 – 1 fi ___ – _____ 2 y = ______ dx 1 + x 1 + x2
IF = e
( )
y (x y) sin __ x = c
Hence, the solution is y p (x y) sin __x = __ 2 3. The given differential equation is
dy x4 + 2xy – 1 ___ = ___________ dx 1 + x2
\
fi
\
(
(
)
fi
– 1
) (
)
cos 2x cos 2x y ◊ _________ = Ú _________ cos 2x dx + c 1 + cos 2x 1 + cos 2x
2 = Ú 1 – _____ 2 dx + c x +1
cos 2x _________ 1 + cos 2x
Hence, the solution is
y x2 – 1 fi _______ = Ú _______ dx + c 2 (x + 1) (x2 + 1)
tan(2x) – Ú ______ dx cos2 x =
IF = e
= (x – 2 tan x) + c
which is the required solution. 2. The given differential equation is y y y cos __ x (x dy – y dx) + x sin __ x (x dy + y dx) = 0
(
)
cos 2x 1 y ◊ _________ = __ Ú cos 2x dx + c 1 + cos 2x 2 sin 2x = _____ 4+c
__
3÷3 p when x = __ , y = ____ then c = 0 8 6 Hence, the required curve is
( )
( ) y y dy __y y dy __y fi __ x cos ( __ x ) ( ___ – x ) + sin ( __x ) ( ___ + x ) = 0 dx dx
1 y = __ ◊ tan (2x) ◊ cos 2x 2 dv dv ___ ___ 4. The given differential equation is fi v cos v v + x – v + sin v v + x + v = 0 dx dx d y + ___ (x y) = x (sin x + log x) y dx __ Let x = v dy dv dv fi x ___ + 2y = x sin x + x log x fi xv cos v ___ + sin v 2v + x ___ = 0 dx dx dx dy 2y dv fi ___ + ___ x = sin x + log x fi x (v cos v + sin v) ___ = – 2v sin v dx dx
(
)
(
( )
)
(
dv 2v sin v fi x ___ = – _____________ dx (v cos v + sin v)
(v cos v + sin v) 2dx fi _____________ dv = – ____ x v sin v
2dx 1 fi cot v + __ v dv = – ____ x
(
)
Integrating, we get
log |sin v| + log |sin |v| = log c – 2 log x
c fi v sin v = __ 2 x y__ y__ c fi x sin x = __ 2 x
( ) ( )
)
which is a linear differential equation. \
2dx ____ IF = e Ú x = x2
Hence, the solution is
y ◊ x2 = Ú (x2 sin x + x2log x) dx + c
fi
2sin x cos x y = – cos x + _____ x + 2 _____ x2 x x c + __ log x – __ + __ . 3 9 x2
5. The given differential equation is
x (1 – x2) dy + (2 x 2 y – y – 5 x3) dx = 0
Differential Equation
dy (2x 2 – 1) y _______ 5 x3 fi ___ + __________ = 2 dx x (1 – x ) x (1 – x2)
dy fi ___ – dx
2
fi
dy (2x – 1) y 5 x2 fi ___ + __________ = _______ 2 dx x(1 – x ) (1 – x2)
which is a linear differential equation \
IF =
(2 x2 – 1) Ú ________ dx 2 e x(1 – x ) =
Hence, the solution is
1 _______ _____ x÷1 – x2
5x 1 _____ y ◊ _______ = Ú ________ dx + c 2 3/2 2 (1 – x ) x÷1 – x
y 5 _____ fi _______ = ______ _____ + c 2 x÷1 – x ÷1 – x2
6. The equation of the tangent at P (x, y) is dy Y – y = ___ (X – x) dx
(
It cuts the line y = x, so the point of intersection
)
dy dy y – x ___ y – x ___ dx dx = _______ , ________ dy dy ___ ___ 1 – 1 – dx dx
fi
dy dy y – x ___ = 1 – ___ dx dx
fi
dy (1 – x) ___ = (1 – y) dx
dy dx fi ______ = ______ (y – 1) (x – 1)
Integrating, we get
log |x – 1| = log |y – 1| + log c
fi
x–1 log _____ = log c y–1
x–1 fi _____ = c y–1
| |
( )
which is the required equation of the curve. 7. The given differential equation is
fi
_______
÷ ( )
y y2 __ x = 1 + __ x
_____ y dv v + x ___ – v = ÷1 + v2 , Let v = __ x dx _____ dv fi x ___ = ÷ 1 + v2 dx dv dx ______ fi _______ = ___ x 2 + v ÷ 1
fi
|
_____
(
_____
|
fi ( v + ÷ 1 + v2 ) = c x
fi y + ÷ y 2 + x2 = c x2
(
______
)
which is the required solution. 8. The given differential equation is y y x e y/x – y sin __ x dx + x sin __ x dx = 0
(
( ) ) ( ) y y __ x sin ( __ x ) – e dy fi ___ = ___________ y dx __ sin ( x ) dv v sin – e fi v + ___ = _______ , ( Let dx sin v y __x
v
fi
dv v sin – ev ___ = _______ –v dx sin v
fi
dv v sin – ev – v sin v ___ = _______________ dx sin v
( ) )
y v = __x
fi e– vsin v dv = – dv Integrating, we get e– vsin v _______ (– sin v – cos v) = – x – c (1 + 1)
e– vsin v fi _______ (sin v + cos v) = x + c (1 + 1)
( ) ( ()
y __y e – x sin __x y y _________ fi sin __ x + cos __x = x + c 2 which is the required solution. 9. The given differential equation is
( ) )
(x + 2y) (dx – dy) = dx + dy
fi
(x + 2y – 1) dx = (x + 2y + 1) dy
dy fi ___ = dx
(x + 2y – 1) ___________ (x + 2y + 1)
...(i)
Let
x + 2y = v
fi
dy dv 1 + 2 ___ = ___ dx dx
______
dy x ___ = – y = ÷x 2 + y2 dx
)
log v + ÷1 + v2 = log c + log |x|
It is given that dy y – x ___ dx ________ =1 dy 1 – ___ dx
4.79
4.80 Integral Calculus, 3D Geometry & Vector Booster
(
(
)
v+1 1 dv __ ___ – 1 = _____ 2 dx v–1
dv fi ___ = dx
2v – 2 ______ +1 v+1
dv 3v – 1 fi ___ = ______ v+1 dx
v + 1 fi ______ dv = dx 3v – 1
fi
fi
(x + 2y) + 4 log |3x + 6y – 1| = 3x + c
fi
2 (y – x) + 4 log |3x + 6y – 1| = c
which is the required solution. 10. The given differential equation is
x2 = v dx dv 2x ___ = ___ dy dy
(
)
dy
1 – ___ IF = e Ú y = e– log y = __ y
( )
1 1 __ v ◊ __ y = – Ú 1 + y2 dy + c
( )
v 1 fi __y = y – __ y + c
x2 1 fi __ y = – y – y + c
x2 + 1 fi ______ + y = c y
(
)
which is the required solution. 12. The given differential equation is
– 1
(1 + y ) dx = (tan y – x) dy.
tan– 1y dx x fi ___ + _____ 2 = ______ dy 1 + y 1 + y2 dy
Ú _____ 2 IF = e 1 + y = etan– 1y
x dy – y dx = x y3 (1 + log x) dx
x dy – y dx fi _________ = x y (1 + log x) dx y2
fi
which is a linear differential equation.
( )
x __x 2 2 – __ y d y = x dx + x log x dx
Integrating, we get
Hence, the solution is tan– 1y etan– 1y y = Ú ___________ dy + c 1 + y2
tan– 1
x ◊ e
)
Hence, the solution is
v + 4 log |3v – 1| = 3x + c
(
x2 1 __ y = – y + __ y Let
\
fi
\
dx 2x ___ – dy
dv v 1 fi ___ – __ = – y + __ y dy y which is a linear differential equation.
2
fi
( ) 3v – 1 + 4 fi ( _________ dv = 3 dx 3v – 1 ) 3 fi ( 1 + ______ dv = 3 dx 3v – 1 )
)
dy dv fi ___ = 1/2 ___ – 1 dx dx
= Ú t e dt + c, t = tan y
= (t – 1) e t + c
= (tan– 1 y – 1) etan– 1y + c
t
– 1
fi
( ) __ x3 + __ x3 log x – __ x9 + C 2x x 1 x – __ ( __ = ___ + __ log x + C 2 y) 3 3 1 x2 – __ __ = 2 y 2
3
3
3
3
3
which is the required solution. 13. The given differential equation is
sec2x ◊ tan y dx + sec2y ◊ tan x dy = 0
which is the required solution. 11. The given differential equation is
sec2 y sec2 x fi _____ dx + _____ dy = 0 tan x tan y
dy 2x y ___ = _________ 2 dx x – 2y – 1
Integrating, we get log tan x + log tan y = log c
2 2 dx x – y – 1 fi ___ = __________ 2x y dy
fi
dx 2x y ___ = x2 – y2 – 1 dy
fi
dx 2x y ___ – x2 = – (y2 + 1) dy
fi (tan x tan y) = c which is the required solution. 14. The given differential equation is dy x+y–1 ________ ___ = _________ dx ÷x + y + 1 Let
x + y + 1 = v2
...(i)
Differential Equation
( )
c dy dv log |v + ev| = log __ y 1 + ___ = 2v ___ fi dx dx c fi (v + ev) = __ y dy dv ___ ___ fi = 2v – 1 dx dx x c fi __ y + ex/y = __ y fi
(
2
dv v –2 fi 2v ___ –1 = _____ v dx
dv v2 – 2 + v fi ___ = _________ dx 2v2
fi
2v2 fi _________ 2 dv = dx v + v – 2
– 1
( (
)
which is a linear differential equation. dy
)
5 dx 1 (2v + 1) 1 fi 1 – __ ________ + __ ◊ ___________ dv = ___ 2 v2 + v – 2 2 (v + 2) (v – 1) 2
| |
v–1 15 x 1 fi v – __ log|v2 + v – 2| + ___ log _____ = __ + C 2 2 v+2 2 ________
_________
1 fi ÷x + y + 1 – __ log |(x + y + 1) + ÷ (x + y + 1) – 2| 2
|
_________
÷(x + y + 1) – 1 15 x _________ = __ fi + ___ log ______________ + C 2 2 ÷(x + y + 1) + 2 which is the required solution. 15. The given differential equation is
(
)
x (1 + ex/y) dx + 1 – __ y ex/y dy = 0
(
)
x __ y – 1 ex/y dx fi ___ = __________ dy (1 + ex/y) fi fi
dv (v – 1) e x v + y ___ = ________ , v = __ y dy 1 + ev v
dv (v – 1) e y ___ = ________ – v dy 1 + ev v
v
ve – e – v – ve = _______________ 1 + ev ev + v = – _____v 1+e dy 1 + ev fi _____v dy = – ___ y v+e
Integrating, we get
\
log |v + ev| = log c – log y
2 Ú _______
– 1
IF = e (1 + y ) = e tan y
Hence, the solution is
( etan y ) = Ú _______ dy +c 1 + y2 2
– 1
tan– 1y
x ◊ e
( etan y ) fi x ◊ e = _______ + C 2 which is the required solution. 18. The given differential equation is – 1
tan– 1y
2
y dx – x (1 + xy) dy = 0
fi
y dx – x dy = x2y dy
y dx – x dy fi _________ = y dy x2 y fi – d __ x = y dy fi
v
v
– 1 dx (1 + y2) ___ + (x – e tan y) = 0 dy
etan y dx x fi ___ + _______ 2 = _______ dy (1 + y ) (1 + y2)
dx 1 (2v + 1) – 5 fi 1 – __ ___________ dv = ___ 2 v2 + v – 2 2
)
which is the required solution. 16. Do yourself. 17. The given differential equation is – 1 dy (1 + y2) + (x – etan y) ___ = 0 dx
dv v2 – 2 v2 – 2 + v fi 2v ___ = ______ + 1 = _________ v v dx
|
4.81
( ) x d ( __ y ) + y dy = 0
Integrating, we get y y2 __ x + __ = c 2
( )
which is the required solution. 19. It is given that dy x+y y ___ = – _____ = – 1 – __ x x dx
(
fi fi
)
dv v + x ___ = – 1 – v, dx dv x ___ = – 1 – 2v dx
dv dx fi ______ = – ___ x 2v + 1
(Let v = y/x)
4.82 Integral Calculus, 3D Geometry & Vector Booster
( ) ( )
dy 2 fi ___ = dx
Integrating, we get
log |2v + 1| + log (x2) = log c
fi
(2v + 1) x2 = c
(
x2 __ 2 y
dy x fi ___ = ± __y dx
)
2y 2 fi ___ x + 1 x = c fi x2 + 2xy = c which is passing thorugh the point (2, 1), so, c = 8 Hence, the equation of the curve is
fi
y dy
fi
x dx ± y dy = 0
x dx = 0
Integrating, we get
20. The equation of tangent at (x, y) is dy Y – y = ___ (X – x) dx
x2 ± y2 = a2 which is the required solution. 22. The given differential equation is dy 2 dy ___ – (ex + e– x) ___ + 1 = 0 dx dx
Putting
fi
p2 – (ex + e– x) p + 1 = 0
fi
(ex + e– x) ± ÷(e x + e– x)2 – 4 p = ________________________ 2
x2 + 2xy = 8
( )
X = 0, we get
_____________
dy Y = y – x ___ dx It is given that,
_________
dy 2 y – x ___ = xy dx
(ex + e– x) ± ÷(e x – e– x)2 = ____________________ 2
dy ___ fi y – x ___ = ÷ x y dx
(ex + e– x) ± (ex – e– x) = ___________________ 2
= ex, e– x
( (
fi
) )
dy ___ x ___ = y – ÷x y dx
dy fi ___ = dx
y __ x –
dy dy fi ___ = ex, ___ = e– x dx dx Integrating, we get
__
÷
y __ x
(
)
__ dv __y fi v + x ___ = v – ÷ v , Let v = x dx __ dv fi x ___ = – ÷v dx dv__ dx fi ___ = – ___ x ÷v
Integrating, we get __
fi
2 ÷v = c – log |x|
fi
y 2 __ x = c – log |x|
__
÷
which is the required equation of the curve. 21. It is given that
÷ ( ) ( ( ) ) ( ( ) ) ________
______ dy 2 y 1 + ___ = ÷x 2 + y2 dx
fi
dy 2 y2 1 + ___ = (x2 + y2) dx
dy 2 (x2 + y2) fi 1 + ___ = _______ dx y2
x2 = __ 2 + 1 y
y = ex + c1, y + ex = c2
fi y – ex – c1 = 0, y + ex – c2 = 0 Hence, the solution is
(y – ex – c1) (y + ex – c2) = 0
23. The given differential equation is
( )
dy 2 dy ___ + 2x ___ = 3x2 dx dx
fi
p2 + 2px – 3x2 = 0
fi
(p + 3x) (p – x) = 0
fi
(p + 3x) = 0, (p – x) = 0
(
) (
)
dy dy fi ___ + 3x = 0, ___ – x = 0 dx dx fi dy = – 3x dx, dy = x dx Integrating, we get 3x2 x2 y + ___ – c1 = 0, y – __ – c2 = 0 2 2 Hence, the solution is
(
) (
)
3x2 x2 y + ___ – c1 y – __ – c2 = 0 2 2
Differential Equation
24. The given differential equation is
{ ( ) } 2
dy dy xy ___ – 1 = (x2 – y2) ___ dx dx
fi
xy (p2 – 1) = (x2 – y2) p
fi
xy p2 – xy = x2 p – y2 p
fi
xy p2 – x2 p – xy + y2 p = 0
fi
xp (yp – p) + y (yp – x) = 0
fi
(xp + y) (yp – x) = 0
fi
(xp + y) = 0, (yp – x) = 0
dy dy x ___ + y = 0, y ___ = x dx dx dy ___ dx fi ___ y + x = 0, y dy – x dx = 0 fi
which is passing through (– 2, 1), so, c = 1 Hence, the equation of the curve is (y + 3) = (x + 4)2. 27. It is given that dy ___ = x + x y dx dy ___ – x y = x dx which is a linear differential equation. fi
\
x2 – __
IF = e – Ú x dx = e 2
Hence, the solution is x2 – __
x2 – __
y ◊ e 2 = Ú x ◊ e 2 dx + c
dy ___ dx fi ___ y + x = 0, x dx – y dy = 0
fi
y ◊ e 2 = – e 2 + c
Integrating, we get
which is passing through (0, 1), so c = 2 Hence, the equation of the curve is
log |y| + log |x| = log c, x2 – y2 = a2
fi xy = c, x2 – y2 = a2 Hence, the solution is
(xy – c) (x2 – y2 – a2) = 0.
25. It is given that dy y ___ = x dx fi
x2 – y2 = c which is passing through (0, – 2), so c = – 4 Hence, the equation of the curve is
x2 – y2 = – 4
dy 26. Clearly, the slope of the tangent is ___ dx Also, the slope of the line segment joining the points (x, y) and (– 4, – 3) is y+3 _____ . x+4 It is given that dy y+3 ___ = 2 _____ x+4 dx
( )
x2 – __
x2 – __
x2 – __
y ◊ e 2 = – e 2 + 2
fi
y = 2e2 – 1
x2 __
28. The given differential equation is
x dx – y dy = 0
Integrating, we get
x2 – __
( )
d2y y3 ___ 2 – y = 1 dx
d2y 1 fi ___2 = y + __ 3 dx y
( )
d dy 1 fi ___ ___ = y + __ 3 dx dx y Integrating, we get
( )
y2 __ – 2
dy ___ = dx
As
y = 1, dy/dx = 0, so c = 0
( ) (
dx fi ___ = dy
y2 dy fi _____ = y4 – 1
Integrating, we get log |y + 3| = 2 log |x + 4| + log c
fi
)
1 1 __ y2 – __ 2 + c 2 y
)
dx ___ 2
2y2 dy fi _____ = d x y4 – 1
(y + 3) = (x + 4)2
1 ___ 2 = 2y
1 1 __ y2 – __ 2 2 y
dy 2 dx fi ______ = ______ (y + 3) (x + 4)
fi
(
2y2 dy _____________ 2 = d x (y – 1) (y2 + 1)
4.83
4.84 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 1 fi _____ 2 + _____ dy = d x y – 1 y2 + 1
p x = 0, y = 1, c1 = __ 4 Hence, the curve is y–1 p 1 __ log _____ + tan– 1y = x + __ 2 y+1 4
Integrating, we get 1 __ log |2v + 1| + log|v – 1| = log c – 3 log x 2 y y c 1 fi __ log 2 __ x + 1 + log __ x – 1 = log __ 3 2 x ________ y__ y__ c fi x – 1 2 x + 1 = __ 3 x which is the required equation of the curve. 30. The equation of the normal at P (x, y) is dx Y – y = – ___ (X – x) dy
29. The equation of the normal at P (x, y) is
Putting
Y = 0, we get
dx – y = – ___ (X – x) dy
Integrating, we get
| ( ) | | ( ) |
| |
y–1 1 __ log _____ + tan– 1y = x + c1 2 y+1
( ( ) ) ( ÷ ( ) ) ( )
When
| |
Putting
dx Y– y = – ___ xy (X – x) Y = 0, we get
dx – y = – ___ xy (X – x)
fi
dy X = y ___ + x dx
fi
dy X = y ___ + x dx
Thus,
dy Q = y ___ + x, 0 dx
Thus,
dy B = y ___ + x, 0 dx
(
)
Also, it is given that,
)
PQ = k
( ( ( (
) )
dy 2 fi y ___ + y2 = k2 dx
Here, A = (x, 0) It is given that
dy x+y x + y ___ – x = _____ 2 dx
fi
dy x + y y ___ = _____ 2 dx
dy 2 fi y ___ = k2 – y2 dx
) )
dy 2 k2 – y2 fi ___ = ______ 2 dx y
÷
______
dy k2 – y2 fi ___ = ± ______ dx y2
dy x + y fi ___ = _____ 2y dx
(
)
fi
y dv 1 + v v + x ___ = _____ , Let v = __ x fi v dx
fi
dv 1 + v – 2v2 x ___ = __________ 2v dx
______
÷k 2 – y2 = c 31. It is given that dy ___ = dx dy fi ___ y = Integrating, we
v dv dx fi _____________ = – ___ x (2v + 1) (v – 1) (v – 1 + 1) dv dx fi _____________ = – ___ x (2v + 1) (v – 1)
(
)
dv 2dv dx 1 dv fi _______ + __ _____ – ______ = – ___ x (2v + 1) 3 v – 1 2v + 1 dv dv 3 dx _______ + _____ = – ____ x (2v + 1) v – 1
y dy ______ _______ = ± dx ÷k 2 – y2
Integrating, we get
v dv dx fi __________ = ___ x 1 + v – 2v2
fi
(
x
2y ___ x 2dx ____ x get
log |y| = log |x2| + log c
fi
y = c x2
which is passing through (1, 1), so, c = 1 Hence, the equation of the curve is y = x2
( )
Differential Equation
32. The equation of the normal at P (x, y) is Here,
dx Y – y = – ___ (X – x) dy dy dx A = x + y ___ , 0 , B = 0, y + x ___ dx dy
(
) (
)
P = (x, y)
It is given that, PB ___ = PA
(
) (
)
dy dx = x + y ___ ◊ y + x ___ dx dy
fi
dy dy 1 + ___ = x + y ___ dx dx
fi
dy (y – 1) ___ + (x – 1) = 0 dx
Integrating, we get 1 __ 2
(x – 1)2 + (y – 1)2 = c
÷ ( )
___________
dx 2 x2 + x ___ dy ____________ fi __________ = dv 2 y ___ + y2 dx
÷( )
which is passing through (5, 4), so c = 25 Hence, the equation of the curve is
1 __ 2
(x – 1)2 + (y – 1)2 = 25.
( ) ( ( ) )
fi
dv 2 dx 2 1 x2 + x ___ = __ y ___ + y2 4 dy dx
fi
x2 x2 + __ 2 = p
1 __ (p2y2 + y2) 4
y2 x2 fi __2 (p2 + 1) = __ (p2 + 1) 4 p 2 x2 y fi __2 = __ 4 p
(Tougher Problems for Jee-Advanced) 1. We have
x
y = C1 cos (2x + C2) – C83C 63 + C6 sin (x – C7)
fi y = C1 cos (2x + C2) – C93x + C6 sin (x – C7) Since the above equation has 5 arbitrary constants, so the order of the differential equation is 5. 2. The Equation of the family of parabolas is (y – k)2 = 4a (x – h),
fi
y2p2 = 4x2
fi
y p = ± 2x
where h and k are arbitrary constants
fi
y dv = ± 2x dx
Differentiating w.r.t. x, we get dy 2 (y – k) ___ = 4a dx
Integrating, we get y2 fi __ = c1 ± x2 2 fi
fi
y2 = 2 c1 ± 2x2
which is passing through (0, 4) so c1 = 8 Hence, the eqution of the curve is 2
y = 16 ± 2x
fi
y2 = 16 + 2x2
___
1 1 ____ + ___ = 1 OM ON
dy ___ = 0 dx
( )
d 2y dy 2 2a ___2 + ___ = 0 dx dx which is the required differential equation. 3. Given condition is
dy ___ = dx
2y ___ x
dy dx ___ fi ___ y = 2 ◊ x Integrating, we get
OM + ON = OM ◊ ON dy dx fi x + y ___ + y + x ___ dx dy
) (
d 2y (y – k) ___2 + dx
From Eqs (ii) and (iii), we get
Clearly, it is passing through (÷10 , – 6). 33. The equation of the normal at P (x, y) is dy (Y – y) = ___ (X – x) dx dy dx Thus, OM = x + y ___ and ON = y + x ___ xy dx Also, it is given that,
(
dy (y – k) ___ = 2a dx
)
...(i)
...(ii)
Again differentiaing w.r.t. x, we get
2
fi
4.85
log |y| = 2 log |x| + log |c|
fi
y = c x2
...(iii)
4.86 Integral Calculus, 3D Geometry & Vector Booster which is passing through the curve (1, 1), so c = 1 Hence, the equation of the curve is y = x2 4. The equation of the tangent is dy Y – y = ___ (X – x) ...(i) dx dx Equation (i) meets the x-axis at x – y ___ , 0 dy dy and the y-axis at 0, y – x ___ respectively. dx
(
)
(
)
dx x – y ___ = 2x dy dy y fi ___ = – __x dx dy ___ d x fi ___ y + x = 0
fi
x y = c
2
)
(
x2 + y2 Let _______ = v x
fi
fi
fi
dv v + x ___ = ev + v dx
fi
dv x ___ = ev dx dx e– v dv = ___ x
– e
fi
– e– v = log |c x|
fi fi
)
x2 + y2 = v x
fi
dz ___ dx dx dx y ___ = ___ dy dz
1 __ x = z
1 Let __ x = v
fi
dv – ___ + v = z dz
1 dx – __2 ___ = x dz
which is a linear differential equation. ...(i)
\
IF = e – Ú dz = e– z
Hence, the solution is
v ◊ e– z = Ú – z e– z dz + c
fi
v ◊ e– z = – (z – 1) e– z + c
fi
1 __ x = (ln y + 1) + c y
Hence, the required integral curve is 1 __ x = (ln y + 1) + e y fi
(
dy 1 ___ fi __ y dx =
( )
– e– v
c x =
1 which is passing through e, __ e , so, c = e
= log |x| + log c
c x = e
ln y = z
dy v = (z + 1) + c ez dv fi 2x + 2y ___ = v + x ___ dx dx 1 ln y fi __ x = (ln y + 1) + c e
Integrating, we get
Let
dv fi ___ – v = – z dz
x______ +y dy x2 + y2 2y ___ = e x + ______ – 2x x dx
– v
...(i)
fi
which is passing through (2, 4), so c = 8 Hence, the equation of the curve is xy = 8. 5. The given differential equation is
fi
dx 2 y ___ xy + x = x ln y
which is a linear differential equation.
log y + log x = log c
fi
1 dx fi __2 ___ + x dz
fi
2
dy x (1 – x ln y) ___ + y = 0 dx
dx fi ___ + x = x2z dz
Given condition is
(
)
2 x______ + y2 e– e– x
which is the required solution. 6. The given differential equation is
((ln y + 1) + e y) x = 1
7. The given differential equation can be written as
(
)
2 (x dx – y dv y dx – xdy ___________ = – ________ 2 2 (x – y ) (x2 + y2)
( ( ) )
d (x2 – y2) y fi ________ = – d tan– 1 __ x 2 (x2 – y )
dv ___ dz
Differential Equation
dt tan x ___ = (2t 2 + t + 2) dx dt dx fi ___________ = ___ x 2 (2t + t + 2)
Integrating, we get
fi
( ( ) )
y log |(x2 – y2)| + tan– 1 __ x = c
which is the required solution. 8. The given differential equation is d y + ___ (xy) = x (sin x + log x) dx
dt dx 1 fi __ ______________ = ___ x 2 2 15 1 t + __ + ___ 4 16 Integrating, we get 1 t + __ 2 4 ___ ____ tan– 1 _____ ___ = log |sin x| + C ÷15 15 ___ 16
{ ( )
fi
dy y + y + x ___ = x (sin x + log x) dx
fi
dy x ___ + 2y = x (sin x + log x) dx
(
which is the required solution. 11. The given differential equation is y y y cos __ x (xdy – ydx) + x sin __ x (xdy + ydx) = 0
dx ___
y ◊ x2 = Ú x2 (sin x + log x) dx + c
fi
y ◊ x2 = – x2 cos x + 2x sin x + 2 cos x
( )
( ) y y dy __y y dy __y fi __ x cos ( __ x ) ( ___ – x ) + sin ( __ x ) ( ___ + x ) = 0 dx dx y ( Let __x = u )
x3 x3 + __ log x – __ + c 3 9
(
2
(xdy + ydx) sin (xy) + (x ydx + xy dx) cos (xy) = 0 fi d (xy) sin (xy) + xy (xdy + ydx) cos (xy) = 0 fi d (xy) sin (xy) + xy d (xy) cos (xy) = 0 fi sin (xy) d (xy) + xy cos (xy) d (xy) = 0
)
Integrating, we get
On Integrating, we get
– cos (xy) + (xy) sin (xy) + cos (xy) = c
fi
(xy) sin (xy) = c
which is the required solution. 10. The given differential equation is 2
dy 2y cos x + y sin 2x + 2 cos x ◊ sin x ___ = ____________________________ dx sin2x
(
log |v sin v| = – 2log |x| + log c c fi v sin v = __ 2 x y__ y__ c fi x sin x = __ 2 x
( ) ( ) y c y sin ( __ x ) = __ x
fi
2
)
dy y2 y fi ___ = 2cosx _____ 2 + ____ + 1 dx sin x sin x
) ( dv fi (v cos v + sin v) ( x ___ ) + 2v sin v = 0 dx 2dx fi (____________ v cos v v sin v+ sin v ) dv + ____ x = 0
dv dv fi v cos v v + x ___ – v + sin v v + x ___ + v = 0 dx dx
9. The given differential equation is 2
)
4y + sin x 2 ___ fi ____ tan– 1 _________ ___ = log |sin x| + C ÷15 ÷15 sin x
\ IF = e2 Ú x = e2 log x = e log x2 = x2 Hence, the solution is
}
( ÷ )
dy 2y fi ___ + ___ x = (sin x + log x) dx which is a linear differential equation.
y Let ____ = t sin x
p p when x = 1, y = __ , then c = __ 2 2 Hence, the required solution is y p y sin __ x = ___ 2x 12. The given differential equation is
( )
÷ ÷
______________________________
dy x4y2 – x6 + 2x4y – x6y2 – 2x6y + x4 ___ ______________________________ = dy dt dx y2 – x2y2 + x3y2 – x5y2 ___ = t cos x + sin x ___ dx dx fi
4.87
dt t cos x + sin x ___ = 2 cos x (t2 + t + 1) dx
______________________________
x4 (y2 – x2 + 2y – x2y2 – 2x2y + 1) = _____________________________ y2 (1 – x2 + x3 – x5)
4.88 Integral Calculus, 3D Geometry & Vector Booster
÷
______________________________
x4 {(y2 + 2y + 1) – x2 (y2 + 2y + 1)} = ______________________________ y2 (1 – x2 + x3 – x5)
÷ ÷
___________________
(x + y)2 + 3 (x + y) + 1 = ____________________ (x + y)2 – 3 (x + y) + 2
dv v2 + 3v + 1 fi ___ – 1 = __________ , dx v2 – 3v + 1
+ 1)2 – x2 (y + 1)2} x2 {(y ___________________ = __ y (1 + x2) (1 + x3)
2 2 x2 {(y + 1) (1 – x )} = __ y _______________ (1 – x2) (1 + x3)
x2 (y – 1) = ________ _____ y ÷1 + x3
v2 – 3v + 1 fi __________ dv = 2 dx v2 + 1
_______________
)
x2 dx 1 fi 1 – _____ dy = _______ _____ y+1 ÷ 1 + x3 Integrating, we get _____
2 y – log |y + 1| = __ ÷ 1 + x3 + c 3
which is the required solution. 13. Given
x
log t f (x) = Ú ________ 2 dt, x ≥ 1 1 1 + t + t
Now,
fi
2v2 + 2 = __________ v2 – 3v + 1
2v 3 _____ fi 1 – __ dv = 2 dx 2 v2 + 1 Integrating, we get 3 v – __ log |v2 + 1| = 2x + C 2 3 fi y – __ log |(x + y)2 + 1| = x + C 2 which is the required solution. 15. Given f (0) = – 1 Put x = 0, y = 1, then f (1) = 1
Now,
f (x + h) + f (x) ◊ f (h) = f (xh + 1)
log t 1 f __ x = Ú ________ dt 1 1 + t + t2
fi
(f (x + h) – f (x)) + (f (x) ◊ f (h) – f (x))
1 1 Let t = __ y fi dt = – __2 dy y 1 y log __ y 1 1 __ __________ f x = Ú – __2 dy 1 1 1 1 + __ + __ y y y2
f (x + h) – f (x) ___________ f (x) f (h – 1) fi _____________ + h h f (xh + 1) – f (1) = ______________ ◊ x xh
1/x
( )
(
)
( )( ) ( )
( )
( Ú (
) )
y
dv v2 + 3v + 1 fi ___ = __________ + 1 dx v2 – 3v + 1
( ( ) )
y dv x2 dx fi _____ = _______ _____ y+1 ÷ 1 + x3
(
(Let v = x + y)
– log (y) = Ú ________ 2 (– 1) dy 1 y + y + 1 y
fi
= f (xh + 1) – f (1)
(
)
(
fi
f ¢ (x) + f (x) ◊ f ¢ (0) = x f ¢ (1)
fi
dy ___ + p y = q x dx
log (t) = Ú ________ 2 dt 1 t + t + 1
which is a linear differential equation.
= f (x)
Hence, the solution is
(
)
Hence, the result. 14. The given differential equation is dy ___ = dx
x2 + y2 + 3x + 3y + 2xy + 1 ________________________ x2 + y2 – 3x – 3y + 2xy + 2
\
)
)
f (xh + 1) – f (1) = lim ______________ x h Æ 0 xh
log (y) = _________ 2 dy 1 y + y + 1 x
(
(f (x + h) – f (x) f (h) – 1 lim _____________ + f (x) ________ h Æ 0 h h)
IF = e Úp dx = ep x
y ◊ epx = q Ú x epxdx + c
fi
y ◊ epx = q Ú x epxdx + c
fi
y ◊ ex = Ú x ex dx + c
4.89
Differential Equation
fi
y ◊ ex = (x + 1) ex + c
fi
y = (x – 1) + c e– x
fi
( ) ( )
when x = 1, y = 1, then c = 0 Thus, the equation of the given curve is
Integrating, we get
y = x – 1.
16. Put
x = 1, y = 1, then f (1) = 0.
Now,
f (x) (1 + h)) = x f (1 + h) + (1 + h) f (x)
fi
f (x + xh) – f (x) = xf (1 + h) + hf (x)
fi
f (x + xh) – f (x)
= x (f (1 + h) – f (1)) + hf (x)
(
)
f (xh + x) – f (x) fi ______________ ◊ x xh
( ) ( )
y y x x sin __ y d __ y + cos __ x d __ x dy + dx + ___ 2 = 0 y
( )
( )
y x 1 – cos __ y + sin __ x + x – __ y = c
which is the required differential equation. 18. The given differential equation is dy ___ = dx
(1 + y2) _________ x y (1 + x2)
y dy dx fi _____ 2 = ________ 1+y x (1 + x2)
(
)
y dy x 1 fi _______ = __ x – _____ 2 dx 2 (1 + y ) 1 + x x ( f (1 + h) – f (1)) hf (x) = ________________ + _____ h h Integrating, we get f (xh + x) – f (x) log |1 + y2| = 2 log |x| – log |1 + x2| + 2 log c fi lim ______________ x h Æ 0 xh fi (1 + y2) (1 + y2) = c2x2 f (1 + h) – f (1) = lim _____________ x + f (x) which is the required solution. h Æ 0 h 19. The given differential equation is
(
)
(
fi
x f ¢ (x) = x f ¢ (1) + f (x)
fi
x f ¢ (x) = x + f (x)
)
dy x ___ = x + y dx dy y fi ___ = 1 + __ x dx dy 1 ___ fi + – __ x y = 1 dx fi
Let
x y = v
fi
y dy __ 1 dv __x + ___ = x ___ dx dx
dv ___________ 1 ___ 1 fi __ 4 x dx = sin v + cos2 v
( )
which is a linear differential equation. \
y dy ________________ 1 __x = ___ = 4 dx sin (xy) + cos2 (xy)
dx ___ 1 IF = e – Ú x = e– log x = __ x
dx (sin4 v + cos2 v) dv = ___ x dx 1 1 fi __ (2sin2v)2 dv + __ (2cos2v) dv = ___ x 4 2 fi
Hence, the solution is y dx __x = Ú ___ x + c y __ fi x = log |x| + c
dx 1 1 fi __ (1 – cos2v)2 dv + __ (1 + cos2v) dv = ___ x 4 2 1 1 fi __ (1 – cos2v) dv + __ (1 + cos4v) dv 4 8
when x = 1, y = 0, then c = 0 Thus, the equation of the given curve is y = x log |x|
17. The given differential equation can be written as
(
)
( )
dx xdy x ___ y – ___ 2 sin __ y y
(
dy + ___ x –
)
( )
ydx y dy ___ 2 cos __ x + dx + ___ 2 = 0 x y
dx 1 + __ (1 + cos2v) dv = ___ x 2
Integrating, we get
(
)
(
)
sin4v 1 1 fi __ (v – sin2v) + __ v + _____ 4 8 4 fi
sin 2v 1 + __ v + _____ = log |x| + c 2 2
(
)
sin (4xy) 1 1 __ (xy – sin (2xy)) + __ xy + _______ 4 8 4
4.90 Integral Calculus, 3D Geometry & Vector Booster
(
)
which is a linear differential equation.
sin 2(xy) 1 + __ xy + _______ = log |x| + c 2 2
\
which is the required solution. 20. The given differential equation can be written as
(y2 + 1) dy + x2 dy + 2xy dx = 0
fi
(y2 + 1) dy + d (x2y) = 0
Hence, the solution is
y3 __ + y + x2y = c 3 which is the required solution. 21. The given differential equation is
Putting
(
fi
dy 1 y ___ + x = ___ 2 (x2 + y2)2 dx 2x
dy Y = y – x ___ dx It is given that,
( (
dy __ fi y – x ___ = ÷ xy dx dy __ fi x ___ + ÷ xy = y dx dy fi ___ + dx
2
d (x + y ) ___ dx fi _________ = 2 2 2 2 (x + y ) x
fi
Integrating, we get 1 1 – _______ 2 = c – __ x 2 (x + y )
fi
) (
y __ x
__
= – ÷v
Integrating, we get
(y + sin x cos2 (xy)) dx + x dy = 0
(
÷
y __ x =
dv__ ___ dx fi ___ + x = 0 v ÷
which is the required solution. 22. The given differential equation is
dy fi y + x ___ + sin x cos2 (xy) = 0 dx
__
__ dv __v v + x ___ + ÷v = v, v = x dx __ dv x ___ = ÷ v = 0 dx
1 _______ 1 fi __ = c x – (x2 + y2)
) )
dy 2 y – x ___ = xy dx
ydy + xdx ___ dx fi _________ = 2 2 2 2 (x y ) 2x 2
X = 0, we get
)
2 2 2 dy 1 x +y y ___ + x = __ ______ x 2 dx
y ◊ e p (x) x = Ú ( ep (x)x ◊ q (x) ) dx + c
24. The equation of tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx
Integrating, we get
IF = e Ú p (x) dx = e p (x) x
)
__
2 ÷v + log |x| = c
fi
y 2 __ x + log |x| = c
__
÷
which is the required solution. 25. We have
dy dv Let xy = v fi y + x ___ = ___ x x dx dx x Ú (1 – t) f (t) dt = Ú tf (t) dt, f (1) = 1 dv fi ___ + sin x cos2v = 0 0 0 dx Applying Newton and Leibnitz formula, we get dv x fi ___ = – sin x cos2v dx x (x – 1) f (x) + Ú (1 – t) f (t) dt = x f (x) fi sec2v dv = – sin x dx 0 Integrating, we get fi
x
fi
tan (xy) – cos x = c
which is the required solution. 23. The given differential equation is dy ___ + p (x) y = q (x) dx
(x – x – x) f (x) + Ú (1 – t) f (t) dt = 0 2
0
x
fi
(– x2) f (x) + Ú (1 – t) f (t) dt = 0 0
2
fi
(– x ) f ¢ (x) – 2x f (x) + (1 – x) f (x) = 0
fi
(x2) f ¢ (x) + (1 – 3x) f (x) = 0
Differential Equation
fi fi fi
dy (x2) ___ + (1 – 3x) y = 0 dx
( )
dy 1______ – 3y ___ dx = 0 y + x2 dy 1 3 ___ y + __ 2 – __ x dx = 0 x
(
)
Integrating, we get
j (y/x) dy y ______ fi ___ = __ + dx x j¢ (y/x) Thus, j (v) dv v + x ___ = v + _____ dx j¢ (v) j (v) dv fi x ___ = _____ dx j¢ (v)
1 log |y| – 3 log |x| – __ x = c
j¢ (v) dx fi _____ dv = ___ x j (v)
fi
y log __ 3 – x
Integrating, we get
when
x = 1, y = 1, then c = – 1
| |
1 __ x = c
Hence, the solution is
| | | |
y 1 log __ 3 – __ x + 1 = 0 x
fi
y 1–x 1 log __ 3 = __ x – 1 = _____ x x
y 1–x fi __3 = _____ x x
= e ( y=
fi
f (x) = x3e (
)
1_____ –x x
26. The given differential equation is
( )
dy 2 dy y ___ + (x – y) ___ – x = 0 dx dx
fi
dy yp2 + (x – y) p – x = 0, where ___ = p dx yp2 – yp + xp – x = 0
fi
yp (p – 1) + x (p – 1) = 0
fi
(yp – 1) (p – 1) = 0
fi
(yp + x) = 0, (p – 1) = 0
fi
x d x + y d y = 0, dy = dx
fi
Integrating, we get
x2 + y2 = a2, y = x + b
which is passing through (3, 4), so a2 = 25 and b = 1 Hence, the equations of the curves are
x2 + y2 = 25, y = x + 1
27. The given differential equation is
y y¢ = __ x +
fi
log |j (v)| = log |c x|
fi
|j (v) = c x
fi
y j __ x = c x
( )
which is the required solution. 28. Clearly, f (0) = 1
(
j (y/x) ______ j¢(y/x)
)
x+h f ______ – f (x) ◊ f (h) = 0 1 + xh
[ (
( )
1_____ –x x3e x
fi
log |j (v)| = log |x| + log c
)
Let y = vx
We have,
1–x _____ x
4.91
) )
] (
h – x2h f 1 + ______ – f (x) ◊ 1 + xh h – x2h fi ______________________ ______ 2 1 + xh h______ –xh ◊ h 1 + xh f (x) ◊ f (h) – f (x) = ______________ h
(
fi
f ¢ (x) ◊ (1 – x2) = f (x) ◊ f ¢ (0)
fi
f ¢ (x) ◊ (1 – x2) = f (x)
dy fi ___ (1 – x2) = y dx dy _______ dx fi ___ y = (1 – x2) Integrating, we get fi
| |
1–x log |y| = log _____ + log c 1+x
( )
1–x y = c _____ 1+x
when f (0) = 1, then we get, c = 1 1–x Thus, y = _____ 1+x which is the required solution.
( )
29. Let
x = 1 = y, then f (1) = 0
)
4.92 Integral Calculus, 3D Geometry & Vector Booster Now,
f (x (1 + h))
Integrating, we get = x f (1 + h) + (1 + h) f (x)
f (x + xh) – f (x) ______________ x {f (1 + h) – 0} _____ hf (x) fi x ◊ ______________ = + xh h h f (x + xh) – f (x) fi lim x ◊ ______________ x Æ 0 xh
(
) (
)
log |x| + log |y| = log |c|
fi
log |xy| log c
fi
x y = c
which is the required equation of the curve. 31. It is given that
x {f (1 + h) – 0} _____ hf (x) = lim ______________ + x Æ 0 h h
fi x f ¢ (x) = x f ¢ (1) + f (x)
÷ ( ) ( ( ) ) ( ( ) ) ( ( ) ) ( ) ( ) ( ) ( ) ________
______ dy 2 y 1 + ___ = ÷ x 2 + y2 dx
dy 2 y2 1 + ___ = (x2 + y2) dx
fi x f ¢ (x) = x + f (x) ( f ¢ (1) = 1) dy fi x ___ = x + y dx dy y fi ___ = 1 + __ x dx y dv fi v + x ___ = 1 + v, v = __ x dx dv fi x ___ = 1 dx dx fi dv = ___ x Integrating, we get
fi
fi v = log |x| + c
x2 ± y2 = a2 32. It is given that dy 1 ___ = __ dx y
dy 2 x 2 fi 1 + ___ = __ y + 1 dx dy 2 fi ___ = dx
dy x fi ___ = ± __ y dx fi
when x = 1, y = 0, then c = 0 y Thus, __ x = log |x|
fi ydy = dx Integrating, we get
fi y = x log |x| which is the required solution. 30. The equation of tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx
( (
) )
dx Y = 0, then A = x – y ___ , 0 dy
dy X = 0, then B = 0, y – x ___ dx It is given that, dx dy x – y ___ y – x ___ dy dx ________ = x and ________ = y 2 2 dy dx fi x – y ___ = 2x and y – x ___ = 2y dy dx dy dx fi y ___ = – x and x ___ = – y dy dx and
dx fi ___ x +
dy ___ y = 0
xdx ± y dy = 0
Integrating, we get
y fi __ = log |x| + c
Put
x 2 __ y
y2 __ = x + c 2 which is passing through (2, 2), so, c = 0 Hence, the eqution of the curve is
y2 = 2x
33. The tangent at P (x, y) is dy Y – y = ___ (X – x) dx If p be the length of perpendicular from the origin, then
| | | |
dy x ___ – y dx ________ p = __________ dy 2 ___ 1 + dx It is given that
÷ ( )
dy x ___ – y dx _________ __________ = x dy 2 ___ 1 + dx
÷ ( )
Differential Equation
) ( ( ) )
(
fi
2 dy dy 2 x ___ – y = x2 1 + ___ dx dx
fi
dy y – 2xy ___ – x2 = 0 dx 2
Hence, the result. dy y2 – x2 Also, ___ = ______ 2 xy dx y 2 __ x – 1 dy fi ___ = _______ x dx 2 __ y
( )
fi fi
( )
2
dv v – 1 v + x ___ = _____ , 2v dx dv v – 1 x ___ = ______ – v 2v dx
)
fi
fi
xdx + ydy _________ ______ = dx ÷ x 2 + y2
fi
2xdx + 2ydy ______ __________ = 2dx ÷ x 2 + y2
fi
d ÷x 2 + y2 = 2 dx
fi
fi x2 + y2 = c x which is the required equation of the curve. 34. The given curve is
y dy a = __ ___ 2 dx
______ dy x + y ___ = ÷ x 2 + y2 dx
fi ...(i)
...(ii)
fi
fi
(
)
dy y dy y2 = 2y ___ x + __ ___ 2 dx dx
( )
dy dy 2 y2 = 2xy ___ + y2 ___ dx dx
Replacing dy/dx by – dx/dy, we get
( ) ( )
dx dx 2 2 ___ y2 = 2xy – ___ xy + y xy
fi
dy 2 dy y ___ + 2xy ___ – y2 = 0 dx dx 2
( )
______
)
( )
______
)
| |
dy 2 xy ___ = y2 – x2 dx dy y2 – x2 ___ = ______ 2xy dx dy (y/x)2 – 1 ___ = _________ 2 (y/x) dx
(
)
fi
y dv v2 – 1 v + x ___ = _____ Let __ = v h 2v dx
fi
dv v2 – 1 x ___ = ______ – v 2v dx
fi
dv v2 + 1 x ___ = – _____ 2v dx
Eliminating a between Eqs (i) and (ii) we get
(
÷ ( )
(v2 + 1) x = c y 2 fi __ x + 1 x = c
dy 2y ___ = 4a dx
fi
______
dy 0 – y + x ___ dx ________ ___________ = x 2 dy 1 + ___ dx
fi
fi
(yp + x) = ÷ x 2 + y2
which is the required orthogonal trajectory. 35. The equation of the tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx It is given that
log |v2 + 1| + log |x| = log c
y2 = 4a (x + a)
fi
(
2v dx fi _____ dv + ___ x = 0 v2 + 1 Integrating, we get
(yp + x)2 = x2 + y2
÷x 2 + y2 = 2x + c
v2 + 1 = – ______ 2v
( ( )
fi
Integrating, we get
2v dx fi _____ dv = – ___ x 2 v +1
dy y2p2 + 2x y p – y2 = 0, p = ___ dx
(Let y = v x)
2
fi
4.93
2v dx fi _____ dv = – ___ x 2 v +1 Integrating, we get log |v2 + 1| = log c – log |x| c fi (v2 + 1) = __ x fi x2 + y2 = c x which is the required solution.
4.94 Integral Calculus, 3D Geometry & Vector Booster
Questions Asked in Past IIT-JEE Examination
3. The given differential equation is
dy ___ = sin (10x + 6y) dx
Let
fi
(
(
)
)
1 dv fi __ ___ – 10 = sin v 6 dx dy fi ___ = 6 sin v + 10 dx dv ___________ = dx Ú (6 sin v + 10) Ú
dv 1 fi __ Ú __________ = x + c 2 (3 sin v + 5)
(
)
2
sec (v/2) dv 1 fi __ Ú _______________________ = x + c 2 (5 + 5 tan2 (v/2) + 6 tan (v/2)) dt = x + c, Ú ____________ (5 + 5t2 + 6t)
(Let tan (v/2) = t))
dt 1 fi __ Ú ___________ = x + c 5 6 2 t + __ t + 1 5
(
)
dt 1 fi __ Ú ______________ = x + c 5 3 2 4 2 t + __ + __ 5 5
( ( ) ( ) ) 5t + 3 1 5 __ fi × __ tan ( _____ = x + c 5 4 4 ) 5 tan (5x + 3y) + 3 1 fi __ tan ( _______________ ) = x + c 4 4
3 3 4 fi (5x + 3y) = tan– 1 __ tan 4x + tan– 1 __ – __ 4 5 5
(
(
)
( ) )
( (
( ) ) )
( )
du equations ___ = p (x) u = f (x) and dx dv ___ + p (x) v = g (x), where p (x), f (x) and g (x) dx are continuous functions. If u (x1) > v(x1) for some x1 and f (x) > g (x) for all x > x1. Prove that any point (x, y) where x > x1 does not satisfy the equation y = u (x) and y = v (x). [IIT-1997]. 7. The equation of the tangent at any point P (x, y) is dy Y – y = ___ (X – x) dx dy It meets the axes at A x – y ___ , 0 and dx
(
(
)
) ( (
) (
)
(
dy B 0, y – x ___ respectively. dx
It is given that
( )
3 1 which is passing through origin so c = __ tan– 1 __ . 4 4 Thus, the equation of the curve is 5 tan (5x + 3y) + 3 3 1 1 __ tan– 1 _______________ = x + __ tan– 1 __ 4 4 4 4
6. Let u (x) and v (x) satisfy the differential
))
dy dx 1 1 \ Mid-point AB is __ x – y ___ , __ y – x ___ . 2 dy 2 dx
– 1
)
4. A curve y = f (x) passes through the point P (1, 1). The normal to the curve at P is a (y – 1) + (x – 1). If the slope of the tangent at any point on the curve is propor tional to the ordinate of the point, determine the equation of the curve. Also, otain the area bounded by the y-axis, the curve and the normal to the curve at P. [IIT-1996] 5. A spherical rain drop evaporates at a rate proportional to its surface area at any instant t. The differential equation giving the rate of change of the radius of the rain drop is....... [IIT-1997]
– 1
(
5 tan (5x + 3y) + 3 3 fi _______________ = tan 4x + tan– 1 __ 4 4
dv 1 fi __ Ú _________________ = x + c 2 _______________ 3 ◊ 2tan (v/2) 1 + tan2 (v/2) + 5
fi
dy dv 10 + 6 ___ = ___ dx dx dy 1 ___ dv fi ___ = __ – 10 dx 6 dx
fi
( )
)
5 tan (5x + 3y) + 3 3 fi tan– 1 _______________ = 4x + tan– 1 __ 4 4
...(i) 10 x + 6 y = v
(
(
)
dy dx 1 1 __ x – y ___ = x and __ y – x ___ = y 2 2 dy dx fi
dx x – y ___ = 2x dy
fi
dx – y ___ = x dy
Differential Equation
dy dx ___ fi ___ y = – x
( ( ) ) ( )
y 2 1 – __ x _________ = y 2 __ x
dy
fi
dx Ú ___y = – Ú ___ x
fi
log |y| = log c – log |x|
fi
x y = c
4.95
fi
dv 1 – v2 v + x ___ = – _____ , 2v dx
which is passing through (1, 1), so c = 1 Hence, the equation of the curve is
fi
dv 1 – v2 – 1 + v2 – 2v2 x ___ = – _____ – v = ___________ 2v 2v dx
x y = 1 8. The given curve is y = (c1 + c2) cos (x + c3) – c4ex + c5
fi
dv 1 + v2 x ___ = – _____ 2v dx
2v dx fi _____ 2 dv = – ___ x 1+v
fi
y = (c1 + c2) cos (x + c3) – c4ec 5◊ ex
fi
y = A cos (x + c3) – B ◊ ex
fi
Since the given curve has 3 arbitrary constants, so, the order of the differential equation represented by the given curve is 3.
2v dx dv = – Ú ___ Ú _____ x 1 + v2
fi
log |1 + v2| = log c – log |x| c (1 + v2) = __ x
9. The equation of the normal at any point P (x, y) is
dx Y – y = – ___ (X – x) dy
| |
...(i)
fi
fi x2 + y2 = c x which is passes through (1, 1), so c = 2 Hence, the equation of the curve is
It is given that
dx y + x ___ dy _________ __________ = |y| dx 2 ___ 1 + dy
10. The given differential equation is
÷ ( ) ( ) ( ( ) ) ( ) ( )
dx dx fi y + x ___ = y2 1 + ___ dy dy 2
2
dx 2 dx fi ___ (x2 – y2) + 2xy ___ = 0 dy dy
dx dx fi ___ = 0, (x2 – y2) ___ + 2xy = 0 dy dy
( )
2xy dx = 0, ___ = _______ dy (x2 – y2)
x2 + y2 = 2x
( ) ( ) ( ) ( )
dy 2 dy ___ – x ___ + y = 0 dx dx fi fi
dy y = x ___ + dx
dy 2 ___ dx
dy y = px + p2, where p = ___ dx
which is a Clairauts differential equation Thus, the solution is
y = c x + c2
fi
y = 2 x + 4
11. The given curve is
__
y2 = 2c (x + ÷c ) dy 2y ___ = 2c dx dy c = y ___ dx
...(i)
dx Now, ___ = 0 fi x = c dy
fi
which is passing through (1, 1), so c =1 Hence, the equation of the curve is x = 1
fi
2xy dx Also, ___ = _______ 2 2 dy (x – y )
From Eq. (i) and (ii), we get
2
2
(x – y ) = _______ 2xy
(Let y = vx)
...(ii)
( ÷ ) ____
dy dy y = 2y ___ x + y ___ dx dx 2
(
)
2 y dx fi __ ___ – x = 2 dy
dy y ___ dx
4.96 Integral Calculus, 3D Geometry & Vector Booster
(
y fi __ – 2
) ( )
dy 2 dy 3 x ___ = y ___ dx dx
Thus, the order of the differential equation is 3. 12. The given expression is x2 + y2 = 1 Differentiating w.r.t. x, we get fi fi
dy 2x + 2y ___ = 0 dx dy x + y ___ = 0 dx
Again differentiating, w.r.t. x, we get
( )
dy 2 d2y 1 + ___ + y ___2 = 0 dx dx
fi
1 + y¢2 + y ◊ y¢¢ = 0
fi
y ◊ y¢¢ + (y¢)2 + 1 = 0
13. The given differential equation is dy (1 + t) ___ – ty = 1 dt
dy fi _____ = y+1
– cos x _______ dx 2 + sin x
dy
fi
– cos x = _______ dx Ú _____ y + 1 Ú 2 + sin x
fi
log |y + 1| = log c – log |2 + sin x|
fi
(y + 1) (2 + sin x) = c
Put x = 0, y = 1, then c = 4 Hence, the equation of the curve is
(y + 1) (2 + sin x) = 4
p When x = __ , then 2 fi
4 y + 1 = __ 3 4 1 __ y = – 1 = __ 3 3
16. Given, dy (x + 1)2 + (y – 3) ___ = _______________ (x + 1) dx
( )
dy – t 1 fi ___ + ____ y = ____ t+1 t+1 dt which is a linear differential equation. t – Ú ____ dt t + 1 =
Thus, IF = e
(t + 1) – 1 – Ú ________ dt e t + 1
= e log (t + 1) – t = (t + 1) e– 1 Therefore, the solution is
( )
dy 2 + sin x ___ _______ = – cos x y + 1 dx
y ◊ (t + 1) e– 1 = Ú e– 1 dt + c – 1
– t
fi y ◊ (t + 1) e = – e + c Put t = 0, y = 1, then c = 2 Hence, the equation of the curve is y ◊ (t + 1) e– t = – e– t + 2 Put t = 1, then
y 3 = (x + 1) + ______ – ______ (x + 1) (x + 1)
dy y 3 fi ___ – ______ = (x + 1) – ______ (x + 1) dx (x + 1) which is a linear differential equation. dx – _____ 1 Thus, IF = e Ú x + 1 = e– log (x + 1) = ______ (x + 1) Therefore, the solution is y 3 _____ = 1 – _______ dx x+1 Ú (x + 1)2
(
)
y 3 fi _____ = x + _____ + c y+1 x+1 Put
x = 2 and y = 0, then c = – 3
2y ◊ e– 1 = – e– 1 + 2
Hence, the equation of the curve is y 3 _____ = x + _____ – 3 x+1 x+1
fi
2y = – 1 + 2e
fi
1 y = e – __ 2 14. A right circular cone with radius R and height H contains a liquid which evaporates at a rate propeortional to its surface in contact with air (proportonality constant k > 0). Find the after which the cone is empty fi
15. The given differential equation is
y = x2 + x + 3 – 3x – 3 = x2 – 2x.
Differential Equation ______
÷ 1 – y2 fi _______ dy = dx y
Hence, the required area 2
= Ú [0 – (x2 – 2x)] dx 0
(
______
)
3 2
x = x2 – __ 3 0
( )
Put fi
x cos y + y cos x = p x = 0, then y = p dy dy cos y – x sin y ___ – y sin x + cos x ___ = 0 dx dx dy Put x = 0, y = p, then ___ = – cos (p) = 1 dx dy dy dy 2 – sin y ___ – sin y ___ – x cos y ___ dx dx dx
( )
( ) ( )
d2y dy – x sin y ___ 2 – y cos x – sin ___ dx dx
dy d2y – sin x ___ + cos x ___ 2 = 0 dx dx
( )
( )
dy Put x = 0, y = p and ___ = 1, then dx d2y (– sin p – sin p – 1) + ___ 2 = 0 dx 2 d y fi ___2 = 1 dx 18. The equation of tangent at any point (x, y) to the curve y = f (x) is Y – y = f ¢ (x) (X – x)
(
)
dx It meets the x-axis at x – y ___ , 0 . dy It is given that
AP = 1
fi
AP 2 = 1
fi
dx 2 2 – y ___ xy + y = 1
( )
( ) ( ) ( ) ÷
dx y2 ___ = 1– y2 dy 2 dx 2 1 – y fi ___ = _____ dy y2 fi
2
fi
1 – y2 ÷_______ dy = Ú dx Ú y
fi
y ÷1 – y2 ________ =x+c Ú y2 dy
fi
t dt = x + c, Ú _____ t2 – 1
fi
1 2 dt = x + c Ú 1 + t_____ –1
fi
t–1 1 t + __ log ____ = x + c 2 t+1
_____
8 4 = 4 – __ = __ sq.u. 3 3 17. The given relation is
÷
______
y2 = ______ 1 – y2
2
(
(Let t2 = (1 – y2))
)
| |
|
_____
|
÷1 – y2 – 1 1 _____ = x + c fi ÷1 – y2 + __ log __________ 2 + 1 ÷ 1 – y2 _____
19. The given differential equation is
(x2 + y2) dy = xy dx
dy xy fi ___ = ______ dx x2 + y2
( ) ( )
y __ x dy fi ___ = _______ y 2 dx __ 1 + x
...(i)
which is a homogeneous differential equation.
(
)
dy dv Let y = v x fi ___ = v + x ___ dx dx fi
dv v v + x ___ = _____ dx 1 + v2
fi
dv v x ___ = _____ – v dx 1 + v2
fi
dv v – v – v3 v3 x ___ = _________ = – _____ 2 dx 1+v 1 + v2
( ) Ú ( )
1 + v2 dx fi ______ 3 dv = – ___ x v fi
_____
1 – y2 dx fi ___ = _____ dy y2
4.97
1 + v2 dx _____ dv = – Ú ___ x v3
fi
1 – ___ 2 + log |v| = c – log |x| 2v
fi
y x2 – ___2 + log __ x = c – log |x| 2y
| |
4.98 Integral Calculus, 3D Geometry & Vector Booster fi
x2 – ___2 + log |y| = c 2y
1 x = 1, y = 1, then c = – __ 2 Thus, the equation of the curve is When
x2 1 – ___2 + log |y| = – __ 2 2y
When
x = x0, y = e, then
x20 1 – ___2 + 1 = – __ 2 2e
fi
x20 3 1 ___ – 2 = – 1 – __ = – __ 2 2 2e
fi x20 = 3e2 fi
___
x0 = ÷3 e
20. The given differential equation is 2
y dx + y
= x dy
fi
ydx – xdy = – y2dy
ydx – xdy fi _________ = – dy y2 x fi d __ y = – dy fi
( ) Ú d ( __ xy ) = – Ú dy
x fi __y = c – y When
x = 1, y =1, then c = 2
Thus, the equation of the curve is x __y = 2 – y When
x = – 3, then
3 2 – y = – __ y
fi
3 y – 2 = __ y
fi
y2 – 2y – 3 = 0
fi
(y – 3) (y + 1) = 0
fi
y = 3, – 1
Since y > 0, so the value of y = 3. 21. The equation of any tangent at P (x, y) is
dy Y – y = ___ (X – x) dx
( )
)
dx It meets the x-axis at A x – y ___ , 0 dy dy and B 0, y – x ___ dx
(
Since P (x, y) divides the line AB in the ratio 3 : 1, we get dy 3 dx 1 fi __ y – x ___ = y, __ x – y, ___ = x 4 4 dx dy
(
)
(
)
dy x ___ = – 3y dx dy dx ___ fi ___ y = – 3 x fi
dy
fi
dx y = – 3 Ú ___ Ú ___ x
fi
log |y| = log c – 3 log |x|
fi
x3y = c
As f (1) = 1, we get c = 1 Thus, the equation of the curve is x3y = 1 1 fi y = __ 3 x which passes through (2, 1/8). dy 3 Also, ___ = – __4 dx x dx m = ___ = – 3 dy (1, 1)
( )
\ The equation of normal at (1,1) is 1 y – 1 = __ (x – 1) 3 fi 3y – 3 = x – 1 fi x – 3y + 2 = 0 22. The given differential equation is dy 2 ___ = _____ dx x + y Let x + y = v dy dv fi 1 + ___ = ___ dx dx dy dv fi ___ = ___ – 1 dx dx dv 2 fi ___ – 1 = __ v dx dv v + 2 fi ___ = _____ v dx
(
)
v fi _____ dv = dx v+2 fi
2 dv = Ú dx Ú ( 1 – _____ v + 2)
Differential Equation
fi
v – 2 log |v + 2| = x + C
fi
x + y – 2 log |x + y + 2| = x + C
fi
y – 2 log |x + y + 2| = C
When x = 1, y = 1, then fi
( )
e c = 1 – 2 log 4 = log ___ 16
Thus, the equation of the curve is fi
( ) ( )
e y – 2 log |x + y + 2| = log ___ 16 e y = log |x + y + 2|2 + log ___ 16
(
( ) )
e fi (x + y + 2)2 ___ = ey 16 16 fi |(x + y + 2)2 e– y)| = ___ e
( )
23. The given differential equation is ______
dy ÷1 – y2 ___ = _______ y dx y dv _____ fi ______ = dx ÷1 – y2 fi
y dv
_____ = Ú dx Ú ______ 2
÷1 – y
Integrating, we get _____
fi (x + c)2 + y2 = 1 which represents a circle with the centre (c, 0) and the radius 1. 24. Given t2 f (x) – x2f (t) lim ____________ = 1 t Æ x t–x
(
Hence, the solution is y dx __2 = – Ú ___4 + c x x y 1 fi __2 = ___ 3 + c x 3x As f (1) = 1, then c = 1 – 1/3 = 2/3 Therefore, the equation of the curve is y 3 1 __2 = ___ 3 + __ 2 x 3x fi
3x2 1 y = ___ + ___ 3x 2
25. The given differential equation is _____
_____
_____
_____
x ÷x 2 – 1 dy – y ÷y 2 – 1 dx = 0
fi
x ÷x 2 – 1 dy = y ÷y 2 – 1 dx
dy dx _____ = ________ _____ fi ________ 2 y ÷y – 1 x ÷x 2 – 1 dy
fi
dx _____ _____ = Ú _______ Ú ________ 2 2
fi
sec– 1 (y) = c + sec– 1 (x)
When
2 x = 2, y = ___ __ , then 3 ÷
y ÷y – 1
x÷x – 1
( )
2 c = – sec–1 (2) + sec–1 ___ __ 3 ÷ p p p fi c = – __ + __ = – __ 3 6 6 Thus, the equation of the curve is p sec– 1 (y) = sec– 1 (x) – __ 6 p – 1 fi y = sec sec (x) – __ 6 Thus, the statement (i) is trure Also, we can write Eq. (i) as fi
– ÷1 – y2 = x + c
)
(
)
2t f (x) – x2 f ¢ (t) fi lim ______________ = 1 t Æ x 1 fi
2x f (x) = x2 f ¢ (x) = 1
fi
dy 2x y – x2 ___ = 1 dx
(
( ) ( ) p 1 1 __ __ fi __ y = cos ( cos ( x ) – 6 )
p 1 1 cos– 1 __ y = cos– 1 __ x – __ 6 – 1
( ( ) ) ( ) p 1 + sin ( cos ( __ x ) ) sin ( __ ) 6
p 1 1 – 1 __ __ fi __ y = cos cos x cos 6
which is a linear differential equation.
3 __ ÷ 1 __ 1 ___ 1 1 __ fi __ y = x 2 + 2 1 – x2
\
IF =
dx ___ e– 2 Ú x =
1 e– 2 log x = __ 2 x
...(i)
)
dy x2 ___ – 2x y = – 1 dx dy 2 1 fi ___ – __ y = – __2 dx x x fi
4.99
– 1
__
÷
_____
Thus, the statement (ii) is false.
4.100 Integral Calculus, 3D Geometry & Vector Booster 26. Given differential equation is y¢ = y +1
Hence, the solution is y __x = – Ú x dx + c
dy fi ___ = y + 1 dx dy fi ___ y + 1 = dx
y x2 fi __x = c – __ 2 When
dy
x = 1, y = 1, then c = 3/2
Hence, the equation of the curve is
fi
= dx Ú y_____ +1 Ú
fi
log |y + 1| = x + c
y 3 __ x2 __x = __ – 2 2
when
x = 0, y = 1, then c = log 2
When x = – 3, then
Hence, the equation of the curve is
log |y + 1| = x + log 2
When
x = ln 2, then
y+1=4fiy=3
29. The given differential equation is
(x – 3)2 y¢ + y = 0
dy (x – 3)2 ___ + y = 0 dx dy y fi ___ = – _______ dx (x – 3)2 fi
Thus.
fi fi
1 C + _____ x – 3 =
1 _____
1 _____
e x – 3 ◊ ec = A ◊ ex – 3
Thus, the domain of the above function is R – {3} 28. The equation of tangent at any point (x, y) to the curve y = f (x) is Y – y = f ¢ (x) (X – x) Intercept on y-axis is y – x f ¢ (x) It is given that,
y – x f ¢ (x) = x3
dy y – x ___ = x3 dx dy y fi ___ – __ x = – x2 dx fi
which is a linear differential equation. Thus
IF =
dx ___ e – Ú x =
1 e– log x = __ x
y ◊ e g (x) = Ú (g (x) Ú g¢ (x)) e g (x) dx
fi
y ◊ e g (x) = e g (x) (g (x) – 1) + c
When x = 0, y = 0, then c = 1 Thus, the curve is
dy dx Ú ___y = – Ú (x_______ – 3)2 1 log |y| = C + _____ x–3 y = e
IF = e Ú g¢ (x) dx = e g (x)
Therefore, the solution is
dy dx _______ fi ___ 2 y = – (x – 3) fi
y ¢ (x) + y (x) g ¢ (x) = g (x) g ¢ (x)
dy fi ___ + g ¢ (x) y = g (x) g ¢ (x) dx which is a linear differential equation.
27. The given differential equation is
9 27 ___ 3x x3 18 y = ___ – __ = – __ + ___ = = 9 2 2 2 2 2
y ◊ e g (x) = eg (x) (g (x) – 1) + 1
When
x = 2, then y = (0 – 1) + 1 = 0.
30. Given equation is x
6 Ú f (t) dt = 3xf (x) – x3 – 5 0
fi
6f (x) = 3f (x) + 3x f ¢ (x) – 3x2
fi
3f (x) = 3x f ¢ (x) – 3x2
fi
f (x) = x f ¢ (x) – x2
dy y = x ___ – x2 dx dy fi x ___ – y = x2 dx dy y fi ___ – __x = x dx fi
which is a linear differential equation
dx ___ 1 IF = e – Ú x = e– log x = __ x
Hence, the solution is y __x = Ú dx + c
Differential Equation
y fi __x = x + c When x = 1, y = 2, then c = 1 Hence, the equation of the curve is y __x = x + 1 when x = 2, then y = 6. Thus, the value of f (2) is 6.
fi
dv v + x ___ = v + sec v dx
fi
dv x ___ = sec v dx
dv dx fi ____ = ___ x sec v fi
dv dx = Ú ___ Ú ____ x sec v
fi
dx Ú cos v dv = Ú ___ x
dy fi ___ – tan x ◊ y = 2x sec x dx
fi
sin (v) = log |x| + C
which is a linear differential equation.
fi
y sin __ x = log |x| + C
31. The given differential equation is
Thus,
y¢ – y tan x = 2x sec x
– Ú tan x dx
IF = e
log cos x
= e
= cos x
( )
( )
p which is passing through 1, __ , so, c = 1/2 6
Hence, the solution is
y ◊ cos x = Ú 2x ◊ sec x ◊ cos x dx + c
Hence, the equation of the curve is
fi
y ◊ cos x = Ú 2x ◊ dx + c
fi
y ◊ cos x = x2 + c
33. The given differential equation is
when x = 0, y = 0, then c = 0 Thus, the equation of the curve is
p p 2 1__ ___ When x = __ , then y ◊ ___ = 4 ÷2 16 p 2 y = ____ __ 8÷2
When
p 2p 2 x = __ , then y = ____ 3 9
dy xy x4_____ + 2x fi ___ – _____ 2 = ______ dx 1 – x ÷1 – x2 which is a linear differential equation.
( ) (
\
( )
__ 2p 2 dy fi ___ = ÷ 3 ◊ ____ + 9 dx
dy 2p 2 ___ 4p fi ___ = ____ __ + 3 dx 3÷3
( ) (
)
)
y ◊ ÷1 – x2 = Ú (x4 + 2x) dx + c
fi
x5 y ◊ ÷1 – x2 = __ + x2 + c 5
_____
_____
dy p 2 p fi ___ = ____ __ + ___ __ dx 8÷2 ÷ 2
x5 y ◊ ÷1 – x2 = __ + x2 5
fi
x5 x2 y = _______ _____ + ______ _____ 5÷1 – x2 ÷1 – x2
...(i)
which is a homogeneous differential equation. Let y = v x dy dv fi ___ = v + x ___ dx dx
__
3 ÷ ___ 2 x5 x2 Now, __ _______ _____ + ______ _____ dx 2 2 3 5 1 – x ÷ 1 – x ___ – 2
Ú
(
÷
÷
_____
= ÷ 1 – x2
which is passing through (0, 0), so c = 0 Hence, the equation of the curve is
dy p p 2 p __ Again, ___ – tan __ ◊ ____ __ = 2 ◊ __ ◊ ÷2 4 8÷2 4 dx
( )
1 __ log (1 – x2)
e 2
_____
4p ___ 3
dy y y 32. Given ___ = __ + sec __ x dx x
x – Ú _____ dx 1 – x2 =
IF = e
Hence, the solution is
dy p 2p 2 p Also, ___ – tan __ ◊ ____ = 2 ◊ __ ◊ 2 3 9 3 dx
( )
y 1 sin __ x = log |x| + __ 2
dy xy x4 + 2x ___ + _____ 2 = _______ _____ dx x – 1 ÷ 1 – x2
y ◊ cos x = x2
fi
4.101
)
4.102 Integral Calculus, 3D Geometry & Vector Booster __
(
Ú
)
÷
__
2–4 Clearly, g (0) = _____ < 0 22
3 ÷ ___ 2 x2 = __ ______ _____ dx 2 3 1 – x ÷ ___ – 2 3 ÷ ___ 2
(
)
2
x = 2 ◊ Ú ______ _____ dx 0 ÷ 1 – x2 =
p __ 3 2 ◊ sin2q dq, 0
Ú
p __ 3 (1 0
Ú
=
sin 2q 3 = q – _____ 2 0
(Let x = sin q)
– cos 2q) dq
( ) p 1 2p = ( __ – __ sin ___ 3 2 ( 3 ))
(
__
) )
g (0) ◊ g (– 1) < 0
(
)
Hence, g (x) has a root in between (– 1, 0). 35. Let the family of circles be
x2 + y2 – a x – a y + c = 0
...(i)
( )
d 2y dy 2 d 2y 2 + 2y ___2 + 2 ___ – a ___2 = 0 ...(ii) dx dx dx Eliminating a between Eqs (i) and (ii), we get
)
( )
dy 2x + 2y ___ d2y 2 2 d y dy dx ___ 2 + 2y ___2 + 2 ___ – __________ 2 = 0 dx dy dx dx 1 + ___ dx
3 p ÷ = __ – ___ 3 4
( )
dy (1 + ex) ___ + y ◊ ex = 1 dx
(
( (
Again differentiating w.r.t. x, we get
34. We have (1 + ex) y ¢ + y ex = 1 fi
)
Differentiating w.r.t. x, we get dy dy 2x + 2y ___ – a – a ___ = 0 dx dx
p __
(
and
3 1 2 1 + __ e – __ 1 – __ e e ___________ _______ g (– 1) = = > 0 1 2 1 2 1 + __ e 1 + __ e
)
(
(
)
(x + 2)2 4 \ _______ > __ > 1 3 3
)
dy ex 1 fi ___ + _______ ◊ y = _____ x dx (1 + ex) 1+e
2x + 2y y¢ fi 2 + 2y y¢¢ + 2 (y¢)2 – _________ y¢¢ = 0 1 + y¢
which is a linear differential equation.
x + y y¢ fi 1 + y y¢¢ + (y¢)2 – _______ y¢¢ = 0 1 + y¢
Thus,
IF =
ex x dx Ú _______ e (1 + e ) =
elog |1 + ex| = (1 + ex)
Multiplying both sides of Eq. (i) by IF and integrating, we get fi
y ◊ (1 + ex) = Ú dx + c
fi
y ◊ (1 + ex) = x + c
When
x = 0, y = 2, then c = 4
(
)
fi (y – x) y¢¢ + (1 + y¢ + y¢ 2) y¢ + 1 = 0 Thus,
P = (y – x) and Q = (1 + y¢ + y¢2)
and P + Q = (– x + 1 + y + y¢ + y¢2) 36. Given curve is dy (x2 + xy + 4x + 2y + 4) ___ – y2 = 0, x > 0 dx
Hence, the solution is
y ◊ (1 + ex) = x + 4
fi
dy {(x + 2)2 + y (x + 2)} ___ – y2 = 0 dx
fi
x+4 y = _______ (1 + ex)
fi
dy {(x + 2)2 + y (x + 2)} ___ = y2 dx
Thus,
y (– 4) = 0
fi
(1 + ex) – (x + 4) ex y¢ = _________________ (1 + ex)2 x
Let
dy ___________________ y d (x + 2) – (x + 2) dy fi ___ y = (x + 2)2 x
(1 + e ) – (x + 4) e g (x) = _________________ (1 + ex)2
(
)
dy y ______ fi ___ y = – d (x + 2) Integrating, we get
Differential Equation
y log |y| = c – ______ (x + 2)
When
x = 1, y = 3, then c = 1 + log3
Thus,
y y log __ + ______ = 1 3 (x + 2)
For
y = (x + 2), then
( )
( ) ( )
x + 2 log _____ = 0 3 x+2 fi _____ = 1 3
fi
x=1
For
y = (x + 2)2, then
(x + 2)2 log _______ + (x + 2) = 1 3
(
(
)
)
(x + 2)2 4 Now for x > 0, _______ > __ > 1 3 3
(
)
(x + 2)2 log _______ + (x + 2) > 2 3
So, it has no solution.
4.103
Chapter
5
Vectors
Concept Booster 1.1 Introduction A study of motion involves the introduction of a variety of quantities that are used to describe the physical world. For examples, distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories—vectors and scalars. A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is described by its magnitude only. The emphasis of this unit is to understand some fundamentals about vectors and to apply the fundamentals in order to understand motion and forces that occur in two dimensions.
For example, the rotation of a rigid body through finite angles have both magnitude and direction but do not satisfy the vector law of addition. Therefore, it is not a vector.
1.3 Representation
of
vectors
Directed line segment Any given protion of a given straight line where the two end-points are distinguished as the initial and the terminal points is called a directed line segment. The directed line segment with the initial point P and the ___›
terminal point Q is denoted by the symbol PQ or PQ .
1.2 Physical quantities The physical quantity is divided into two categories: (a) scalar quantities (b) vector quantities (a) Scalar quantities A quantity which has only magnitude and not related to any fixed direction in the space is called the scalar quantity. For example, mass, length, volume, density, time, temperature, etc., are all scalar quantities. If the unit of measurement is defined, a real number is sufficient to represent a scalar quantity. Thus, in this chapter, we shall represent scalars by real numbers. (b) Vector quantities A quantity which has magnitude as well as direction is called a vector. For exmaple, force, velocity, acceleration, displacement, momentum, etc. are all vector quantities. Notes: Quantities having magnitude and direction but not obeying the vector law of addition will not be treated as vectors.
The two end-points P and Q are not interchangeable. A directed line segment is called a vector if it has three following characteristics.
_ __›
(i) Length: The length of PQ is denoted by the symbol ___›
|PQ | . (ii) Support: The line of unlimited length of which a directed line segment is a part is called the support. (iii) Sense: The sense of the directed line segment is from its initial point to its terminal point. We genrally denote the vector by bold letter or by a single letter with an arrow or by a letter with a bar over its head, _ __› _› _ i.e. a, a , a denote the vector PQ .
5.2 Integral Calculus, 3D Geometry & Vector Booster
1.4 types
of
vectors
Like vectors
(i) Zero or Null vector A vector whose initial or terminal points are identical or coincident, is called a zero vector. It is also known as null vector. It is denoted by 0 and its direction is indeterminate. ___› _ ___› __› Thus AA , BB and CC are zero vectors. Geometrically, it represents a point.
Collinear vectors are called like vectors if their direction are the same.
(ii) Proper vector
Collinear vectors are called unlike vectors if their directions are opposite.
Any non-zero vector is called a proper vector. (iii) Unit vector If the magnitude of a vector is unity, it is called a unit vector. _› If |a | = 1, then a is called a unit vector. It is generally a denoted as and is defined as = ___ _› . |a | (iv) Co-initial vectors Two or more vectors are said to be co-initial vectors, if they have the same initial point.
_ __›
_ __›
Here, PQ , and QR vectors are like vectors. Unlike Vectors
___›
___›
Here, PQ and QR are unlike vectors. Difference between parallel and collinear vectors Every collinear vectors are parallel, whereas every parallel vectors need not be a collinear. Notes 1. Two non-zero vectors a and b are collinear (or parallel) if a = l b, " l ŒR. 2. If a = a1i + a 2 j + a3k and b = b1i + b2j + b3k are two collinear (or parallel) vectors, then a1 __ = b1
__› _ ___› ___› ___› _ __›
Here, OP , OQ , OR , OS , OT are co-initial vectors. (v) Co-terminal vectors Two or more vectors are said to be co-terminal vectors if they have the same terminal point.
a2 __ = b2
a3 __ . b3
(vii) Coplanar vectors Three or more vectors are said to be coplanar if they lie in the same plane, or they are parallel to the same plane.Three vectors a, b and c are coplanar if any one of their is a linear combination of the other two vectors. i.e
a = x b + y c
or
b = x a + y c
or
c = x a + y b
(viii) Free vector If the origin of a vector is not specified, it is called a free vector.
_ __› _ __› __› _ __› _ __› _
Here, PM , DM , CM , BM , AM are co-terminal vectors. (vi) Collinear or parallel vectors Two vectors are said to be collinear vectors if their supports are parallel or the same irrespective of their direction. Collinear vectors are also called parallel vectors.
___›
__› _ › _
_› ___›
_ ›
Here, OA , = a , OB = b , OC = c › _ › _
_ ›
Thus a , b , and c are free vectors.
Vectors
5.3
(ix) Localized vector For a vector of given magnitude and direction, if its initial point is fixed in space, the vector is called a localized vector.
It is also stated as, If two vectors are represented by the two sides of a triangle taken in order, their sum is represented by the third side of the triangle, taken in reverse order.
(x) Position vector
Parallelogram law of addition of vectors
Let O be the origin and P be a point in the space. The position vector of P is OP.
Let the parallelogram OACB, where _ __›
_›
_ __›
___›
___›
___›
___›
___›
_ ›
OA = a and OB = b
If a and b are the position vectors of the points A and B, then ___›
___›
___›
AB = OB – OA = b – a. (xi) Negative of a vector The vector, which has the same magnitude as the vector a but has the direction opposite to that of a, is called the negative of a and is written as – a. (xii) Equality of two vectors Two vectors are (a) the same (b) the same (c) the same
1.5 Algebra
of
= OA + OB
= a + b .
Properties of addition of vectors
vectors
(i) Addition of vectors is commutative, i.e. _›
1.5.1 Addition of vectors Let a and b be two given vectors. Take a point O in the space. ___›
_ ›
_›
This method of addition of two vectors is called the parallelogram law of addition vectors.
said to be equal if they have length or parallel supports sense.
It is possible to develop an algebra of vectors which is useful in the study of geometry, mechanics and other branches of applied mathematics.
___›
Thus, OC = OA + AB
_›
Let OA , = a, AB _ ›= b, so that the terminal point of a is the initial point of b .
_ ›
_ ›
_›
a + b = b + a (ii) Addition of vectors is associative, i.e. _ ›
_›
_ ›
_›
_ ›
_ ›
(a + b ) + c = a + (b + c )
(iii) Additive identity exists, i.e. _›
_ ›
_ ›
_›
_›
a + 0 = a = 0 + a (iv) Additive inverse also exists, i.e. _ ›
_ _›
_›
_ _›
_ ›
a + (– a ) = 0 (– a ) + a
Polygon law of vector addition Let OABCDE be a polygon where
___›
The vector OB is defined as the vector sum (or resultant) _› _ › of a and b . ___›
___›
___›
_ ›
_›
Thus, OB = OA + AB = a + b which is known as the triangle law of vector addition.
___›
__› _ › _
_› _ __›
__› _ › _
_›
___›
_ ›
OA = a , AB = b , BC = c , CD = d and DE = e .
5.4 Integral Calculus, 3D Geometry & Vector Booster Applying the triangle law, we have.
(i) If the rotation from OX to OY is in the anti-clockwise direction and OZ is directed upwards, the system is called the right handed system. (ii) If the rotation from OX to OY is in the clockwise direction and OZ is directed upwards, the system is called the left handed system.
_ _ __› _ __› _ › ___› › OB = OA + AB = a + b
___›
___›
___›
_ ›
_›
_ ›
___›
___›
___›
_›
_›
_›
___›
___›
___›
_›
_›
_›
OC = OB + BC = (a + b ) + c
_›
OD = OC + CD = (a + b + c ) + d _›
_›
OE = OD + DE = (a + b + c + d ) + e Thus, to find the sum of more than two vectors, a polygon is formed. Therefore, this method is known as the polygon law of vector addition. Note: If the initial point of the first vector and the final point of the last vector are the same, the sum of the vectors is a zero vector.
1.5.2 Subtraction of vectors If a and b be any two vectors, the subtraction of b from a is defined as the addition of – b to a and is written as a + (– b) = a – b, i.e. _›
_ ›
_ ›
_›
a + (– b ) = a – b
1.7 Position vector
of a
point
in a
space
Let O be a fixed point, known as the origin, and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis, respectively, in such a way that they form a right handed system. The plane XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane, respectively. Let P be a point in a space such that its distances from yz-, zx- and _ › › _ › _ xy-plane be a, b, c respectively and i , j , k are the vectors along x, y and z axes, respectively. Let
OA = a, OB = b and OC = c
Now, and
OP = OL + LP = OA + AL + LP = OA + OB + OC. = a i___________ + b j + c k. |OP| = ÷ a 2 + b2 + c2 .
1.8 Linear Combination 1.5.3 Multiplication of a vector by a scaler Let_ m be any scaler and a be any vector. Then its product › m a is called_ the multiplication of a vector by a scaler. › _› If a and be two vectors and m, n are scalers, then
_›
_›
_›
(i) m (a ) = m a = (a ) m _›
_›
(ii) m (n a ) = (m n) a (iii) (iv)
_› (m + n) a = _ › _› m (a + b ) =
1.6 Left
and
_› ma + _›
m a + m b
right handed orientation
__›
Z
Z
X
Y Z
Right Handed System
__›
__›
A system of vectors a1 , a2 , ..., an is said to be linearly independent if_there exist scalars x1, x2,...., xn (all zero) such _› _› › that x1a 1 + x2a 2 + ... + xna n = 0
X
O Y
X
x1a1 + x2a2 + ... + xnan = 0.
1.10 Linearly independent vectors
Y
O
1.9 Linearly dependent vectors A system of vectors a1, a2, ..., an is said to be linearly dependent if there exist scalars x1, x2, ..., xn (not all zero) such that
_› n a _ ›
Y
A vector r is said to be a linear combination of the vectors a, b, c, ..., if there exist scalars x, y, z, ... such that r = x a + y b + z c + ...
1.11 Section formulae Z X
Left Handed System
1.11.1 Internal section If a point R (r) divides the line segment joining the points P(r1) and Q (r2) internally in the ratio m : n, then
Vectors
m r2 + n r1 r = __________ . m+n
From the figure, we have by using triangle rule,
5.5
1.11.4 Centroid The point of intersection of the medians of a triangle is called its centroid. Let A (r1), B (r2) and C (r3) be the vertices of a triangle ABC and G (r) be its cen troid. Then the position vector r of the centroid is given by
r1 + r2 + r3 r = __________ . 3
m PR __ ____ = n RQ
OR – OP __ m fi _________ = n OQ – OR fi
r – r1 __ m ______ r 2 – r = n
fi
n (r – r1) = m (r2 – r)
fi
r (m + n) = m r2 + n r1
fi
m r2 + n r1 r = __________ m+n
1.11.2 External section If a point R(r) divides the line segment joining the points P(r1) and Q(r2) externally in the ratio m : n, then
1.11.5 Centroid of a tetrahedron Let _the position vectors of the points A, B, C and D are _› _› › _› a , b , c and d respectively. The position vector r of the centroid is
a + b + c + d r = ____________ 4
m r2 – n r1 r = _________ m – n . It can be proved by using triangle rule.
1.11.6 Incentre The point of intersection of the angle bisectors of a triangle is called its incentre.
1.11.3 Mid-Point formula If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio 1 : 1, then
r2 + r1 r = ______ . 2
Let the position vectors of the points A, B, C are r1, r2, r3, respectively, and r be the position vector of the incentre I. The position vector r of the incentre is given by
5.6 Integral Calculus, 3D Geometry & Vector Booster
a r1 + b r2 + c r3 r = ______________ a+b+c
1.11.7 Circumcentre The point of intersection of the perpendicular bisector of a triangle is called its circumcentre. Let the position vectors of the points A, B, C are r1, r2, r3, respectively, and r be the position vector of the circumcentre O. The position vector r of the incentre is given by (sin 2A) r1 + (sin 2B) r2 + (sin 2C) r3 r = ______________________________ sin 2A + sin 2B + sin 2C
1.13 straight line 1.13.1 Equation of a line passing thorugh a point and parallel to a vector The equation of a line passing through a point A with position vector r1 and parallel to a vector m is given by
r = r1 + l m
Now,
OP = OA + AP
fi
OP = OA + l M
fi
r = r1 + l m
1.11.8 Orthocentre The point of intersection of the perpendiculars of a triangle is called its orthocentre. Let the position vectors of the points A, B, C are r1, r2, r3, respectively and r be the position vector of the orthocentre H. The position vector of the orthocentre is
1.13.2 Equation of a line passing through two points A (r1) and B (r2) is
(tan A) r1 + (tan B) r2 + (tan C) r3 r = ___________________________ tan A + tan B + tan C
1.12 Bisector
of
angle
between
vectors A
and
r = r1 + l (r2 – r1)
Now,
OP = OA + AP
fi
OP = OA + l AB
fi
OA + l (OB – OA)
fi
r = r1 + l (r2 – r1).
b
(i) If a is not parallel to b, the vectors along the bisectors (internal) of angle between a and b is given by
___› OM =
(
_ ›
a l ___ _› + |a |
__›
)
b ___ _› |b |
(ii) If a is not parallel to b, the vectors along the bisectors (external) of the angle the between a and b is given by
(
_›
___› a ON = l ___ _› – |a |
_›
)
b ___ _› |b |
1.14 Plane 1.14.1 Vector Equation The vector equation of a plane passing through a point having position vector a and the normal to a vector n is given by r ◊ n = d
Vectors
N n
P
r
5.7
Let a and b be two non-zero vectors and q the angle between them. Its scalar product is denoted as a ◊ b and is defined as a ◊ b = ab cos (q ), 0 £ q £ p.
O
Let
OP = r and ON = n
It is given that ON ^ NP NP ◊ ON = 0 (OP – ON) ◊ ON = 0 (r – n) ◊ n = 0 r ◊ n – n ◊ n = 0 r ◊ n = n◊ n r ◊ n = d.
fi fi fi fi fi fi
2.2 Geometrical interpretation
of a ◊ b
1.14.2 Equation of a plane in normal form The vector equation of a plane normal to a unit vector and at a distance from the origin is r ◊ = d.
We have a ◊ b = a b cos (q ) = a [b cos (q )] = a OM a◊b Thus, OM = ____ a a◊ b Projection of b on a is ____ a .
Let O be the origin and ON be the perpendicular from O to the given plane such that ON = d . Let P be a point on the plane with position vector r so that OP = r Now, ON ^ NP
Thus, geometrically product of two vectors a and b is defined as the product of the magnitude of the first vector and the projection of the 2nd vector on the 1st vector.
2.3 Properties
of
dot product
NP ◊ ON = 0
fi
(OP – ON) ◊ ON = 0
= ba cosq
fi
(r – d ) ◊ d = 0
= ab cos q
fi
(r ◊d – d ◊ d ) = 0
fi fi
(r ◊ – d) = 0 r ◊ = d
(ii) If q acute, then a ◊ b > 0
Scalar product of two vectors 2.1 Definition If the product of two vectors is a scalar, it is known as scaler product or dot product of two vectors.
two vectors
(i) Dot product is commutative, i.e. a ◊ b = b◊ a
fi
which is the required equation of the plane.
of
Now
(iii) (iv) (v) (vi) (vii) (viii) (ix)
b ◊ a = ba cos (– q)
= a ◊b
If q obtuse, then a ◊ b < 0 If a and b are perpendiculars, then a ◊ b = 0 If a and b are like vectors, then a ◊ b = ab If a and b are unlike vectors, then a ◊ b = – a b Maximum value of a, b is ab Minimum value of a ◊ b is – ab a ◊ a = |a|2 = a2
5.8 Integral Calculus, 3D Geometry & Vector Booster a ◊b (x) Projection of a on b = ____ . |b|
Eliminating x, y and z from Eqs. (i), (ii) and (iii), we get
a. b (xi) Projection of b on a = ____ . |a| (xii) Orthonormal triad vectors If three vectors which are mutually perpendicular to each other, they are called orthonormal triad vectors. Let , , be three unit vectors which are mutually perpendicular to each other. Thus . = . = . and
a b c a⋅a a⋅b a⋅c = 0 b⋅a b⋅b b⋅c
2.4 Component vector A
of a
vector B
along and
perpendicular
to
. = 0 = ◊ = ◊
Therefore,
0 0
Here,
0
1 0
0
0 1
b = OM + MB Component of a vector b along a = OM
(xiii) If
a = a1 + a2 + a3
and
b = b1 + b2 + + b3 ,
then
a ◊ b = a1b2 + a2b2 + a3b3
|a| = ÷ a 21 + a22 + a 23
___________
|b| = ÷
Thus,
a1b1 + a2b2 + a3b3 ___________ ___________ cos (q) = ________________________ ÷a 21 + a22 + a23 ÷b 21 + b22 + b 23
(xiv) Any vector a can be written as a = (a ◊ ) + (a ◊ ) + (a◊ )
(xv) |a + b + c| 2 = |a|2 + |b|2 + |c| 2 + 2 S (a ◊ b) (xvi) If a, b, c are coplanar vectors, then
x a + y b + z c = 0
= (OB cos q) = (b cos q) b a◊ b = _____ a b a ◊ b = ____ a a◊ b __ a = ____ a a a ◊ b = ____ 2 a. |a| Component of a vector b perpendicular to a vector a = MB
( ) ( )
= OB – OM = b – OM
a◊ b = b – ____ 2 a |a|
( )
2.5 Physical significance two vectors
a b c a⋅a a⋅b a⋅c = 0 b⋅a b⋅b b⋅c
As we know that three coplanar vectors are always linearly dependent. Thus,
= (OM)
( ) ( )
___________ b21 + b22 + b23
and
OB = OM + MB
1
of the
dot prodcut
F
Taking dot product a and b with Eq. (i), we get x (a ◊ a) + y (a ◊b) + z (a ◊ c) = 0
...(ii)
x (b ◊ a) + y (b ◊ b) + z (b ◊ c) = 0
...(iii)
A force acting on a particle is said to be work done if the particle is displaced in a direction which is not perpendicular to the force applied.
...(i)
of
q A
B
Vectors
Let a force F acts on a particle P and the particle P gets displaced from the position A to B, where AB is not perpendicular to the applied force F. Let AB = d. \ Work done = F d (cos q) = F ◊ d = (Force) ◊ (Displacement). Note: The work done by the resultant of the number of forces F1, F2, ..., Fn in a displacement d of a particle is equal to the sum of the work done by the forces separately, i.e. Work done = F1 ◊ d + F2 ◊ d + ... + Fn ◊ d.
= (F1 + F2 + ... + Fn) ◊ d
= R ◊ d
5.9
Area of a parallelogram OACB = (base) (height) = (OA) (BM) = (a) (b sin q) = (ab sin q) = |a × b| Thus, |a × b| reprtesents the area of a parallelogram with adjacent sides represented by the vectors a and b.
3.3 Properties
of
vector product
of
two vectors
(i) Vector product is not commutative, i.e. a × b π b × a (ii) a × b = – b × a (iii) If a and b are parallel vectors, then a × b = 0 (iv) If a and b are collinear vectors, then a × b = 0
where R = F1 + F2 + ... + Fn
CROSS PRODUCT OF TWO VECTORS 3.1 Introduction If the product of two vectors is a vector, it is known as the vector product or cross product or outer product of two vectors.
(v) (m a) × b = a × (m b) = m (a × b), where m is any scalar. (vi) a × (b + c) = (a × b) + (a + c). (vii) Unit vector perpendicular to the plane of a and b is a×b = ± ______ . |a × b| (viii) A vector of the magnitude l and perpendicular to the plane of a and b is
(a × b) ± l _______ . |a × b|
(ix) If q be the angle between two vectors a and b, then |a × b sin (q) = ______ . |a||b| Let a and b be two_ vectors. The cross product of _two _› _ › › › vectors is denoted as a × b and is defined as a × b = (ab sin q) , _ › _› where q the angle between a _ and b and is unit vector › perpendicular to the plane of a and b. Here a, b, form a right handed system.
3.2 Geometrical Interpretation
of
×
Let OACB be a parallelogram with O origin. Let OA = a and OB = b and –AOB = q
(x) The area of DOAB is given by 1 __ |a × b|, 2 where OA = a, OB = b 1 (xi) The area of DABC = __ |a × b + b × c + c × a|, 2 position vectors of the points where a, b and c are the A, B and C, respectively. 1 (xii) The area of a parallelogram = __ |d1 × d2|, 2 where d 1 and d 2 are the diagonals of the parallelogram. (xiii) If A, B and C are three collinear, points then (a × b + b × c × c × a) = 0, where a, b and c are the position vectors of the points A, B and C, respectively. (xiv) A unit vector perpendicular to the plane of
5.10 Integral Calculus, 3D Geometry & Vector Booster –ABC is given by (a × b + b × c + c × a) = ± _____________________ . |a × b + b × c + c + a|
(xv) If a, b and a × b form a right handed system, any vector r is a linear combination of a, b and a × b, i.e. r = x a + y b + z (a × b), where x, y and z are scalars. (xvi) Orthonormal Triad Vectors If three vectors which are mutually perpendicular to each other, they are called orthonormal triad vectors. Let , , be three unit vectors which are mutually perpendicular to each other. Thus
×
and
×
=0= = ,
= abc sin q cos j
= (a b sin q) (c cos j)
= |a × b| (c cos j)
= (area of the IIgm OALB) × (height)
,
= Volume of a parallelopiped.
–
4.2 Properties
× ×
,
, 0
=
×
= ,
×
=
, – 0
– 0
(xvii) If a = a1 + a2 + a3 and b = b1 + b2 + b3 , then iˆ a × b = a1 b1
We know that [a, b, c] = a ◊ (b × c)
Therefore, ×
Here OA, OB and OC are the coterminous edges of a parallelopiped represented by the vectors a, b and c respectively.
kˆ a3 b3
ˆj a2 b2
( xviii) Lagranges identity For any two vectors a and b, a.a a.b (a × b)2 = (ab)2 – (a ◊ b)2 = . a.b b.b
Scalar Triple Product of Vectors
(i) Let
of
scalar triple product
vectors
a = a1i + a2 j + a3k,
b = b1i + b2 j + b3k
and
c = c1i + c2j + c3k
Then
[a b c] = a ◊ (b × c)
of
a1 = b1 c1
a2 b2 c2
a3 b3 c3
(ii) a ◊ (b × c) = b ◊ (c × a) = c ◊ (a × b) i.e.
[a b c] = [b c a] = [c a b]
(iii) a ◊ (b × c) = – b ◊ (a × c) i.e.
[a b c ] = – [b a c]
(iv) [ka b c] = k [a b c], where k ΠR
4.1 Introduction If the product of three vectors is a scalar, then it is known as scalar triple or box product or mixed product of three vectors. Let a, b and c are three vectors. Its scalar triple product is denoted as a ◊ (b × c) and is defined as a ◊ (b × c) = abc sin q cos j _ ›
_›
where q the angle between a and b and j the angle between c and a × b. Geometricel interpretation of [a, b, c]
(v) [la mb nc] = lmn [a b c] [a + b + c, d] = [a, c, d] + [b, c, d] If any two vectors are same, then [a, b, c] = 0 If any two vectors are parallel or collinear, then [a, b, c] = 0 If any one of them be a zero vector, then [a, b, c] =0 (x) If a, b, c are coplanar vectors, then [a, b, c] = 0
(vi) (vii) (viii) (ix)
Vectors
If a, b, c form a right handed system, then [a, b, c] > 0 If a, b, c form a left handed system, then [a, b, c] < 0 For any three vectors a, b and c, [a + b, b + c, c + a] = 2 [a, b, c] If a, b, c are coplanar vectors, then a + b, + b + c, c + a are also coplanar vectors If a and b are non-zero and non-collinear vectors, then [a, b, i] i + [a, b, j] j + [a, b, k] k = a × b (xvi) For any three vectors a, b, c, (xi) (xii) (xiii) (xiv) (xv)
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a
c⋅b
c⋅c
(xvii) If a, b, c, m, n are non-zero vectors, then a b c [a, b, c] ◊ [m × n] = a ⋅ m b ⋅ m c ⋅ m a ⋅n b⋅n c ⋅n
(xix) Any vector r can be expressed as a linear combination of three non-coplaner vectors a, b, c, then
r = ma + n b + p c
i.e.
[rbc] [rca] [rab] r = _____ a + _____ b + _____ c [abc] [abc] [abc]
(xx) If four points a, b, c and d are coplanar vectors, then [a b c] = [a b d] + [b c d] + [c a d]
VECTOR TRIPLE PRODUCT 5.1 Introduction If the product of three vectors is also a vector, then it is known as vector triple product. Let a, b and c be any three vectors, then its vector triple product is denoted as a × (b × c) and is defined as a × (b × c) = (a ◊ c) b – (a ◊ b) c
5.2 Geometrical r r r r
0 = l (a ◊ b) + m (a ◊ c)
l (a ◊ b) + m (a ◊ c) = 0
l (a ◊ b) = – m (a ◊ c)
(a ◊ b) (a ◊ c) _____ = – _____ = l (say) m l Therefore,
a × (b × c) = l [(a◊c) b – (a◊b) c]
Let a = i, b = i + j and c = i + j + k Now, i j k (b × c) = 1 1 0 = i – j 11 1
| |
2
Let fi fi fi
significance of
a × (b × c)
= a × (b × c) ^ a and r ^ (b × c) ^ a and r lies in the plane of b and c is orthogonal to a and coplanar with b and c
5.3 Explanation
of
Vector Triple Product
Since r is coplanar with b and c, then we can write r = l b + m c Thus,
a × (b × c) = l b + m c
a ◊ (a × (b × c)) = l (a ◊ b) + m (a ◊ c)
5.11
and a × (b × c) = i × (i – j) = – i × j = – k Also, (a ◊ c) b – (a ◊ b) c = i + j – i – j – k = – k Thus,
– k = l × – k
fi Hence,
l=1 a × (b × c) = (a◊c) b – (a◊b) c
5.4 Properties
of
vector triple product
(i) A unit vector perpendicular to a and coplanar with b and c is given by
a × (b × c) ± ___________ . |a × (b × c)|
If a, b and c be three non-zero vectors, then a × (b × c) + b × (c × a) + c × (a × b) = 0 If a, b and c be three non-zero vectors, then a × (b × c) + b × (c × a) + c × (a × b) are coplaner. The vector product of three vectors is not associative, i.e. a × (b × c) π (a × b) × c (v) If a, b and c be three non-zero vectors, then [(a × b), (b × c), (c × a)] = [a, b, c]2 (ii) (iii) (iv)
Proof: We have
(b × c) × (c × a)
= d × (c × a) (Let d = (b × c)]
= (d ◊ a) c – (d ◊ c) a
= (b × c ◊ a) c – (b × c ◊ c) a
= [a, b, c] c – 0
= [a, b, c,] c
Thus, [(a × b), (b × c), (c × a)]
= (a × b) ◊ ((b × c) × (c × a))
5.12 Integral Calculus, 3D Geometry & Vector Booster
= (a × b) ◊ [a, b, c] c
= ((a × b) ◊ c) [a, b, c]
= [a, b, c] [a, b, c]
= [a, b, c]2
(vi) If a, b and c be coplanar vectors, then (a × b), (b × c), (c × a) are also coplanar. Proof: As we know that, if a, b, c are coplanar vec-
tors, then [a, b, c] = 0 We have [(a × b), (b × c), (c × a)]
= [a, b, c]2 = 0
Thus, (a × b), (b × c), (c × a) are coplanar. (vii) For any vector a, i × (a × i) + j × (a × j) + k × (a × k) = 2a Proof: We have i × (a × i) = (i ◊ i) a – (i ◊ a) i = a – (i◊a) i = a – a1i, where a = a1i + a 2 j + a3k Similarly, j × (a × j) = a – a2 j and
k × (a × k) = a – a3k
Thus,
i × (a × i) + j × (a × j) + k × (a × k)
= a – a1i + a – a2 j + a – a3k
= 3 a – (a1i + a2 j + a3 k) = 3 a – a = 2 a (viii) If a and b be two non-zero vectors, then i × ((a × b) × i) + j × ((a × b) × j)
+ k × [(a × b) × k] = 2 (a × b).
Proof: Replace a by a × b in property (vii), we get the required result. (ix) If a × (b × c) = (a × b) × c, then (c × a) × b = 0 Proof: We have
a × (b × c) = (a × b) × c
fi
a × (b × c) = – c × (a × b
fi
(a ◊ c) b – (a ◊ b) c = – (c ◊ b) a + (c ◊ a) b
fi
– (a ◊ b) c = – (c ◊ d) a
fi
(c ◊ b) a – (a ◊ b) c = 0
fi
(c × a) × b = 0
(x) If a, b, c and d are four coplanar vectors then, [a, b, c] = [a, b, d] + [b, c, d] + [c, a, d]
Proof: Let the position vectors of the points A, B, C and D are a, b, c and d, respectively. Thus, AB = b – a, AC = c – a, AD = d – a Given A, B, C and D are coplanar, so, AB ◊ AC, AD are coplanar Therefore, AB ◊ (AC × AD) = 0 fi (b – a) ◊ ((c – a) × (d – a)) = 0 fi (b – a) ◊ (c × d – c × a – a × d) = 0 fi (b ◊ c × d) – (b ◊ c × a) – b ◊ (a × d) – (a ◊ c × d) = 0 fi [b, c, d] – [b, c, a] – [b, a, d] = [a, c, d] fi [b, c, d] – [b, a, d] – [a, c, d] = [b, c, a] fi [b, c, a] + [a, b, d] + [c, a, d] = [b, c, a] fi [b, c, a] = [b, c, d] + [a, b, d] + [c, a, d] fi [a, b, c] = [a, b, d] + [b, c, d] + [c, a, d] Hence, the result.
5.5 Scaler product
of
four vectors
If a, b, c and d are four non-zero vectors, then the scalar product of (a × b) and (c × d) is called scalar product of four vectors and is denoted as (a × b) ◊ (c × d) and is defined as (a ⋅ c) (a ⋅ d) (b ⋅ c) (b ⋅ c)
(a × b)◊(c × d) =
We have
(a × b) ◊ (c × d) = (a × b) ◊ p, [Let p = (c × d)]
= a ◊ (b × p) = a ◊ (b × (c × d))
= a ◊ ((b ◊ d) c – (b ◊ c) d)
= (a ◊ c) (b ◊ d) – (a ◊ d) (b ◊ c)
=
(a ⋅ c) (a ⋅ d) (b ⋅ c) (b ⋅ c)
5.5.1 If a and b lie in a plane normal to the plane containing c and d, then (a × b) ◊ (c × d) Proof: it is given that (c × d) is perpendicular to the plane containing c and d. But a and b are normal to the plane containing c and d Thus, c and d lie in the plane containing a and b. Also (a × b) is perpendicular to the plane containing a and b. Therefore, (a × b) is perpendicular to (c × d) also. Hence,
(a × b) ◊ (c × d) = 0
Vectors
5.5.2 If a, b, c and d are four non-zero vectors, then (a × b) ◊ (c × d) + (b × c) ◊ (a × d) + (c × a)◊(b × d) =0 Proof: We have, (a × b) ◊ (c × d) + (b × c) ◊ (a × d) + (c × a) ◊ (b × d) =
(a ⋅ c) (a ⋅ d) (b ⋅ a) (b ⋅ d) (c ⋅ b) (c ⋅ d) + + (b ⋅ c) (b ⋅ d) (c ⋅ a) (c ⋅ d) (a ⋅ b) (a ⋅ d)
= (a ◊ c) (b ◊ d) – (a ◊ d) (b ◊ c)
+ (b◊a) (c◊d) – (c◊a) (b◊d)
+ (c◊b) (a◊d) – (a◊b) (c◊d)
= 0
the plane containing a and b and the other one parallel to the plane containing c and d. 5.7.1 If a, b, c and d are four non-zero vectors, then a × [b × (c × d)] = (b ◊ d) (a × c) – (b◊c) (a × d) Proof: We have,
a × [b × (c × d)]
= a × [(b ◊ d) c – (b ◊ c) d]
= (b ◊ d) (a × c) – (b ◊ c) (a × d) Hence, the result. 5.7.2 If a, b, c and d are coplanar vectors then (a × b) × (c × d) = 0 Proof: As we know that (a × b) is perpendicular
5.6 Vector product
of
four vectors
If a, b, c and d are four non-zero vectors, the vector product of (a × b) and (c × d) is called the vector product of four vectors and is denoted as (a × b) × (c × d) and is defined as (a × b) × (c × d) = [a, b, d] c – [a, b, c] d
5.13
= [a, c, d] b – [b, c, d] a
to a and b vectors and (c × d) is also perpendicular to c and d vectors. Since a, b, c and d are coplanar vectors, so (a × b) and (c × d) are perpendicular to the same plane. Thus, (a × b) and (c × d) are parallel. Therefore, (a × b) × (c × d) = 0 5.7.3 If b, c and d are three non-coplanar vectors, then (a × b) × (c × d) + (c × d) + (a × d) × (b × c) is parallel to a.
Proof: We have,
Proof: We have,
(a × b) × (c × d)
(a × b) × (c × d) = [a, b, c] c – [a, b, c] d Again, (a × c) × (d × b) = – (d × b) × (a × c) = – [d, b, c] a + [d, b, a] c = – [b, c, d] a – [a, b, d] c Also, (a × d) × (b × c) = – (b × c) × (a × d)
= p × (c × d), [Let a × b = p]
= (p ◊ d) c – (p ◊ c) d
= ((a × b) ◊ d) c – ((a × b) ◊ c) d
= [a, b, d] c – [a, b, c] d
Also,
(a × b) × (c × d)
= (a × b) × q, [Let q = (c × d)]
= (a◊q) b – (b◊q) a
= [a◊(c × d)] b – [b◊(c × d)] a
= [a, c, d] b – [b, c, d] a Hence, the result.
5.7 Geometrical interpretation
of
(a × b) × (c × d )
(a × b) × (c × d) is a vector perpendicular to (a × b) and (c × d). But (a × b) is perpendicular to the plane containing a and b. Thus, (a × b) × (c × d) represents a vector coplanar with a and b. Similarly (a × b) × (c × d) represents a vector coplanar with c and d. Hence, (a × b) × (c × d) represents a vector parallel to the line of intersections of two planes, one being parallel to
= – [b, c, d] a + [b, c, a] d
= – [b, c, d] a + [a, b, c] d
...(i)
...(ii)
...(iii)
Adding Eqs. (i), (ii) and (iii), we get (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) = – 2 [b, c, d] a Therefore, (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) is parallel to a.
5.8 Reciprocal system
of
vectors
Let a, b and c be three non-zero vectors such that [a, b, c] π 0. The reciprocal system of vectors are denoted as a¢, b¢, c¢ and are defined as b×c c × a a×b a¢ = _______ , b¢ = _______ and c¢ = _______ . [a, b, c] [a, b, c] [a, b, c]
5.14 Integral Calculus, 3D Geometry & Vector Booster
Properties
of
reciprocal system
of
v ectors
If a, b, c and a¢, b¢, c¢ be reciprocal system of vectors, then (i) a ◊ a¢ = 1 = b ◊ b¢ = c ◊c¢ Proof: We have,
(
)
(b × c◊a) c – (b × c ◊ c) a = ____________________ [a, b, c]2
a ◊ (b × c) = _________ [a, b, c]
[a, b, c] c – 0 = ____________ [a, b, c]2 c = _______ [a, b, c] a Similarly, b¢ × c¢ = _______ [a, b, c]
[a, b, c] = ________ = 1 [a, b, c] Similarly, b ◊ b¢ = 1 and c ◊ c¢ = 1 (ii) a ◊ a¢ + b ◊ b¢ + c ◊ c¢ = 0 (iii) a ◊ b¢ = 0 = a ◊ c¢, b ◊ c¢ = 0 = b ◊ a¢ and c ◊ a¢ = 0 = c ◊ b¢
(
)
c×a a ◊ b¢ = a ◊ _______ [a, b, c]
(
)
(
(
)
)
[a¢, b¢, c¢] = a¢◊(b¢ × c¢)
)
c×a b×c a×b = _______ ◊ ________ × _______ [a, b, c] [a, b, c,] [a, b, c] 1 = ________ [a × b, b × c, c × a] [a, b, c]3 [a, b, c,]2 = _________ [a, b, c]3 1 = _______ [a, b, c] Thus, [a¢, b¢, c¢] [a, b, c] = 1 (v) a × a¢ + b × b¢ + c × c¢ = 0 a+b+c (vi) a¢ × b¢ + b¢ × c¢ + c¢ + a¢ = _________ [a, b, c] Proof: We have,
b×c c×a a¢ × b¢ = _______ × ________ [a, b, c] [a, b, c,]
Hence,
a + b + c a¢, × b¢ + b¢ × c¢ + c¢ × a¢ = _________ [a, b, c]
(x) If a, b and c be three non-coplanar vectors and a¢, b¢ and c¢ be its reciprocal system of vectors, then any vector r can be expressed as r = (r ◊ a¢) a + (r ◊ b¢) b + (r ◊ c¢) c
Proof: We have,
(
b c¢ × a¢ = _______ [a, b, c]
(ix) If a¢, b¢ and c¢ be the reciprocal system of a, b and c, then a, b and c is the reciprocal system of a¢, b¢ and c¢,
a×b a◊c¢ = a ◊ _______ [a, b, c]
a ◊ (a × b) = _________ = 0 [a, b, c] (iv) [a, b, c] [a¢, b¢, c¢] = 1
and
(vii) (a + b + c) ◊ (a¢+ b¢ + c¢) = 3. (viii) The system of unit vectors i, j, k be its own reciprocal.
a ◊ (c × a) = _________ = 0 [a, b, c] Also,
d × (c × a) = __________ [a, b, c]2 (d ◊ a) c – (d ◊ c) a = ______________ [a, b, c]2
b×c a ◊ a¢ = a ◊ _______ [a, b, c]
Proof: We have
(b × c) × (c × a) = _______________ [a, b, c]2
or r = (r ◊ a) a¢ + (r ◊ b) b¢ + (r ◊ c) c¢ Proof: Since a, b, c be three non-coplanar vectors, any vector r can be expressed as a linear combination of a, b and c. Thus, r = x a + y b + z c, where x, y, z Œ R Taking dot product of (b × c), we have r ◊ (b × c) = x a ◊ (b × c) + y b ◊ (b × c) + z c ◊ (b × c) fi r ◊ (b × c) = x a ◊ (b × c) + 0 + 0 = x a ◊ (b × c) fi r ◊ (b × c) = x a ◊ (b × c) r ◊ (b × c) r ◊ (b × c) fi x = ________ = ________ a ◊ (b × c) [a, b, c] fi x = r ◊ a¢ Similarly, y = r◊b¢, = z = r ◊ c¢ (xi) If a, b and c be three non-coplanar vectors, any vector r can be expressed as a linear combination of a, b and a × b, i.e. r = x a + y b + z (a × b).
Vectors
5.15
Exercises (Problems based on Fundamentals)
DOT PRODUCT OF VECTORS 1. If a = 3 i – 2 j + k and b = 2 i + 3 j, find the angle between a and b. 2. If a = i + 2 j – 3 k and b = 3 i + j + 2 k, prove that (a + b) is perpendicular to (a – b). 3. If |a| = 3, |b| = 4, |c| = 5 such that (a + b + c) = 0, find the value of (a ◊ b + b ◊ c + c ◊ a). 4. If a, b and c are mutually perpendicular unit vectors, find |a + b – c|. 5. Find the projection of the vector a = 3 i + j + 2 k on the vector b = j – 2 j = i – j. 6. Find the angle between the vectors a and b such that 2a + b = i + j and a + 2 b = i – j. 7. Find the unit vector perpendicular to each of the vectors i – 2 j + k and 2 i + j – 3 k. 8. Find a unit vector in xy-plane that makes an angle of 45° with the vector i + j and an angle of 60° with the vector 3 i – 4 j. 9. If a, b and c be mutually perpendicular vectors of equal magnitude, show that (a + b + c) is equally inclined to a, b and c. ___ 10. Find a vector of magnitude ÷51 which makes equal angles with the vectors 1 1 a = __ (i – 2 j + 2 k), b = __ (– 4 i – 3 k) and c = j. 3 5 11. Find a vector of magnitude 4, which is equally inclined to vectors a = i + j, b = j + k and c = k + i. 12. Find a unit vector in the plane of (i + 2 j + k) and (i + j + 2 k) and perpendicular to (2 i + j + k). 13. If and are unit vectors and q the angle between q 1 them, show that sin __ = __ | – |. 2 2 14. If , and are unit vectors perpendicular to each other, find the value of ( – ) ◊ ( – ). 15. Let and be two unit vectors and q the angle between them. If ( + ) is a unit vector, find sin (q). 16. If and be two unit vectors inclined to x-axis at angles 30° and 120° respectively, find the value of | + |. 17. If a and b are two vectors and their lengths are a a b 2 a–b2 and b, show that __ 2 – __ 2 = _____ ab a b 18. Let u = i + j, v = i – j, and w = i + 2 j + 3 k. If is a unit vector such that u ◊ = 0 and v ◊ = 0, then find |w ◊ |.
( )
(
) ( )
19. If a, b and c be unit vectors such that a is perpendicular to the plane of b and c and the angle between p b and c is __ , find |a + b + c|. 3 20. If |a| = 3, |b| = 4, |c| = 5 and a ^ (b + c) b ^ (c + a) and c ^ (a + b), find |a + b + c|. 21. Let and are two unit vectors and the angle between them is 60°, find the value of | + / – |. ___
17 ÷ 22. If |a| = 1, |b| = 4, |c| = ____ such that (a + b + 3 c) 3 = 0, find the value of (a ◊ b + b ◊ c + c ◊ a). 23. Constant forces F1 = i – j + k, F2 = – i + 2 j – k and F3 = j – k act on a particle at a point A. Find the work done when the particle is displaced from the point A to B where A = 4 i – 3 j – 2 k and B = 6 i + j – 3 k. 24. Prove that, by vector method, b2 + c2 – a2 cos (A) = ___________ . 2bc 25. 26. 27.
Prove that, by vector method, a = b cos (C) + c cos (B). Prove that, by vector method, cos (A – B) = cos A ◊ cos B + sin A ◊ sin B Prove that, by vector method,
cos (A + B) = cos (A) cos (B) – sin (A) sin (b)
Cross product of two vectors 28. If a = 2 i – j + k, and b = i – 3 j – 5 k, find a × b. 29. Find the area of a triangle whose adjacent sides are determined by the vectors a = 3 i + 4 j and b = 3 i – 3 j + k. 30. Find the area of a parallelogram whose adjacent sides are determined by the vectors a = i + 2 j + 3 k and b = 3 i – 2 j + k. 31. Find the area of a parallelogram whose diagonals are given by the vectors a = i + j + k and b = 3 i – 4 j + k. 32. Find the area of a triangle whose vertices are (3, – 1, 2), (1, – 1, – 1) and (4, – 3, 1). 33. Find a unit vector perpendicular to each of the vectors 2 i – j + k and 3 i – 4 j – k. 34. Prove that a × (b + c) + b × (c + a) + c × (a + b) = 0. 35. If a × b = c × d and a × c = b × d, show that (a – d) is parallel to (b – c). 36. If (a + b + c) = 0, prove that a × b = b × c = c × a.
5.16 Integral Calculus, 3D Geometry & Vector Booster 37. Prove that |a × |2 + |a × |2 + |a × |2 = 2a2. 38. Let , and be three unit vectors such that p a ◊ b = a ◊ c = 0 and the angle between b and c is __ , 6 prove that a = ± 2 (b × c). 39. If a × b = a × c, a π 0 and b π c, prove that b = c + a, " Œ R. 40. Let A = 2 i + k, B = i + j + k and C = 4 i – 3 j + 7 k. Find a vector R satisfying R × b = R × C and R ◊ A = 1. 41. Find the number of vectors of unit length perpendicular to the vectors a = i + j and b = j + k. 42. Let a = x i + y j + z k and b = j, find a vector c for which a, b and c form a right handed system. 43. Let a = i + j and b = 2 j – k. Find the point of intersection of the lines r × a = b × a and r × b = a × b. 44. If a and b are two non-zero vectors such that r ◊ a = 0 and r × a = b, find r. 45. Let a = i + j + k, c = j – k such that a ◊ b = 3 and a × b = c, find b. 46. Find the angle between the vectors [a – (a ◊ b) b] and (a × b). 47. Find the area of a square whose one diagonal is (3 i + 4 j). 48. A constant force F = 3 i + 2 j – 4 is applied at the point (1, – 2, 2). Find the vector moment of F about the point (2, – 1, 3). 49. Forces (5 i + k) and (– 5 i – k) act at the points P (9, – 1, 2) and Q (3, – 2, 1), respectively. Find the moment of the couple. 50. Find the torque about the point 2 i + j + k of a force 4 i + k acting through the point (i – j + 2 k). 51. In a D ABC, a = BC, b = CA and c = AB, by using vector method, prove that a b c ____ = ____ = _____ . sin A sin B sin C
Find [a, b + c, a + b + c]. If a, b, and c be three non-coplanar vectors, find [a + b + c, a + b, a + c]. Find the value of [a – b, b – c, c – a]. If a, b and c are three non-zero and non-coplanar vectors such that a ◊ (b × c) = |a| |b| |c|, find the angle between a and b. 60. For non-zero vectors a, b and c, if a ◊ (b × c) = 4, 1 find the value of __ (b × c) ◊ (a + b + c). 2 56. 57. 58. 59.
61. If the vectors 2 i – j + l k, i – j + 2 k and 3 i – 2 j + k are coplanar vectors, find the value of l. 62. Find the volume of a parallelopiped whose coterminous edges are a = 3 i – 2 j + k, b = i + j + k and c = i + j – 2 k. 63. Find the volume of a parallelopiped whose coterminous edges are a + b = i + j – k, b + c = 2 i + 3 j – 4 k and c + a = 3 i + 4 j – 7 k. 64. The volume of a parallelopiped determined by the vectors a, b and c is 5. Find the volume of the parallelopiped determined by the vectors 3 (a + b), (b + c) and 2 (c + a). 65. If the volume of the parallelopiped formed by the vectors i + a j + k, j + a k and a i + k, a > 0, is minimum, find the value of a. 66. Let V1 be the volume of the parallelopiped formed by three non-zero vectors a, b and c and V2 be the volume of the parallelopiped formed by the vectors p = a – 2 b + 2 c, q = 3 a – b + c and r = a – b + V1 2 c. Find ___ . V2
( )
67. If the vectors a i + j + k, i + b j + k and i + j + c k, (a, b, c π 1) are coplanar, then find the value 1 1 1 of _____ + _____ + _____ . 1–a 1–b 1–c 68. For any three vectors a, b and c, prove that
52. Prove, by the vector method,
sin (A – B) = sin A cos B – cos A sin B.
53. Prove, by vector method,
sin (A + B) = sin A cos B + cos A sin B
SCALER TRIPLE PRODUCT OF VECTORS 54. For any three vectors a, b and c, prove that [a + b, b + c, c + a] = 2 [a, b, c]. 55. If ,
and
be unit coplanar vectors, find
[2a – b, 2b – c, 2c – a].
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a c⋅b c⋅c 2
69. The edges of a parallelopiped are of unit length and parallel to non-coplaner unit vectors a, b and c such 1 that a ◊ b = b ◊ c = c ◊ a = __ . Find the volume of the 2 parallelopiped. 70. If a and b be non-zero and non-collinear vectors, prove that [a, b, i] i + [a, b, j] j + [a, b, k] k = a × b. 71. If a, b, c, m and n are non-zero vectors, prove that
Vectors
a b c [a, b, c] (m × n) = a ◊ m b ◊ m c ◊ m a ◊n b◊n c ◊n 72. If a vector r can be expressed as a linear combination of three non-coplanar vectors a, b and c, prove that [r b c] r = ______ a + [a b c]
[r c a] ______ b+ [a b c]
[r a b] ______ c [a b c]
73. If a and b be unit vectors such that 1 [a, b, c] = __ , find the angle between a and b. 4 74. If a = x (a + b) + y (b × c) + z (c × a) and [a, b, c] 1 = __ , find (x + y + z). 8 75. Find the volume of the tetrahedron with edges a = i + j + k, b = i – j + k and c = i + 2 j – k. 76. If a, b and c are unit vectors satisfying |a – b|2 + |b – c|2 + |c – a|2 = 9, find |2a + 5b + 5c|. 77. The three vectors i + j, j + k, k + i taken two at a time form three planes. The three unit vectors drawn perpendicular to three planes form a parallelopiped. 4 Prove that its volume is ___ __ c.u. 3 ÷ 78. If a, b and c are three non-coplanar vectors such that a + b + c = a d and b + c + d = b a, find a + b + c + d + 4.
VECTOR TRIPLE PRODUCT 79. Find a unit vector which is orthogonal to 3 i + 2 j + 6 k and is coplanar with the vectors 2 i + j + k and i – j + k. 80. If a = i + j + k, a ◊ b = 1, a × b = j – k, find b. 81. If a, b and c are three non-parallel unit vectors such 1 that (a × (b × c)) = __ b, find the angles which a 2 makes with b and c. 82. If a, b and c are three non-coplanar unit vectors such that 1 (a × (b × c)) = ___ __ (b + c), 2 ÷ where b and c are non parallel, find the angle between a and b. 83. If a × (a × c) + b = 0 such that |a| = 1 |b| and |c| = 2, find the angle between a and c.
If q be the angle between b and c, find the value of sin q. 86. If a and b be mutually perpendicular unit vectors satisfying r ◊ a = 0, r ◊ b = 1 and [r, a, b] = 1, find r. 87. If |a| = 7, |b| = 1 = |c| and a × (b × c) + b × (c × a) 1 = __ a, where a and b are non-collinear vectors, find 2 |a × c|. 88. Simplify: (d + a) ◊ a × b × (c × d)]. 89. Solve: p x + (x × a) = b. 90. If [x a b] = 0, x ◊ a = 0, x ◊ b = 1, a ◊ b = 0, find x. 91. Find the value of [a × (b + c), b × (c – 2a), c × (a + 3b)] if [a, b, c] = 2. 92. Find the value of [(a – b), (a – b – c), (a + 2b – c)] if [a, b, c]. 93. Find the value of [(2a × 3b), (3b × 4c), (4c × 2a)] if [a, b, c].
Vector equations 94. Solve: x × b = a × b, where a and b are non-zero and non-collinear vectors. 95. Solve: x × a = b and x ◊ a = 0. 96. Solve: x × b = a × b and x ◊ c = 0, where c is not perpendicular to b. 97. Solve: x a + y b + z c = d. 98. Solve: x + y = a, x × y = b, x ◊ a = 1
84. If a × (b × c) is perpendicular to (a × b) × c, prove that (a ◊ c) = 0. 85. Let a, b and c be three non-zero vectors such that
1 (a × b) × c = __ |b| |c| |a. 3
5.17
_›
Mixed Problems (More than one options are correct) _ ›
› _› _
1. If a and b are two vectors such that a , b < 0 and › _› _
_›
_ ›
_›
_ ›
|a ◊ b | < |a × b | , the angle between a and b is p (a) p (b) __ 3 p 3p __ ___ (c) (d) 4 _ 4 _› › _› 2. Let a , b and c are three vectors of equal magnitudes. p The angle between each pair of vectors is __ such that 3 _ __ › _› _› |a + b + c| = ÷ 6 , the value of (|a | + 2) is (a) 2 (b) 3 (c) 1 (d) 0 3. If a, b and c are three unit vectors inclined to each other at angle q, the maximum value of q is p p (a) __ (b) __ 3 2 2p 5p ___ ___ (c) (d) 3 6 4. The number of vectors of unit length perpendicular _ › _› to vectors a = (1, 1, 0) and b = (0, 1, 1) is
5.18 Integral Calculus, 3D Geometry & Vector Booster
(a) 1 (b) 2 (c) 3 (d) infinite 5. Let a, b and g be distinct real numbers. The points with position vectors a , b , g , b , g , a and g , a , b (a) are collinear (b) form an equilateral triangle (c) form a scalene triangle (d) form a right-angle triangle. _›
_›
6. If P and Q have position vectors a and___ b relative to › the origin O such that X and Y divide PQ internally and externally respectively in the ratio 2 :1, the ___› vector XY is _ 3 _ › _› 4 _› › (b) __ (a – b ) (a) __ (b – a ) 3 2 _ 5 _› _› 4 › _› __ __ (c) (b – a ) (d) (b – a ) 3 6 7. If _the_› three points with position vectors _› _ _ › › _› › (1, a , b )_, (a , _›2, b ) and (a , b , 3) are collinear, the › value of a + b is (a) 1 (b) 4 (c) 5 (d) none. 8. The acite angle between the medians drawn from the acute angle of an isosceles right-angled triangle is 3 2 (a) cos– 1 __ (b) cos– 1 __ 3 4 4 2 – 1 __ – 1 __ (c) cos (d) cos 3 5
( ) ( )
_›
( ) ( )
_›
_›
_›
__›
12. Let u = + , v = – and w = – 2 + 3 _ __› › If is a unit vector such that u ◊ , the value of |w ◊ | is (a) 1 (b) 2 (c) 3 (d) 0 _›
_ ›
_ ›
_ ›
_›
_ ›
13. If a + b _+ c = _0, |a | = 3, |b | = 5, |c | = 7, the angle › › between a and b is p 2p (a) __ (b) ___ 3 6 5p p (c) ___ (d) __ 3 3 _› _›
_›
14. Let a , b and c be three vectors of the lengths 3, 4 _› _› _› _› and 5 respectively. Let a _be perpendicular to b _+ c , b › › _› _› _› _› _› _› to a + c and c to a + b , the value of |a + b + c | is __ __ (a) 2 ÷5 (b) 2 ÷2 __
__
(c) 10 ÷5
(d) 5÷2
_›
15. If a vector a of length 50 is collinear with the _› vector b = 6 – 8 – and makes an acute angle with the positive z-axis, then _›
_›
_›
_›
(a) a = 4b
(c) b = 4 a
_›
(b) a = – 4 b (d) none.
_ › 16. Let and b be two non-parallel unit vectors in a _ › _› plane. If (a a + b ) bisects the internal angle between _› _› a and b , then a is _› a
9. If e 1 and e 2 are two unit vectors and q the angle q between them, then cos __ is 2 _› _› 1 _› 1 _› __ (a) |e 1 + e 2| (b) __ |e 1 – e 2| 2 2 _› 1 _› 1 _› _› __ __ (c) |e 1 × e 2| (d) |e 1 ◊ e 2| 2 2
(a) 1/2
(b) 1
(c) 2
(d) – 1
10. If the vectors (3 p + q ) , (5 p – 3 q ) and (2 p + q ) , _› _› (4 p – 2 q ) are pairwise mutually perpendicular vectors _› _› such that q the angle between p and q , the value of sin (q ) is
( )
_›
___
÷55 (a) ____ 4 3 (c) ___ 16
_›
_›
_›
_›
_›
___
÷55 (b) ____ 8 ____ 247 ÷ (d) _____ 16
11. The set of values of c for which the angle between the vectors c x – 6 + 3 , x – 2 + 2c x is acute for every x in R is 4 4 (a) 0, __ (b) 0, __ 3 3 11 4 4 (c) ___ , __ (d) 0, __ 9 3 3
( ) ( )
[ ] [ )
17. Let , and are three unit vectors such that ( + + ) is also a unit vector. If pairwise angle between , , are q1, q2 and q3, spectively, the value of cos (q1) + cos (q 2) + cos (q3) is (a) 3 (c) 1
_› _›
(b) – 3 (d) – 1 _›
18. Given three vectors a , b and_ c each two of which are › _› _› non-collinear. Further (a + b ) is collinear with c and _› _› _› _› _› (c +__ b ) is collinear with_ a and each length of a , b _ › _› _› › _› _› is ÷ 2 . The value of (a ◊ b + b ◊ c + c ◊ a ) is
(a) 3
(b) – 3
(c) 0
(d) not defined.
19 ___ (a) – _____ 5÷43
(b) 0
19.
_ _ _ › _ › _ › _› _› _› › _› _› › If p =_ 3a – 5b , q = 2a + b , r = a + 4b and s = _ _ _ › › › › – a + b are four vectors such that sine of ( p and q ) _ _ › › is 1 and sine of (r and s ) is also 1, the value of _ › _ › cosine of (a and b ) is
Vectors
19 ___ (d) _____ 5÷43
(c) 1
_› _ › 20. If a and b are non-zero and non-collinear and the _› _› _ _ › › linear combination (2x – y) a + 4b = 5 a + (x – 2y) b
holds good for all real x and y, then the value of x + y is (a) – 3 (b) 1 (c) 17 (d) 3. _ _ › › 21. If p the position vector of the orthocentre and q the position vector of the centroid of _the triangle ABC _ › › and origin O is the circumcentre. If p = k q , the value of k is (a) 3 (b) 2
(c) 1/3
(d) 2/3
(c) 275 (d) 300 28. The set of values of m for which the vectors + m , + + (m + 1) and coplanar, is
–
5.19 +
+ m are non-
(a) R
(b) R – {1}
(b) R – {– 2}
(d) j
_ › 29. Given a = x + y + z , _ › _ _ › › _ › b = – + , c = + 2 , (a ◊ c ) = _› _ › p between a and b is __ , then 2 _› _› _› 2 _› _› _› (a) [a , b , c ] = |a | (b) [a , b , _ _ › _› _ › 2 › _› (c) [a , b , c ] = 0 (d) [a , b ,
4 and the angle _ _› 2 › c ] = |a | _ _ › › c ] = |a | 2
22. If the vectors – 2x – 3y and +3x + 2y are orthogonal to each other, the locus of the point is (a) a/an circle (b) ellipse (c) a/an parabola (d) straight line.
30. Let a, b, c be three distinct non-negative numbers. If the vectors a + a + a , + and c + c + b lie in a plane, then c is
23. A tangent is drawn to the curve x2 y = 8 at a point A (x1, y1) where x1 = 2. The tangent cuts the x-axis
(a) AM of a and b
(b) GM of a and b
(c) HM of a and b
(d) equal to zero.
___› ___›
at a point B. The value of (AB ◊ OB ) is
(a) 3 (c) 6
(b) – 3 (d) – 6.
_ _ › › 24. If e 1 and e 2 be two unit vectors and q the angle
( )
q between them, then sin __ is 2
_› 1 _› (a) __ |e 1 + e 2 | 2 _ › 1 _ › (c) __ |e 1 × e 2| 2
_› 1 _› (b) __ |e 1 – e 2 | 2 1 _ › _ › (d) __ |e 1 ◊ e 2| 2
_ › _ _› _› _ _ _ _ › › › › › 25. If u = a – b , v = a + b and |a | = 2 = |b | _› _› _› _› = (a ◊ b ) , the value of |u × v | is
(a) 0 (c) 4
(b) 2 (d) 16.
(a) (5, 2, 2)
(
)
2 5 __ 2 (c) __ , __ , 3 3 3
( ) 5 2 2 __ (d) ( __ , __ , . 3 3 3)
5 2 __ 2 (b) __ , __ , 3 3 3
_›
_›
__›
_›
_›
__›
_›
__›
_›
_›
_›
__›
(a) u ◊ (v × w )
(b) (v × w ) ◊ u
(c) v ◊ (u × w )
(d) (u × v ) ◊ w
32. The vector perpendicular to the vectors _› _› a = (2, – 3, 1), b = (1, – 2, 3) and satisfies the _›
_›
condition c ◊ ( + 2 + 7 ), the vector c is
(a) (7, 5, 1)
(b) (– 7, – 5, 1)
(c) (1, 1, – 1)
(d) (1, 2, 3)
_ › _› _› 33. If a = ( + + ) and b = ( + 2 + ), the vector c _ › _ _› _› › _› such that a ◊ c = 2 and a × c = b is
1 1 (a) __ ( 3 + 2 + 5 ) (b) __ ( – + 2 + 5 ) 3 3 1 1 (c) __ ( + 2 + 5 ) (d) __ ( 3 + 2 + ) 3 3 34. The altitude of a parallelopiped whose coterminous
_›
(a) 225
__›
and C = ( +
_ ›
_ ›
+ 3 ) with A and B as the sides of
the base of the parallelopiped, is
(b) 250
_›
edges are the vectors A = ( + + ) B = (2 + 4 – )
_ _ › › _› _› 27. If |a | = 1, |b | = 2 and and the angle between a and b _› _› _ _ › › 2p is ___ , the value of {(a + 3 b ) × (3 a – b ) }2 is
3
31. For three vectors which of the following expressions is not equal to any of the following three?
_› _ › 26. If a = (1, 1, 1) are given vectors, then a vector b _› _ _ _ › › › _› satisfying the conditions a × b = c and a ◊ b = 3
is
_› _ › __› u , v , w ,
2 ___ (a) ____ ÷19
4 ___ (b) ____ ÷19
3 ___ (c) ____ ÷19
2÷2 ___ (d) ____ . ÷19
__
5.20 Integral Calculus, 3D Geometry & Vector Booster 2. Let a and b are two vectors satisfying the conditions a × b = 0. Does it imply that one of the vectors a + a + b are coplanar and a, b and c are distinct, and b must be a null vector. Give one example in then support of your answer. (a) a3 + b3 + c3 = 1 (b) a + b + c = 1 [Roorkee-JEE, 1984] 3. Constant forces P = 2 i – 5 j + 6 k and Q = – i + 1 __ 1 __ 1 (c) __ (d) a + b + c = 0. 2 j – k act on a particle. Determine the work done a + b + c = 1 when the particle is displaced from a point A with _› _› __› _› _› __› If u , v , w are vectors such that u + v + w = 0, then position vector 4 i – 3 j – 2 k to a point B with a _› _› _› __› __› _› position vector 6 i + j – 3 k. Îu + v , v + w , w + u ˚ is [Roorkee-JEE, 1984] (a) 1 (b) 0 4. Prove that for any two vectors a and b (c) –1 (d) None. _ (a × b)2 = a2b2 – (a ◊ b)2 _› › _› If a , b and c are three mutually perpendicular [Roorkee-JEE, 1985] _ _ _ › _ › _ › › _ › _› › _ › _ › vectors, then (r ◊ a ) a + (r ◊ a ) b + (r ◊ a ) c is 5. If (a × b) = (b × c) π 0, show that a + c = k b, where _› _› k is a scalar. (a) r (b) 2 r [Roorkee-JEE, 1985] _› _› (c) – r (d) – 2 r 6. Find the value of the constant S such that the scalar _› _› product of the vector (i + j + k) with the unit vector Let a = + and b = 2 – . The point of intersec_ _ _ parallel to the sum of the vectors (2i + 4 j – 5k) and › › › _› _› _› _› _› tion of the lines r × a = b × a and r × b = a × b (S i + 2 j + 3k) is equal to 1. [Roorkee-JEE, 1985] is 7. Find a such that the vectors (2 i – j + k), (i + 2 j – 3k) (a) (3, – 1, 1) (b) (3, 1, – 1) and (3i + a j + 5k) are coplanar. (c) (– 3, 1, 1) (d) (– 3, – 1, – 1) [Roorkee-JEE, 1986] _› _› _› _› _› _› _› 8. If a and b are non-null vectors and |a + b| = |a – b|, If |a | = 1 = |b | and |c | = 2 and a × (a × c ) + b _ show that a and b are perpendicular to each other. _› › _› = 0 , the acute angle between a and b is [Roorkee-JEE, 1986] ___ p p 9. Find a vector of magnitude 51 which makes equal ÷ (a) __ (b) __ 4 6 1 angles with the three vectors a = __ (i – 2 j + 2k), b p __ 3 (c) (d) p 2 1 __ = (– 4i – 3k) and c = j. _ › _› 5 If b and _c are two mutually perpendicular unit vec[Roorkee-JEE, 1987] › tors and a is any vector, then _ › 10. Three vectors a = (12, 4, 3), b = (8, – 12, – 9) and _› _ _ ( › _› _› › _ › _ › a ◊ b × c ) ( _› _ › ) c = (33, – 4, – 24) define a parallelopiped. Evaluate b × c is (a ◊ b ) b + (a ◊ c ) c + _________ _› _ ( b × c › 2 ) the length of its edges, area of the faces and its volume. _› (a) 0 (b) a [Roorkee-JEE, 1988] _ › _› __ a (c) 2 a (d) b × c c × a 2 11. It is given that x = _______ , y = _______ and [a, b, c] [a, b, c] a × b z = _______ , where a, b and c are non-coplanar [a, b, c]
35. If the vectors a + b + c , b + c + a and c
36. 37. 38.
39.
40.
1. Three force vectors P, Q and R of 15 KN each acts along AB, BC and CA, respectively. The position vectors OA, OB and OC are given in metres as OA = 2 i – 4 j + 3 k, OB = 5 i + 3 j – 2k and OC = – 2 i + 2 j + 3 k. Find the resultant force vector S of the vectors P, Q and R. [Roorkee-JEE, 1983]
vectors. Show that x, y and z also form a non-coplanar system. Find the value of x ◊ (a + b) + y ◊ (b + c) + z ◊ (a + b).
[Roorkee-JEE, 1989]
12. It is given that r × b = c × b, r ◊ a = 0, a ◊ b π 0 What is the geometrical meaning of these equations
Vectors
separately? If the above three statements hold good simultaneously, determine the vector r in terms of a, b and c. [Roorkee-JEE, 1989] 13. Let a, b, c are three unit vectors, a + b + c = 0. Then a ◊ b + b ◊ c + c ◊ a is (a) – 3/2 (b) 0 (c) – 1 (d) 1 [Roorkee-JEE, 1990] 14. Let a = x i + y j + z k, b = j. For what c, a, b, c form a right handed system? (a) 2 i – x k (b) 0 (c) – z i × x k (d) y j [Roorkee-JEE, 1990] 15. Let a, b, c are in the same plane. Which of the following is correct? (a) a ◊ b × c = 0 (b) a ◊ b × c = 1 (c) a ◊ b × c = 2 (d) a ◊ b × c = 3
[Roorkee-JEE, 1990]
16. Given that the vectors A, B, C form a triangle such that A = B + C. Find__a, b, c and d such that the area of the triangle is 5 ÷6 , where A = a i + b j + c k, B = d i + 3 j + 4 k, C = 3 i + j – 2 k. [Roorkee-JEE, 1990] 17. If a and b be two unit vectors, the vector (a + b) × (a × b) is (a) perpendicular to (a – b) (b) parallel to (a – b) (c) equal to 2 (a – b) (d) equal to 2 (a – b) [Roorkee-JEE, 1991] 18. If the position vectors of points A and B with respect to origin O are a = 2 i – j + 2 k and b = i + 2 j – 2k, respectively, the projection of the vector a + b + a × b on a line perpendicular to the plane OAB is (a) 15 (b) 10
___ (c) ÷75
___ (d) ÷ 65
[Roorkee-JEE, 1991] 19. If two vectors a and b are such that a ◊ b = 0, the solution of x + a = b for all x (a) is unique (b) is possible only, when a = b (c) does not exist (d) will be infinitely many. [Roorkee-JEE, 1991]
5.21
20. Let be a unit vector and b be a non-zero vector not parallel to . Find the angles of the triangle, two sides of each are represented by the vectors __
÷ 3 ( × b) and b – ( ◊ b) [Roorkee-JEE, 1991] 21. Unit vector in the xy-plane that makes an angle of 45° with the vector i + j and an angle of 60° with the vector 3 i – 4 j is
(a) i
i – j __ (c) ____ 2 ÷
i+j __ (b) _____ 2 ÷ (d) none
[Roorkee-JEE, 1992] 22. Find the value of l that the vectors 2 i – j + k, i + 2 j – 3k and 3i + l j + 5k are coplaner (a) – 4 (b) 0 (c) 2 (d) 4 [Roorkee-JEE, 1992] 23. In parallelogram ABCD, the interrior bisectors of the consecutive angles B and C intersect at P. Find –BPC. [Roorkee-JEE, 1992] 24. If the sides of an angle are given by the vectors a = i – 2 j + 2k and b = 2i + j + 2k, the internal bisector is 1 (a) 3i + j + 4k (b) __ (3i – j + 4k) 3 1 (c) __ (– i – 3j) (d) 3i – j – 4k 3 [Roorkee-JEE, 1993] 25. Let a, b and c are three vectors such that a × b = c, b × c = a, then (a) a = 1, b = c (b) c = 1, a = b (c) b = 2, c = 2a (d) b = 1, c = a
[Roorkee-JEE, 1993]
26. Let a, b and c are in the same plane. Which of the following is correct? (a) a ◊ b × c = 0 (b) a ◊ b × c = 1 (c) a ◊ b × c = 2 (d) a ◊ b × c = 1 [Roorkee-JEE, 1994] 27. If two vectors a and b are such that a ◊ b = 0, the solution of x × a = b for all x. (a) is unique (b) is possible only, when a = b (c) does not exist (d) will be infinitely many.
[Roorkee-JEE, 1994]
5.22 Integral Calculus, 3D Geometry & Vector Booster 28. Solve the following system of simultaneous equations for vectors x and y x + y = a, x × y = b, x ◊ a = 1
(a) p – 4q
7 1 (b) __ p + __ r 5 5
(c) 2p – 3q – r
(d) – 4p – 2r
[Roorkee-JEE, 1998] 37. The vector c directed along the bisectors of the angle between the vectors a__= 7i – 4 j – 4k and b = – 2i – j + 2k, where |c| = 3 ÷6 is given by
[Roorkee-JEE, 1994]
29. The vector – i + j – k bisects the angle between vector c and 3 i + 4 j. Determine the unit vector along a.
1 (a) – ___ (3i – 8jz + 12k) 15 1 (b) – ___ ( 11i + 10j + 2k ) 15 1 (c) – ___ ( 3i + 8j + 12k ) 15 (d) none.
[Roorkee-JEE, 1995]
30. Find the scalers a and b if a × (b × c) + (a ◊ b) b = (4 – 2b – sin a) b + (b 2 – 1) c and (c ◊ c) a = c, where b and c are non-collinear. [Roorkee-JEE, 1995] 31. |(a × b)◊ c| = |a| |b| |c|, where a, b, c are non-zero vectors holds, if (a) a ◊ b = 0 (b) b ◊ c = 0
(c) c ◊ a = 0
(d) a ◊ b = b ◊ c = c ◊ a [Roorkee-JEE, 1996]
32. Let x, y and z be unit vectors such that x + y + z 3 a, x × (y × z) = b (x × y) × z = c, a ◊ x = __ , a ◊ y = 2
= 7 __ 4
and |a| = 2. Find x, y and z in terms of a, b, and c. [Roorkee-JEE, 1996] __
33. Vectors x, y and z each of magnitude ÷ 2 makes an angle of 60° with each other. If x × (y × z) = a, y × (z × x) = b and (x × y) = c, find x, y, z in terms of a, b and c. [Roorkee-JEE, 1997] 34. Which of the following is a true statement? (a) (a × b) × c is coplaner with c (b) (a × b) × c is perpendicular to a (c) (a × b) × c is perpendicular to b (d) (a × b) × c is perpendicular to c. 35.
[Roorkee-JEE, 1998] If A, B, C and D are any four points, then AB ◊ CD + BC ◊ AD + CA◊ BD is equal to (a) 1 (b) 0 (c) – 1 (d) None.
[Roorkee-JEE, 1998] 36. If p = 2a – 3b, q = a – 2b + c, r = – 3a + b + 2c, where a, b, c are non-zero are non-coplanar vectors, then the vector – 2a + 3b – c is equal to
(a) i – 7 j + 2k
(b) i + 7 j – 2k
(c) – i + 7 j – 2k
(d) i – 7 j – 2k
[Roorkee-JEE, 1998] 38. a, b, c three non-coplanar vectors such that r1 = a – b + c, r2 = b + c – a, r3 = c + a – b, and r = 2a – 2b + 4c, if r = x1r1 + x2r2 + x3r3, then
7 (a) x1 = __ 3 (c) x1 + x2 + x3 = 4
(b) x1 + x3 = 3 (d) x2 + x3 = 2
[Roorkee-JEE, 1998] 39. If x × y = a, y × z = b, x ◊ b = c, x ◊ y = 1 and y ◊ z = 1, find x, y and z in terms of a, b and c. [Roorkee-JEE, 1998] 40. If a, b, c are non-coplanar vectors and d is a unit vector, find the value of |(a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)| independent of d. [Roorkee-JEE, 1999] 41. If a = i + j – k, b = – i + 2 j + 2k and c = – i + 2 j – k, find a unit vector normal to the vectors a + b and b – c. [Roorkee-JEE, 2000] 42. Given that the vectors a and b are perpendicular to each other, find vector v in terms of a and b satisfying the equations v ◊ a = 0, v ◊ b = 1 and [v, a, b] = 1 [Roorkee-JEE, 2000] 43. a, b, c are three unit vectors such that a × (b × c) 1 = __ (b + c). Find the angle between vectors a and b 2 given that vectors b and c are nonparallel. [Roorkee-JEE, 2000] 44. The diagonals of a parallelogram are given by the vectors 2i + 3 j – 6k and 3i – 4 j – k. Determine its sides and the area also. [Roorkee-JEE, 2001] 45. Find the value of l that a, b, c are all non-zero and (– 4i + 5 j) a + (3i – 3 j + k) b (i + j + k) c = l (a i + b j + c k)
[Roorkee-JEE, 2001]
Vectors
46. Find the vector r which is perpendicular to a = i – 2 j + 5k, b = 2i + 3 j – k and r ◊ (2i + j + k) + 8 = 0.
_›
_›
5.23
_›
9. Let a and b are two unit vectors such that |a | = 1, _›
_› _›
_›
_›
_›
_›
[Roorkee-JEE, 2001]
|b | = 4, a ◊ b =_›2. If _c = (2a × b ) – 3b , find the › angle between b and c .
47. Two vertices of a triangle are – i + 3 j and 2i + 5 j and its orthocentre is at i + 2 j. Find the position vector of the third vertex.
10. Given that a , b , p and q are four vectors such
where m is a scaler, prove that
[Roorkee-JEE, 2001]
_›
_›
_›
(Tougher Problems for JEE-Advanced)
_›
_›
_› _› _ ›
_›
that a + b = m p , b ◊ q = 0 and (b ) 2 = 1, _› _ › _ ›
_›
_› _ › _ ›
_› _ ›
|(a ◊ q ) p – (p ◊ q ) a | = |p ◊ q | _›
11. Find a vector v , which is coplanar with with the
_
› _› vectors + – 2 and + – 2 and is orthogonal 1. If a and b be two unit vectors, find the range of _› to the vector 2 + + . _› 3 _› _› __ _ › |a + b | + 2 |a – b | . It is given that the projection of v along the vector 2 __ p 2. Find the unit vector which makes an angle __ – + is equal to 6÷3 . 4 2 1 _› 1 with axis of z and is such that ( + + ) is a unit _› _› 12. Let a = , b = 1 and c = 1 . 0 vector. 0 –1 – 3 3. If r and s are non-zero vectors and the sacalar b is Find the numbers a, b and g such that _› _› chosen in such a way that |r + b s | is minimum, find –2 _› _ _ › › _› _› a a = b b + g c = –5 the value of |r + b s |2 + |b s |2. _ 6 › _› u = – 2 + 3 , 13. Let a 3-dimentional vector V satisfies the condition _› __› _ _ › › 4. Given that v = 2 – + 4 and w = + 3 + 3 , 2V + V × ( + 2 ) = 2 + 2 . and _ › __ _ _ _ _ _ _ _ › › › › › › If 3 |V | = ÷ m , m Œ N, find m. (u ◊ R – 10) + (v ◊ R 20) + (w ◊ R – 20) = 0 _› __› _› _› 14. Let A = – 2 + 3 , B = 2 + – and C = + . Find the vector R .
( )
[ ] [ ]
[ ]
[ ]
5. Suppose the_ vectors a,_ b, c on a_ › plane satisfy the _ › › › conditions |a | = |b | = |c | = |a + b | = 1; c ◊ a = 0 _› _ ›
_›
_›
and _b ◊ c > 0. Find the angle between (2a + b ) › and b .
6. Find the minimum area of the triangle whose vertices are A (– 1, 1, 2), B (1, 2, 3) and C (t, 1, 1), where t is a real number. 7. Suppose the vectors a, b, c on a plane satisfy the _›
_›
_›
_›
and b ◊ c > 0. If the vector c is a linear combination _›
_›
l a + m + b , find the ordered pair (l ◊ m).
_›
_ ›
8. Given that a and b are two unit vectors such that the
( )
_› _ _ › › 1 angle between a and b is cos– 1 __ . If c be a vector 4 _ _ › › _› _› _› in the plane of a and b such that |c | = 4 c × b = _› _› _ _ _ › › › 2a × b and c = l a × m b , find the
__›
_ ›
_ _›
__›
15. Find the angle between any edge and face, which is not containing the edge of a regular tetrahedron and also find the angle between the two faces of a regular tetrahedron.
INTEGER TYPE QUESTIONS
_›
_›
conditions |a | = |b | = |c | = |a + b | = 1; c ◊ a = 0 _› _ ›
_ ›
If the vector B × C = x A + y B + z C , where x, y and z are scalars, find the value of (100x + 10y + 8z).
1. If [a, b, c] = 1, find the value of
a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) _________ + + (c × a)◊ b (a × b)◊ c (b × c)◊ a
2. If a, b and c are non-coplanar vectors and p, q, r are defined as
b×c c×a a×b p = ______ , q = ______ , r = ______ , [b c a] [c a b] [a b c]
(i) values of l
find the value of (a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r
(ii) sum of the values of m
(iii) product of all values of m.
and c = l +
_›
3. Let a = _›
_›
– 2 – 3 , b = 2 + 3 – _›
+ (2 l – 1) . If c
5.24 Integral Calculus, 3D Geometry & Vector Booster _›
_›
is parallel to the plane of the vectors a and b , find the value of (l2 + 1).
_› _› _› _› 4. Let r be _a vector perpendicular to ( a + b + c ) _ › _› › _› _› _› _› _› where [a , b , c ] = 2. If r = P (b × c ) + Q (c × a ) + _› _ › R (a × b ) , find the value of (P + Q + R + S). _› _› _› _› _› _› 5. Let |a | = 1, |b | = 1, |c | = 2 and a ^ (b + c ), _ _ _ › › › _› _› _› _› _› _› b ^ (c + a ) , c ^ (a + × b ) such that m = |a + b + c | ,
Linked Comprehension Type (For Jee-Advanced Examination Only) Passage I _ _ _ › _ › › › Let the vectors x , y and z are coplanar such that x = a + _ ›
– , y =
2
find the value of (m + 1).
6. If a, b and c are the p th, q th and r th terms of an _› HP and u = (q – r) + (r – b) + (p – q) and _› v =
__ a + __ + b value of (m +
_› _› __ c such that m = |(u ◊ v ) + 2|, find the 4). _›
_›
7. Let a = – 2 + 3 , b = 2 + 3 – + (2 l – 1) . _›
_›
If c is parallel to the plane of a and
_›
and c = l + _› b ,
find the value
of (l + l + 2). › _› _
_›
8. Let a , b and c be three vectors of magnitudes 3, 4 _›
_›
› _› _
_›
_›
and 5 respectively and a ^ b + c , b ^ c + a and _ › _ _› › c ^ a + b such
(
that m =
_ › _ _› › |a + b + c | ,
find the value
)
m of ______ __ . 2 + 3 ÷
_ ›
_ ›
+ , v =
__›
–
and w =
+ 2 – 3 . If
is a unit vector such that ◊ = 0 and v ◊ = 0, find the value of_ | ◊ |. › _› 11. If a and b are non-zero and non-collinear and _› _› _› the linear combination (2x – y) a + 4 b = 5 a + _› (x – 2y) b holds for real x and y, find the value of (x + y + 2). _ ›
_ ›
12. Given the vectors u = 2 – – , v = – – 2 __› and w = – 3 . If the volume of the parallelopiped _ __› › _ › having – c u , v and c w are concurrent edges is 8, find the positive integral value of c. _› p =
_›
13. Consider three vectors + + , q = 2 + 4 _› – and r = + + 3 and s_ be a unit vector. If _› _› _› _› _› › (p × q ) × r = u p + v q + w r , find the value of (u + v + w + 4). _›
_›
14. Consider_ three vectors p = _ + + , q = 2 + 4 › › – and r = + + 3 and s be a unit vector. Find the value of _› _›
_›
_›
_› _›
_›
_›
(
_› _›
_›
_›
|(p ◊ s ) (q × r ) + (q ◊ s ) (r × p ) + (r ◊ s ) (p × q )|
+
–c .
)
1 1 1 (i) The value of _____ + _____ + _____ is 1–a 1–b 1–c
(a) 1
(b) 2
(c) 0
(d) – 1
(
)
a b c (ii) The value of _____ + _____ + _____ is 1–a 1–b 1–c
(a) – 1 (c) – 3
(b) – 2 (d) 0 _› _ ›
_› _ ›
_›
(iii) The value of Îx + y , y + z , z + z ˚ is
(a) 1 (c) 0
(b) 2 (d) –1
Passage II
_ ›
_ ›
Consider the three vectors_ p = + + , q = 2 + 4 – _ › › and r = + + 3 also s be a unit vector.
9. If the three points with position vectors (1, a, b), (a, 2, b) and (a, b, 3) are collinear in the space, find the value of (a + b).
10. Let u =
_ ›
and z =
_›
2
+b –
_› _› _›
(i) p , q , r are (a) (b) (c) (d)
linearly dependent vectors can form the sides of a possible triangle _› _› _› (q – r ) is orthogonal to p any one can be expressed as a linear combination of the other two.
_›
_›
_›
_›
_›
_›
(ii) If (p × q ) × r = u p + v q + w r , the value of (u + v + w + 2) is (a) 6 (b) 4 (c) 0 (d) 2 (iii) The value of _› _› _› _› _› _› _› _› _› _› _› _› (p ◊ s ) (q × r ) + (q ◊ s ) (r × p ) + (r ◊ s ) (p × q ) is
|
|
(a) 4 (c) 18
(b) 8 (d) 2.
Passage III _› _› _› _› Consider the three vectors p , q and r such that p = _›
_ _› › (p × r )
+ , q = – + and 2, then _› _› _› (i) the value of Î p , q , r ˚ is
=
_› q +
4 (a) – __ 3
8 (b) – __ 3
(c) 2
(d) 1
_› c p and
_› _› (p ◊ q )
+ =
Vectors _›
_› _ ›
_› _ ›
_›
(ii) the value of Î p + q , q + r , r + p ˚ is
4 (a) – __ 3
8 (b) – __ 3
16 (c) – ___ 3
(d) 2
(iii) the value of Î p × q, q × r, r × p˚ is
16 (b) ___ 9
4 (a) __ 3 64 (c) ___ 9
(d) 1
Passage IV The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors , , such that 1 ◊ = ◊ = ◊ = __ . 2
(i) The value of Î × , ×
× ,
1 (a) __ 2
1 (b) __ 4
1 (c) __ 3
1 (d) __ 9
× ˚ is
(ii) The volume of the parallelopiped with coterminous edges are given by , , is
1 (a) __ 3
1__ (b) ___ 3 ÷
1__ (c) ___ 2 ÷
1 (d) __ . 2
(iii) The volume of the tetrahedron formed by the vectors , , is
1__ (a) ____ 6÷2
1__ (b) ____ 3÷2
1__ (c) ____ 4÷2
1 __ (d) _____ . 12÷2
Passage V _›
If a + b + g = 2 and a = a + b + g and also _ _ › › × ( × a ) = a . (i) The value of (a 2 + 1) is (a) 2 (b) 1 (c) –1 (d) 0. 2 (ii) The value of (b + b + 1) is (a) 1 (b) 2 (c) –1 (d) 0 (iii) The value of (g 2 – g) is (a) – 2 (b) 0 (c) 4 (d) 2
Passage Vi _›
_›
__›
Let u , v , w be three _› _› _› 3 _› _› = a , a ◊ u = __ , a ◊ v = 2 (i) The value of (a) 3/4 (c) 1/2 (ii) The value of (a) – 3/4 (c) – 1/4 (iii) The value of (a) 1/2 (c) – 1/2 Passage VII _›
_›
_›
_›
5.25 __›
unit vectors such that u + v + w _› 7 __ and |a | = 2. 4
_› __› a ◊ w is
_ › __› u ◊ w is
_› _› u ◊ v is
(b) 1/4 (d) –1/2 (b) 1/2 (d) 1/6 (b) 3/4 (d) – 3/4.
_›
Let a , b , and c be three vectors with magnitude 4 such _› _ _› _› _› _› › 1 that a ◊ b = b ◊ c = c ◊ a = __ . 2 (i) The height of the parallelopiped whose adjacent › _› _ _› edges by the vectors a , b and c is
÷
÷ 2 (d) 4 × __ ÷ 3
__
__
2 (a) __ 3
2 (c) 3 × __ 3
2 (b) 2 × __ 3
÷
__
__
(ii) The volume of_ the prism whose adjacent edges by _› › _› the vectors a , b and c is __
__
(a) 2 ÷2
(b) 3 ÷2
(c) 4 ÷2
(d) 16 ÷2
__
__
(iii) The volume of the_tetrahedron whose adjacent edges _› › _› by the vectors a , b and c is __
__
4÷2 (a) ____ 3
8÷2 (b) ____ 3
16÷2 (c) _____ 3
32÷2 (d) _____ 3
__
Passage VIII _›
_›
__
_›
Let a , b and c are three non-parallel unit vectors such __ _› _› _› 1 › _› that a × ( × ) = __ (b + c ) . 2 _ _ › › (i) The angle between a and c is (a) 90° (b) 60° (c) 30° (d) none. _ ›
_›
(ii) The angle between a and b is (a) 90° (b) 60° (c) 120° (d) 0°
5.26 Integral Calculus, 3D Geometry & Vector Booster _ ›
_›
_ ›
_ ›
(a – b) × c – (a + b) (d) y = __________________ 2 _› (iii) The vector z is
(iii) The value of |a × b | 2 + |a × c | 2 is
(a) 1 (c) 0
(b) 2 (d) 4.
› _› _
_›
_›
_›
_›
Let a , b and c be three vectors such that |a | = 1, |c | , |b | = _›
_›
___
4 and |b × c | = ÷ 15 and also
(i) The angle between the
( ) ( )
_›
_› _› b – 2 c = l a . _ › _› vectors b and c is
( ) ( )
1 1 (a) sin–1 __ (b) cos–1 __ 4 4 1 1 (c) tan–1 __ (d) cos–1 __ 4 4 (ii) The value of l is (a) ± 4 (b) ± 3 (c) ± 2 (d) ± 6 (iii) The angle between the vectors a and b is
( ) ( ) ( ) ( )
7 (a) p – cos– 1 __ 8 7 (b) p + cos– 1 __ 8 7 (c) – p + cos– 1 __ 8 1 (d) p – cos– 1 __ . 8
Passage X
If x × (y × z) = a, y × (z × x) = b and (x × y) = c.
_›
(a + b) × c – (a + b) (a) x = ___________________ 2 (a + b) × c + (a + b) (b) x = ___________________ 2 (a – b) × c + (a + b) (c) x = __________________ 3 (a + b) × c – (a – b) (d) x = __________________ 3 _ ›
(ii) The vector y is
b + a + (a – b) × c (d) z = _________________ 2.
Passage XI Let x, y and z be three unit vectors such that _›
_›
_›
_›
_›
8 4 (c) x = a + __ b – __ c 3 3 8 4 (d) x = – a + __ b – __ c 3 3 _ ›
(ii) The vector y is (a) y = a – 4c
(b) y = – 4c
(c) y = a + 4c + 3b
(d) y = 2b – 4c _›
(iii) The vector z is 2 (a) z = __ (c – b) 3 1 (b) z = __ (c – b) 3 4 (c) z = __ (c – b) 3 5 (d) z = __ (c – b). 3
(a – b) × c + (a + b) (a) y = __________________ 2
Passage XII
(a + b) × c + (a + b) (b) y = __________________ 2
(a – b) × c + (a – b) (c) y = __________________ 2
_›
_› _› _› _› _› _› _› 3 _› _› 7 (x × y ) × z , c , a ◊ x = __ , a ◊ y = __ and |a | = 2. 2 4 _ › (i) The vector x is 8 4 (a) x = a + __ b + __ c 3 3 8 4 (b) x = a – __ b + __ c 3 3
_› _›
x + y + z = a , x × (y × z ) = b
__
(i) The vector x is
Vectors x, y and z each of magnitude ÷ 2 makes an angle of 60° with each other.
b – a + (a – b) × c (a) z = _________________ 2 b + a – (a + b) × c (b) z = _________________ 2 b – a + (a + b) × c (c) z = __________________ 2
Passage IX
_ ›
_ ›
_ › _ ›
_ ›
_› _ › _ ›
If x + y = a , x × y = b , x ◊ a = 1.
_ ›
(i) The vector x is _› 1 _› (a) x = __ 2 ( a + a _› 1 _› (b) x = __ 2 ( a – a
_ › _› a × b
)
_›
_ ›
a × b )
Vectors
_ _› 1 _› _› › (c) x = – __ 2 ( a + a × b ) a
_ _ › › 1 ( _ › (a) y = a + __ 2 a + a _› _› 1 _› (b) y = a – __ 2 ( a + a
_› _ › a × b ) _› a ×
_› b
)
_ _› _› 1 ( _› _› › ) (c) y = – a + __ 2 a + a × b a
_ _› _› 1 ( _› _› › ) (d) y = a + __ a – a × b . _›
a2
_›
+
+
_› (b) – a _› (d) 3a .
_›
and b =
Column I (A) The projection of
30°
_›
_ › _› a on b is _›
_
_ ›
_ ›
__
_
180°
120°
4. Match the following Columns: Column I
Column II
(A) If a be any vector, the value of (P) _›
_›
_›
2 |a |
_›
|(|a ◊ |2 + |a ◊ |2 + |a ◊ |2)1/2 | is
+ 2 + 2
› (B) The projection of b on a is
Column II (P) (Q)
÷
___
26 ___ 2 ___
÷26 ____ 4
__ _ ›
(B) If a be any vector, the value of | (Q) ÷2 |a | _›
× (a × )|2 × )|
_›
_›
– (a × )|2 – (a __ _›
formed (R) (C) The area of a _triangle _› › by the vectors a and b is
1 __ 3
|a | (C) If a be any vector, the value (R) ÷3 _ _ _ › › › 2 2 of (|(|a × | + |a × | + |a × |2)1/2 |
(D) The area of a parallelogram (S)
1 ___ __ 3 ÷
(D) If a be any vector, the value of (S)
having diagonals
(B) If a, b and c be unit vectors (Q) such that a is _perpendicular to _› _› › b and c and |a + b + c | = 1, the angle between b and c is
› _› › › (D) If |_a ◊ b | = ÷3 | a × b | , the angle (S) between a and b is
1. Match the following Columns:
Let a =
60°
_ ›
Matrix Match (For Jee-Advanced Examination Only)
(A) If a, b and c be three vec- (P) _› unit _ _ › › tors such that a + _b – c _=› 0, › the angle between a and b is
_
_› (a) a _› (c) 2 a
Column II
+ b + c = 0 and |a | = (R) (C) If a _› _› 1, |b | = _5 |c | =_› 7, the angle › between a and b is
_›
(iii) The vector (x + y ) is
3. Match the following Columns: Column I
_ _› 1 _› › _› (d) x = __ 2 ( a + b × a ) a _› (ii) The vector y is
5.27
_ › _› a and b is
_›
_›
_›
|a |
_›
|(a ◊ ) + (a ◊ ) + (a ◊ ) | is
2. Match the following Columns: Column I
_›
_›
_›
_›
Column II
(A) If |a + b | = |a – b |, the angle (P) _ › _› between a and b is _
_
› › _› › (B) If |_a + 2b | = |a – 2b | , the angle (Q) _ › _› between a and b is
_ › _› |a + b |
(C) If the angle
_
_ _› › _› = 2b | and |a | = |b | , _› _ › between a and b is _› |a –
_
(R)
› › _› _› › (D) If |_a + 2b | = |a – 3b | , and |a | = 1, (S) _ › _› the angle between a and b is
60°
5. Match_ the following Columns: _› › _› If a , b and c are three non-coplaner vectors such › _ _› _ ›
that [a , b , c ] = 2. 45° 0°
90°
Column I
Column II
(A) The value of _ _ [ _a › + b › , b › + _c › , _c › + _a › ] is
(P)
12
(B) the value of
(Q)
16
[ _a › +
_› 2 b ,
_› 2 b +
_›
_›
_›
3 c , 3 c + a ] is
5.28 Integral Calculus, 3D Geometry & Vector Booster (C) The value of _ ›
_›
_›
_ › _ ›
_ ›
[ a + b , b + c , c + a ] is
(D) The value of
_ › _› _ _ _ › › _ › › a × 2 b , 2 b × c , c × a is
[
(R)
9
(S)
4
__›
+ 4 and C = 3 + triangle is 5 ÷6 sq.u. Column I
]
6. Match the following Columns: If a, b, c and a¢, b¢, c¢ be the reciprocal systems of vectors. Column I
Column II
(A) The value of (a + b + c) is
(P)
15
(B) The value of (a + c + d) is
(Q)
11
(C) The value of (b + c + d) is
(R)
19
(D) The value of (a + b + c + d) is
(S)
14
10. Match_ the following Columns: _ _ › › › If a , b and c be unit vectors satisfying
(B) The value of (Q) a¢ (a + b) + b¢ ◊ (b + c) + c¢ ◊ (c + a) is
0
|a – b | 2 + |b – c | 2 + |c – a | 2 = 9.
(C) The value of (a + b + c) ◊ (a¢ + b¢ + c¢) is
(R)
1
(A) The value of |2 a + 5 b + 4 c (P) | is
1
(D) The value of (a × a¢ + b × b¢ + c × c¢) is
(S)
3
› › (B) The value of |2 a + 3 b + 3c (Q) | is
5
› › (C) The value of |2 a + b + c | is
(R)
4
(D) The value of |2 a + 7 b + 7 c (S) | is
3
(P)
_›
Column I
Column II (P)
1
(B) The value of (b 2 + 1) is
(Q)
2
(C) The value of (l + 1) is
(R)
0
(D) The value of (b + g ) is
(S)
3
_› c =
Let + , + , and a a + b b. If the _› vectors – 2 + , 3 + 2 – and the vector c are coplanar. Column I
Column II
(A) The value of ( a + 3b + 2 ) is
(P)
1
a (B) The value of __ + 4 is b
(Q)
4
b 2 The value of 3 __ + __ is a 3 (D) a b The value of 3 __ + __ + 11 is b a
(R)
6
(
(C)
)
( )
( )
_›
(S)
_›
__›
_›
_›
_›
_›
_
_›
_
_ ›
_›
Column II
_
_›
_›
11. Match the following Columns: Column II
(A) The volume of the parallelopiped (P) › _ _ › _ › determined by the vectors a , b , c is 2. The volume of the parallelopiped › _ › _ determined by the vectors 2 (a × b ) , _› _› _› _› 3 (b × c ) and (c × a ) is
100
(B) The volume of the parallelopiped (Q) _› determined by the vectors a , _› _ › b , c is 5. The volume of the parallelopiped determined by the
30
_›
_›
_›
_›
vectors 3 (a + b ), (b + c ) and _› _› 2 (c + a ) is
_›
_›
24
_›
(2 a + 3 b ) and (a – b ) is
2
_›
__›
_›
(C) The area of a triangle with (R) adjacent sides determined by _› _› the vectors a and b is 20. The area of the triangle with adjacent sides determined by the vectors
Given that A , B , C form a triangle such that A = B + C . If A = a +
_›
_›
_›
9. Match the following Columns: _›
_›
Column I
8. Match the following Columns: _› b =
_ ›
_
(A) The value of (a + 1) is
_› a =
_ ›
Column I
7. Match the following Columns: _ › If _a + b + g = 2 and a = a + b + g , + ( × a) › = 0 .
Column II
2
(A) The value of a ◊ a¢ + b ◊ b¢ + c ◊ c¢ is
– 2 and the area of the
__
+ b + c and B = d + 3
(D) The area of a parallelogram with (S) adjacent_sides determined by the _› › vectors a and b is 30. The area of the parallelogram with adjacent _› sides by the vectors (a _› determined _› + b ) and a is
60
Vectors
Questions asked in Previous Years’ JEE-Advanced examinations 1. The value of A {(B + C) × (A + B + C)} is (a) 0 (b) [ABC] + [BCA] (c) [ABC] (d) none [IIT-JEE, 1981]
2. Let A, B, C be there unit vectors. Suppose that A B = 0 = A C and the angle between B and C is p /6. Then A = ± 2 (B × C). Is it true or false? [IIT-JEE, 1981] 3. For non-zero-vectors a, b, c |(a × b) c| = |a||b||c| holds if and only if (a) a b = 0, b c = 0 (b) c a = 0, b c = 0 (c) c a = 0, a b = 0 (d) a b = b c = c a = 0 [IIT-JEE, 1982] 4. A1, A2, ..., An are the vertices of a regular polygon with n sides and O its centre. Show that n – 1
S (OAi × OAi + 1) = (1 – n) (OA2 × OA1 + 1). i = 1
[IIT-JEE, 1982]
5. Find all values of l such that (x, y, z) π (0, 0, 0) and x (i, j, 3k) + y (3i – 3j + k) + z (– 4i + 5j) = l (x i + y j + z k), where i, j, k are unit vectors along the co-ordinate axes. [IIT-JEE, 1982] 6. The points with position vectors 60i + 3j, 40i – 8j, a i – 52j are collinear if (a) a = – 40 (b) a = 40 (c) a = 20 (d) None. [IIT-JEE, 1983] 7. If X A = 0 = X B = X C for some non-zero vector X, then [A B C] = 0. Is it true or false? [IIT-JEE, 1983] 8. The volume of the parallelopiped whose sides are given by OA = 2i – 3j, OB = i + j – k and OC = 3i – k is (a) 4/13 (b) 4 (c) 2/7 (d) None. [IIT-JEE, 1983] 9. If c be given non-zero scalar and A and B be given non zero vectors such that A ^ B. Find the vector X which satisfies the equations A X = c, A × X = B [IIT-JEE, 1983] 10. A vector A has components A1, A2, A3 in a right handed rectangular cartesian co-ordinate system
5.29
OXYZ. The co-ordinate system is rotated about the p x-axis through an angle __ . Find the components of 2 A in the new co-ordinate system in terms of A1, A2, A3 [IIT-JEE, 1983] 11. The points with position vectors a + b, a – b and a + k b are collinear for all real values of k. Is it true or false? [IIT-JEE, 1984] 12. Let a = a1i + a2 j + a3k, b = b1i + b2 j + b3k and c = c1i + c2 j + c3k be three non-zero vector such that c is a unit vector perpendicular to both the vectors a and b. If the angle between a and b be p /6, then
| |
a 1 b 1 c1
a 2 b 2 c2
a3 b 3 is equal to c3
(a) 0 (b) 1 1 (c) __ a21 + a22 + a 23 b21 + b 22 + b 23 4 3 (d) __ a21 + a22 + a23 b21 + b22 + b 23 c21 + c22 + c 23 4 [IIT-JEE, 1986] 13. A vector a has components 2p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If, with respect to the new system, a has component p + 1 and 1, then (a) p π 0 (b) p = 1 or p = – 1/3 (c) p = – 1 or p = 1/3 (d) p = 1 or p = – 1 [IIT-JEE, 1986] 14. The position vectors of the points A, B, C and D are 3i – 2j – k, 2i + 3j – 4k, – i + j + 2k and 4i + 5j + l k respectively. If the points A, B, C and D lie on a plane, find the value l. [IIT-JEE, 1986] 15. The number of vectors of unit length perpendicular to vectors a = (1, 1, 0) and b = (0, 1, 1) is (a) 1 (b) 2 (c) 3 (d) infinite [IIT-JEE, 1987]
(
) (
)
(
) (
) (
)
16. If the vectors a i + j + k, i + b j + k and i + j + c k, (a, b, c π 1) are coplanar, the value of 1 1 1 _____ + _____ + _____ is ... 1–a 1–b 1–c [IIT-JEE, 1987]
5.30 Integral Calculus, 3D Geometry & Vector Booster 17. Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively, are given by ... [IIT-JEE, 1987] 18. If A, B, C, D are four points in the space, prove that |AB × CD × BC × AD × CA × BD| = 4 (area of D ABC) [IIT-JEE, 1987] 19. Let a, b, c be three non-coplanar vectors and p, q r are vectors defined by the relations b×c c×a a×b p = _______ , q = _______ and r = _______ , [a, b, c] [a, b, c] [a, b, c] the value of the expression (a + b) p + (b + c) q + (c + a) r is equal to
(a) 0 (c) 2
(b) 1 (d) 3
[IIT-JEE, 1988] 20. The components of a vector a along and perpendicular to a non-zero vector b are ... and ... respectively. [IIT-JEE, 1988] 21. Let OACB be a parallelogram with O at the origin and OC a diagonal. Let D be the mid-point of OA. Using vectors method, prove that BD and CO intersect in the same ratio. Determine the same ratio. [IIT-JEE, 1988] 22. For any three vectors a, b, c, (a – b) {(b – c) × (c – a)} = 2a (b × c). Is it true or false? [IIT-JEE, 1989] 23. If vectors a, b and c are coplanar, show that
|
|
b c a a a a c = 0 a b b a b b b c
[IIT-JEE, 1989]
24. In a triangle OAB, E is the mid-point of BO and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : OD, using vector method. [IIT-JEE, 1989] 25. Let A = 2i + k, B = i + j + k and C = 4i – 3j + 7k. Determine a vector R satifying R × B = C × B and R A = 0. [IIT-JEE, 1990] 26. Determine the value of c so that for all real x, the vectors c x i – 6j – 3k and x i + 2j + 2c x k make an obtuse angle with each other. [IIT-JEE, 1991]
27. Given that a = (1, 1, 1), c = (0, 1 –1,) a b = 3 and a × b = c, then b = ... [IIT-JEE, 1991] 28. A unit vector coplanar with i + j + 2k and i + 2 j + k and perpendicular to i + j + k is ... [IIT-JEE, 1992] 29. Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie on a plane, then c is (a) the AM of a and b (b) the GM of a and b (c) The HM of a and b (d) equal to zero. [IIT-JEE, 1993] 30. Let a = 2i – j + k, b = i + 2j – k, and c = i + j – 2k be three vectors. A vector in the plane of b and c, __ 2 whose projection on a is __ , is 3 (a) 2i + 3j – 3k (b) 2i + 3j + 3k
÷
(c) –2i – j + 5k
(d) 2i + j + 5k
[IIT-JEE, 1993] 31. In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find BP the ratio ___ , using vector method. PE [IIT-JEE, 1993] 32. Let p and q be the position vectors of P and Q, respectively, with respect to O and |p| = p, |q| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular, then
(a) 9p2 = 4q2
(b) 4p2 = 9q2
(c) 9p = 4q
(d) 4p = 9q
[IIT-JEE, 1994] 33. Let a, b and g be distinct real numbers. The points with position vectors a i + b j + g k, b i + g j + a k and g i + a j + b k (a) are collinear (b) form an equilateral triangle (c) form a scalene triangle (d) form a right angled triangle. [IIT-JEE, 1994] 1 34. The vector d = __ (2i – 2j + k) is 3
(a) a unit vector.
Vectors
p (b) makes an angle __ with the vector. 3 (2i – 4j + 3k) 1 (c) parallel to the vector – i + j – __ k 2
(
)
(d) perpendicular to the vector (3i + 2j – 2k) [IIT-JEE, 1994] 35. A unit vector perpendicular to the plane determined by the points P (1, –1, 2), Q (2, 0, –1) and R (0, 2, 1) is ... [IIT-JEE, 1994] 36. If the vectors a, b, c, d are not coplanar, prove that the vector (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) is parallel to a. [IIT-JEE, 1994] 37. Let a = i – j, b = j – k, c = k – i. If d be a unit vector such that a d = 0 = [b, c, d], then d equals
( (
)
(
)
i + j – 2k __ (a) ± _________ ÷6
i + j – k __ (b) ± ________ 3 ÷
i + j + k __ (c) ± ________ 3 ÷
(d) ± k
)
[IIT-JEE, 1995] 38. If , b and c be non-coplanar unit vectors such that 1 a × (b × c) = ___ __ (b + c), the angle between a and 2 ÷ b is 3p p (a) ___ (b) __ 4 4 p (c) __ (d) p 2 [IIT-JEE, 1995] 39. Let u, v and w be the vectors such that u + v + w = 0. If |u| = 3, |v| = 4, |w| = 5, the value of u v + v w + w u is (a) 47 (b) – 25 (c) 0 (d) 25 [IIT-JEE, 1995] 40. If a, b and c are three non-coplaner vectors,
(a + b + c) ◊ {(a + b) × (a + c)} is
(a) 0
(b) [a, b, c]
(c) 2 [a, b, c]
(d) – [a, b, c]
[IIT-JEE, 1995] 41. A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the line determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is ... [IIT-JEE, 1996]
5.31
42. If b and c any two non-collinear unit vectors and a be any vector, then a (b × c) (a b) b + (a c) c + _________ (b × c) = ... |(b × c)| [IIT-JEE, 1996] 43. The position vectors of the vertices A, B, C of a tetrahedron ABCD are i + j + k, i and 3i, respectively. The altitude from vertex D to the opposite face ABC, meets the median line through A of the D ABC at E. If the length of the side AD is 4 and the volume of __ 2÷2 ____ the tetrahedron is . Find the position vector of 3 E for all its possible positions. [IIT-JEE, 1996] 44. Let p, q and r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation p × [(x – q) × p] + q × [(x – r) × q] + r × [(x – p) × r] = 0, then x is given by 1 (a) __ (p + q – 2r) 2 1 (c) __ (p + q + r) 3
45.
46.
47. 48. 49.
50.
1 (b) __ (p + q + r) 2 1 (d) __ (2p + q – r) 3 [IIT-JEE, 1997] Let OA = a, OB = 10a + 2b and OC = b where O, A and C are non-collinear points. Let p denotes the area of the quadrilateral OABC and q denotes the area of the parallelogram with OA and OC as adjacent sides. If p = kq then k = ... [IIT-JEE, 1997] Let a, b, and c be three vectors having magnitudes 1, 1 and 2, respectively. If a × (a × c) + b = 0, the acute angle between a and c is ... [IIT-JEE, 1997] If A, B and C are vectors such that |B| = |C|. Prove that |(A + B) × (A + C)| × (B × C) (B + C) = 0 [IIT-JEE, 1997] Let a, b and c be non-coplaner unit vectors, equally inclined to one another at an angle q. If a × b + b × c = p a + q b + r c, find scalars p, q and r in terms of q. [IIT-JEE, 1997] If a = i + j + k, b = 4i + 3j + 4k and c = i + a j __ + b k are linearly independent vectors and |c| = ÷ 3 , then (a) a = 1, b = 1 (b) a = 1, b = ± 1 (c) a = –1, b = ± 1 (d) a = ± 1, b = 1 [IIT-JEE, 1998] For the vectors u, v and w, which of the following expressions is not equal to any of the remaining three?
5.32 Integral Calculus, 3D Geometry & Vector Booster (b) (v × w) ◊ u (d) (u × w) ◊ w [IIT-JEE, 1998] 51. Which of the following expressions are meaningful?
(a) u ◊ (v × w) (c) v ◊ (u × w)
(a) u ◊ (v × w)
(b) (u ◊ w) ◊ w
(d) u × (v ◊ w) [IIT-JEE, 1998] 52. For any two vectors u and v, prove that (i) (u ◊ v)2 + |u × v|2 = (u2 v2)
(c) (u ◊ v) × w
2
2
(ii) (1 + |u| ) (1 + |v| )
53.
= |u + v + (u × v)|2 + (1 – (u ◊ v))2 [IIT-JEE, 1998] Let a and b be two non-collinear unit vectors. If u = a – (a b) b and v = a × b, then |v| is (a) |u| (b) |u| + |u a| (c) |u| + |u b| (d) |u| + u (a + b) [IIT-JEE, 1999]
planes determined by the pair of vectors a, b and c, d respectively, then the angle between P1 and P2 is
(a) 0
p (c) __ 3
[IIT-JEE, 2000] 59. If a, b and c are unit coplanar vectors, then the value of [2a – b 2b – c 2c – a] is
(a) 0
(c) – ÷3
that a ◊ c = |c|, |c – a| = 2 ÷2 and the angle between (a × b) and c is 30°, the value of |(a × b) × c| is (a) 2/3 (b) 3/2 (c) 2 (d) 3. [IIT-JEE, 1999] 55. Let a = 2 i + j + k and b = i + 2 j – k and a unit vector c be coplanar. If c is perpendicular to a, then c is
1__ (a) ___ (– j + k) 2 ÷
1__ (b) ___ (– i – j – k) 2 ÷
1__ (c) ___ (– i – 2j) ÷5
1__ (d) ___ (i – j – k) 3 ÷
[IIT-JEE, 1999] 56. Let u and v be unit vectors. If w be a vector such that 1 w × (w × u) = v, prove that |(u × v) w| £ __ and that 2 the equality holds if and only if u is perpendicular to v. [IIT-JEE, 1999] 57. If the vectors a, b and c from the sides BC, CA and AB, respectively, of a triangle ABC, then (a) a ◊ b + b ◊ c + c ◊ a = 0
(b) a × b + b × c = c × a (c) a ◊ b = b ◊ c = c ◊ a
(d) a × b + b × c + c × a = 0
[IIT-JEE, 2000] 58. Let the vertices a, b, c and d be such that (a × b) × (c × d) = 0. Let P1 and P2 be the two
__
(b) 1
__
(d) ÷ 3
[IIT-JEE, 2000]
60. If a, b and c are unit vectors, then |a – b|2 + |b – c|2 + |c – a|2 does not exceed
(a) 4 (c) 8
(b) 9 (d) 6
[IIT-JEE, 2000] 61. Let a = i – k, b = x i + j + (1 – x) k and c =
54. Let a = 2i + j – 2k and b = i + j. If c be a vector such __
p (b) __ 4 p (d) __ 2
y i + x j + (1 + x – y) k, then [a b c] depends on (a) only x (b) only y (c) neither x nor y (d) both x and y. [IIT-JEE, 2001]
62. Find 3-dimensional vectors v1, v2, v3 satisfying v1 ◊ v1 = 4, v1 ◊ v2 = – 2 and v1◊v3 = 6, v2 v2 = 2, v2 ◊ v3 = 5 and v3◊v3 = 29. [IIT-JEE, 2001] 63. Let A (t) = f1 (t) i + f2 (t) j and B (t) = g1 (t) i + g2 (t) j, t Œ [0, 1], where f1, f2, g1, g2 are continuous functions. If A (t) and B (t) are non-zero vectors for all t and A (0) = 2i + 3j, A (1) = 6i + 2j, B (0) = 3i + 2j, B (1) = 2i + 6j, show that A(t), B(t) are parallel for some t. [IIT-JEE, 2001] 64. Let V be the volume of the parallelopiped formed by the vectors a = a1i + a2 j + a3k, b = b1i + b2 j + b3k and c = c1i + c2 j + c3k. If ar, br, cr where r = 1, 2, 3, are non-negative real numbers and 3
S (ar + br + cr) = 3L, r = 1
Show that V £ L3.
[IIT-JEE, 2002] 65. If a and b are two unit vectors such that a + 2b and 5a – 4b are perpendicular to each other, the angle between a and b is p p (a) __ (b) __ 4 3
Vectors
( )
( )
1 (c) cos–1 __ 3
2 (d) cos–1 __ 7 [IIT-JEE, 2002] 66. Let v = 2 i + j – k, w = i + 3 k and u is a unit vector, the maximum value of [u v w] is
(a) –1
(c) ÷59
__ ___ (b) ÷ 10 + ÷6 ___ (d) ÷ 60
___
[IIT-JEE, 2002] 67. The value of a be such so that the volume of a parallelopiped formed by the vectors i + a j + k, j + a k, a i + k becomes minimum, is (a) –3 (b) 3 __ __ (c) 1/÷3 (d) ÷ 3 [IIT-JEE, 2003] 68. If u, v, w are three non-coplaner unit vectors and a, b, g are the angle between u and v, v and w and w and u respectively, and x, y, z are unit vectors along the bisectors of the angle a, b, g respectively. Prove that [x × y, y × z, z × x]
( )
( ) ( )
g b a 1 = ___ [u v w]2 sec2 __ sec2 __ sec2 __ 2 2 2 16 69. 70. 71.
[IIT-JEE, 2003] If a, b, c and d are distinct vectors satisfying relation a × b = c × d and a × c = b × d. Prove that a ◊ b + c ◊ d + a ◊ c + b ◊ d [IIT-JEE, 2004] If a = i + j + k, a. b = 1 and a × b = j – k, then b is (a) i – j + k (b) 2 j – k (c) i (d) 2 i – k [IIT-JEE, 2004] The unit vector which is orthogonal to the vector 5 i + 2 j + 6 k and is coplanar with the vectors 2 i + j + k and i – j + k is 1 ___ (a) ____ (2i 41 ÷ 1 ___ (b) ____ (2i ÷29 1 ___ (c) ____ (3j 10 ÷ 1 ___ (d) ____ (2i ÷69
– 6j + k)
73. Let a, b, c be three non-coplanar vectors and a ◊ b a ◊ b b1 = b – ____ a, b2 = b + ____ 2 a, 2 |a| |a| c ◊ a c ◊ b c1 = c – ____ 2 a + ____ 2 b1, |a| |c| b1 ◊ c c ◊ a c2 = c – ____2 a – ____ 2 b1, |a| |b1| c ◊ a b ◊ c c3 = c – ____ 2 a + ____ 2 b1, |c| |c| c ◊ a b ◊ c and c4 = c – ____ 2 a – ____ 2 b1, |c| |b1| the triplet of pairwise orthogonal vectors is (a) (a, b1, c1) (b) (a, b1, c2) (c) (a, b2, c2) (d) (a, b1, c3) [IIT-JEE, 2005]
74. Let a = i + 2 j + k, b = i – j + k and c = i + j – k. A vector in__ the plane of a and b, whose projection on c is 1/÷3 , is
(a) 4 i – j + k
(b) 3 i + j + k
(c) 2 i + j + 2k
(d) 4 i + j – 4 k
[IIT-JEE,2006]
75. The number of distinct real values of g which the vectors – g 2 i + j + k, i – g 2 j + k and i + j – g 2 k are coplanar is (a) 0 (b) 1 (c) 2 (d) 3 [IIT-JEE, 2007] 76. Let a, b and c be unit vectors such that a + b + c = 0, Which one of the following is correct? (a) a × b = b × c = c × a = 0 (b) a × b = b × c = c × a π 0 (c) a × b = b × c = a × c π 0 (d) a × b, b × c, c × a are mutually perpendicular. [IIT-JEE, 2007]
– 5j) – k) – 8j + k)
[IIT-JEE, 2004] 72. Let v be a unit vector along the incident ray, w be a vector along the reflected ray and a be a unit vector along the outward normal to the plane mirror at the point of incidence P. Express w in terms of a and v. [IIT-JEE, 2005]
5.33
77. Let the vectors PQ, QR, RS, ST, TU and UP represent the sides of a regular hexagon. Statement I: PQ × (RS + ST) π 0 Statement II: PQ × RS = 0 and PQ × ST π 0
[IIT-JEE, 2007]
78. Let two non-collinear unit vectors a and b form an acute angle. A point P moves so that at any time t, the position vector OP (where O is the origin) is given by a cos t + b sin t, where P is the farthest from origin O, let M be the length of OP and u be the unit vector along OP. Then
5.34 Integral Calculus, 3D Geometry & Vector Booster
a+b (a) u = ______ and M = (1 + a ◊ b)1/2 |a + b| a–b (b) u = ______ and M = (1 + a ◊ b)1/2 |a – b| a+b (c) u = ______ and M = (1 + 2 a ◊ b)1/2 |a + b|
a–b (d) u = ______ and M = (1 + 2 a ◊ b)1/2 |a – b| [IIT-JEE, 2008] 79. The edges of a parallelopiped are of length and parallel to non-coplanar unit vectors a, b, c such 1 that a ◊ b = b ◊ c = c ◊ a = __ . The volume of the 2 parallelopiped is 1__ 1__ (a) ___ (b) ____ 2 2÷2 ÷
___
(a) 8/9
(b) ÷ 17 / 9
(c) 1/9
(d) 4÷5 / 9
__
[IIT-JEE, 2010] 85. Let non-zero vectors a, b and c such that a ◊ b = 0, (b – a) ◊ (b + c) = 0 and 2 |b + c| = |b – a|. If a = m b + 4 c, the possible values of m are ...
__
3 ÷ (c) ___ 2
1__ (d) ___ 3 ÷
[IIT-JEE, 2008] 80. If a, b, c and d are unit vectors such that 1 (a × b) ◊ (c × d) = 1 and (a × c) = __ , 2 then (a) a, b, c are non-coplanar (b) b, c, d are non-coplanar (c) b, d are non-parallel (d) a, d are parallel and b, c are parallel
[IIT-JEE, 2009]
81. The volume of the parallelopiped with its edges represented by the vectors i + j, i + 2j and i + j + p k is... [IIT-JEE, 2009] 82. Angle between vectors a and b where a, b and c are unit vectors satisfying a + b + R3c = 0 is ... [IIT-JEE, 2009] 83. Let P, Q, R, S be the points on the plane with position vectors – 2 i – j, 4 i, 3 i + 3j and – 3 i + 2j respectively. The quadrilateral PQRS must be a
(a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c) rectangle but not a square (d) rhombus but not a square.
[IIT-JEE, 2010]
84. Two adjacent sides of a parallelogram ABCD are given by AB = 2 i + 10 j + 11 k and AD = – i + 2 j + 2 k. The side AD is rotated by an acute angle a in the plane of the parallelogram so that AD becomes AD ¢. If AD ¢ makes a right angle with the side AB, the cosine of the angle a is given by
[IIT-JEE, 2010] 86. If a and b be vectors in the space given by i – 2 j 2i + j + 3k ___ a = ______ __ and __________ , then the value of 14 ÷ ÷5 (2a + b) ◊ {(a × b) × (a – 2 b)} is [IIT-JEE, 2010] 87. Let a = i + j + k, b = i – j + k and c = i – j – k be three vectors. A vector v in the plane of a and b, 1 whose projection on c is ___ __ , is given by 3 ÷ (a) i – 3 j – 3 k (b) – 3 i – 3 j + k (c) 3i – j + 3k (d) i + 3 j – 3 k [IIT-JEE, 2011] 88. The unit vector(s) which is/are coplanar with vectors i + j + 2 k, i + 2 j + k and perpendicular to the vector i + j + k is given by (a) j – k (b) – i + j (c) i – j (d) – j + k [IIT-JEE, 2011] 89. Let a = i – k, b = – i + j and c = i + 2 j + 3 k be three given vectors. If r be a vector such that r × b = c × b and r ◊ a = 0, the value of r ◊ b is ... [IIT-JEE, 2011] __
__
90. If a = j + ÷3 k, b = – j + R3 k and c = 2 ÷3 k form a triangle, the internal angle of the triangle between a and b is. [IIT-JEE, 2011] 91. If a, b and c are unit vectors satisfying |a – b|2 + |b – c|2 + |c – a|2 = 9, then |2a + 5b + 5c| is. [IIT-JEE, 2012]
___
92. If a and b are vectors such that |a + b| = ÷ 29 and a × (2i + 3j + 4k) = (2i + 3j + 4k) × b, a possible value of (a + b) ◊ (– 7i + 2j + 3k) is
(a) 0 (c) 4
(b) 3 (d) 8.
[IIT-JEE, 2012] 93. Let PR = 3i + j – 2k and SQ = i – 3j – 4k determine diagonals of a parallelogram PQRS and PT = i + 2j + 3k be another vector. The volume of the parallelopiped determined by the vectors PT, PQ and PS is
Vectors
(a) 5 (c) 10
(b) 20 (d) 30
[IIT-JEE, 2013] 94. The volume of the parallelopiped determined by the vectors a, b and c is 2. The volume of the parallelopiped determined the vectors (a × b), 3 (b × c) and (c × a) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 95. The volume of the parallelopiped determined by the vectors a, b and c is 5. The volume of the parallelopiped determined the vectors 3 (a + b), (b + c) and 2 (c + a) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 96. The area of a triangle with adjacent sides determined by the vectors a and b is 20. The area of the triangle with adjacent sides determined by the vectors (2a + 3b) and (a – b) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 97. The area of a parallelogram with adjacent sides determined by the vectors a and b is 30. The area of the parallelogram with adjacent sides determined by the vectors (a + b) and a is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013]
__
98. Let x, y and z be three vectors each of magnitude ÷ 2 p and the angle between each pair of them is __ . If a 3 be a non-zero vector perpendicular to x and (y × z) and b is another non-zero vector perpendicular to y and (z × x), then (a) b = (b ◊ z) (z – x) (b) a = (a ◊ y) (y – z) (c) (a ◊ b) = (a ◊ y) (b ◊ z) (d) a = (a ◊ y) (z – y) [IIT-JEE, 2014]
99. If a, b and c are three non-coplaner unit vectors such p that the angle between every pair of them is __ . If 3 a × b + b × c = pa + qb + rc, where p, q and r are scalears, find the value of
(
)
p2 + 2q2 + r2 ____________ is... q2
[IIT-JEE, 2014] _›
› ___› _
___› _ ›
1. (d) 6. (d) 11. (d) 16. (b)
2. (b) 7. (b) 12. (c) 17. (d)
3. (c) 8. (c) 13. (d) 18. (b)
4. (b) 9. (a) 14. (d) 19. (d)
5. (b) 10. (b) 15. (b) 20. (b)
___›
100. Let PQR be a triangle. a = QR , b = RP , c = PQ _›
_›
_› _ ›
__
If |a | = 12, |b | = 4÷3 , and b ◊ c = 24, which of the following is (are) true? _
| c › | 2 – |_a › | = 12 (a) ____ 2
_›
| c | 2 + |_a › | = 30 (b) ____ 2 _ __ › _› _› _› (c) |a × b + c × a | = 48÷3
› _ › _
(d) a ◊ b = – 72
[IIT-JEE, 2015] 2
101. In R , if the magnitude of the projection vector of __
__
the 3 + is ÷ 3 and a = 2 + __ vector a + b on ÷ ÷ 3 b, find the possible values of |a |. [IIT-JEE, 2015] _› _› _› 102. Suppose that p , q and r are three non-coplanar vectors in R3. _› _› _› Let the components of a vector s , along p , q _› and r be 4,_3 and 5, respectively. If _the components _› _› _› _› _› › › of _a vector s along ( – p + q + r ), ( p – q + r ) and _ _ ( – p › – q › + r › ) are x, y and z, respectively, find the value of 2x + y + z. [IIT-JEE, 2015] 103. Let
__›
= u1 + u2 + u3 be a unit vector in R3 and w
_ › 1 = ___ __ ( + + ). Given that there exists a vector v ÷6 _› _› _› _› in R3 such that |u × v | = 1 and . (u × v ) = 1.
Which of the following statement (s) is (are) correct? _ › (a) There is exactly one choice for such v . _› (b) There are infinitely many choices for such v . _› (c) If u lies in the xy-plane, then |u1| = |u2|. _›
(d) If u lies in the xz-plane, then 2 |u1| = |u3|. [IIT-JEE, 2016]
Answers LEVEL II
5.35
21. (a) 26. (b) 31. (c) 36. (b)
22. (a) 27. (c) 32. (a) 37. (a)
23. (**) 28. (a) 33. (b) 38. (b)
24. (b) 29. (d) 34. (d) 39. (a)
25. (a) 30. (c) 35. (d) 40. (b)
5.36 Integral Calculus, 3D Geometry & Vector Booster
LEVEL-IV
COMPREHENSIVE LINK PASSAGES
1. [– 5, 5]
2. – __ – __ + ___ __ 2 2 ÷ 2
3. |r | 2
4. – + 2 + 5 p 5. __ 2
_›
__
3 ÷ 6. ___ . 2
1 2__ 7. ___ __ , ___ 3 ÷ 3 ÷
8. (i) 2
5p 9. ___ 6
(
)
(ii) –1 (iii) 12
(ii) (b) (ii) (b) (ii) (c) (ii) (c) (ii) (a) (ii) (c) (ii) (d) (ii) (c) (ii) (a) (ii) (b) (ii) (b)
(iii) (c) (iii) (a) (iii) (c) (iii) (a) (iii) (d) (iii) (b) (iii) (c) (iii) (a) (iii) (a) (iii) (c) (iii) (c)
P-XII :
(i) (a)
(ii) (b)
(iii) (a)
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
12. a = – 1, b = – 2, g = 3. 13. m = 6 14. 101 1 1 15. cos–1 ___ __ , cos–1 __ 3 3 ÷
( )
( )
INTEGER TYPE QUESTIONS 1. 3 6. 6 11. 3
2. 3 7. 2 12. 2
3. 3 8. 8 13. 2
4. 4 9. 4 14. 4
5. 7 10. 3
Hints
1. Let q the angle between a and b.
Now,
a ◊ b = a b cos (q)
fi
a ◊ b cos (q) = ____ a b
fi
6 ___ – 6 +___0 cos (q) = _________ =0 14 ÷ 13 ÷
fi
p q = __ 2
p Hence, the angle between a and b is __ . 2 2. We have, a + b = (i + 2j – 3k) + (3 i + j + 2k)
(i) (a) (i) (c) (i) (b) (i) (a) (i) (b) (i) (a) (i) (d) (i) (b) (i) (b) (i) (a) (i) (a)
MATRIX MATCH
11. 9 (– + )
P-I : P-II : P-III : P-IV : P-V : P-VI : P-VII : P-VIII : P-IX : P-X : P-XI :
= (4 i + 3j – k)
and
(A)Æ(R), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(P), (A)Æ(S), (A)Æ(S), (A)Æ(S),
(B)Æ(S), (B)Æ(S), (B)Æ(R), (B)Æ(P), (B)Æ(P), (B)Æ(S), (B)Æ(P), (B)Æ(P), (B)Æ(P), (B)Æ(P),
(C)Æ(P), (C)Æ(P), (C)Æ(P), (C)Æ(Q), (C)Æ(S), (C)Æ(S), (C)Æ(S), (C)Æ(P), (C)Æ(Q), (C)Æ(P),
(D)Æ(P) (D)Æ(P) (D)Æ(Q) (D)Æ(S) (D)Æ(Q) (D)Æ(Q) (D)Æ(Q) (D)Æ(P) (D)Æ(R) (D)Æ(Q)
11. (A)Æ(R), (B)Æ(S),
(C)Æ(P),
(D)Æ(Q)
solutions Also, a – b = (i + 2j – 3k) – (3i + j + 2k) Now,
= (– 2 i + j – 5k) (a + b) ◊ (a – b) = – 8 + 3 + 5 = 0
Thus, (a + b) is perpendicular to (a – b). 3. Given,
(a + b + c) = 0
fi | (a + b + c) |2 = 0 fi |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 9 + 16 + 25 + 2(a ◊ b + b ◊ c + c ◊ a) = 0 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 50 fi (a ◊ b + b ◊ c + c ◊ a) = – 25
Vectors
|(a + b – c)|
Now,
= |a|2 + |b|2 + |c|2 + 2 (a ◊ b – a ◊ c + b ◊ c)
fi
=1+1+1+0
fi
=3 __ |a + b – c| = ÷ 3
4. We have, 2
fi and
5. As we know that the projection of a on b
a ◊ b = ____ . |b| 3_________ –2+2 3 = __________ = ___ __ = 1 + 4 + 1 ÷6 ÷
÷
__
3 __ 2
6. Given 2 a + b = i + j and a + 2b = i – j Solving, we get
1 1 a = __ (i + 3j) and b = __ (i – 3j) 3 3
Since,
a ◊ b = a b cos (q )
fi
a ◊ b cos (q ) = ____ a b 1 __ (1 – 9) 9 8 4 ________ cos (q ) = = – ___ = – __ 10 10 5 ___ 9
fi
fi
( )
4 (q ) = cos–1 – __ 5
7. Let r = a i + b j + c k. r We have to find = __ . |r| Here, r is perpendicular to the given vectors, so a – 2 b + c = 0 and 2a + b – 3c = 0 Solving, we get a b c _____ = _____ = _____ 6–1 2+3 1+4
r ◊ a = r a cos (45°) _______
__ 1 __ x+y=÷ x 2 + y2 ◊ ÷2 ◊ ___ 2 ÷ ______
x+y=÷ x 2 + y2
...(i)
r ◊ b = r b cos (60°) ______
1 3 x – 4y = ÷x 2 + y2 ◊ 5 ◊ __ 2 ______ 2 fi __ (3 x – 4 y) = ÷ x 2 + y2 5 From Eqs. (i) and (ii), we get fi
x i + y j r _______ = __ = ________ |r| ÷x 2 + y2
13 y i + y j _________ = __________ ÷169y 2 + y2
9. Given Let
...(ii)
x = 13 y
Thus,
13 i + j ____ = ______ . 170 ÷ a ◊ b = 0 = a ◊ c = b ◊ c |a| = |b| = |c| = l
Now, (a + b + c) ◊ a = |a + b + c||a| cos (q1) fi |a + b + c||a| cos (q1) = (a ◊ a + a ◊ b + a ◊ c) fi |a + b + c||a| cos (q1) = |a|2 l |a| fi cos (q1) = _________ = _________ |a + b + c| |a + b + c| l |b| Similarly, cos (q2) = _________ = _________ |a + b + c| |a + b + c| and
l |c| cos (q3) = _________ = _________ |a + b + c| |a + b + c|
Thus,
cos (q1) = cos (q2) = cos (q3)
fi
(q1) = (q2) = (q3).
___
a b __ c fi __ = __ = 5 5 5
10. Let r = x i + y j + z k such that |r| = ÷ 51 . It is given that
fi
Thus,
a = b = c = l (say) l (i + j + k) (i + j + k) r __ __ = __ = __________ = _________ . |r| l ÷3 ÷3
8. Let the vector be r = x i + y j. r We have to find = __ . |r|
Let
a = i + j and b = 3i – 4j
5.37
cos (q1) = cos (q2) = cos (q3)
a ◊ r b ◊ r c ◊ r fi ____ = ____ = ____ |a||r| |b||r| |c||r| a ◊ r b ◊ r ___ c◊r fi ____ = ____ = |a| |b| |c| fi
a ◊ r = b ◊ r = c ◊ r ( |a| = 1 = |b| = |c|)
x – 2y + 2z ________ y – 4x – 3z __ fi __________ = = 3 1 5
5.38 Integral Calculus, 3D Geometry & Vector Booster x – 2y + 2z ________ y – 4x – 3z __ fi __________ = = = l (say) 3 1 5 Solving, we get Also,
___
x + y + z = 51
fi
25l2 + l2 + 25l2 = 51
fi
51 l2 = 51
fi Also,
fi
l =1
fi
l = ± 1
fi
r = ± (– 5i + j + 5k)
fi
cos (q1) = cos (q2) = cos (q3)
c ◊ r ___ |c|
a ◊ r = b ◊ r = c ◊ r
fi
2x + y + z = 0 z __ 5 z __ = l (say) 1
= ± ( j + k)
13. We have, 1 |a| = ___ __ = |b| = |c| |( – )|2 2 ÷ = | | + | |2 –2 ( ◊ ) fi x+y=y+z=x+z fi
(
x + y = y + z = x + z = l (say)
Also, fi
= [1 + 1 – 2 ( ◊ )]
= 2 – 2 ( ◊ )
l x = – __ = y = z 2
= 2 – 2 cos q
= 2 (1 – cosq)
|r| = 4
2
2
2
x +y +z =4
l2 l2 __ l2 fi __ + __ + = 4 4 4 4 3l2 fi ___ = 4 4 16 fi l2 = ___ 3 fi
4__ l = ± ___ 3 ÷
4__ Therefore, r = ± ___ (i + j + k) 3 ÷ 12. Let
)
Solving, we get
a = (i + 2j + k),
( ) q fi | – | = 2 sin ( __ ) 2 q 1 fi sin ( __ ) = __ | – | 2 2 14. Given
...(i)
r ◊ c = 0
y x __ = ___ = 0 – 5 y x fi __ = ___ = 0 – 1 r Therefore, = __ |r|
a ◊ r b ◊ r ____ c ◊ r fi ____ = ____ = |a| |r| |b|r| |c| |r|
fi
3x – y – z = 0
Solving Eqs. (i) and (ii), we get
11. Let r = x i + y j + z k such that |r| = 4. It is given that
b ◊ r ____ = |b|
| |
r ◊ (a × b) = 0
x y z fi 1 2 1 = 0 112
2
a ◊ r fi ____ = |a|
c = (2i + j + k)
2
fi
and
such that r = (x i + y j + z k). It is given that r, a and b lie in the same plane, so
|r| = ÷ 51 2
b = (i + j + 2k)
and r be a vector
x = – 5l, y = l, z = 5l
2
q = 2 × 2 sin2 __ 2
◊ = 0 = ◊ = ◊
and | | = 1 = | | = | | We have,
( – ) ◊ ( – )
= ( ◊ – ◊ – ◊ + ◊ )
= (0 – 0 – ◊ + 0)
...(ii)
Vectors
= – | |2
= – 1.
15. Given
| |=1=| |=| + |
Also, fi fi
| + |=1
fi
| + |2 = 1 2
Also,
| | + | + 2 ( ◊ ) = 1
fi
1 + 1 + 2 ( ◊ ) = 1
fi
2 cos (q ) = 1 – 2 = – 1
fi
1 cos (q ) = – __ 2
Thus,
sin (q ) = ÷ 1 – cos2 (q )
fi
3 ÷ 1 ___ = sin (q ) = 1 – __ 4 2
fi ÷x 2 + y2 + z2 = 1
__________
÷
__
______
__
3 ÷ 16. It is given that, a ◊ i = |a| |i| cos (30°) = ___ 2 1 and b ◊ i = |b| |i| cos (120∞) = – __ 2 Clearly, the angle between a and b is 90°.
=1+1+0
=2
__
17. Given,
)
a2 __ 4 + a
b2 __ 4 – b
2 (a ◊ b) ______ 2 2 a b
1 = __ 2 + a
1 __ 2 – a
2 (a ◊ b) ______ 2 2 a b
)
Now,
|(w ◊ )| = |– 3| = 3.
19. Given
a ◊ b = 0 = a ◊ c
Also,
p 1 b ◊ c = |b||c| cos __ = __ 3 2
u ◊ = 0 x+y=0 x = – y
( )
|a + b + c|2 = |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) 1 = 1 + 1 + 1 + 2 0 + __ + 1 2 =4 |a + b + c| = 2.
(
...(i)
)
2 (a ◊ b + b ◊ c + c ◊ a) = 0 |a + b + c|2
= |a|2 + |b|2 +|c|2 + 2 (a ◊ b + b ◊ c + c ◊ a)
= 9 + 16 + 25 + 0
Thus,
= 50 2 |a + b + c| = 5.
| |
2
| |2 + | |2 + 2( ◊ ) = _________________ | |2 + | |2 – 2 ( ◊ )
1 + 1 + 2 cos (60°) = ________________ 1 + 1 – 2 cos (60°)
1 + 1 + 1 = _________ 1+1–1
= 3.
= x i + y j + z k such that | | = 1.
Now, fi fi
r = ± k
| + | + 2 _______ _____ = – | – |2
a–b 2 = _____ a b 18. Let
Thus,
21. We have,
a2 + b2 – 2 (a ◊ b) = _______________ a2 b2
(
0 + 0 + z2 = 1 z = ± 1
20. Given Now,
fi | + | = ÷ 2
b 2 __ 2 = b
fi fi
Thus,
| + |2 = | |2 + | |2 + 2( ◊ )
(
x2 + y2 + z2 = 1
We have,
a __ 2 – a
fi
Now,
( ◊ ) = 0
| |=1 __________
2
fi
Thus,
...(ii)
From Eqs. (i) and (ii), we get x=0=y
We have,
v ◊ = 0 x–y=0 x = y
5.39
5.40 Integral Calculus, 3D Geometry & Vector Booster
| |
__ + Thus, _____ = ÷ 3 –
fi b2 + c2 + 2 b c cos (p – A) = a2
22. Given,
fi 2 b c cos (A) = b2 + c2 – a2 (a + b + 3c) = 0
fi
(a + b + c) = (– 2c)
fi b2 + c2 + 2 b c cos (A)= a2
We have,
|a + b + c|2 = |(– 2c)|2
b2 + c2 – a2 fi cos (A) = ___________ 2 b c Hence, the result. _ __›
__› _› _
___›
___›
_ ›
_ __›
_›
25. Let BC = a , CA = b and AB = c
fi |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 4|c|2 68 17 fi 1 + 16 + ___ + 2 (a ◊ b + b ◊ c + c ◊ a) = ___ 9 9 68 – 17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = _______ – 17 9 51 fi 2 (a ◊ b + b ◊ c + c ◊ a) = ___ – 17 9 17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = ___ – 17 3 34 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – ___ 3
We have,
17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – ___ 3 23. Let the resultant force be F such that
fi b + c = – a
F = F1 + F2 + F3
i.e.
F = 2j – k
Now,
AB = 2 i + 4 j – k
= (2 j – k) ◊ (2i + 4j – k)
= 8 + 1 = 9.
24.
_›
_›
_ ›
_›
_ ›
_ ›
_›
fi a + b + c = 0 › _ › _
_ ›
_ › _ ›
_ › _ ›
fi a ◊ b + a ◊ c = – a ◊ a fi
a b cos (p – C ) + a c cos (p – B ) = – a2
fi
– a b cos (C) – a c cos (B) = – a2
fi
\ Work done = (Force) ◊ (Displacement)
___›
BC + CA + AB = 0
a = b cos (C) + c cos (B)
26.
_ › ___› ___› _ _ › ___› › Let BC = a , CA = b and AC = c
_ __›
Here, OP = a + b ___›
and OQ = c + d
___› ___› ___› We have BC + CA + AB = _ › _ _ › › fi a + b + c = 0 _ › _› _› fi b + c = – a _›
_ ›
_ ›
fi ( b + c ) = (– a ) 2 2
fi b2 + c2 + 2 b ◊ c = a2
___› ___›
0
___
___
| › | | › |
Now, OP ◊ OQ = OP OQ cos (– POQ) Now,
(a + b ) ◊ (c + d )
______
______
= ÷a 2 + b2 ÷c 2 + d2 cos (A – B) fi
cos (A – B)
Vectors
( a + b ) ◊ ( c + d ) ______ ______ = __________________ ÷ c 2 + d2 ÷ a 2 + b2 a c + b d ______ ______ = _______________ 2 ÷ c 2 + d2 ÷ a + b2 d a c b ______ ______ = _______ ______ ◊ _______ + ________ _______ ◊ _______ 2 2 2 2 2 2 2 ÷ c + d ÷a + b ÷c + d2 ÷ a + b = cos A ◊ cos B + sin A ◊ sin B 27.
30. Now,
5.41
| |
i a × b = 1 3
j 2 – 2
k 3 1
= 8 i + 8 j – 8 k
= 8 (i + j – k)
Area of the parallelogram = |a × b| __
= 8÷3 sq.u.
31. Now,
| |
i a × b = 1 3
j 1 – 4
k 1 1
= 5 i + 2 j – 7 k
Area of the parallelogram Here, and
___› OP =
___› OQ =
1 = __ |a × b| 2
___________ 1 = __ × ÷ 25 + 4 + 49 2
78 ÷ = ____ sq.u. 2
32. Let and Now, and
a = 3 i – j + 2 k, b=i–j–k c = 4 i – 3 j + k. a b = – 2 i – 3 k a c = i – 2 j – k
\
i j a b × a c = – 2 0 1 – 2
a + b c + d
___
Now, ___› ___›
___
___
| › | | › |
OP ◊ OQ = OP OQ cos (– POQ) fi
a c – b d = (OP) (OQ) cos (A + B)
fi
a c – b d cos (A + B) = _________ (OP) (OQ)
a c b d = ___ ◊ ____ – ___ ◊ ____ OP OQ OP OQ
= cos (A) cos (B) – sin (A) sin (B)
28. We have,
| |
i a × b = 2 1
j – 1 – 3
k 1 –5
= 8 i + 11 j – 5 k
1 29. Area of a triangle = __ |a × b| 2 Now,
| |
i a × b = 3 5
j 4 7
k 0 0
= (21 – 20) k =k
1 Thus, the area of the triangle = __ sq.u 2
| |
k – 3 – 1
= – 6 i – 5 j + 4 k
1 Area of the triangle = __ |a b × a c| 2 ___________ 1 = __ ÷36 + 25 + 16 = 2 33. Let a = 2 i – j + k and b = 3 i – 4 j – k. A unit vector perpendicular to a and b (a × b) = ± _______ |a × b| Now,
| |
i (a × b) = 2 3
j – 1 – 4
k 1 – 1
= 5i + 5j – 5k
= 5 (i + j – k)
___
77 ÷ ____ sq.u 2
5.42 Integral Calculus, 3D Geometry & Vector Booster (a × b) Thus, the unit vector = ± ______ |a × b|
and Therefore,
1__ = ± ___ (i + j – k). 3 ÷
|(a × k)|2 = a12 + a22
|
|2
| a × |2 + a × + | a × | 2
34. We have,
= (a22 + a32 + a12 + a32 + a12 + a22)
a × (b + c) + b × (c + a) + c × (a + b)
= 2 (a12 + a22 + a32)
= 2a2
= a × b + a × c + b × c + b × a + c × a + c × b = a × b – c × a + b × c – a × b + c × a – b × c = 0 35. Given
a × b = c × d
...(i)
and
a × c = b × d
...(ii)
(b × c) = ± ______ |b × c|
Subtracting Eq. (ii) from Eq. (i), we get
38. Given a ◊ b = 0 = a ◊ c Thus, a is perpendicular to b and c. A unit vector perpnedicular to b and c
a×b–a×c=c×d–b×d
fi
a × (b – c) = (c – b) × d
fi
a × (b – c) – (c – b) × d = 0
fi
a × (b – c) + (c – b) × d = 0
fi
a × (b – c) – d × (b – c) = 0
fi
(a – d) × (b – c) = 0
(b × c) = ± __________ p |b| |c| sin __ 6 (b × c) = ± ______ 1/2
( )
= ± 2 (b × c).
(a + b + c) = 0
fi
b × (a + b + c) = b × 0 = 0
fi
b×a+b×b+b×c=0
fi
b×a+b×c=0
39. Given, fi \ fi fi 40. Given, fi \ fi fi Also,
fi
b × c = – b × a
– 6 l – 6l = 1
fi
fi
b × c = a × b
fi
1 l = – ___ 12
Thus,
1 R = – __ (–3 i + 4 j – 6k). 2
Thus, (a – d) is parallel to (b – c). 36. Given, (a + b + c) fi
a × (a + b + c) = a × 0 = 0
fi
a×a+a×b+a×c=0
fi
a×b+a×c=0
fi
a × b = – a × c
fi
a × b = c × a
...(i)
Also,
...(ii)
From Eqs. (i) and (ii), we get
a×b=b×c=c×a
37. Let
a = a1i + a2j + a3k
Now,
(a × i)
= (a1i × i + a2j × i + a3 k × i)
= (– a2k – a3j)
Thus,
|(a × i)|2 = a22 + a32
41. Now,
2
2
Similarly, |(a × j)| = a1 + a3
2
a×b=a×c a × (b – c) = 0 (b – c) is parallel to a (b – c) = l a b – c + l a, l ŒR R× R× R is R= R=
B=R×C (B – C) = 0 parallel to (B – C) l (B – C) l (– 3 i + 4 j – 6 k)
R ◊ A = 1
| |
i a × b = 1 0
j 1 1
k 0 1
=i–j+k
Unit vector perpendicular to a and b
...(i)
Vectors
a×b = ± ______ |a × b|
= ± (i – j + k) Thus, the number of vectors of unit length = 2. 42. As we know that, if a, b and c form a right handed system, then
| |
c=a×b
fi
i c = x 0
43. Given,
j y 1
k z = – z i + x k 0
5.43
Comparing, we get
(z – y) = 0, (x – z) = 1, (y – x) = – 1
Solving, we get fi
5 2 x = __ , y = __ = z 3 3
Thus,
1 b = __ (5 i + 2 j + 3 k) 3
46. The vector [a – (a ◊ b) b] lies in the plane of a and b and the vector (a × b) is perpendicular to each of a and b Thus, the angle between (a – (a ◊ b) b) and (a × b) = 90°.
r × a = b × a and r × b = a × b
fi
r × a = – a × b and r × b = a × b
fi
r × a = – r × b
fi
r×a+r×b=0
fi
r × (a + b) = 0
\
r is parallel to (a + b)
fi
r = l (a + b)
fi
r = l (a + b)
fi
r × b = l (a × b + b × b)
fi
r × b = l (a × b)
Here,
fi
a × b = l (a × b)
Given a + b = 3 i + 4 j
fi
l=1
Now,
|a + b|2 = |a|2 + |b|2 + 2 (a ◊ b)
Therefore, r = (a × b)
fi
|a + b|2 = |a|2 + |b|2 = 2 |a|2
fi
2 |a|2 = |a + b|2 = |a + b|2 = (32 + 42) = 25
fi
25 |a|2 = ___ 2
= 3 i + j – k
44. Given that
r × a = b and r ◊ a = 0
fi
r×a=b
fi
a × (r × a) = a × b
fi
(a ◊ a) r – (a ◊ r) a = a × b 2
47.
|a| = |b| and (a ◊ b) = 0
25 \ Area of a square = |a|2 = ___ . 2
48. Let
A = (1, – 2, 2) and B = (2, – 1, 3)
Thus,
r = BA = OA – OB = – i – j – k
fi
|a| r – 0 = a × b
Therefore, Momentum about the point B
fi
a×b r = _____ |a|2
=r×F
45. Given
a × b = c and a ◊ b = 3
Let
b = x i + y j + z k
i j k = – 1 – 1 – 1 3 2 – 4
Thus,
x+y+z=3
= 6 i – 7 j + k
49. Here,
PQ = – 6 i – j – k
| |
i and 1 x fi
j 1 y
k 1 z
(z – y) i + (x – z) j + (y – x) k = j – k
| |
Thus, moment of the couple
=M
= PQ × F
5.44 Integral Calculus, 3D Geometry & Vector Booster
| |
___›
i j k = – 6 – 1 – 1 5 0 1
Here, OP = a + b
= – i + j + 5 k
Now,
OP × OQ = (a d – b c)
50. Here,
r = (i – j + 2 k) – (2 i + j + k)
fi
[(OP) (OQ) sin (–POQ)] = (a d – b c)
fi
r = i – 2 j + k
fi
(a d – b c) – sin (A – B) = _________ (OP) (OQ)
a d ___ b c = ___ ◊ ____ ◊ ◊ ____ OP OQ OP OQ
= cos A ◊ sin B – sin A ◊ cos B
fi
sin (A – B) = sin A ◊ cos B – cos A ◊ sin B
___›
and OQ = c + d
Thus, torque about the point
| |
=r×F
i = 1 4
= – 2 i – 3 j + 8 k.
51. Given
D ABC, a = BC, b = CA, c = AB
j k 1 – 2 0 1
Hence, the result. 54. We have, [a + b, + b + c, c + a] = (a + b) ◊ (b + c) × (c + a) = (a + b) ◊ (b × c + b × a + c × c + c × a) = (a + b) ◊ (b × c + b × a + c × a) = (a ◊ b × c + a ◊ b × a + a ◊ c × a)
_ ›
_›
+ (b ◊ b × c + b ◊ b × a + b ◊ c × a)
= [a, b, c] + [a, b, c]
_ ›
Clearly, a + b + c = 0
= 2 [a, b, c]
55. We have,
_ _ › › _ _ _ _› › › › fi a × b = b × c = c × a
_ _ › › _ _ _ _› › › › fi a × b = b × c = c × a
|
| |
|
|
|
fi | a b sin (p – C) | = | b c sin (p – A) | = | c a sin (p – B) |
fi
a b sin C = b c sin A = c a sin B
a b sin C _______ b c sin A _______ c a sin B fi _______ = = a b c a b c a b c sin C ____ sin A ____ sin B fi _____ = a = c b sin C sin A ____ sin B _____ fi ____ = = c a b Hence, the result. 52.
[2 a – b, 2b – c, 2c – a] = (2 a – b) ◊ (2b – c) × (2c – a) = (2 a – b) ◊ (4 b × c – 2 b × a + c × a) = (2 a – b) ◊ (4 b × c + 2 a × b + c × a) = (8 a ◊ b × c – b ◊ c × a) = 7 [a, b, c] = 0, a, b and c are coplaner vectors 56. We have, [a, b + c, a + b + c] 57.
= a ◊ (b + c) × (a + b + c) = a ◊ (b × a + b × c + c × a + c × b) = (a ◊ b × c + a ◊ c × b) = (a ◊ b × c – a ◊ c × b) = 0. We have, [a + b + c, a + b, a + c] = (a + b + c) ◊ (a + b) × (a + c) = (a + b + c) ◊ (a + c + b × a + b + c) = (a ◊ b × c + b ◊ a × c + c ◊ b × a)
Vectors
58.
= (a ◊ b × c – a ◊ b × c + c ◊ b × a) = [a, b, c] – [a, b, c] + [a, b, c] = [a, b, c] We have, [a – b, b – c, c – a] = (a – b) ◊ (b – c) × (c – a) = (a – b) ◊ (b × c – b × a + c × a) = (a – b) ◊ (b × c + a × b + c × a) = (a ◊ b × c – b ◊ c × a) = [a, b, c] – [a, b, c] = 0.
59. We have,
a ◊ (b × c) = |a| |b| |c|
fi
(a × b) ◊ c = |a| |b| |c|
fi
|a| |b| |c| sin (q ) cos (q ) = |a| |b| |c|
fi
sin (q ) cos (j) = 1
fi
sin (q ) = 1, cos (j) = 1
fi
p q = __ , j = 0 2
p Hence, the angle between a and b is q = __ 2 60. We have, 1 __ (b × c) ◊ (a + b + c) 2 1 = __ ((b × c) ◊ a + (b × c) ◊ b + (b × c) ◊ c) 2 1 = __ ((b × c) ◊ a + 0 + 0) 2 1 = __ (b × c) ◊ a 2 1 = __ [a, b, c] 2 1 = __ × 4 = 2 2 61. Since the given vectors are coplanar, so
| |
2 1 3 fi fi
–1 – 1 – 2
l 2 = 0 1
6 – 5 + l (– 2 + 3) = 0 1+l=0
fi l = – 1 62. The volume of the parallelopiped
| |
= [a, b, c]
3 = 1 1
– 2 1 1 1 1 – 2
= 3 (– 3) + 2 (– 3) + 0
= |– 3| = 3 c.u.
5.45
63. We have,
2 (a + b + c) = 6 i + 8 j – 12 k
fi
(a + b + c) = 3 i + 4 j – 6 k
Thus, a = i + j – k, b = k, c = 2 i + 3 j – 5 k Hence, the volume of the parallelopiped
| |
1 = 0 2
1 0 3
64. Given
[a, b, c] = 5
Now,
[3 (a + b), (b + c), 2 (c + a)]
– 1 1 = 1 c.u. – 5
= 3.2 [(a + b), (b + c), 2 (c + a)]
= 6 [(a + b), (b + c), 2 (c + a)]
= 6 × 2 [a, b, c]
=6×2×5
= 60. 65. The volume of the parallelopiped,
| |
1 V = 0 a
a 1 0
1 a 1
= 1 + a (a2 – 1)
= a3 – a + 1
dV fi ___ = 3a2 – 1 da For maximum or minimum, dV ___ = 0 da gives fi
3a2 – 1 = 0
fi
1__ a = ± ___ 3 ÷ –
+
+
1__ 1__ ___ – ___ 3 ÷3 ÷
1 By sign scheme, the point of minima is a = ___ __ . 3 ÷ 67. Since the given vectors are coplanar, so
| | |
1 1 b 1 = 0 1 c – a 1 –a a 1 C Æ C 2 – C1 fi 1 b = 0, 2 0 – 1 C3 Æ C3 – C1 1 c–1 0
a 1 1
|
(
)
5.46 Integral Calculus, 3D Geometry & Vector Booster __
fi a (b – 1) (c – 1) (a – 1) (c – 1)
3 ÷ __ = ____ 2÷2
fi (1 – b) (1 – c) + (1 – a) (1 – c)
70. Let
a × b = a1i + a2j + a3k
+ (1 – a) (1 – b) = 0 Dividing both the sides by (1 – a) (1 – b) (1 – c), we get a 1 1 _____ + _____ + _____ = 0 1–a 1–b 1–c a 1 1 fi 1 + _____ + _____ + _____ = 1 1–a 1–b 1–c 1 1 1 fi _____ + _____ + _____ = 1 1–a 1–b 1–c
We have,
[a, b, i] i + [a, b, j] j + [a, b, k] k
68. Let a = a1i + a2j + a3k,
= a1i + a2j + a3k
b = b1i + b2j + b3k and c = c1i + c2 j + c3k We have, [a, b, c]2
= (a × b) Hence, the result. 71. Let a = a1i + a2 j + a3k,
+ (1 – b) (1 – b) = 0.
(
)
| | | |
= [a, b, c] [a, b, c]
a 1 b 1 = c1
a 2 b 2 c2
a3 b3 × c3
a 1 b 1 c1
a 2 b 2 c2
a 3 b 3 c3
a12 + a2 2 + a32 a1b1 + a2 b2 + a3 b3 a1c1 + a2 c2 + a3 c3 b1a1 + b2 a 2 + b3 a3 b12 + b2 2 + b32 b1c1 + b2 c2 + b3 c3 c1a1 + c2 a2 + c3 a3 c1b1 + c2 b2 + c3 b3 c12 + c2 2 + c32
|
|
a ◊ a a ◊ b a ◊ c = b ◊ a b ◊ b b ◊ c c ◊ a c ◊ b c ◊ c
|
69. As we know that,
|
=
1 2 1 2
1 2
1 2
1
1
1 2
= a1i
Similarly, [a, b, j] j = a2j and
[a, b, k] k = a3k
Thus,
b = b1i + b2 j + b3k and c = a = c1i + c2 j + c3k Also, let m = m1i + m2 j + m3k and
n = n1i + n2 j + n3k
Now,
[a, b, c] (m × n)
| | | |
b a c = a ◊ m b ◊ m c ◊ m a ◊ n b ◊ n c ◊ n
|
a 2 b 2 c2
a3 b 3 × c3
a 1 = b 1 c1
i m 1 n1
|
j m 2 n2
k m 3 n3
r ◊ (b × c) = x a ◊ (b × c) + y b ◊ (b × c) + z c ◊ (b × c)
1
( ) (
72. Given r = x a + y b + z c Taking scalar product of (b × c)
a ◊ a a ◊ b a ◊ c [a, b, c]2 = b ◊ a b ◊ b b ◊ c c ◊ a c ◊ b c ◊ c 1
[a, b, i] i = ((a × b) ◊ i) i
)
1 1 1 __ 1 = 1 – __ + __ __ – 2 2 4 2
fi
r ◊ (b × c) = x a ◊ (b × c)
fi
[r, b, c] = x [a, b, c]
fi
[r, b, c] x = _______ [a, b, c]
[r, c, a] [r, a, b] Similarly, y = _______ & z = _______ [a, b, c] [a, b, c]
3 1 1 __ = __ – __ = 2 8 8
Hence,
3 ÷ Thus, [a, b, c] = ____ __ 2÷2
73. Given,
__
Therefore, the volume of the parallelopiped
[r b c] r = ______ a + [a b c]
[r c a] ______ b + [a b c]
1 [a b, a × b] = __ 4
fi
1 (a × b) ◊ (a × b) = __ 4
[r a b] ______ c [a b c]
Vectors
fi fi fi fi fi 74. Given fi fi
1 |(a × b)|2 = __ 4 1 __ |(a × b)| = 2 1 a b sin (q ) = __ 2 1 sin (q ) = __ 2 p q = __ . 6 = x (a × b) + y (b × c) + z (c × a) ◊ c = x (a × b) ◊ c + y (b × c) ◊ c + z (c × a) ◊ c
Now, |2a + 5b + 5c| = |2a + 5 (b + c)| = |2a + 5 (– a) = |– 3 (a)| = |– 3| |a| = 3. 77. Let Now,
= x (a × b) ◊ c = x [a b c] ◊ c x = _______ [a, b, c]
a = i + j, b = j + k and c = k + i
| |
i (a × b) = 1 0
j 1 1
k 0 = (i – j + k) 1
(a × b) 1 = _______ = ___ __ (i – j + k) |(a × b)| ÷3
◊ c ◊ b Similarly, y = _______ , z = _______ [a, b, c] [a, b, c]
similarly,
(b × c) 1 = _______ = ___ __ (i + j – k) |(b × c)| ÷3
Thus,
x+y+z
and
(c × a) 1 = _______ = ___ __ (i + j – k) |(c × a)| ÷ 3
◊ c ◊ a ◊ b = _______ + _______ + _______ [a, b, c] [a, b, c] [a, b, c]
Volume of a parallelopiped
◊ (a + b + c) = ____________ [a, b, c]
1 1 1 1__ ___ = ___ __ ◊ ___ ◊ __ 1 ÷3 – 1 3 ÷3 ÷
◊ (a + b + c) = ____________ 1/8
= 8 a ◊ (a + b + c).
1 1__ = ____ 1 3÷3 – 1
2 = ____ __ (1 + 1) 3÷3
75. Volume of the tetrahedron
|
| | 0 2 0
1 = __ [a, b, c] 6
1 1 = __ × 1 6 1
4__ = ____ c.u 3÷3
78. Given
a + b + c = a d
and
b + c + d = b a
1 1 1 – 1 2 – 1
|
– 1 1 1 – 1 1 1
1 (C2 Æ C2 + C1) – 1 1
| |
1 = __ (– 1 + 2 + 3) 6
fi
d = – b – c – b a
fi
a d = – a b – a c – a b a
2 = __ c.u. 3
fi
a + b + c = a b – a c – a b a
76. Given |a – b|2 + |b – c|2 + |c – a|2 = 9 fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3 fi (a2 + b2 + c2) + 2 (a ◊ b + b ◊ c + c ◊ a) = 3 – 3 fi |a + b + c|2 = 0 fi (a + b + c) = 0
5.47
Comparing, the co-efficients of the vectors, we get
a = – 1, a b = – 1
fi
a = – 1, b = 1
Now. a + b + c + d + 4 = a d + d + 4 = – d + d + 4 = 4. 79. Let a = 3 i + 2 j + 6 k, b = 2 i + j + k
5.48 Integral Calculus, 3D Geometry & Vector Booster c = i – j + k.
and
a × (b × c) We have to find ± ___________ . |a × (b × c)|
| | | |
Now,
i (b × c) = 2 1
j 1 – 1
i a × (b × c) = 3 2
j 2 – 1
(21 i – 7 k) ________ = ± _________ ÷441 + 49
80. We have, a × (a × b) = (a ◊ b) a – (a ◊ a) b = a – |a|2 b = a – 3 b
| | j 1 1
k 1 – 1
= – 2 i + j + k Thus, from Eq. (i), we get
3 b = (i + j + k) – (– 2 i + j + k)
fi
3 b = 3 i
fi
b=i
81. We have, fi
1 (a × (b × c)) = __ b 2 1 (a ◊ c) – (a ◊ b) c = __ b 2 1 = __ , (a ◊ b) = 0 2
1 1 (a ◊ c) = ___ __ , (a ◊ b) = ___ __ 2 2 ÷ ÷
fi
1 1 a c cos(q1) = ___ __ and a b cos (q2) = ___ __ 2 2 ÷ ÷
fi
1 1 cos (q1) = ___ __ and cos (q2) = ___ __ 2 2 ÷ ÷
fi
(3 i – 7 k) ___ = ± ________ 10 ÷
i a × (a × b) = 1 0
fi
fi
(21 i – 7 k) ___ = ± _________ 7÷10
Now,
1 cos (q1) = __ and cos (q2) = 0 2 p p (q1) = __ , (q2) = __ 3 2
82. We have, 1 (a × (b × c)) = ___ __ (b + c) 2 ÷ 1 fi (a ◊ c) b – (a ◊ b) c = ___ __ (b + c) 2 ÷
k 6 – 3
3 b = a – a × (a × b)
fi
Hence, the result.
= 21 i – 7 k Thus, a × (b × c) ± ___________ |a × (b × c)|
fi
1 a c cos (q1) = __ and a b cos (q2) = 0 2
fi
k 1 1
= 2 i – j – 3 k Also,
fi
...(i)
p p q1 = __ and q2 = __ 4 4 p q2 = __ 4
p Hence, the angle between a and b is __ 4 83. Given, a × (a × c) + b = 0 fi a × (a × c) = – b fi |a × (a × c)| = |– b| p fi |a| |(a × c)| sin __ = 1 2 fi |a| |(a × c)| = 1
( )
fi
|(a × c)| = 1
fi
a c sin q = 1 1 sin q = __ 2 p q = __ 6
fi fi
p Hence, the angle between a and c is __ . 6 84. We have, a × (b × c) ◊ (a × b) × c = 0 fi {(a ◊ c) b – (a ◊ b) c}◊{(c ◊ a) b – (c ◊ b) a} = 0 fi ((a ◊ c) b)2 – (a ◊ c) (c ◊ b) (b ◊ a) – (a ◊ b) (c ◊ a) (c ◊ b) + (a ◊ b)(c ◊ b) (c ◊ a) = 0 fi ((a ◊ c) b)2 – (a ◊ c) (c ◊ b) (b ◊ a) = 0 fi (a ◊ c) {(a ◊ c) b2 – (a ◊ b) (b ◊ c) = 0 fi (a ◊ c) = 0, {(a ◊ c) b2 – (a ◊ b) (b ◊ c)} = 0 Hence, (a ◊ c) = 0
Vectors
85. Given, 1 (a × b) × c = __ |b| |c| a 3 1 fi c × (a × b) = – __ |b| |c| a 3 1 fi (c ◊ b)a – (a ◊ a) b = – __ |b| |c| a 3 1 __ fi (c ◊ b) = – |b| |c| a 3 1 fi |c| |b| cos (q) = – __ |b| |c| 3 1 fi cos q = – __ 3 __ _____ 1 2÷2 fi sin q = 1 – __ = ____ 9 3 Hence, the result. 86. Given,
÷
a ◊ b = 0
r = m a + n b + p (a × b) ...(i) r ◊ a = m (a ◊ a) + n (a ◊ b) + p (a ◊ (a × b)) m=0 r ◊ b = m (a ◊ b) + n (b ◊ b) + p (b ◊ (a × b)) 1 = 0 + n (b ◊ b) + 0 n=1 r ◊ a × b = m (a ◊ a × b) + n (b ◊ a × b)
Let Now, fi Also, fi fi Again,
+ p [(a × b) ◊ (a × b)]
fi
1 = 0 + 0 + p |(a × b)|2
fi
p=1
Hence, from Eq. (i),
r = b + (a × b)
87. We have,
1 a × (b × c) + b × (c × a) = __ a 2
1 fi (a ◊ c) b – (a ◊ b) c + (b ◊ a) c – (b ◊ c) a = __ a 2 1 fi (a ◊ c) = 0, (a ◊ b) = 0, (b ◊ a) = 0, (b ◊ c) = – __ 2
Thus,
|(a × c)|
( )
p = a c sin __ 2 = 7. 1. 1
= 7.
88. We have, (d + a) ◊ [a × b × (c × d)]
= (d + a) ◊ (a × [(b ◊ d) c – (b × c) d]
= (d + a) ◊ [(a × c) (b ◊ d) – (a × d) (b ◊ c)]
= [d ◊ (a × c) (b ◊ d) + a ◊ (a × c) (b ◊ d)
– d ◊ (a × d) (b ◊ c) – a ◊ (a × d) (b ◊ c)]
= [a, c, d] (b ◊ d)
89. Given, fi
px + (x × a) = b p (a ◊ x) + a ◊ (x × a) = a ◊ b
fi
p (a ◊ x) + 0 = a ◊ b
fi
p (a ◊ x) = a ◊ b (a ◊ b) p = _____ (a ◊ x)
fi
px + (x × a) = b p (a × x) + a × (x × a) = a × b
fi
p (a × x) + (a ◊ a) x – (a ◊ x) a = a × b
fi
(a ◊ a) x = (a ◊ x) a – a × b – p (a × x)
fi
(a ◊ b) (a ◊ a) x = _____ p a – a × b – p (a × x)
fi
(a ◊ b) (a ◊ a) x = _____ p a – a × b – p (p x – b)
fi
(a ◊ b) (a2 + p2) x = _____ p a – a × b – p b
fi
(a ◊ b) _____ p a – a × b – p b x = ___________________ (a2 + p2)
90. Given fi fi
[x a b] = 0 x, a, b are coplanar vectors x = m a + n b
(a ◊ c) = 0
fi
a c cos (q1) = 0
fi
x ◊ a = m (a ◊ a) + n (b ◊ a)
fi
cos (q1) = 0
fi
x ◊ a = m (a ◊ a) + n (b ◊ a)
fi
0 = m |a|2 + 0
fi
m=0
Also,
x = m a + n b
fi
x ◊ b = m (a ◊ b) + n (b ◊ b)
fi
( )
p = cos __ 2
( )
p q1 = __ 2
...(i)
Also, fi
Now,
5.49
5.50 Integral Calculus, 3D Geometry & Vector Booster fi
1 = 0 + |b|2
fi
(x × a) × a = b × a
fi
1 n = ___ 2 |b|
fi
a × (x × a) = – b × a = a × b
fi
(a ◊ a) x – (a ◊ x) a = a × b
Thus,
b x = m a + n b = ___ 2 = |b|
fi
|a|2 x = a × b
fi
a × b x = _____ 2 . |a|
b ___ |b|
91. We have, [a × (b + c), b × (c – 2a), c × (a + 3b)] = [a × b + a × c, b × c – 2b × a, c × a + 3c × b] = [a × b – c × a, b × c + 2a × b, c × a – 3b × c] = [x – z, y + 2x, z – 3y] where x = a × b, y = b × c, z = c × a
= (x – z) ◊ (y + 2x) × (z – 3y)
96. Given, fi
r×b=a×b (r – a) × b = 0
fi fi
(r Рa) = l b, l ΠR r = a + l b
fi r ◊ c = (a ◊ c) + l (b ◊ c)
= (x – z) ◊ (y × z – 6 x × y + 2x × z) = x ◊ (y + z) + 6 z ◊ (x + y)
fi
(a ◊ c) l = – _____ (b ◊ c)
Thus,
(a ◊ c) r = a – _____ b. (b ◊ c)
= 7 x ◊ (y × z) = 7 [x y z] 2
= 7 [a b c] 92.
97. Given,
= 28 We have, [(a – b), (a – b – c), (a + 2 b – c)] = (a – b) ◊ (a – b – c) × (a + 2 b – c)
x a + y b + z c = d Taking scalar product of (b × c), we get x a ◊ (b × c) + y b◊ (b × c) + z c◊ (b × c) = d ◊ (b × c)
= (a – b) ◊ [2 a × b – a × c – b × a + b × c – c × a – 2c × b] = a ◊ (b × c) – 2a ◊ (c × b) + b ◊ (a × c) + b ◊ (c × a) = a ◊ (b × = 3 [a, b, = 15 93. We have,
c) + 2a ◊ (b × c) – b ◊ (c × a) + b ◊ (c × a) c]
94. Given,
= 576 × 4 = 2304
[(2a × 3b), (3b × 4c), (4c × 2a)] = [x × y, y × z, z × x] where 2a = x, 3b = y, 4c = z = [x, y, z]2 = [2 a, 3 b, 4 c]2 = 4. 9. 16 [a, b, c]2
x×b=a×b fi (x – a) × b = 0 fi (x – a) = l b, fi x = a + l b. 95. We have, x×a=b
( )
fi
x a ◊ (b × c) + 0 + 0 = d ◊ (b × c)
fi
x a ◊ (b × c) = d ◊ (b × c)
fi
d ◊ (b × c) [d b c] x = ________ = ______ a ◊ (b × c) [a b c]
Similarly, we can easily find that [d c a] [d a b] y = ______ and z = ______ [a b c] [a b c] 98. We have, x+y=a fi
(x + y) ◊ a = a ◊ a
fi
x ◊ a + y ◊ a = a ◊ a
fi
1 + y ◊ a = |a|2
...(i)
Also, x × y = b
where l ΠR
fi
a × (x × y) = a × b
fi
(a ◊ y) x – (a ◊ x) y = a × b
fi
(|a|2 – 1) x – y = a × b [from Eq. (i)] ...(ii)
Again x + y = a Adding Eqs. (ii) and (iii), we get
(|a|2) x = (a × b + a)
...(iii)
Vectors
(a × b + a) x = __________ |a|2
fi
Thus,
From Eq. (iii), we get (a × b + a) y = a – x = a – __________ |a|2
Hence,
| | j 3 3
k 0 4
= 12 i – 8 j + 6 k π 0
(a × b + a) (a × b + a) x = __________ , y = a – __________ 2 |a| |a|2
a = 2 i + 3 j and b = 3 j + 4 k i a × b = 2 0
5.51
3. Resultant of P and Q
AB = 3 i + 7 j – 5 k
= (2 i – 5 j + 6 k) + (– i + 2 j – k) = (i – 3 j + 5 k). Displacement, = AB = (6 i + j – 3 k) – (4 i + 3 j – 2 k) = (2 i + 4 j – k)
BC = – 7 i – j + 5 k
Workdone = Force ◊ Displacement
CA = 4 i – 6 j
= (i – 3 j + 5 k) ◊ (2 i + 4 j – k) = |2 – 12 – 5| = |– 15| = 15units. 4. We have,
1. We have,
and
Unit vectors along A B, B C, C A are 1 ___ ____ (3i + 7j – 5k) 83 ÷ 1 ___ ____ (– 7i + j + 5k) ÷75
(a × b)2 = (a b sin q )2
= a2 b2 sin2 q
1 ___ and ____ (4 i – 6 j). ÷52
= a2 b2 (1 – cos2q )
= a2 b2– a2 b2 cos 2q
Since the forces P, Q, R are of 15 KN each, so
= a2 b2 – (a ◊ b)2
15 P = ____ ___ (3 i + 7 j – 5 k) 83 ÷
15 Q = ____ ___ (– 7 i – j + 5 k) ÷75
= (b × c)
= – (c × b)
15 and R = ____ ___ (4 i – 6 k) ÷52 Thus, the resultant S is given by
fi
(a × b) + (c × b) = 0
fi
(a × c) × b = 0
15 15 = ____ ___ (3 i + 7 j – 5 k) + ____ ___ (– 7 i – j + 5 k) 83 ÷ ÷75 15 ___ (4 i – 6 j) + ____ ÷52
) (
)
3 6 7 7 4 1 = 15 ____ ___ – ____ ___ + ____ ___ i + ____ ___ – ____ ___ + ____ ___ j 83 ÷75 83 ÷75 ÷52 ÷52 ÷ ÷
(
5 ___ + – ____ + 83 ÷
(a × b) = (b × c) π 0
fi a × c = k b, where k is a scalar. 6. Sum of the vectors
= P + Q + R
[ (
5. Given,
)]
= (2 i + 4 j – 5 k) + (S i + 2 j + 3 k)
= (2 + S) i + 6 j – 2 k Unit vector parallel to the sum of the vectors
(2 + S) i + 6 j – 2 k _______________ = _________________ ÷(2 + S)2 + 36 + 4
(2 + S) i + 6 j – 2 k = _________________ ___________ 44 ÷ S 2+ 4S +
5 ____ ___ k ÷75 Also, it is given that
2. As a × b = 0 does not imply that the vectors a and b are null vectors, let
(2 + S) i + 6 j – 2 k ___________ (i + j + k) ◊ ________________ = 1 44 ÷ S 2 + 4S +
5.52 Integral Calculus, 3D Geometry & Vector Booster ___________
fi
(2 + S) + 6 – 2 = ÷ S 2 + 36 + 44
Thus, the vector r is
fi
(S + 6)2 = (S2 + 4S + 44)
fi
S2 + 12 S + 36 = (S2 + 4S + 44)
10. We have |a| = ÷ 144 + 16 + 9 = 13
fi
8S = 8
___________
fi S=1 7. Since the vectors (2i – j + k),
____________
____
|b| = ÷ 64 + 144 + 81 = ÷ 289 = 17
and
|c| = ÷ 1089 + 16 + 576 = ÷ 1681 = 41
_______________
_____
Thus, the length of the edges are 13, 17 and 41 respectively. \ Area of the faces = |a × b|, |b × c|, |c × a|
(i + 2j – 3k) and (3i + aj + 5k)
| |
are coplanar, so,
2 –1 1 1 2 –3 = 0 3 a 5 fi
= ± (– 5i + j + 5k).
3 (3 – 2) – a (– 6 – 1) + 5(4 + 1) = 0
= 220, 435, 455
Volume of the parallelopiped = [a, b, c]
|
fi
3 + 7a + 25 = 0
fi
7a = – 28
fi
a = – 4.
8. Given,
|a + b| = |a – b|
fi
|a + b|2 = |a – b|2
fi
a2 + b2 + 2a ◊ b = a2 + b2 – 2a ◊ b
= 3696 11. Since a, b and c are non-coplanar vectors, so [a, b, c] π 0.
fi
4 a ◊ b = 0
fi
a ◊ b = 0
|
12 4 3 = 8 – 9 –12 33 – 4 –24
Now,
[x, y, z] = x ◊ (y × z)
c×a b×c a×b = _______ ◊ _______ × _______ [a, b, c] [a, b, c] [a, b, c]
...(i)
Let the vector r makes an angle q with the vectors a, b and c.
1 = ________ [b × c, c × a, a × b] [a, b, c]3
1 = ________ 3 [a, b, c]2 [a, b, c]
1 = _______ π 0. [a, b, c]
Thus, a and b are perpendicular to each other. 9. Let r = x i + yj + z k fi
x2 + y2 + z2 = 51
r ◊ a r ◊ b r ◊ c cosq = ____ = _____ = ____ |r| |a| |r| |b| |r| |c| r ◊ a r ◊ c r ◊ b r ◊ c fi ____ = ____ and _____ = ____ |r| |a| |r| |c| |r| |b| |r| |c| Thus,
1 1 __ (– 4x – 3z) __ (x – 2y + 2z) y y 3 5 ___ ___ ___ fi _____________ = ____ ___ ; ___________ = ____ ÷51 ÷51 ÷51 ÷51 fi
x – 5y + 2z = 0; 4x + 5y + 3z = 0
Solving, we get x ___ = – 5
y __ = 1
z __ = l (say) 5
From Eqs. (i) and (ii), we get
25l2 + l2+ 25l2 = 51
fi
51l2 = 51
fi fi
l2 = 1 l = ± 1
...(ii)
(
)
Thus, x, y and z are non-coplanar vectors. Now, x ◊ (a + b) + y ◊ (b + c) + z ◊ (a + b) (b × c) ◊ (a + b) + (c × a) ◊ (b + c) + (a × b) × (c + a) = ______________________________________________ [a, b, c] (b × c) ◊ a + (c × a) ◊ b + (a × b) ◊ c = _______________________________ [a, b, c] [a, b, c] + [a, b, c] + [a, b, c] = __________________________ [a, b, c] 3[a, b, c] = _________ [a, b, c] = 3
Vectors
12. Given,
r×b=c×b
fi
(r – c) × b = 0
fi (r – c) is parallel to b. or r is coplanar with the vectors b and c. Also, r ◊ a = 0 fi r is perpendicular to a. Again a ◊ b π 0 fi neither a nor b is a null vector or a is not perpendicular to b. 13. Given, a+b+c=0 fi (a + b + c)2 = 0 fi a2 + b2 + c2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 1 + 1 + 1 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3 – 3 fi (a ◊ b + b ◊ c + c ◊ a) = ___ 2 14. As we know that, if a, b, c form a right handed system, so a×b=c fi c=a×b fi
| |
i j k c = x y z 01 0 = x k – z i = – z i + x k
15. Since a, b and c are lie in the same plane, so they are coplanar. Thus, [a, b, c] = 0 fi
a ◊ (b × c) = 0
16. Given,
A=B+C
= 10 i – (2d – 12) j + (9 – d) k __
Given the area of the triangle is 5÷6 . __ 1 fi __ |A × B| = 5 ÷6 2 __
fi |A × B| = 10 ÷6
________________________
fi 100 + (2d + 12)2 + (9 – d)2 = 600 fi (2d + 12)2 + (9 – d)2 = 500 fi 4d 2 + 48d + 144 + 81 – 18d + d 2 = 500 fi 5 d 2 + 30 d – 275 = 0 fi d 2 + 6 d – 55 = 0 fi (d + 11) (d – 5) = 0 fi d = –11, 5 Hence,
a = 8, b = 4, c = 2, d = 5
or
a = – 8, b = 4, c = 2, d = –11
17. We have, (a + b) × (a × b) = a × (a × b) + b × (a × b)
= (a ◊ b)a – (a ◊ a) b + (b ◊ b)a – (b ◊ a)b
=a–b+a–b
= 2 (a – b) 18. Ans. (c) 19. Ans. (d) 20. Also, Let , be unit vectors on form_ a right handed system › Let b = b1 + b2 + b3 C
fi ai + bj + ck = di + 3j + 4 k + 3i + j – 2k
A
R
B
fi a = d + 3, b = 4, c = 2.
We have,
Then P = ÷ 3 ( × b )
| | | |
i j k = a 4 2 d3 4
= 10 i – (4a – 2d ) j + (3a – 4d) k
= 10 i – (4d + 12 – 2d) j + (3d + 9 – 4d)k
__
2 fi ÷100 + (2d + 12) + (9 – d)2 = 10÷6
Q
i j k (A × B) = a b c d3 4
5.53
P
_›
_›
__ __
= ÷ 3 (b2 – b3 ) __›
_›
_›
and Q = b – ( ◊ b ) = (b2 + b3 ) _› __›
P Q Then cos A = _____ _ › _ _› =0 |P ||Q |
A = 90°
_›
_›
__›
Also, R = P – Q
so that , ,
5.54 Integral Calculus, 3D Geometry & Vector Booster __
__
23. Let BP as x-axis and BY as y-axis.
= (÷3 b 2 – b3) – (÷3 b 2 – b2) Now, cos B =
_› _› P R _____ _ › _ › |P | |R |
(
__
) ( ) __ 3 ( b22 + b23 ) – 2 ÷3 b2 b3 – 1 __________________ B = cos __ 2 2 ÷3 ( b2 + b 23 )
Let i and j be unit vectors along x-axis and y-axis, respectively.
C = 180° – 90° – B = 90° – B
Here,
3 b22 + b 23 – 2 ÷3 b2 b3 = __________________ __ 2 2 ÷3 b2 + b23 Thus, Also,
(
)
Hence, the result,
p a + b = __ 2 Let q the angle between BP and CP fi
Question asked in Past IIT-JEE Examinations 21. Let x be a unit vector such that x = ai + bj Thus, |x| = 1
Now,
fi
|a i + b j| = 1
fi
÷a 2 + b2 = 1
fi
a2 + b2 = 1
_______
...(i)
Also (ai + bj) ◊ (i + j) = |(ai + bj)| ◊ |(i + j)| cos (45°) fi fi Again,
__ 1 a + b = 1 ◊ ÷2 ◊ ___ __ 2 ÷
a + b = 1
...(ii)
(ai + bj) ◊ (3i – 4j) = |(ai + bj)| ◊ |(3i – 4j)| cos (60°) 1 fi 3a – 4b = 1.5 ◊ __ 2 5 fi 3a – 4b = __ 2 On solving (ii) and (iii), we get 13 1 a = ___ and b = ___ 14 14 1 Thus, x = ___ (13i – j) 14
| |
22. Since the given vectors are coplanar, so 2 –1 1 1 2 – 3 = 0 3 l 5 fi 3(3 – 2) – l (– 6 – 1) + 5 (4 + 1) = 0 fi 3(3 – 2) – l (– 6 – 1) + 5 (4 + 1) = 0 fi 3 + 7l + 25 = 0 fi 7l + 28 = 0 fi l = – 4
2a + 2b = p
...(ii)
CP = i cos (a + b) + j sin (a + b)
( )
( )
p p = i cos __ + j sin __ = j 2 2 p Thus, BP ◊ CP = i ◊ j = __ . 2 Therefore, the angle between the angle bisectors of angles B and C is 90°. 24. As we know that, a vector in the direction of the bisector of the angle between two vectors a and b is a b 1 1 __ + ___ = __ (i – 2j + 2k) + __ (2i + j + 2k) |a| |b| 3 3
1 = __ (3i – j + 4k) 3
25. We
a=b×c = b × (a × b)
= (b ◊ b) a – (b ◊ a) b
= |b|2a – (b ◊ a) b
Comparing, we get
|b|2 = 1, (b ◊ a) = 0
fi Also,
|b| = 1, a ^ b c=a×b
= (b × c) × b
= (b ◊ b) c – (b ◊ c) b
= |b|2 c – (b ◊ c) b Comparing, we get |b|2 = 1,(b ◊ c) = 0 fi
|b| = 1, b ^ c
Thus,
|b| = 1 and a = c
26. Ans. (a) 27. Ans. (d)
Vectors
28. Given,
x×y=b
fi
(x × y) × a = b × a
fi
a × (x × y) = – b × a
31. Given
p b = and a = (4n + 1) __ , n ŒI 2 |(a × b) ◊ c| = |a| |b| |c|
fi
|a| |b| |c| sinq cosj = |a| |b| |c|
fi
sinq cosj = 1
fi
sinq = 1 and cosj = 1 p q = __ and j = 0 2
fi
=a×b
fi
(a ◊ y) x – (a ◊ x) y = a × b
fi
(a ◊ y) x – y = a × b
...(i)
fi
Also,
x + y = a
...(ii)
Similarly, |(a × b) ◊ c| = |a| |b| |c|
fi
x ◊ a + y ◊ a = a ◊ a
fi
1 + y ◊ a = a2
fi
y ◊ a = a2 – 1
...(iii)
From Eqs. (i) and (iii), we get
2
(a – 1) x – y = a × b
...(iv)
Adding and subtracting Eq. (ii) from Eq. (iv), we get
1 x = __ 2 (a + a × b) a 1 y = a – __ 2 (a + a × b). a
fi
|(b × c) ◊ a| = |a| |b| |c|
fi
b^c
Again
|(a × b) ◊ c| = |a| |b| |c|
fi
|(c × a) ◊ b| = |a| |b| |c|
fi
a ^ c
Thus,
a ^ b, a ^ c and b ^ c
32. Given,
x+y+z=a
fi
(x + y + z) ◊ a = a ◊ a
fi
(x ◊ a + y ◊ a + z ◊ a) = |a|2
30. We have,
fi
3 __ + 2
a × (b × c) + (a ◊ b) b = (4 – 2b – sina) b + (b 2 – 1)c
fi
fi (a ◊ c) b – (a ◊ b) c + (a ◊ b) b
Also,
and 29. Ans. (b)
2
= (4 – 2b – sina) b + (b – 1) c
Comparing, we get
(a ◊ c + a ◊ b) = 4 – 2b – sina
...(i)
2
and
(b – 1) = – (a ◊ b)
fi
b 2 = 1(a ◊ b)
Also
(c ◊ c) a = c
fi
(c ◊ c) (a ◊ c) = (c ◊ c)
...(ii)
fi (a ◊ c) = 1 From Eqs. (i) and (iii), we get
...(iii)
...(iv)
1 + (a ◊ b) = 4 – 2b – sina
Adding Eqs. (ii) and (iv), we get
(
)
7 __ + z ◊ a = 4 4 13 3 (a ◊ z) = 4 – ___ = __ 4 4
...(i)
x+y+z=a
fi
x ◊ x + y ◊ x + z ◊ x + a ◊ x
fi
3 1 + y ◊ x + z ◊ x = __ 2
fi
1 y ◊ x + z ◊ x = __ 2
...(ii)
3 Similarly, x ◊ y + z ◊ y = __ 4
...(iii)
and
1 x ◊ z + y ◊ z = – __ 4
...(iv)
Adding Eqs. (ii), (iii) and (iv), we get
1 x ◊ y + y ◊ z + z ◊ y = __ 2
4 – 2b – sina + b 2 = 2
Hence,
fi
b2 – 2b + 2 = sina
3 1 x ◊ y = __ , y ◊ z = 0, z ◊ x = – __ 4 4
Now,
x × (y × z) = b
fi (b – 1)2 + 1 = sina It is possible only when
fi
(x ◊ z) y – (x ◊ y) z = b
fi
3 1 – __ y – __ z = b 4 4
(b – 1) = 0 and sina = 1
5.55
...(v)
5.56 Integral Calculus, 3D Geometry & Vector Booster Also,
(x × y) × z = c
fi
fi
z × (x × y) = – c
Solving Eqs. (iii) and (vi), we get
fi
(z ◊ y) x – (z ◊ x) y = – c
fi
1 0x + __ y = – c 4
fi
y = – 4c
...(vi)
From Eqs. (v) and (vi), we get fi
3 c – __ z = b 4 4 z = __ (c – b) 3
Finally, x + y + z = a fi
x=a–y–z
fi
8 4 x = a + __ b + __ c 3 3 __ __ 1 x ◊ y = ÷2 ◊ ÷2 cos (60°) = 2 ◊ __ 2
Thus,
x ◊ y = 1 = y ◊ z = z ◊ x
fi
x ◊ x = 2 = y ◊ y = z ◊ z
Now,
x × (y × z) = a
fi
(x ◊ z) y – (x ◊ y) z = a
fi
y – z = a
Also,
y × (z × x) = b
fi
(y ◊ z) z – (y ◊ z) x = b
fi
z–x=b y – x = a + b
Again,
(x × y) = c
fi
x × (x × y) = x × c
fi
(x ◊ y) x – (x ◊ x) y = x × c
fi
x – 2y = x × c y × (x × y) = y × c
fi
(y ◊ y) x – (y ◊ x) y = y × c
fi
2x – y = y × c
Subtracting Eq. (iv) from Eq. (v), we get
x + y = (y – x) × c
(a + b) × c – (a + b) x = __________________ 2
(a + b) × c + (a + b) y = __________________ 2
and
b – a + (a + b) × c z = _________________ 2
34. Let
r = (a × b) × c
fi
r ^ (a × b) and r ^ c
fi
[(a × b) × c] ^ c
–2a + 3b – c = lp + mq + nr
fi ...(i)
where l, m, n ŒR – 2a + 3b – c = l (2a – 3b) + m (a – 2b + c) + n (– 3a + b + 2c) = (2l + m – 3n) a
+ (– 3l – 2m + n) b + (m + 2n) c
Comparing, we get ...(ii) ...(iii)
(2l + m – 3n) = – 2,
(– 3l – 2m + n) = 3
and
(m + 2n) = –1
Solving, we get ...(iv)
Finally, x × y = c fi
and A = i, B = i + j, C = i + j + k Now, AB ◊ CD + BC ◊ AD + CA ◊ BD = –1 + 0 + 1 = 0. 36. Since a, b and c are non-zero and non-coplanar vectors, so any vector can be expressed as a linear combination of these three vectors, i.e.
Adding Eq. (i) and (ii), we get
...(vi)
35. Let D be the origin.
33. We have,
x + y = (a + b) × c
7 1 Therefore, –2a + 3b – c = – __ q + __ r. 5 5 37. According to the question,
...(v)
7 1 l = 0, m = – __ , n = __ 5 5
fi fi
(
a c = l __ + |a|
( (
)
b ___ , |b|
where l ŒR
)
7i – 4j – 4k – 2i – j + 2k c = l ___________ + ___________ 9 3 i – 7j + 2k c = l __________ 9
)
Vectors
It is given that, __ |c| = 3÷6 fi
|c|2 = 54
fi
1 + 49 + 4 54 = l2 _________ 81
(
l = 81
l = ± 9
fi
y + y × a x = _________ |y|2
(
)
i – 7j + 2k Therefore, c = ± 9 __________ 9
|y|2 x = y + y × a
)
2
fi
fi
= ± (i – 7j + 2k)
38. Given r = x1r1 + x2r2 + x3r3
fi (2a – 3b + 4c)
( ) ( ) ( ) a×b a×b ( _____ ) – ( _____ ) × a c c = ____________________ a_____ ×b ( c ) b× a b× a – _____ c – ____ × a c ____________________ = b×a2 _____ c
2
c ((a × b) + (a × b) × a) = ______________________ (a × b)2
= (a – b + c) x1 + (b + c – a) x2 + (c + a – b) x3
Finally, y × z = b
= (x1 – x2 + x3) a + (– x1 + x2 – x3) b
fi
y × (y × z) = y × b
+ (x1 + x2 + x3) c
fi
(y ◊ z)y – (y ◊ y)z = y × b
Comparing, we get
fi
y – |y|2 z = y × b
fi
|y|2z = y – y × b
fi
y – y × b z = _________ |y|2
x1 – x2 + x3 = 2
fi
– x1 + x2 – x3 = – 2
fi x1 + x2 + x3 = 4 On solving, we get,
7 1 x1 = __ , x2 = 1, x3 = – __ 2 2
39. Given,
x×y=a
fi
b × (x × y) = b × a
fi
(b ◊ y)x – (b ◊ x) y = b × a
Also,
y×z=b
fi
y ◊ (y × z) = y ◊ b
fi
0 = y ◊ b
fi
y ◊ b = 0 – (b ◊ x) y = b × a
fi
b×a b×a y = – _____ = – _____ c (b ◊ x)
Again
x×y=a
fi
y × (x × y) = y × a
fi
(y ◊ y) x – (y ◊ x) y = y × a
fi fi
...(ii)
...(iii)
c ((a × b) – (a × b) × b) = ______________________ . (a × b)2
r = a (b × c) + b (c × a) + g (a × b) where a = (a ◊ d), b = (b ◊ d), g = (c ◊ d)
Now,
r ◊ a = a [a, b, c]
r ◊ b = b [a, b, c]
r ◊ c = g [a, b, c]
Thus, r ◊ (a + b + c) = (a + b + g ) [a, b, c] fi r ◊ (a + b + c) = (a ◊ d + b ◊ d + c ◊ d) [a, b, c] fi r ◊ (a + b + c) = (a + b + c) ◊ d [a, b, c] fi (r – d [a, b, c]) (a + b + c) = 0
|y| x – (y ◊ x)y = y × a |y| x – y = y × a
fi
2 2
) ( ) ( )
40. Let r = (a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)
From Eqs. (i) and (ii), we get
(
[\ (y ◊ z) = 1]
a×b a × b _____ c – _____ × b c ____________________ = a×b 2 _____ c
...(i)
5.57
[ (y ◊ x) = 1]
fi (r – d [a, b, c]) = 0 [ coplanar]
a, b, c are non-
5.58 Integral Calculus, 3D Geometry & Vector Booster fi r – d [a, b, c] = 0 fi r = d [a, b, c] fi |r| = |d [a, b, c]|
(a × b) b v = ___ + ________ 2 |b | |(a × b)|2
Therefore, 43. Given,
r ◊ (a + b) = 0 and r ◊ (b – c) = 0
1 a × (b × c) = __ (b + c) 2 1 fi (a ◊ c) b – (a ◊ b) c = __ (b + c) 2 Comparing, we get 1 1 a ◊ c = __ , a ◊ b = – __ 2 2 1 1 fi cos q1 = __ and cos q2 = – __ 2 2 p 2p __ ___ fi q1 = and q2 = 3 3 2p Thus, the angle between a and b is ___ . 3 44. Area of a given parallelogram 1 = __ |d1 × d2| 2 i j k 1 = __ 2 3 – 6 2 3 – 4 –1
fi
3y + z = 0 and 3z = 0
fi
y = 0 and z = 0
fi |r| = |[a, b, c]| ( d is a unit vector) which is independent of d. Thus, |(a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)|
= [a, b, c]
41. Let
r = xi + yj + zk
Now,
a + b = (i + j – k) + (– i + 2j + 2k) = (3j + k)
and b – c = (– i + 2j + 2k) – (– i + 2j – k) = 3k It is given that,
Since r is a unit vector, so
x2 + y2 + z2 = 1
2
| | | |
|
1 = __ (– 27i – 16i – 17k) 2 1 _______________ = __ ÷729 + 256 + 289 2 _____ 1 = __ × ÷ 1274 2 39.5 = ____ = 19.75 2
fi
x =1
fi
x = ± 1
Thus,
r = ± i
45. We have,
fi
rˆ = ± i
(– 4i + 5j) a + (3i – 3j + k) b (i + j + 3k)c
Hence, the result. 42. Here, v, b and a × b are coplanar.
= l (ai + bj + ck)
Thus,
v = a b + b (a × b)
(– 4a + 3b + c)i + (5a – 3b + c)j + (b + c)k = l (ai + bj + ck) Comparing, we get
fi
v ◊ b = a b ◊ b + b (a × b) ◊ b v ◊ b = a |b|2 + 0
(– 4a + 3b + c) = l a,
fi fi
Now,
1 = a |b|2 + 0 1 a = ___ 2 |b| [v, a, b] = 1
fi
v ◊ (a × b) = 1
fi
l3 + 4l2 – 25l = 0
fi
(a b + b (a × b)) ◊ (a × b) = 1
fi
l (l2 + 4l – 25) = 0
fi
(a b ◊ (a × b) + b (a × b) ◊ (a × b)) = 1
fi
l = 0 and (l2 + 4l – 25) = 0
fi
(a 0 + b (a × b)2) = 1 1 b = ________ |(a × b)|2
fi
fi
– 4 ± ÷116 l = 0 and l = _________ 2
fi
l = 0 and l = – 2 ± ÷29
fi
(5a – 3b + c) = l b and (b + 3c) = l c.
|
|
3 – 4 – l 1 Thus, –3 – l 5 1 0 1 3–l
____
___
Vectors
46. Let Given,
r = xi + yj + zk
x – 2y + 5z = 0
r ◊ a = 0, r ◊ b = 0
fi
2x + 3y – z = 0 y z x Thus, ______ = ______ = ____ 2 – 15 10 + 1 3+ 4 y z x ____ fi = ___ = __ = l (say) – 13 11 7 Also,
r ◊ (2i + j + k) + 8 = 0
fi
2x + y + z + 8 = 0
...(i)
...(ii)
– 26l + 11l + 7l + 8 = 0
fi
– 8l + 8 = 0
fi
l=1
(Tougher Problems for Jee Advanced) 1. We have,
› _› _
|a + b | 2 = a2 + b2 + 2 (a ◊ b )
= 1 + 1 + 2cosq
= 2 (1 + cosq)
q = 2.2 cos2 __ 2
2
( ) q = 4 cos ( __ ) 2
( ) 3 q __ | a + b | = 3 cos ( __ ) 2 2 _ › _› q |a + b | = 2 cos __ 2 _›
_›
fi Also,
Hence, r = – 13i + 11j + 7k 47. Let the position vector of the third vertex be x i + y j.
_ ›
_›
fi
From Eqs (i) and (ii), we get
_›
_ ›
› _› _
|a – b | 2 = a2 + b2 – 2 (a ◊ b )
= 1 + 1 – 2 cosq
= 2 (1 – cosq)
( ) q |a – b | = 2 sin ( __ ) 2 q 2 |a – b | = 4 sin ( __ ) 2 q = 4 sin2 __ 2
_›
fi
_›
fi
_›
_›
Thus, Here,
AC = (x + 1)i + (y – 3)j
fi
BO = i + 3j
Since
AC ^ BO
fi
AC ◊ BO = 0
fi
(x + 1) + 3(y – 3) = 0
fi
x + 3y = 8
Also,
BC = (x – 2)i + (y – 5) j
and
OA = 2i – j
Again,
BC ◊ OA
fi
2 (x – 2) – (y – 5) = 0
fi
2x – y = 1
|
|
( )
|
= [– 5, 5]. _ ›
...(i)
...(ii)
5 17 x = __ , y = ___ 7 7 5 17 = __ i + ___ j. 7 7
|
_ › _› 3 _› _ › Hence, the range of __ a – b + 2 |a – b | 2
Hence, the position vector of the thrid vertex is
( )
_› _› q q 3 _› _› __ a + b + 2|a – b | = 3cos __ + 4sin __ 2 2 2
On solving Eqs (i) and (ii), we get
5.59
2. Let a = x + y + z , where It is given that, _› _› 1 a ◊ k = ___ __ 2 ÷ 1 fi z = ___ __ 2 ÷ _›
x2 + y2 + z2 = 1
Also,
|a +
+ |=1
fi
|(x + 1) + (y + 1) + z | = 1
fi
(x + 1) + (y + 1)2 + z2 = 1
fi
1 (x + 1)2 + (y + 1)2 + __ = 1 2 1 2 2 (x + 1) + (y + 1) = __ 2
fi
2
5.60 Integral Calculus, 3D Geometry & Vector Booster _ _ › › _› p Thus, the angle between (2a + b ) and b is __ . 2 ___› 6. Now, AB = 2 + +
1 x2 + y2 + 2 (x + y) + 2 = __ 2 1 1 __ __ + 2 (x + y) + 2 = 2 2
fi fi
_ __›
fi
x + y = –1
...(i)
Also,
1 x2 + y2 = __ 2
...(ii)
and AC = (t + 1) – Thus,
1 ___› ___› ) ar (D ABC) = __ ( AB × AC 2
|
|
| | |
From Eq. (i) and (ii), we get 1 1 x = – __ = y, z = ___ __ 2 ÷2
1 = __ 1 1 2 2 (t + 1) 0 –1
_› kˆ 1 Thus, a = – __ ( + ) + ___ __ 2 2 ÷
1 = __ ( – + (t + 3) – (t + 1) ) 2
3.
_ _› _› It is given that |r + bs | is minimum __› __› _ › Hence, the value of |r + bs | 2 + |bs | 2
when b = 0
_›
4. Let R = a + b + c .
› _› _ (u ◊ R –
10) +
› _› _ (v ◊ R –
1 = __ ÷ 2
1 = __ ÷2t 2 + 8t + 11 2
Let 20) +
___________________ 2 1 + (t + 3) + (t + 1)2 ____________
_ ›
= |r | 2
It is given that
|
__› _› (w ◊ R )
=0
z = 2t2 + 8t + 11
dz fi __ = 4t + 8 dt
fi (a – 2b + 3c) + (2a + b + 4c)
fi
fi
d 2z ___ 2 = 4 > 0 dt z is minimum
fi
Area is minimum.
+ (a + 3b + 3c) = 10 – 20 – 20
Comparing the co-efficients, we get a – 2b + 3c = 10, 2a + b + 4c = – 20 and a + 3b + 3c = – 20 Solving, we get a = –1, b = 2, c = 5 _›
Hence, R = – + 2 + 5 .
5. It is given that,
fi
_ › _› |a + b | = 1 _ › _ › |a + b | 2 = 1
fi
a2 + b2 + 2 (a ◊ b ) = 1
fi
1 + 1+ 2 (a ◊ b ) = 1
_› _›
_› _›
fi
_› _› (a ◊ b )
Now,
(2a + b ) ◊ b = (2a ◊ b + b ◊ b )
_›
For maximum or minimum, dz __ = 0 gives 4t + 8 = 0 dt t = – 2 Thus, the minimum area is
1 = – __ 2
_›
_ ›
_›
_ ›
|a + b | = 1
fi
|a + b | 2 = 1
fi
|a | 2 + |b | 2 + 2 (a ◊ b ) = 1
fi
1 + 1 + 2 (a ◊ b ) = 1
› _› _
› _› _ 2(a ◊ b )
( )
_ › _ ›
+
_› |b | 2
1 = 2 – __ + 1 2
= –1 + 1
= 0.
_›
_ ›
› _ › _
_› _›
_
_› › 1 (a ◊ b ) = – __ . 2 It is also given that,
fi
_ › _ ›
=
7. Given,
3 1 ___________ ___ = __ ÷8 – 16 + 11 = __ 2 2 ÷
_ ›
_ ›
_›
c = l a + m b _›
› _› _
fi
|c | 2 = l2 + m2 + 2l m (a ◊ b )
fi
1 1 = l2 + m2 + 2l m – __ 2
( )
= l2 + (2l)2 – l (2l)
|
Vectors
fi fi fi Again
fi
2 m = 2l = ___ __ 3 ÷
10. Given, _› _ _ › › a + b = m p
fi
(
)
1 2__ (l, m) = ___ __ , ___ 3 ÷ 3 ÷ 8. It is given that,
Thus,
( )
1 cosq = cos (cos–1) __ = 2
Now.,
1 __ 4
___›
__›
_ ›
_›
_ ›
› _ › _
( )
_›
_ › __› (2a × b )
_› _ › Now, b ◊ c =
_› _› a ◊ c
_› – 3 |b | 2
_›
=
= – 6
2 2
2
= 4a b sin q + 144
= 64 sin2q + 144
3 = 64 × __ + 144 = 48 + 144 = 192 4
_› _›
11. Let a =
+
–2 +
_› b = _› and c =
–2
| | | | | |
– 2 +
+
_›
____
__
|c | = ÷ 192 = 8÷3
_› _ › Again, b ◊ c =
– 48
= – 3 – 3 – 3
Also,
|
_ › _› _› a × (b × c ) = 1 1 – 2 –3 – 3 – 3
| |
= 3 1 1 – 2 –1 –1 –1
= 3 1 1 – 2 –1 –1 –1
= – 48
_› _› _ _ › › |c | 2 = |(2a × b ) |2 + 9 |b | 2 – 0
_›
= |(p ◊ q ) |
– 3b
› _› _ – 3(a ◊ b )
_ ›
9. Given, c =
fi
_› _›
_ ›
Sum of the roots = –1 and the product of the roots = 12 Hence, sum of the values of m = –1 and the product of all values of m = 12.
Also,
_ › _ ›
l = 2, m = m
1 16 = 4 + m2 + 4m __ 4 m2 + m – 12 = 0
_ › _ › _ ›
= |(p ◊ q ) | |b |
Now,
fi
_ › _ › _ ›
_ ›
|(a ◊ q ) p – (p ◊ q ) a | = |(p ◊ q ) | |m p – a |
|c | 2 = 4 |a | 2 + m2|b | 2 + 4m (a ◊ b )
_ › _ ›
_› _ ›
_›
fi
_› _ ›
_ › _ ›
fi a ◊ q + b ◊ q = m (p ◊ q )
_› _› b × c = 1 – 2 1 – 2 1 1
\ c = 2a + mb
fi
__
3 ÷ – 48 __ 3__ cosj = ________ = – ____ = – ___ 2 4 × 8÷3 2÷3 5p j = ___ 6
_› _ ›
_ _ _› _ › › _› × b Also, c × b = 2a _ _ _› _ › › _› fi (c – 2a ) × b = 0 _› __› _› \ (c – 2a ) is parallel to b _› __› _› fi (c – 2a ) = m b _› __› _› fi c = 2a + m b _› _ › = l a + m b
Thus,
|b | |c | cosj = – 48
fi a ◊ q = m (p ◊ q )
_› _›
a ◊ b = ab cos (q) fi
_› _ ›
3l2 = 1 1 l2 = __ 3 1 l = ___ __ 3 ÷
fi
5.61
|
= 3 1 1 – 2 0 0 – 3
12. It is given that,
= – 9 ( – )
[ ]
– 2 _ › _› _› a a + b b + g c = – 5 6
fi
– 2 1 2 1 a 0 + b 1 + g 1 = – 5 0 – 3 –1 6
( ) ( ) ( ) ( )
5.62 Integral Calculus, 3D Geometry & Vector Booster fi
a + 2b + g = – 2
fi
b + g = – 5
fi
– 3a – g = 6
and 3x – y + z = 2 Solving, we get 1 1 x = 1, y = __ , z = – __ 2 2 Hence, the value of
Solving, we get a = 1, b = – 2, g = 3.
13.
_› Let V =
a + b + c
_› X =
+2
Now,
fi
| |
_›
= 2 (a – c) + (2b + c) + (2a – b + 2c)
2 (a – c) + (2b + c) + (2a – b + 2c)
= (2 + )
Comparing the coefficients of , , and , we get a – c = 1, 2b + c = 0, 2a – b + 2c = 1. Solving, we get Now,
7 1 2 a = __ , b = __ , c = – __ 9 9 9 _›
_ ›
9 |V | 2 = m
fi
49 + 1 + 4 m = 9 × __________ 81
= 2 ( –
= 3
2. We have,
(a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r (a + b) ◊ (b × c) (b + c) ◊ (c × a) = _____________ + _____________ [a, b, c] [a, b, c] (c + a) ◊ (a × b) + _____________ [a, b, c]
_›
3. It is given that, c is parallel to the plane of the _ › _› vectors a and b , i.e.
c ^ (a × b) fi (a × b) ◊ c = 0
+ ) x ( – 2 + 3 ) + y (2 +
= 2 ( –
= 1 + 1 + 1
__› _ __› › z C = B × C
[a, b, c] ________ [a, b, c] [a, b, c] = ________ + + _______ [a, b, c] [a, b, c] [a, b, c]
= 3
14. It is given that, _› y B +
1. We have,
[a, b, c] _______ [a, b, c] ________ [a, b, c] = _______ + + [a, b, c] [a, b, c] [a, b, c]
( ) 54 = 9 × ( ___ ) = 6 81
= 101
a ◊ (b × c) + 0 b ◊ (c × a) + 0 c ◊ (a × b) + 0 = _____________ + ____________ + ____________ [a, b, c] [a, b, c] [a, b, c]
__
_› x A +
3 |V | = ÷ m
fi
= 100 + 5 – 4
a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) _________ + + (c × a) ◊ b (a × b) ◊ c (b × c) ◊ a
2V + V × X
fi
_›
= –2c + c + (2a – b) _›
(100 x + 10 y + 8 z)
Integer type Questions
_› _› V × X = a b c 120
15. Do yourself
_ ›
and Y = 2 +
– 2x + y + z = – 2
– ) + z ( + )
+ )
|
|
3 1 –2 fi 2 3 = 0 – 1 l 1 (2l – 1) fi (6l – 3 + 1) + 2 (4l – 2 + l) + 3(2 – 3l) = 0
Comparing the co-efficients of the unit vectors, we get
fi (6l – 2) + 2 (5l – 2) + 3 (2 – 3l) = 0
fi
x + 2y = 2
fi (6l – 2) + (10l – 4) + 3(6 – 9l) = 0 7l = 0
Vectors
fi l=0 Hence, the value of (l2 + 2) is 2. 4. We have, r ◊ (a + b + c) = 0 fi r ◊ a + r ◊ b + r ◊ c = 0 Now, r ◊ a = P [b, c, a]
Thus, m = 0 Hence, the value of (m + 4) = 4.
b×c c×a a×b 7. Clearly, p = _______ , q = _______ , r = _______ [a, b, c] [a, b, c] [a, b, c] _›
_›
_›
_›
_›
Hence, the value of (a + b + c) ◊ (p + q + r )
r ◊ b = Q [c, a, b]
and r ◊ c = R [a, b, c] Adding, we get
a ◊ b × c _______ b ◊ c × a _______ c ◊ a × b = _______ + + [a, b, c] [a, b, c] [a, b, c]
[a, b, c] [a, b, c] [a, b, c] = _______ + _______ + _______ [a, b, c] [a, b, c] [a, b, c]
=3
P [b, c, a] + Q [c, a, b] + R [a, b, c] = 0
fi
(P + Q + R) = 0
Hence, the value of (P + Q + R + 5) = 5. 5. It is given that _ ›
_›
_›
_ ›
_›
_› _ _›
_›
8. We have, _›
_ ›
a ^ (b + c ) , b ^ (c + a ) , c ^ (a + × b ) _› _› _› fi a ◊ (b + c ) = 0, _ › _ _ › › b ◊ (c + a ) = 0 _ › _› _› and c ◊ (a + b ) = 0 _ › › _ _› _ _› _› › Thus, a ◊ b + b ◊ c + c ◊ a = _ › _› _› Now, |a + b + c | 2
_›
_›
_›
_›
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi – 2 (a ◊ b + b ◊ c + c ◊ a) = 3 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3
0
Now,
|a + b + c|2
= a2 + b2 + c2 + 2 (a ◊ b + b ◊ a + c ◊ a)
=3–3=0
= a2 + b2 + c2
Thus,
|a + b + c| = 0
= 1 +1 + 4
fi
a + c = – a
2
2
Hence, the value of
=6 Hence, the value of (m2 + 1) = 7.
6. In an AP,
_›
1 tp = A + (p – 1) d = __ a
( ) 1 c–b q – r = __ ( _____ d cb )
(
Now,
= 0
= |3a|
=3
+a +b ,
9.
_› Let p =
a + 2 + b
_ › and r = a + b + 3 __› Now, pq = (a – 1) + (2 – a) __› and qr = + (b – 2) + (3 – b)
)
(
(a – 1) + (2 – a) = l{(b – 2) + (3 – b) }
)
= {(q – r) + (r – p) + (p – q) } ◊ __ a + __ + __ b c 1 1 1 __ 1 1 __ 1 1 = _____ __ – __ + – __ + – __ abcd b c c a a b
(
Since the vectors are collinear, so
1 a–c r – p = __ _____ ac d u ◊ v
_›
_› q =
1 b–a p – q = __ _____ d ab
Thus,
_ ›
|2a + 5b + 5c | = |2a + 5 (b + c)| = |2a – 5a|
1 tq = A + (q – 1) d = __ b 1 tr = A + (r – 1) d = __ c
_›
|a – b |2 + |b – c |2 + |c – a | 2 = 9
› _› _ _ _ › _ › › _ › = a + b + c + 2 (a ◊ b + b ◊ c + c ◊ a ) 2
5.63
)
fi
(a –1) = 0, (3 – b) = 0
fi a = 1, b = 3 Hence, the value of (a + b) = 4. _›
10. Let n = a + b + c .
5.64 Integral Calculus, 3D Geometry & Vector Booster _ ›
Thus, _›
n = ___ _› |n |
Comparing the co-efficients, we get
Now, u ◊ = 0 fi a + b = 0 fi
_› v ◊
=0fia–b=0
Hence, the value of
(u + 4v + w) = 17
_ ›
_› | ◊
(u + 2v + w) = 7
and (u – v + 3w) = – 8 Solving, we get
Solving, we get a = 0, b = 0 and c = 1 Thus, n =
u = – 3, v = 5 and w = 0
Hence, the value of (u + v + w + 4) = – 3 + 5 + 0 + 4 = 6.
| = 3.
11. Clearly,
2x – y = 5
14. Let s = x (p + q ) + y (q + r ) + z (r + p )
and
x – 2y = 4
Now, p ◊ s = y (p ◊ (q × r ) ) = y [p , q , r ]
_›
3x = 6
fi
x=2
and so
y = – 1
_›
_ › › _
Solving, we get
_›
_›
_ › _ ›
_›
_ ›
_›
_›
_ › _ › _ ›
› _› _
p ◊ s y = ________ _› _› _› [p , q , r ]
fi
_ › › _
_ _
Hence, the value of (x + y + 2) = 3.
› › q ◊ s r ◊ s Similarly, x = ________ _ › _ › and z = ________ _ › _ › _ _ › › [p , q , r ] [p , q , r ]
12. It is given that,
Thus,
|– cu , v , cw | = 8
fi
– 2 1 1 c2 1 2 = 8 –1 1 0 –1
fi
– 2 1 –1 c2 1 –1 3 = 8 1 0 0
| | | |
fi
c2(3 – 1) = 8
fi
c2 = 4
fi
c=2
(C3 Æ C1 + C3)
_ ›
_› _›
_›
_›
_› _›
_›
_›
› _› _
_›
_›
(r ◊ s ) (p × q ) + ____________ _› _› _ › [p , q , r ] › _› _
_›
_›
_› _›
_›
_›
_›
+ (q ◊ s ) (r × p ) + (r ◊ s ) (p × q )
_›
_› _›
_›
_›
+ (r ◊ s ) (p × q ) |
_ ›
_ ›
_› _ › _ › _ ›
= |s [p , q , r ] |
= |s | |[p , q , r ] |
=1×4
=4
_›
_› _› _›
Questions asked in Past Iit-Jee Examinations
_ ›
1. We have,
A ◊ {(B + C) × (A + B + C)} = A ◊ {B × A + B × B + B × C + C × A + C × B
_ ›
(p × q ) × r = up + vq + wr
7 + 17 – 8 = ( +
_›
fi s [ p , q , r ] = (p ◊ s ) (q × r )
_› _›
= 7 + 17 – 8
_ ›
_›
fi |(p ◊ s ) (q × r ) + (q ◊ s ) (r × p )
(p × q) × r = – 5 3 2 1 13
It is given that,
_› _›
_›
_› _ › _ › _ ›
(p × q) = 1 1 1 = – 5 + 3 + 2 2 4 – 1
Thus,
_›
| | | |
13. Now,
_› _›
_ (p ◊ s ) (q × r ) (____________ q ◊ s ) (r × p ) › fi s = ____________ + _› _› _ _› _› _ › › [p , q , r ] [p , q , r ]
__› _ › _ _›
+ C × C} = A ◊ {B × A + B × C + C × A + C × B}
+ ) u
+ (2 + 4 + ) v + ( +
+ 3 ) w
= (u + 2v + w) + (u + 4v + w) + (u – v + 3w)
= A ◊ (B × A) + A ◊ (B × C) + A ◊ (C × A)
+ A ◊ (C × B)
Vectors
= 0 + A ◊ (B × C) + 0 + A ◊ (C × B)
fi
x + 3y – 4z = l x
= [A B C] – [A B C]
x – 3y + 5z = l y
and
3x + y = l z
fi
(1 – l)x + 3y – 4z = 0
x – (3 + l) y + 5z = 0
and
3x + y – l z = 0
= 0
2. We have,
( )
p 1 |(B × C)| = BC sin __ = __ . 2 6
Now,
(B × C) A = ± ________ = ± 2 (B × C). |(B × C)|
3. We have |a| |b| |c| = |(a × b) ◊ c| = Volume of a parallelopiped having three adjacent sides as a, b, c. = Volume of a rectangular parallelopiped having a, b, c as adjacenat sides.
|
Eliminating x, y, z, we get
|
3 1 – l – 4 – (l + 3) 5 = 0 1 3 – 5 1 fi
l3 + 2l2 + l = 0
fi
l (l2 + 2l + 1) = 0
fi
l (l + 1)2 = 0
fi a ◊ b = b ◊ c = c ◊ a = 0
fi
l = –1, 0
OP = p = 60i + 3j,
4. Let |OAi| = R, for every i
5.65
6. Let
OQ = q = 40i – 8j
and
OR = r = a i – 52 j
Since the points are collinear, so
Let p be a unit vector perpendicular to the plane of the regular polygon.
( ) ] [ 2p = [ R sin ( ___ n ) ] p
2p OAi × OAi + 1 = (OAi) (OAi + 1) sin ___ n p
fi
-20 - 11 =0 a - 60 - 55
fi
-20 1 =0 a - 60 5
fi
– 100 – a + 60
fi
a = – 40.
X ◊ A = 0 = X ◊ B = X ◊ C
2
Thus,
( ( ) ) 2p × OA ) = – ( R sin ( ___ n ) )p
n – 1 2p S (OAi × OAi + 1) = (n – 1) R2sin ___ n p
i = 1
Also,
(OA2
1
2
(
fi
60 3 1 R Æ R 2 – R1 -20 -11 0 = 0 2 R3 Æ R3 – R1 a - 60 -55 0
We have, for 1 £ i £ n –1
60 3 1 40 -8 1 = 0 a -52 1
7. Since,
fi
A, B, C are coplanar
n – 1
Thus,
[A, B, C] = 0
i = 1
Thus, S (OAi × OAi + 1) = (1 – n) (OA2 × OA1) 5. Given x (i + j + 3k) + y (3i – 3j + k) + z (– 4i + 5j)
= l (x i + y j + z k)
8. V = [OA OB OC]
2 -3 0 = 1 1 -1 3 0 -1
)
5.66 Integral Calculus, 3D Geometry & Vector Booster 9. Given,
= 2 (–1 – 0) + 3 (–1 + 3) = – 2 + 6 = 4.
Now,
A ◊ B = 0, A ◊ X = c, A × X = B
A × (A × B) = A × B
fi
(A ◊ X) A – (A ◊ A) X = A × B 2
fi
cA – A X = A × B
fi
A2X = cA – A × B
cA – A × B X = ___________ A2 10. Let i, j, k be the unit vectors along the positive directions of x, y and z axes respectively. fi
= |c| |a × b| cos (0) = |a × b| 1 = __ (ab) 2
Thus, a1 b 1 c1
a2 b2 c2
A1 = r sin q cos j,
and
A2 = r sin q sin j A3 = r cos q.
(
(
and
)
)
B3 = r cos q = A3.
11. The given statement is true. Let the position vectors of the points A, B and C are a + b, a – b and a + kb respectively. Then AB = – 2b Similarly, BC = (k + 1)b Clearly, AB is parallel to BC for all k in R Thus, A, B and C are collinear p 1 12. Given (a × b) = (a b) sin __ = __ (ab) 2 6 Now, [c a b] = c ◊ (a × b)
( )
= [a b c]2
1 = __ a2 b2 4 1 = __ (a 21 + a22 + a 23) (b21 + b 22 + b23) 4 13. As the length of the vectors remain the same.
(2p)2 + 1 = (p + 1)2 + 1
fi
4p2 = (p + 1)2
fi
p + 1 = ± 2p 1 fi p = 1, – __ . 3 14. Given A, B, C and D are coplanar, so AB, AC and AD are also coplanar. AB = OB – OA
= (2i + 3j – 4k) – (3i – 2j – k)
= – i + 5j – 3k.
Also,
p When the x-axis is rotated through an angle of __ , the 2 new components of A are B1, B2, B3 respectively. p Thus, B1 = r sin q cos j + __ 2 = – r sin q cos = – A1 p B2 = r sin q sin j + __ = r sin q cos j 2 = A1
2
Now,
Here,
a3 b3 c3
AC = OC – OA
= (– i + j – 2k) – (3i – 2j – k)
= 2i + 3j + 3k.
Finally,
AD = OD – OA
= (4i + 5j + lk) – (3i – 2j – k)
Given,
= i + 7j + (l + 1) k
fi
[AB AC AB] = 0 -1 5 -3 2 3 3 =0 1 7 (l + 1)
fi 1 (15 + 9) – 7 (– 3 + 6) + (1 + l) (– 3 – 10) = 0 fi 24 – 21 – 13 (l + 1) = 0 fi 13 (l + 1) = 3 3 fi (l + 1) = ___ 13 3 10 fi l = ___ – 1 = – ___ 13 13 (a × b) 15. Number of unit vectors = ± ________ |(a × b)|
Vectors
= 2 |a × b + b × c + c × a| Also, area of a triangle D ABC 1 = __ |a × b + b × c + c × a| 2 From, (i) and (ii), we get,
16. Given
a 1 1 1 b 1 =0 1 1 c
fi
a 1- a 1- a C Æ C 2 – C1 1 b -1 0 = 0 2 |AB × CD + BC × AD + CA × BD| C3 Æ C3 – C1 1 0 c -1 = 4 ar (D ABC)
(
)
fi a (b – 1) (c – 1) – (1 – a) (c – 1)
19. (a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r
+ (1 – a) (1 – b) = 0
(a + b) × (b × c) (b + c) ◊ (c × a) = ______________ + _____________ [a b c] [a b c]
fi a (1 – b) (1 – c) + (1 – a) (1 – c) + (1 – a) (1 – b) = 0 Divide both the sides by (1 – a) (1 – b) (1 – c), we get a 1 1 _____ + _____ + _____ =0 1– 1–b 1–c a 1 1 fi 1 + _____ + _____ + _____ = 1 1– 1–b 1–c 1 1 1 fi _____ + _____ + _____ =1 1– 1–b 1–c _›
Also,
4a + 3b = a __ = 3 _›
0 0 b ___ = l – 4
(c + a) ◊ (a × b) + _____________ [a b c] a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) = _________ + + [a b c] [a b c] [a b c] [a b c] _______ [a b c] _______ [a b c] = _______ + + [a b c] [a b c] [a b c]
= (Projection of a on b)
a ◊ b = ____ |b|
Let a = x + y be the required vectors a ◊ b Projection of a along b = ____ = 1 b 4x + 3y ______ _______ = 1 ÷16 + 9 4x + 3y = 5 Also, projection of a along c is 2 a ◊ c ____ = 2 c a x + b y _______ ________ = 2 ÷a 2 + b 2
...(i)
= 3 20. Component of a along b
17. Let c = a + b _› _ › b ◊ c =
5.67
( ) a ◊ b b = ( ____ ) ___ |b| |b|
( )
a ◊ b = ____ 2 b |b|
Component of a perpendicular to b a ◊b = a – ____ 2 b |b|
|b|2 a – (a ◊ b) b = _____________ |b|2
b × (a × b) = __________ . |b|2
...(i)
3x – 4y = 10 ...(ii) Solving (i) and (ii), we get, x = 2, y = – 1 Hence, the required vector is 2 – 18. Let the position vectors of the points A, B, C and D be a, b, c and d respectively w.r.t the origin O Now, AB = b – a, AD = d – a BC = c – b, BD = d – b CD = d – c, CA = a – c Now, |AB × CD + BC × AD + CA × BD|
( )
21. Do Yourself By the help of internal section formula 22. (a – b) ◊ {(b – c) × (c – a)} = (a – b) ◊ (b × c – b × a – c × c + c × a) = (a – b) ◊ (b × c – b × a + c × a) = (a ◊ b × c – a ◊ b × a + a ◊ c × a) – (b ◊ b × c – b ◊ b × a + b ◊ c × a) = (a ◊ b × c) – (b ◊ c × a) = [a b c] – [a b c] = 0
5.68 Integral Calculus, 3D Geometry & Vector Booster 23. Given
[a b c] = 0.
Let
a = a1i + a2 j + a3 k
b = b1i + b2 j + b3k
and
c = c1i + c2 j + c3k
=0
a3 i b3 ¥ a1 c3 b1
j a2 b2
25. Given,
R × B = C × B and R ◊ A = 0.
Now,
(R – B) × C = 0
fi
R = B + l C
Also,
R ◊ A = 0
fi
(B + l C) × A = 0
fi
B ◊ A + l (C ◊ A) = 0
fi
(2 + 1) + l (8 + 7) = 0
k a3 = 0 b3
fi
15l + 3 = 0 1 fi l = – __ 5 Hence, the vector
4 c < 0 and – __ < c < 0 3
27. Given, Now,
( a, b, c are coplanar vectors.) 24. Do yourself By the help of internal section formula and the vector equation of a line.
fi
( )
a b c a a ◊a b◊a a ◊ a a ◊ b a ◊ c = b a ◊ b b ◊ b b◊a b◊b b◊c c a ◊c b◊c a2 b2 c2
c < 0 and c (3c + 4) < 0
4 Hence, the value of c = – __ , 0 3
Then
a1 = b1 c1
fi
1 R = (i + j + k) – __ (4i – 3j + 7k) 5 1 __ = (i + 8j – 2k) 5
a ◊ b = 3 and a × b = c a × (a × b) = a × c
fi
(a ◊ b) a – (a ◊ a) b = a × c
fi
3a – 3b = a × c
fi
3b = 3a – a × c
= (i + j + k) – (– 2i + j + k)
= 3i
fi b=i 28. We have,
i j k (b × c) = 1 1 1 1 1 2 =i–j
Therefore,
i j k a × (b × c) = 1 2 1 1 -1 0 = i + j – 3k
Thus,
29. Given
a × (b × c) ____ 1 = ___________ = ___ (i + j – 3k). |a × (b × c)| ÷ 11 a a c 1 0 1 =0 c c b
fi
a (0 – c) – a(b – c) + c(c) = 0
fi
c2 = ab
26. Let
a = c x i – 6j – 3k
and
b = x i + 2j + 2c x k
Given,
a ◊ b < 0
fi
c x2 – 12 – 6c x < 0
fi
c x2 – 6c x – 12 < 0
= (i + 2j – k) + l(i + j – 2k)
fi
c < 0 and D < 0
= (1 + l)i + (2 + l) j – (2l + 1) k
fi
c < 0 and 36c2 + 48c < 0
fi
c < 0 and 3c2 + 4c < 0
2 Now, the projection of v on a = __ . 3
Thus, c is the GM of a and b. 30. Let v be a vector in the plane of b and c. Then
v = b + lc
÷
__
Vectors
÷
__
(
v ◊ a 2 fi ____ = __ |a| 3 __
– l – 1 = 2
fi
l = – 3
)
1 So, d is parallel to – i + j – __ k . 2
2(1 + l) – (2 + l) – (2l + 1) ___ ÷2 __ fi __________________________ = __ ÷3 ÷6 fi
Now,
d ◊ (3i + 2j – 2k) 1 = __ (2i – 2j + k) ◊ (3i + 2j – 2k) 3 1 = __ (6 – 4 – 2) = 0 3
Therefore, v = – 2i – j + 5k 31. Do yourself By the help of internal section formula and basic Geometry. 32. Given OP = p and OQ = q
Thus, d is parallel to (3i + 2j – 2k). QP × QR 35. Unit vector = ± _________ |QP × QR| 36. We have,
1 = ± ___ __ (2 + ÷6
+ )
3p + 2q 3p – 2q OR = ________ and OS = _______ 1 5
Now,
= p × (c × d), p = (a × b)
OR ^ OS = 0
= (p ◊ d)c – (p ◊ c) d
fi
OR ◊ OS
= ((a × b) ◊ d)c – ((a × b) ◊ c) d
Thus,
(
) (
)
3p + 2q _______ 2p – 2q fi ________ ◊ = 0 1 5 fi
9p ◊ p = 4q ◊ q 2
5.69
(a × b) × (c × d)
= [a b d]c – [a b c] d. Similarly,
(a × c) × (b × d) = [a c d]b – [a b c] d
and (a × d) × (b × c) = [a d c]b – [a d b]c
2
fi
9p = 4q .
Therefore, x = (a × b) × (c × d) + (a × c) × (d × b)
33. Let
OP = a i + b j + g k,
OQ = b i + g j + a k
= [a b d] c – [a b c] d + [a c d] b – [a b c] d
and
OR = g i + a j + b k
+ [a d c] b – [a d b]c = – 2[a b c] d
Now, PQ = OQ – OP
= (b – a) i + (j – b) j + (a – g ) k
= (g – a) i + (a – b) j + (b – g ) k
Again,
QR = OR – OQ
Thus,
= (g – b) i + (a – g )j + (b – a) k |PQ| = |QR| = |PR| _________________________ (b – a)2 + (g – b)2 + (a – g )2
=÷
Therefore, D PQR is an equilateral triangle. 34. Given, 1 d = __ (2i – 2j + k) 3 1 Clearly |d| = __ ◊ 3 = 1 3 Thus, d is a unit vector. 2 1 Also, d = – __ – i + j – __ k 3 2
(
= – 2[b c d] a Thus, x is parallel a.
Also, PR = OR – OP
+ (a × d) × (b × c)
)
37. Given fi Thus,
[b c d] = 0 b, c, d are coplanar vectors d = b + l c
= (j – k) + l (– i + k)
Also,
= (– l) i + j + (l – 1) k a ◊ d = 0
fi
– l – 1 = 0
fi
l = –1
Therefore, d = i + j – 2k Hence,
(i + j – 2k) d __ = ± ___ = ± _________ . |d| ÷6
38. Given
1 a × (b × c) = ___ __ (b + c) 2 ÷
5.70 Integral Calculus, 3D Geometry & Vector Booster fi
1 (a ◊ c) b – (a ◊ b) c = ___ __ (b + c) 2 ÷
Comparing the components of b and c, we get, fi fi fi
1__ 1 (a ◊ b) = – ___ and (a ◊ c) = ___ __ 2 2 ÷ ÷ 1__ a b cos q = – ___ 2 ÷ 1__ cos q = – ___ 2 ÷ 3p q = ___ 4
Volumne of the tetrahedron __
39. Given, u+v+w=0 fi u2 + v2 + w2 + 2(u ◊ v + v ◊ w + w ◊ u) = – 50 fi 2 (u ◊ v + v ◊ w + w ◊ u) = – 50 fi (u ◊ v + v ◊ w + w ◊ u) = – 25 40. Given, (a + b + c) ◊ {(a + b) × (a + c)} = (a + b + c) ◊ {(a × a) + (a × c) + (b × a) + (b × c)} = (a + b + c) ◊ {(a × c) + (b × a) + (b × c)} = a ◊ (b × c) + b ◊ (a × c) + c ◊ (b × a) = [a b c] – [a b c] – [a b c] 41. Let N1 = normal to the plane parallel to i and i + j = i × (i + i) = k
and N2 = normal to the plane parallel to i – j, i + k
= (i – j) × (i + k)
= – i – j + k.
Note that a is parallel to N1 × N2 So, we can consider a = N1 × N2 = k × (– i – j + k)
= (i – j) Let q the angle between a and i – 2j + 2k. Thus, fi
2 cos q = 1 + _____ __ = 2 . 3 ÷ p __ q = 4
1 ___ __ 2 ÷
42. If b is perpendicular to c, the given expression is equal to a. Otherwise can not be calculated. 43. Let DE = p.
__
fi
|BC × BA| p = 4÷2
fi
2÷2 p = 4÷2
fi
p=2
__
__
Here, D ADE is a right-angled triangle. Thus,
AE 2 = AD 2 – DE 2 = 16 – 4 = 12
fi
AE = 2÷3
__
OM = OB + OC Now, _______________ = 2i 2 Thus, fi
= – [a b c]
__
2÷2 1 fi __ |BC × BA| p = ____ 3 6
fi (u + v + w)2 = 0
2÷2 ABCD = ____ 3
AM = OM – OA = 2i – i – j – k = (i – j – k) __
AM = ÷3
If E lies on AM produced, then M is the mid-point of AE. Then 1 __ (e + (i + j + k)) = 2i 2 fi e = 3i – j – k If E lies on MA produced, then A divides the join of E and M in the ratio 2:1. Thus, 2(2i) + e ________ =i+j+k 2 + 1 fi
e = – i + 3j + 3k
44. Let
|p| = |q| = |r| = l
Also,
p ◊ q = 0 = q ◊ r = r ◊ p
Let Now,
x = a p + b q + g r, (where a, b, g, Œ R) p × ((x – q) × p) = (p ◊ p)(x – q) – [p ◊ (x – q)] p = (p ◊ p)(x – q) – (p ◊ x – p ◊ q) p
Vectors
= (p ◊ p)(x – q) – (p ◊ x) p 2
48. Given,
2
= p (x – q) – (a p ) p
= l2 (x – q) – l2 a p
= l2 [(x – q) – a p] Similarly,
...(i)
a ◊a a ◊b a ◊c [a b c] = b ◊ a b ◊ b b ◊ c c◊a c◊b c◊c
q × [(x – r) × q] = l2 [(x – r) – b q]
...(ii)
r × [(x – p) × r] = l2 [(x – p) – g r]
...(iii)
1 = l l
= (1 – l2) – l (l – l2) + l (l2 – l)
= 1 – l2 – l2 + l3 + l3 – l2
= 2l3 – 3l2 + 1.
fi
[a b c]2 = 2l3 – 3l2 + 1
fi
[a b c] = ÷ 2l 3 – 3l2 + 1
2
Adding Eqs. (i), (ii) and (iii), we get fi 45. Given,
0 = l2 (3x – (p + q + r) – x) 1 x = __ (p + q + r). 2 p = ar (quad OABC)
= ar (DOAB) + ar (DOCB) 1 1 = __ |a × (10a + 2b)| + __ |b × (10 a + 2b)| 2 2 1 1 = __ |2a × b| + __ |10b × a| 2 2 = |a × b| + 5|b × a|
= 6 |a × b|
Also,
q = ar (parallelogram OABC) = |a × b|
Here,
__________
Taking dot product of a × b + b × c = pa + pb + rc with a, b and c we get p + q cos q + r cos q = m ...(i)
p cos q + q + r cos q = 0
...(ii)
p cos q + q cos q + r = m
...(iii)
k = 6.
Adding, we get
a × (a × c) = – b
fi fi
|a × (a × c)| = |– b| p |a| |(a × c)| sin __ = 1 2 |a| |(a × c)| = 1
fi
|(a × c)| = 1
fi
a c sin q = 1 1 sin q = __ 2 p __ q = 6
fi
fi fi
( )
47. We have, [(A × C + B × A + B × C) (B × C) (B + C)] = (A × C + B × A + B × C) ◊ [(B × C) × (B + C)] = (A × C + B × A + B × C) ◊ [(B × C) + (C × B)] = (A × C + B × A + B × C) ◊ [(B × C) – (B × C)] = (A × C + B × A + B × C) ◊ 0 = 0.
(where l = cos q)
m = (1 – cos q ) ÷1 + 2 cos q
fi
fi
l l , 1
= m, (say)
p = 6q
a × (a × c) + b = 0
l 1 l
____________
Thus,
46. Given,
5.71
(p + q + r) (1 + 2 cos q )
fi
2m (p + q + r) = ___________ (1 + 2 cos q)
...(iv)
Multiplying Eq. (iv) by cos q and subtracting from Eq. (i), we get,
2m cos q m – __________ = p (1 – cos q) 1 + 2 cos q
m 1 fi p = ___________________ = ___________ __________ (1 + 2 cos q)(1 – cos q) ÷1 + 2 cos q Similarly, – 2 cos q 1 r = ___________ __________ and q = ___________ __________ ÷1 + 2 cos q ÷1 + 2 cos q 49. Given that _› _› a = ( + + ), b = (4 + 3 + 4 ) _› and c = ( + a +_ b ) are linearly independent › _› _› vectors c = l a + m b for some scalers l and m not all zeroes. + a + b = ( + 4m) + (l + 3m) + (l + 4m) l + 4m = 1 l + 3m = a Thus, l + 4m = b
5.72 Integral Calculus, 3D Geometry & Vector Booster On solving, we get, b =__1 Also, given that, |c| = ÷ 3 2 2 1 + a + b = 3 a = ± 1 50. Ans. (c) 51. Ans. (a) 52. We have,
55. Given c is coplanar with a and b. Thus,
(i) (u ◊ v)2 + |(u × v)|2
= u2 v2 cos2q + u2v2 sin2q
= u2v2 (cos2q + sin2q)
= u2v2
1 = __ |a × b| 2 3 = __ , where (a × b) = 2i – 2 j + k 2
(ii) |u + v + (u × v)|2 + (1 – (u ◊ v))2
c = a + l b
= (2i + j + k) + l (i + 2j – k)
= (2 + l) i + (1 + 2l)j + (1 – l) k
Also, fi
c is perpendicular to a c ◊ a = 0
fi
2(2 + l) + (1 + 2l) + (1 – l) = 0
fi
4 + 2l + 1 + 2l + 1 – l = 0
fi
3l + 6 = 0
= (u + v)2 + |(u × v)|2 + 2(u + v) ◊ (u × v)
fi
l+2=0
+ (1 – (u ◊ v))2
fi
l = – 2
So,
c = – 3j + 3k
2
2
2
= (u + v) + |(u ◊ v) | + (1 – (u ◊ v)) 2
2
2 2
2
= u + v + 2u ◊ v + u v sin q 2 2
+ 1 – 2u ◊ v + u v cos q
2
2
2 2
= u + v + u v + 1 = (1 + u2) (1 + v2) = (1 + |u|2) (1 + |v|2). 53. We have |v| = |a × b| = a b sin q = sin q Also,
|u|2 = a2 – 2(a ◊ b) (a ◊ b) + (a ◊b)2 b2
= 1 – 2 (a ◊ b)2 + (a ◊ b)2
= 1 – (a ◊ b)2 = 1 – cos2q = sin2q
fi
|u| = sin q = |v|
Again,
u ◊ b = (a ◊ b) – (a ◊ b)(b ◊ b)
= (a ◊ b) – (a ◊ b) b2
= (a ◊ b) – (a ◊ b) = 0
Thus, 54. Given,
2
|v| = |u| + |u ◊ b| __
|c – a| = 2÷2
__
fi |c – a|2 = ( 2÷2 )2 = 8 2 2 fi c + a – 2c ◊ a = 8 fi c2 + 9 – 2 c = 8 fi c2 – 2c + 1 = 8 fi (c – 1)2 = 0 fi c=1 We have, |(a × b) × c| = |a × b| |c| sin (30°)
1 = ___ __ (– j + k) 2 ÷ 56. Let u and v are not parallel. Let w be a linear combination of u, v and u × v Then w = a u + b v + g (u × v).
Thus,
w × u = b (v × u) + g ((u × v) × u) = b (v × u) + g ((u ◊ u)v – (v ◊ u)u)
Now,
w+w×u=v
fi
a u + b v + g (u × v) + g v – g (v ◊ u)u + b (v × u) = v
fi (a – g (v ◊ u)) u + (b + g ) v + (g – b) (u × v) = v fi (a – g (v ◊ u)) = 0, (b + g ) = 1, (g – b ) = 0 1 1 fi b = g = __ , a = __ (v . u) 2 2 Also, fi
(u × v) ◊ w = a (0) + b (0) + g (u + v) = g (u × v) |(u × v) ◊ w|2 = g 2 |(u × v)|2
= g 2 (u2v2 – (u . v)2)
1 = __ (1 – (u ◊ v)2) 4
fi
1 £ __ 4 1 |(u × v) ◊ w| £ __ 2
and the equality holds iff u ◊ v = 0 ¤ u and v are perpendicular to each other.
Vectors
57. Given
a+b+c=0
fi
a + b = – c
fi
a × (a + b) = – a × c
fi
(a × b) = c × a
Also,
a + c = – b
fi
b × (a + c) = – b × b
fi
b × c = – b × a = a × b
...(i)
...(ii)
From Eqs. (i) and (ii), we get
a×b=b×c=c×a
5.73
=1+x–x =1 = neither depends on x nor y.
62. Thre vectors v1, v2, v3 involve 9 unknown quantities. Since, we are given six equations involving 9 unknown quantities, it is not possible to have uniquely 3 vectors-satisfying these conditions. As v1 ◊ v2 = 4, let us consider v1 = 2k. Also, let v2 = ai + bj + ck and v3 = xi + yj + zk.
58. Let N1 = normal vector to plane P1 = a × b
Now,
v1 ◊ v2 = – 2 fi 2c = – 2 fi c = – 1
and N2 = normal vector to plane p2 = c × d Now,
Now,
v1 . v3 = 6 fi 2z = – 6 fi z = – 3
Now,
v2 ◊ v2 = 2, v2 ◊ v3 = 5, v3 ◊ v3 = 29
\
a2 + b2 + c2 = 2 fi a2 + b2 = 2 – 1 = 1
x2 + y2 + z2 = 29 fi x2 + y2 = 20
and
ax + by + cz = 5
fi
ax + by = 5 – 3 = 2.
Put
b = 0, we get,
a2 = 1 fi a = ±1
and
ax = 2 fi x = ± 2
Now,
x2 + y2 = 20 fi y2 = 20 – 4 = 16
fi
y = ± 4.
N1 × N2 = (a × b) × (c × d) = 0
Thus, N1 and N2 are parallel. Therefore, the angle between P1 and P2 is 0. 59. Given [a b c] = 0 Now, [2a – b 2b – c 2c – a]
= (2a – b) ◊ {(2b – c) × (2c – a)}
= (2a – b) ◊ {4(b × c) – 2(b × a) + (c × a)}
= 8 (a ◊ (b × c)) – (b ◊ (c × a))
= 8 [a b c] – [a b c]
= 7 [a b c]
Thus, one possible set of vectors is v1 = 2k, v2 = i + 2k, v3 = – 2i + 4j + 3k. 63. A (t) is parallel to B (t) for some t in [0, 1]
= 0. 60. We have, |a – b|2 + |b – c|2 + |c – a|2 = 2(a2 + b2 + c2) – 2(a ◊ b + b ◊ c + c ◊ a) = 6 – 2(a ◊ b + b ◊ c + c ◊ a) Also,
...(i)
(a2 + b2 + c2) + 2 (a ◊ b + b ◊ c + c ◊ a)2 = (a + b + c)2 ≥ 0 fi
3 + 2 (a ◊ b + b ◊ c + c ◊ a) ≥ 0
fi
2 (a ◊ b + b ◊ c + c ◊ a) ≥ – 3
fi
– 2 (a ◊ b + b ◊ c + c ◊ a) ≥ 3
...(ii)
|a – b|2 + |b – c|2 + |c – a|2 £ 6 + 3 = 9 1 0 -1 61. [a b c] = x 1 (1 - x ) y x 1+ x - y
Let h(t) = f1(t) g2(t) – f2(t) g1(t)
h(1) = f1 (1) g2(1) – f2(1) g1(1)
= 6.6 – 2.2 = 32 > 0 since h is a continuous function and h(0) h(1) < 0 Then there is some t in [0, 1] for which h(1) = 0 Thus, A(t) is parallel to B (t) for this t. 64. We have, V = [a b c]
(C3 Æ C3 + C1)
h(0) = f1(0) g2(0) – f2(0) g1(0)
= 2.2 – 3.3 = – 5 < 0
From Eqs (i) and (ii), we get
1 0 0 = x 1 1 y x 1+ x
f1 (t) f2 (t) _____ = _____ for some t in [0, 1] g1 (t) g2 (t)
a1 = b1 c1
a2 b2 c2
a3 b3 c3
= a1(b2c3 – b3c2) + a2(b3c1 – b1c3) + a3 (b1c2 – b2c1)
5.74 Integral Calculus, 3D Geometry & Vector Booster = (a1b2c3 + a2b3c1 + a3b1c2) – (a1b3c2 + a2b1c3 + a3b2c1)
We assume that (a1b2c3 + a2b3c1 + a3b1c2) ≥ (a1b3c2 + a2b1c3 + a3b2c1)
i j k = 2 1 -1 = 3i – 7j – k 1 0 3 _________
As we know that, AM ≥ GM
1 fi __ (a1 + b2 + c3) ≥ (a1 b2 c3)1/3 3
Thus, the maximum value of ___
1 fi (a1 b2 c3) ≥ ___ (a1 + b2 + c3)3 27 1 Similarly, (a2 b3 c1) ≥ ___ (a2 + b3 + c1)3 27 1 and (a3 b1 c2) ≥ ___ (a3 + b1 + c2)3 27 Thus,
[u v w] = ÷ 59 .
67. Given
1 a 1 V= 0 1 a a 0 1
= 1 + a (a2 – 1)
(a1 b2 c3) + (a2 b3 c1) + (a3 b1 c2)
= a3 – a + 1
1 £ ___ [(a1 + b2 + c3)3 + (a2 + b3 + c1)3 + (a3 + b1 + c2)3]
dV fi ___ = 3a2 – 1 da
27
1 £ ___ [(a1 + b2 + c3) + (a2 + b3 + c1) + (a3 + b1 + c2)]3 27 Since,
x3 + y3 + z3 £ (x + y + z)3 for x, y, z ≥ 0
3 1 3 = ___ S (ar + br + cr) 27 r =1
1 = ___ (3L) 3 27
= L3
Thus, 65. Given,
V £ L3.
fi
5a2 – 4 (a ◊ b) + 10 (a ◊ b) – 8b2 = 0
fi
5 + 6(a ◊ b) – 8 = 0
fi
6(a ◊ b) = 3
fi
1 (a ◊ b) = __ 2
fi fi fi
(
)
___
r = |r| = ÷ 9 + 49 + 1 = ÷ 59
dV For maximum or minimum, ___ = 0 da fi
3a2 – 1 = 0
fi
1 a2 = __ 3 1__ a = ± ___ 3 ÷
fi
1__ 1__ ___ ___ 3 ÷3 ÷
(a + 2b) ◊ (5a – 4b) = 0
1 Hence, the value of a = ___ __ . 3 ÷ 68. Given x = u + v, y = v + w, z = w + u u + v ˆ _______ v + w ˆ _______ w+u Thus, xˆ = ______ y = , z = . |u + v| |v + w| |w + u| We have,
1 a b cos q = __ 2
( )
p 1 cos q = __ = cos __ 2 3 p q = __ . 2
66. We have, [u v w] = u ◊ (v × w) = u ◊ r
(u + v)2 = u2 + v2 + 2u ◊ v
= 1 + 1 + 2 cos a
= 1 + 1 + 2 cos a
= 2(1 + cos a)
fi
( ) ( )
a = 2 ◊ 2 cos2 __ 2 a |(u + v)| = 2 cos __ 2
= u r cos q
a Similarly, |(v + w)| = 2 cos __ 2
= r cos q, where r = u × v
and
|(w + u)| = 2 cos __
Vectors
Thus,
x = 1/2 (u + v) sec __,
y = 1/2(v + w) sec __
and
z = __(w + u) sec g/2
Now,
[x × y y × z z × x] = [x y z]2
Thus,
a × (b × c) 1 = __________ = ____ ___ (3j – k) |a × (b × c)| ÷ 10
72. Note that a is directed along the internal bisector of the triangle formed by the vectors - v and w Thus,
k a = – v + w
fi
k2a2 = v2 + w2 – 2 (v ◊ w)
fi fi
k2 = 1 + 1 – 2vw cos (p – 2q) k2 = 2 + 2 cos 2q
fi
k2 = 2(1 + cos 2 q) = 2 . 2 cos2q
g 2 b a 1 = __ [u v w] sec __ sec __ sec __ 4 2 2 2
fi
k = 2 cos q
fi
k = 2(a ◊ (– v))
g b a 1 = ___ × sec2 __ sec2 __ sec2 __ [a b c]2 2 2 2 16
Thus, w = v + ka = v – 2(a ◊ (– v)). 73. Ans. (b)
[
( )
( ) ( ) ]
g 2 b a 1 = __ [u + v v + w w + u] sec __ sec __ sec __ 8 2 2 2
[ [
( ) ( ) ] ( ) ( ) ( ) ] [ ( ) ( ) ( ) ] ( )
g b a 1 = __ ◊ 2[u v w] sec __ sec __ sec __ 8 2 2 2
69. Given, fi fi fi fi \ fi fi
2
a × b = c × d and a × c = b × d a×b–a×c=c×d–b×d a × (b – c) = (c – b) × d = d × (b – c) a × (b – c) – d × (b – c) = 0 (a – d) × (b – c) = 0 (a – d) is parallel to (b – c) (a – d) ◊ (b – c) π 0 a ◊ d + c ◊ d π a ◊ c + b ◊ d.
Hence, the result. 70. We have, a × (a × b) = (a ◊ b)a – (a ◊ a)b a × (a × b) = a – a2b = a – 3b 3b = a – a × (a × b) 3b = (i + j + k) – (– 2i + j + k) 3b = 3i b=i a × (b × c) 71. To find = ___________ |a × (b × c)|
fi fi fi fi fi
We have, Now,
i j k b × c = 2 1 1 = 2i – j – 3k 1 −1 1 i j k a × (b × c) = 5 2 6 2 −1 −3
= 27j – 9k
= 9 (3j – k)
5.75
a ◊ b We have a ◊ b1 = a ◊ b – ____ 2 (a ◊ a) = 0. |a|
b1 ◊ c a ◊ c a ◊ c2 = a ◊ c – ____ 2 (a ◊ a) – ____ 2 (b1 ◊ a) |a| |b1| = a ◊ c – a ◊ c = 0
and
b1 ◊ c c ◊ a b1 ◊ c2 = b1 ◊ c – ____ 2 (b1 ◊ a) – ____ 2 (b1 ◊ b1) |a| |b|
= b1 ◊ c2 – 0 – b1 ◊ c = 0
74.
Thus, the required triplet is (a b1 c2). Ans.(c) Let v be a vector. A vector v in the plane of a and b So, v, a, b are coplaner vectors. Thus, v = a + l b fi v = (1 + l) i + (2 – l) j + (1 + l) k fi v = (1 + l) i + (2 – l) j + (1 + l) k
1 the projection of v on c is ___ __ 3 ÷ v ◊ c 1 fi ____ = ___ __ |c| 3 ÷ Also,
(1 + l) + (2 – l) – (1 + l) ___ 1 __ fi ________________________ = __ 3 ÷3 ÷ fi
(2 – l) = 1
fi l=1 Therefore, v = 2i + j + 2k 75. Ans. (c)
Given
−l 2 1 1 1 −l 2 1 =0 1 1 −l 2
5.76 Integral Calculus, 3D Geometry & Vector Booster fi
– l2 (l4 – 1) – 1(l2 – 1) + 1(l2 + 1) = 0
= cos2t + sin2t + (a ◊ b) sin 2 t
fi
– l2 (l4 – 1) + (l2 + 1) + (l2 + 1) = 0
= 1 + (a ◊ b) sin 2 t
fi
6
2
– l + 3l + 2 = 0 6
2
Thus,
________ p |OM| = |OP|max = ÷ 1 + (a ◊ b) , t = __ 4
fi
1 1 1 OM = ___ __ a + ___ __ b = ___ __ (a + b) 2 2 2 ÷ ÷ ÷
fi
l – 3l – 2 = 0
fi
a3 – 3a – 2 = 0, a = l2
fi
a2 (a – 2) + 2a(a – 2) + 1(a – 2) = 0
fi
(a – 2) (a2 + 2a + 1) = 0
Therefore,
fi
(a – 2) (a + 1)2 = 0
79. We have,
fi
(a – 2) = 0, (a + 1)2 = 0
fi
a = 2, a = – 1
fi
2
l = ± ÷2
__
Thus, the number of real values of l is 2. Ans. (b) Given a+b+c=0 fi a + b = – c fi a × (a + b) = – a × c fi (a × b) = c × a Also, a + c = – b fi b × (a + c) = – b × b fi b × c = – b × a = a × b From (i) and (ii), we get, a×b=b×c=c×a
77. We have RS + ST = RT
As PQ and TR are not parallel to each other.
PQ × (RS × ST) π 0
Thus, statement-I is true. Next, since PQ and RS are not parallel, so,
2
2
fi
a⋅a a⋅b a⋅c 1 [a b c] = b ⋅ a b ⋅ b b ⋅ c = 2 c ⋅a b⋅c c ⋅c 1 2
l = 2, l = – 1 l =2
76.
1
2
fi
PQ × RS π 0
Thus, statement-II is false. 78. Ans. (a) Given OP = a cos t + b sin t |OP|2 = |a cos t|2 = |b sin t|2 + 2 (a ◊ b) sin t cos t
OM a+b = _____ = ______ |OM| |a + b|
1 2 1 1 2
1 2 1 2 1
( ) ( ) ( 3 1 __ 1 2 1 = ( __ – __ – = __ = __ 4 8 8) 4 2
)
1 1 1 __ 1 1 1 __ 1 = 1 1 – __ – __ __ – + __ __ – 4 2 2 4 2 4 2
1 [a, b, c] = ___ __ 2 ÷
...(i)
fi
...(ii)
80. Let q1 and q2 be the angles between a and b, and c and d and n1 and n2 are the unit vectors perpendicular to the plane of a and b, and c and d. Now,
a × b = (a b sinq1) n1
and
c × d = (c d sinq2) n2
Let j the angle between n1 and n2. Given
(a × b) ◊ (c × d) = 1
fi
(sin q1) (sin q2) (n1 ◊ n2) = 1
fi
(sin q1) (sin q2) (cos j) = 1
p It is possible only when q1 = __ = q2, j = 0 2 Since j = 0, so n1 and n2 are parallel. Thus, a, b, c and d are coplaner vectors. Also, fi fi
1 a ◊ c = __ 2
( )
p 1 cos q3 = __ = cos __ 2 3 p q3 = __ . 3
( )
p Since the angle between a and c is __ and b and d 3 p is also __ . So b and d are non-parallel. 3
Vectors
81. We have,
1 1 0 V= 1 2 0 1 1 p
= p (2 – 1)
= p.
82. Given,
2
– 2 + 20 + _________ 22 ____________ = _______________________ 4 + 100 + 121 ÷ 1 + 4 + 4 ÷
40 = _____ = 15 . 3
(
8 __ 9
p Note that a = __ – q 2 p Thus, cos a = cos __ – q 2
__
a + b = – ÷3 c
__
(a + b) = (– ÷3 c )
fi
a2 + b2 + 2a ◊ b = 3c2
) (
)
= sin q
÷ ( ) 81 – 64 71 17 = _______ = ___ ÷ 81 ÷ 81 = ÷ ___ 9 .
8 2 = ÷1 – cos q = 1 – __ 9 2
_______
fi
85. Given,
a = mb + 4c
fi
a ◊ b = mb ◊ b + 4b ◊ c
fi
mb2 + 4b ◊ c = 0
83.
________
_________
2
fi
1 + 1 + 2a ◊ b = 3 1 fi a ◊ b = __ 2 1 fi a b cos q = __ 2 p 1 fi cos q = __ = cos __ 2 3 p __ fi q = 3 Let M be the mid-point of PR. Then the position vector of M is
( )
___
___
...(i)
Also,
a ◊ a = m(a ◊ b) + 4(a ◊ c)
fi
|a|2 = 4(a ◊ c)
...(ii)
1 1 __ (– 2i – j + 3i + 3j) = __ i + j 2 2
Given,
(b – a) ◊ (b + c) = 0
Let N be the mid-point of QS. Then the position vector of N is
fi
(b ◊ b) + (b ◊ c) – (a ◊ b) – (a ◊ c) = 0
fi
|b|2 + (b ◊ c) – (a ◊ c) = 0
1 1 = __ (4i – 3i + 2j) = __ i + j 2 2 Thus, PQRS is a parallelogram. ______
___
= ÷36 + 1 = ÷37
And QR = |PR| = |3i + 3j – 4i| = |– i + 3j|
_____
...(iii)
Again,
Also, PQ = |PQ| = |4i – (– 2i – j)| = |6i + j|
5.77
___
= ÷1 + 9 = ÷10
2|b + c| = |b – a|
fi
4|b + c|2 = |b – a|2
fi
4(b2 + c2 + 2b ◊ c) = (b2 + a2 – 2(a ◊ b))
fi
4(b2 + c2 + 2b ◊ c) = (b2 + a2)
fi
3b2 + c2 + 8(b ◊ c) = a2
...(iv)
2
Therefore, PQRS is not a rhombus.
Eliminating (b ◊ c) and a from Eqs (i), (ii), (iii) and (iv), we get
Also, PR = |PR| = |3i + 3j – (– 2i – j)|
(2m2 – 10m) |b|2 = 0
fi
(2m2 – 10m) = 0
fi
m(m – 5) = 0
fi
m = 0, 5.
___
= |5i + 4j| = ÷ 41
And QS = |QS| = |– 3i + 2j – (– 4i)|
_____
__
= |i + 2j| = ÷ 1 + 4 = ÷ 5
So, PQRS is not a rectangle. Therefore, PQRS is a parallelogram, which is neither a rhombus nor a rectangle. 84. Let q be the angle between AB and AD. AB ◊ AD Thus, cos q = ________ |AB| |AD|
86. Given |a| = 1 = |b| and a . b = 0 Now, (2a + b) ◊ {(a × b) × (a – 2b)}
= (2a + b) ◊ {(2b – a) × (a × b)}
= (2a + b) ◊ {(2b × (a × b) – a × (a × b))}
= (2a + b) ◊ {(2(b ◊ b)a – 2(b ◊ a)b – (a ◊ b)a + (a ◊ a)b)}
5.78 Integral Calculus, 3D Geometry & Vector Booster
= (2a + b) ◊ (2b2a + a2b)
90. We have,
= (2a + b) ◊ (2a + b)
|a| = 2 = |b| and a ◊ b = –1 + 3 = 2
= (4a2 + b2 + 4a ◊ b)
Let q be the angle between them.
=4+1+0
=5 87. Given v, a, and b are coplaner, so, v = a + lb = (i + j + k) + l (i – j + k)
= (1 + l) i + (1 – l) j + (1 + l) k
1 Also, the projection of v on c is ___ __ . 3 ÷ v ◊ c 1 fi ____ = ___ __ |c| 3 ÷
|
a2 + b2 – c2 cosq = ___________ 2ab
Then
Thus, 91. Given
4 + 4 – 12 1 = __________ = – __ 8 2 2p q = ___ . 3 |a – b|2 + |b – c|2 + |c – a|2 = 9
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9
|
(1 + l) – (1 – l) – (1 + l) 1 __ fi _______________________ = ___ __ 3 ÷3 ÷
fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3
fi
fi |a + b + c|2 = 0
(1 – l) = 1
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 3 – 3 fi (a + b + c) = 0
fi Thus,
l=2 v = 3i – i + 3k
88. Let
a = i + j + k,
Now,
b = i + j + 2k
= |2a + 5 (– a)|
and
c = i + 2 j + k
= |– 3 (a)|
= |– 3| |a|
Now,
i j k (b × c) = 1 1 2 1 2 1
=3
Also,
= – 3i + j + k i j k a × (b × c) = 1 1 1 –3 1 1
|2a + 5b + 5c| = |2a + 5 (b + c)|
92. Given, a × (2i + 3j + 4k) = (2i + 3j + 4k) × b fi a × (2i + 3j + 4k) = – b × (2i + 3j + 4k) fi (a + b) × (2i + 3j + 4k) = 0 fi (a + b) = l (2i + 3j + 4k)
= – 4 j + 4k
fi |(a + b)| = |l (2i + 3j + 4k)|
= 4(– j + k).
fi ÷29 l2 = ÷29
a × (b × c) Thus, ___________ = (– j + k). |a × (b × c)| 89. We have, r×b=c×b
___
___
fi |l| = 1 fi l = ±1 Thus,
(a + b) = ± (2i + 3j + 4k)
Now,
(a + b) ◊ (– 7i + 2j + 3k)
= ± (– 14 + 6 + 12)
fi
a × (r × b) = a × (c × b)
fi
(a ◊ b) r – (a ◊ r) b = a × (c × b)
fi
(a ◊ b) r = a × (c × b)
fi
– r = a × (c × b) = 3i – 6j – 3k
fi
r = – 3i + 6j + 3k
= ± 4 93. When diagonal are given, then area of a parallelogram 1 PQRS = __ |PQ × SQ| 2
Thus,
r . b = 3 + 6 + 0 = 9.
Volume of the parallelopiped
5.79
Vectors
= ((PQ × PS)) ◊ PT)
1 = 2 × __ × (PR × SQ) ◊ PT 2
= [(PR × SQ)] ◊ PT)
3 1 −2 = 1 −3 − 4 1 2 3
[
]
= 30.
98. It is given that a is in the direction of x × (y × z) fi
a = l1(x × (y × z)) = l2((x ◊ z)y –(x ◊ y)z)
( ( ) ( ) )
1 1 fi a = l1 2 × __ y – 2 × __ z = l1 (y – z) 2 2 fi a ◊ y = l1(y ◊ y – y ◊ z) a = a ◊ y(y – z)
Thus,
= |3 (– 9 + 8) – (3 + 4) – 2 (2 + 3)|
Similarly, b = (b ◊ z)(z – x)
= |– 3 – 7 – 10|
Now,
= 20
= (a ◊ y)(b ◊ z)(1 – 1 – 2 + 1)
94. Given
[a b c] = 2
= – (a ◊ y)(b ◊ z).
99. Given a × b + b × c = pa + qb + rc
Now,
[2(a × b) 3(b × c) (c × a)]
fi
= 6[(a × b) (b × c) (c × a)]
fi
= 6[a b c]2
= 24
95. Given
[a b c] = 5
Now,
[3 (a + b) (b + c) 2(c + a)]
= 6 [(a + b) (b + c) (c + a)]
= 6 × 2 [a b c]
= 60
1 96. Given __ |a × b| 2 fi
a ◊ b = (a ◊ y)(b ◊ z){(y – z) ◊ (z – x)}
|a × b| = 0
a ◊ [(a × b) + (b × c)] = p + q (a ◊ b) + r (a ◊ c)
a ◊ (b × c) = p + q (a ◊ b) + r(a ◊ c) q r fi a ◊ (b × c) = p + __ + __ 2 2 q r fi p + __ + __ = a ◊ (b × c) = [a, b, c] 2 2 p r fi __ + q + __ = 0 2 2 p q __ + __ + r = [a, b, c]2 2 2 Also,
...(i) ...(ii) ...(iii)
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a c⋅b c⋅c 2
1 Now, __ |(2a + 3b) × (a – b)| 2
1
1 = __ |– 2a × b + 3b × a| 2
1 = __ |– 2a × b – 3a × b| 2
1 = __ |– 5a × b| 2
5 = __ |a × b| 2
= 100
fi
97. Given
|a × b| = 30
Now,
|a × (a + b)|
= |a × a + a + b|
= |0 + a × b|
= |a × b|
=
1 2 1 2
1 2 1 1 2
1 2 1 2 1
( ) ( ) ( 3 1 __ 1 2 1 = ( __ – __ – = __ = __ 4 8 8) 4 2
)
1 1 1 __ 1 1 1 __ 1 = 1 1 – __ – __ __ – + __ __ – 4 2 2 4 2 4 2
1 [a, b, c] = ___ __ 2 ÷
From Eqs (i), (ii) and (iii), we get
q = r = – q
(
)
2 p2 + 2q2 + q2 q____________ + 2q2 + q2 Now, ____________ = =4 q2 q2
5.80 Integral Calculus, 3D Geometry & Vector Booster 100. We have,
fi
_› _ _ › › |b + c | = |a | _ › _› _› |b + c | 2 = |a | 2 _ › _› _ _ _ › › › |b | 2 + |c | 2 + 2b ◊ c = |a | 2
fi
48 + |c | 2 + 48 = 144
fi
|c | 2 = 144 – 96 = 48
fi
4÷3
=
_›
_› |c |
_›
_›
_› _›
› _› _ (a ◊ b )
= – 72
Again, _› _› _› _› a + b + c = 0
_›
_›
_ ›
_›
_›
_ ›
|a × b + c × a | = |a × b + a × b | _›
_›
_›
_›
_›
9 7 x = 4, y = __ , z = – __ 2 2
Hence, the value of 2x + y + z = 9 103. We have, _› ◊ ( × v ) _ › fi | | | × v | cos a = 1 fi cos a = 1 _› fi ^ and ^ v
_ › _› _› _› fi a × b = c × a
_›
= 2|a × b |
_ › › 2 = 2 a2b2 – (a ◊ b )
= 2÷144 × 48 – (– 72)
Hence, infinitely many such vectors v exist.
= 2÷144 × 48 – 72 × 72
If
=
fi
=
=
=
=
____________ _
÷________________
2
_________________
Hence, the result. 101. It is given that
|
__
______________ 2÷72(2 × 48 – 72) ___________ 2÷72(96 – 72) _______ 2÷72 × 24 __ 2 × 24 ÷3 __ 48 ÷3
|
__ a ÷3 + b ________ = ÷ 3 2
_›
_›
Solving, we get
_›
_›
+ (x + y + z)r
144 + 48 + 2(a ◊ b ) = 48
_› _›
_›
+ z (– p – q + r ) _ _ › › = (– x + y – z)p + (x – y – z)q
(x – y – z) = 3 (x + y + z) = 5
_›
= |c | 2
_›
= x (– p + q + r ) + y (p – q + r )
fi fi
_›
_›
_ ›
_›
|a | 2 + |b | 2 + 2(a ◊ b ) = |c | 2
fi
_›
_›
(– x + y – z) = 4
fi
Thus,
...(ii)
fi
fi
_ › _› |a + b | 2
fi
a=2+÷ 3 b
_ › 48 Now, – |a | = ___ – 12 = 24 – 12 = 12 2 2 Also,
Given
__
s = 4p + 3q + 5r
__
_ › _› |a + b |
...(i)
102. We have,
_ ›
fi
__
a ÷3 + b = ± 2÷3
From Eqs (i) and (ii), we get a = – 1 or 2 Thus, |a| = 1 or 2
_›
_› |c | = _› |c | 2 ___
__
_›
As it is given that there exists a vector v . fi
must be perpendicular to
_›
= u1 + u2 __›
◊ w = 0
fi fi fi
(u1 + u2) = 0 u1 = – u2 |u1| = |– u2| = |u2|
If
u = u1 + u3 _ › __›
fi u ◊ w = 0 fi
u1 + 2u3 = 0
fi
|u1| = 2|u3|
Chapter
6
3D-Co-ordinate Geometry
Concept Booster 1.1 Introduction In earlier classes, we learnt about points, lines, cicles and conic sections in two-dimensional geometry. In 2D-geometry, a point represented by an ordered pair (x, y) for which both x and y are real numbers. In the space, each body has length, breadth and height, i.e. each body exits in three-dimensional space. Thus, three independent quantities are required to represent any point in a space and so three axes are required to represent these quantities.
1.2 Rectangular
1.3 Position
vector of a point in a space
Let O be a fixed point, known as the origin, and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis, respectively, in such a way that they form a right handed system.
co-ordinate system
The cartesian system of three lines which are mutually perpendicular to each other is called rectangular co-ordinate system. When three mutually perpendicular planes intersect at a point, the mutually perpendicular lines are obtained and these lines also pass through that point. If we assume the point of intersection as the origin, these three planes are known as co-ordinate planes and the three lines are known as co-ordinate axes.
The plane XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane, respectively. Let P be a point in a space such that its distances from yz-, zx- and xy-planes be a, b and c, respectively and i, j and k are the vectors along x, y and z axes, respectively, i.e.
= OA + AL + LP
Octants
= OA + OB + OC.
Every plane bisects the space. Hence three-co-ordinate plane divide the space into eight parts. Each part is called an octant.
= ai + bj + ck.
OA = a, OB = b and OC = c
Now,
OP = OL + LP
__________
and
|OP| = ÷ a 2 + b2 + c2 .
Notes: 1. Equation 2. Equation 3. Equation 4. Equation 5. Equation 6. Equation
of of of of of of
x-axis: y = y-axis: x = z-axis: x = xy-plane: z yz-plane: x zx-plane: y
0, z = 0 0, z = 0 0, y = 0 =0 =0 =0
6.2 Integral Calculus, 3D Geometry & Vector Booster
1.4 Distance
r – r1 m fi _____ = __ n r2 – r
between two points in space
From the figure, it is clear that
PQ = OQ – OP
= r2 – r1
P
Q r2
r1 O
fi
n (r – r1) = m (r2 – r)
fi
r (m + n) = mr2 + nr1
fi
mr2 + nr1 r = ________ m+n
|PQ| = |r2 – r1|
Let
P = (x1, y1, z1) and Q = (x2, y2, z2)
fi
(xi + yj + zk)
Then
PQ = OQ – OP
m (x2i + y2 j + z2k) + n (x1i + y1j + z1k) = _________________________________ m+n
= (x2 – x1) i + (y2 – y1) j + (z2 – z1) k
Thus, the distance between two points, ____________________________
PQ = ÷ (x 2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
1.4.1 Distance from Origin Let O be the origin and P (x, y, z) be any point, then
__________ x2 + y2 + z2 .
OP = ÷
1.4.2 Distance of a Point from Co-ordinate Axes _______
mx2 + nx1 my2 + ny1 mz2 + nz1 fi x = ________ , y = ________ , z = ________ m+n m+n m+n
1.5.2 External Section If a point R(r) divides the line segment joining the points P(r1) and Q(r2) externally in the ratio m : n, then mr2 – nr1 r = ________ . m – n mx2 – nx1 x = ________ , m – n
The distance from a point to x-axis = ÷ b 2 + c2
my2 – ny1 y = ________ , m – n
The distance from a point to y-axis = ÷ a 2 + c2
mz2 – mz1 z = _________ m – n
_______
_______
and the distance from a point to z-axis = ÷ a 2 + b2
1.5.3 Mid-Point Formula If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio 1 : 1, then r2 + r1 r = ______ . 2 Thus, the co-ordinates of R are
1.5 Section
(
)
x1 + x2 y______ 1 + y2 z______ 1 + z2 _______ , , 2 2 2
formulae
1.5.1 Internal Section
1.5.4 Centroid
If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio m : n, then
mr2 + nr1 r = ________ . m+n
PR fi ____ = RQ
m __ n
OR – OP m fi _________ = __ n OQ – OR
The point of intersection of the medians of a triangle is called the centroid of the triangle. Let A (r1), B (r2) and C (r3) be m the vertices of a triangle ABC PR where ___ = __ n and G (r) be its centroid. Then RQ the position vector of the centroid is
r1 + r2 + r3 r = __________ . 3
3D-Co-ordinate Geometry
Let the position vectors of OA, OB and OC are r1, r2 and r3, respectively. Let the position vectors of G be r. Now, the position vector of D is r2 + r3 OD = ______ 2
= OG = r r2 + r3 2 ◊ ______ + 1 r1 2 = ______________ . 1+2
1.6.3 Refult-1 Let OP = r = ai + bj + ck and i, j, k be the unit vectors along the x-axis, y-axis and z-axis, respectively Now,
r i = a
fi
|r| |i| cos a = a a cos a = __ |r| a l = __ |r|
fi
Thus, the co-ordinates of the centroid G are
(
A set of three numbers a, b, c, which are proportional to the direction cosines l, m, n respectively of a line, are called the direction ratios.
fi
r1 + r2 + r3 = __________ 3
1.6.2 Direction Ratios
l __ m __ n Thus, __ a = b = c .
As we know that, the centroid divides the median in the ratio 2 : 1. Thus, the position vector of G,
)
x1 + x2 + x3 y1__________ + y2 + y3 z1_________ + z2 + z3 __________ , , 3 3 3
Similarly,
b c m = __ and n = __ |r| |r|
i.e.
a l = ____________ __________ , 2 c2 ÷ a + b2 +
and
c n = ____________ ___________ . 2 c2 ÷ a + b2 +
Centroid of a Tetrahedron Let the co-ordinates of A, B, C and D are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) and (x4, y4, z4), respectively. Then the co-ordinates of its vertices are
(
6.3
a = ____________ ___________ 2 c2 ÷ a + b2 +
1.6.4 Refult-2
If l, m, n be the direction cosines of a line, then x1 + x2 + x3 + x4 y1_____________ + y2 + y3 + y4 z1_____________ + z2 + z3 + z4 ______________ , , l 2 + m2 + n2 = 1 4 4 4
)
1.6.5 Refult-3 Any vector r can be expressed as Let
r = |r| (li + mj + nk) r = ai + bj + ck.
r fi __ = |r|
1.6 Direction line
cosines and direction ratios of a vector or a
1.6.1 Direction cosines If a line makes an angle a, b, g with the positive direction of the co-ordinate axes. Then cos a, cos b, cos g are known as the direction cosines of the given line and are, generally, denoted as l, m and n, respectively. Thus, l = cos a, m = cos b, n = cos g.
a b c __ i + __ j + __ k |r| |r| |r|
= li + mj + nk
fi r = |r| (li + mj + nk)
1.6.6 Refult-4 If a vector r having direction cosines l, m, n, the projection of r on the co-ordinate axes are given by
l|r|, m|r|, n|r|.
1.6.7 Refult-5 The projection of the segment joining the points P (x1, y1, z1) and Q (x2, y2, z2) on a line having direction cosines l, m, n is given by l (x2 – x1) + m (y2 – y1) + n (z2 – z1).
6.4 Integral Calculus, 3D Geometry & Vector Booster
1.6.8 Refult-6
= (l21 + m 21 + n 21) (l22 + m 22 + n 22)
If the co-ordinates of P and Q are (x1, y1, z1) and (x2, y2, z2), the direction ratios of the line PQ are a = x2 – x1, b = y2 – y1, c = z2 – z1 and the direction cosines of the line PQ are
fi
(l1m2 – l2m1)2 + (m1n2 – n2m1)2
x2 – x1 y2 – y1 z2 – z1 l = ______ , m = ______ , n = ______ |PQ| |PQ| |PQ|
fi
1.6.9 Direction Cosines of the Axes
fi
Since the positive x-axis makes angles 0°, 90°, 90° with the axes of x, y and z, respectively, then the direction cosines of x-axis are (1, 0, 0). Similarly, the direction cosines of y-axis and z-axis are (0, 1, 0) and (0, 0, 1) respectively.
l1 fi __ = l2
m = a1i + b1j + c1k
and
n = a2i +
Then
m ◊ n = m n cos q
2j
+ c2k
a1a2 + b1b2 + c1c2 __________ __________ = ________________________ 2 ÷a 1 + b 21 + c21 ÷a 22 + b 22 + c22 a1a2 + b1b2 + c1c2 = _______________ l1l2
___________
l1 = ÷ a 21 + b 21 + c21 , l2 = ÷ a 22 + b 22 + c22
a1 a2 fi cos q = __ __ + l1 l2
b1 2 __ __ + l1 l2
m1 ___ m = 2
n1 __ n . 2
Straight Line A straight line in a space can be determined uniquely if (i) it passes through a fixed point and is parallel to a fixed line. (ii) it passes through two fixed points. (iii) it is the intersection of two given non-parallel planes.
___________
where
l1m2 = l2m1; m1n2 = n2m1; n1l2 = l2n1
2.1 Definition
m ◊ n _____ m ◊ n fi cos q = ____ mn = |m| |n|
(l1m2 – l2m1) = 0, (m1n2 – n2m1) = 0, (n1l2 – l2n1) = 0
1.6.10 Angle between Two Vectors Let
+ (n1l2 – l2n1)2 = 0
c1 c2 __ __ l1 l2
= l1l2 + m1m2 + n1n2
(i) Condition of perpendicularity
2.2 Equation
of a line passing thorugh a point and
parallel to a vector
The equation of a line passing through a point A with position vector r1 and parallel to a vector m is r = r1 + lm Now, fi
OP = OA + AP = OA + lm
Let r = x i + y j + z k
r1 = x1 i + y1 j + z1 k
and
m = a i + b j + c k r – r1 = l m
Here,
q = 90°
Then
fi
cos q = 0
fi (x – x1) i + (y – y1) j + (z – z1) k
fi
l1l2 + m1m2 + n1n2 = 0
(ii) Condition of parallelism Here,
q = 0°
fi
cos q = 1
fi
lll2 + m1m2 + n1n2 = 1
fi
(l1l2 + m1m2 + n1n2)2 = 1
fi
(l1l2 + m1m2 + n1n2)2
= l (a i + b j + c k)
x – x1 y_____ – y1 z_____ – z1 Thus, ______ = = c . a b Notes: 1. In case of a line, the direction ratios and the direction cosines are the same. 2. Any point on the line can be considered as
(al + x1, bl + y1 + cl + z1)
3D-Co-ordinate Geometry
2.3 Equation of a Line Passing Through Two Points A (r1) and B (r2) is r = r1 + (r2 – r1) OP = OA + AP
Now,
fi OP + OA + l AB fi r = r1 + l (r2 – r1). r = x i + y j + z k
\
r1 = x1 i + y1 j + z1 k
r2 = x2 i + y2 j + z2 k
Then
(r – r1) = l(r2 – r1)
2.5 Skew lines Two lines are said to be skew lines if they are neither parallel nor intersecting. Clearly, the skew lines can never be coplnar.
2.5.1 Shortest Distance Between Two Skew Lines
fi OA + l (OB – OA)
Let
6.5
Let
L1 : r = r1 + l u
and
L1 : r = r2 + m v
Let
OA = r1 and OB = r2
Here,
L1 || u and L2 || v
fi (x – x1) i + (y – y1) j + (z – z1) k
= (x2 – x1) i + (y2 – y1) j + (z2 + z1) k
x – x1 fi ______ = x2 – x 1
y – y1 ______ = y2 – y 1
z – z1 ______ z2 – z 1
fi PQ ^ L1 and PQ ^ L2 fi PQ ^ u and PQ ^ v fi PQ || u × v Thus,
2.4 Angle
between
two straight lines
Let
L1 : r = r1 + l m
and
L2 : r = r2 + m n
Then
m ◊ n = m n cos q
The shortest distance = PQ
= Projection of AB on PQ
AB ◊ PQ = _______ |PQ|
(r2 – r1) ◊ (u × v) = ______________ . |(u × v)|
Notes 1: If two straight lines intersect, the shortest distance between them is zero.
m ◊ n fi cos q = ____ mn
The Plane
Let
m = a1 i + b1 j + c1 k
and
n = a2 i + b2 j + c2 k
3.1 Definition
Thus,
a1a2 + b1b2 + c1c2 __________ __________ cos q = ________________________ 2 2 + c21 ÷a 22 + b22 + c22 ÷ a 1 + b 1
A plane is a surface such that if any two points on it are taken, the line joining them lies completely on it.
1. Condition of perpendicularity When
\
p q = __ 2 a1a2 + b1b2 + c1c2 = 0
2. Condition of parallelism When
q=0
a1 \ __ a2 =
b1 __ + b2
c1 __ c 2
3.2 General Form A first degree equation represents a plane. The general equation of a plane is given by a x + b y + c z + d = 0.
Notes: 1. Equation 2. Equation 3. Equation 4. Equation given by
of xy-plane is z = 0. of yz-plane is x = 0 of zx-plane is y = 0. of a plane passing through the origin is a x + b y + c z = 0.
6.6 Integral Calculus, 3D Geometry & Vector Booster
3.3. Equation
of a plane passing through a point (x, y, z,)
The equation of a plane passing through a point (x1, y1, z1) is a (x – x1) + b (y – y1) + c (z – z1) = 0 Let the equation of the plane be a x + b y + c z + d = 0 ...(i) which is passing through (x1, y1, z1). Thus, a x1 + v y1 + c z1 + d = 0 ...(ii) Subtracting Eqs. (ii) from Eq. we get a (x – x1) + b (y – y1) + c (z – z1) = 0
3.4 Equation of a plane passing non-collinear points
through three
The equation of any plane passing through (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by x – x1 y – y1 z – z1 x – x y z = 0 2 – z1 2 1 2 – y1 x3 – x1 y3 – y1 z3 – z1
|
|
The equation of any plane passing through (x1, y1, z1) is given by a (x – x1) + b (y – y1) + c (z – z1) = 0 ...(i) When it is passing through (x2, y2 , z2) and (x3, y3, z3), then a (x2 – x1) + b (y2 – y1) + c (z2 – z1) = 0 ...(ii) a (x3 – x1) + b (y3 – y1) + c (z3 – z1) = 0 ...(iii)
|
Eliminating a, b, c from Eqs (i), (ii) and (iii), we get x – x1 y – y1 z – z1 x y z = 0 2 – z1 2 – x1 2 – y1 x3 – x1 y3 – y1 z3 – z1 which is the required equation of the plane.
3.5 Coplanarity
|
and
B ◊ b + D = 0
C ◊ c + D = 0
D fi B = – __ b D fi C = – __ c .
Putting the values of A, B and C in Eq. (i), we get D D D __ __ – __ a x – b y – c z + D = 0 y __z x __ fi __ a + b + c – 1 = 0 y __z x __ fi __ a + b + c = 1
which is the required equation of the plane.
3.7 Normal
to a plane
A line perpendicular to a plane is called the normal to the plane. Clearly every line in a plane is perpendicular to the normal to the plane.
3.8 Vector
form
The vector equation of a plane passing through a point having position vector a and normal to a vector n is given by r n = d
of four points
The points P (x1, y1, z1), Q (x2, y2, z2), R (x3, y3, z3) and S (x4, y4, z4) are coplanar, then x – z1 2 – x1 y 2 – y1 z2 x y z = 0 3 – z1 3 – x1 3 – y1 x4 – x1 y4 – y1 z4 – z1
3.6 Intercept
Let the equation of the plane be passing through A (a, 0, 0), B (0, b, 0), and C (0, 0, c). A x + B y + C z + D = 0 ...(i) D So, A ◊ a + D = 0 fi A = – __ a
|
form of a plane
|
The equation of a plane in intercept form is given by y __z x __ __ a , b , c = 1, where a, b, c are the lengths of x, y and z axes, respectively.
Let OP = r and ON = n Now, ON ^ NP fi NP ON = 0 fi (OP – ON) ON = 0 fi (r – n) n = 0 fi r n – n n = 0 fi r n = n n fi r ◊ n = d.
3.9 Equation
of a plane in normal form
The vector equation of a plane normal to a unit vector and at a distance from the origin is given by r ◊ = d Let O be the origin and ON be the perpendicular from O to the given plane such that ON = d .
3D-Co-ordinate Geometry
Let P be a point on the plane with position vector r so that OP = r. ON ^ NP
Now,
fi NP ON = 0 fi (OP – ON) ON = 0 fi (r ◊ d ) ◊ d = 0
(ii) Castesian form: a1b1 + a2b2 + a3b3 __________ __________ cos q = ________________________ ÷a 21 + a 22 + a23 ÷b 21 + b22 + b 23 where and
n1 = a1 i + a2 j + a3 k n2 = b1 i + b2 j + b3 k.
Condition of Perpendicularity
fi (r ◊d – d ◊ d ) = 0 fi (r ◊ – d) = 0
fi r ◊ = d
Condition of Parallelism
which is the required equation of the plane.
a3 a1 a2 __ __ = __ = . b1 b2 b3
3.10 Normal
form of a
Plane
If l, m, n be the direction cosines of the normal to a given plane and p be the length of the perpendicular from the origin to the plane, the equation of the plane is given by l x + m y + n z = p. As we know that the normal form of a plane is given by r = d Let r = x i + y j + z k and = l i + m j + n k The equation of the plane becomes l x + m y + n z = p.
3.11 Theorem Equation of a plane passes through a point A with position vector a and is parallel to the given vectors b and c. b P c r
a O
Let r be the position vector of any point P in the plane, then AP = OP – OA =r–a Since, the vectors r – a, b, c are coplanar So,
[r – a b c] = 0
fi [r b c] – [a b c] = 0 fi [r b c] = [a b c] which is the required equation of the plane.
3.12 Angle
3.13 Angle
a1b1 + a2b2 + a3b3 = 0
between a line and a plane
The angle between a line and a plane is the angle between the line and the normal to the plane. Let the equation of the plane be
a1x + b1y + c1z + d = 0
between two planes
The angle between two planes is defined as the angle between their normals. Let r n1 = d1 and r n2 = d2 and q be the angle between them. Then n1 n2 (i) Vector form: cos q = ______ . |n1| |n2|
...(i)
and the equation of the line be – y1 z – z1 x – x1 y_____ _____ = = _____ a2 c2 b2 Thus, cos (90° – q)
A
6.7
fi
a1a2 + b1b2 + c1c2 __________ __________ = ________________________ 2 ÷a 1 + b21 + c21 ÷a 22 + b22 + c 22
a1a2 + b1b2 + c1c2 __________ __________ sin q = _______________________ ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Condition of Perpendicularity a1 __ b1 __ c1 __ a2 = b2 = c2 Condition of parallelism a1a2 + b1b2 + c1c2 = 0
3.14 Equation
of a plane parallel to a given plane
The equation of a plane parallel to another plane a x + b y + c z + d = 0 is a x + b y + c z + k = 0.
3.15 Equation
of a plane parallel to the axes
3.15.1 Equation of a plane parallel to x -axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of x-axis is x y __z __ = __ = 1 0 0 Since the plane (i) is parallel to the line (ii), so, a ◊1 + b . 0 + c . 0 = 0 fi a=0
...(i) ...(ii)
6.8 Integral Calculus, 3D Geometry & Vector Booster
3.16 Equation of a (x2, y2, z2) and ratios a, b, c
Putting the value of a in Eq. (i), we get b y + c z + d = 0. which is the required equation of the plane.
3.15.2 Equation of a plane parallel to y -axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of y-axis is x y __z __ = __ = 0 1 0 Since the plane (i) is parallel to the line (ii), so, a ◊ 0 + b ◊ 1 + c ◊ 0 = 0 fi b=0 Putting the value of b in Eq. (i), we get a x + c z + d = 0. which is the required equation of the plane.
...(ii)
fi c=0 Putting the value of a in (i), we get a x + b y + d = 0. which is the required equation of the plane.
3.15.4 Equation of a plane parallel to xy -plane The equation of xy-plane is given by z=0 So the equation of any plane parallel to xy-plane is z+k=0
A x + B y + C z + D = 0
\
A (x – x1) + B (y – y1) + C (z – z1) = 0
and
A (x – x2) + B (y – y2) + C (z – z2) = 0
Also,
A ◊ a + B ◊ b + C ◊ c = 0
...(i) ...(ii)
Hence, the result.
3.17 Equation of the plane passing through a point (x1, y1, z1) and parallel to two lines having direction ratios (a1, b1, c1) and (a2, b2, c2) is x – x1 a1 a2
y – y1 b1 b2
z – z1 c1 = 0 . c2
The equation of any plane passing through (x1, y1, z1) is A (x – x1) + B (y – y1) + C (z – z1) = 0 Since the plane (i) is parallel to two lines, so A a1 + B b1 + C c1 = 0 A a2 + B b2 + Cc2 = 0
x
x1
y
y1
z
...(ii) ...(iii)
z1
a1
b1
c1
a2
b2
c2
=0
which is the required equation of the plane.
3.18 Equation
3.15.6 Equation of a plane parallel to zx -plane
...(i)
Eliminating A, B, C from Eqs (i), (ii) and (iii), we get
So the equation of any plane parallel to yz-plane is x+k=0
...(i)
which is passing through (x1, y1, z1) and (x2, y2, z2)
The equation of yz-plane is given by x=0
z – z1 z2 – z1 = 0 . c
Let the equation of the plane be
3.15.5 Equation of a plane parallel to yz -plane
y – y1 y2 – y1 b
Eliminating, A, B and C from above three equations, we get x – x1 y – y1 z – z1 x2 – x1 y2 – y1 z2 – z1 = 0 a b c
3.15.3 Equation of a plane parallel to z-axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of z-axis is x y z __ = __ = __ 0 0 1 Since the plane (i) is parallel to the line (ii), so, a ◊ 0 + b ◊ 0 + c ◊ 1 = 0
x – x1 x2 – x1 a
...(i)
plane passing thorugh (x1, y1, z1), parallel to the line having direction
of a plane passing through the line of
intersection of planes
The equation of zx-plane is given by
The equation of a plane passing through the line of intersection of planes
y=0 So the equation of any plane parallel to zx-plane is
a1x + b1y + c1z + d1 = 0
is a1x + b1y + c1z + d1 + l (a2x + b2y + c2z + d2) = 0.
y + k = 0.
and a2x + b2y + c2z + d2 = 0
3D-Co-ordinate Geometry
3.19 Distance
3.24 Condition
of a point from a plane
The length of the perpendicular from a point P (x1, y1, z1) to the plane a x + b y + c z + d = 0 is ax1 + b y1 + c z1 + d __________ _________________ + c2 ÷ a 2 + b2
3.20 Distance
|
|
between two parallel planes
of coplanarity of two lines
Let
x – x1 y_____ – y1 z_____ – z1 L1: ______ a = = c b1 1 1
and
x – x1 y_____ – y2 z_____ – z2 L2: ______ a = = c b2 2 2
If two lines L1 and L2 are coplanar, then
The distance between two parallel planes a x + b y + c z + d1 = 0 and a x + b y + c z + d2 = 0 is given by d – d2 ____________ ___________ . 2 c2 ÷ a + b2 +
3.21 Sides
and the equation of the plane containing them is
|
|
of a plane
We know that a plane divides the three-dimensional space in two equal parts. Two points A (x1, y1, z1) and B (x2, y2, z2) are on the same side or opposite side of the plane a x + b y + c z + d = 0 if a x1 + b y1 + c z1 + d __________________ > 0 or < 0. a x2 + b y2 + c z2 + d
3.22 Intersection
of a line and a plane
x – x1 y_____ – y1 z_____ – z1 Let L: _____ = = c a b and P: A x + B y + C z + D = 0 Any point on the line (i) can be considered as (x1 + al, y1 + bl, z1 + cl)
x2
...(iii)
A (x1 + al) + B (y1 + bl) + C (z1 + cl) + D = 0 fi (A x1 + B y1 + C z1 + D) + l (Aa + Bb + Cc) = 0
y2
y1
z2
z1
a1
b1
c1
a2
b2
c2
x1
y
y1
z
=0.
z1
a1
b1
c1
a2
b2
c2
=0
or x
...(i) ....(ii)
x1
x
If it lies on the plane (ii), then
3.25 Equation planes
x2
y
y2
z
z2
a1
b1
c1
a2
b2
c2
=0.
of the planes bisecting the angle between the
The equation of the planes bisecting the angle between the planes a1x + b1y + c2z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by
|
| |
|
(A x1 + B y1 + C z1 + D) fi l = – ____________________ (Aa + Bb + C c)
a1x + b1y + c1y + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Substituting the values of l in Eqs (iii), we get the required co-ordinates of the point of intersection.
3.26 Bisector planes
3.23 Condition
Let the equation of the two planes be
If the line
for a line to lie in a plane
x – x1 ______ y – y1 z_____ – z1 L: ______ = m = n l
lies in the plane P: a x + b y + c z + d = 0, then
(i) a x1 + b y1 + c z1 + d = 0
(ii) a l + b m + c n = 0 Since the line L = 0 lies in a plane P = 0, therefore the point (x1, y1, z1) on L = 0 will also lie in the plane P = 0. So,
a x1 + b y1 + c z1 + d = 0
Also, since the line L = 0 lies in the plane P = 0, therefore, the normal to the plane is also normal to the line. Thus,
a l + b m + c n = 0.
6.9
of the acute and obtuse angles between two
a1 x + b1y + c1z + d1 = 0
...(i)
and
a2 x + b2y + c2z + d2 = 0
...(ii)
First we make d1 > 0, d2 > 0 (i) If a1a2 + b1b2 + c1c2 > 0, the origin lies in the obtuse angle between two planes and the equation of the obtuse angle bisector is a1x + b1y + c1z + d1 a2x + b1y + c2z + d2 __________ __________ __________________ = __________________ 2 2 2 + c 1 ÷a 22 + b 22 + c22 ÷ a 1 + b1 and the acute angle bisector is a1x + b1y + c1z + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ 2 2 2 ÷a 1 + b 1 + c1 ÷a 22 + b 22 + c22
6.10 Integral Calculus, 3D Geometry & Vector Booster (ii) If a1a2 + b1b2 + c1c2 < 0, the origin lies in the acute angle between the two planes and the equation of the bisector of the acute angle is
is
x – x1 l1 l2
y – y1 m1 m2
z – z1 n1 = 0 . n2
a1x + b1y + c1z + d1 __________________ a2x + b2y + c2z + d2 __________ __________ __________________ = 2 2 2 ÷a 1 + b1 + c 1 ÷a 22 + b22 + c22
The equation of the plane containing L1 = 0 is a (x – x1) + b (y – y1) + c (z – z1) = 0
...(i)
and the obtuse angle is
where
...(ii)
a1x + b1y + c1z + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ . ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Since L2 = 0 is parallel to the plane (i), therefore the normal to the plane (i) is also normal to the line L2 = 0.
3.27 Foot
Eliminating a, b, c from Eqs (i), (ii) and (iii), we get
of perpendicular of a point w.r.t. a plane
The foot of the perpendicular M of a point P (x1, y1, z1) w.r.t. a plane a x + b y + c z + d = 0 is given by x2 – x1 y______ z2 – z1 (ax1 + by1 + cz1 + d) 2 – y1 ______ = = ______ = – __________________ a c b (a2 + b2 + c2)
3.28 Image
of a point w.r.t. a palne
Thus,
3.30 Equation
a l1 + b m2 + c n2 = 0 x – x1 l1 l2
y – y1 m1 m2
...(iii)
z – z1 n1 = 0 . n2
of the plane containing two given lines
The equation of the plane containing two lines
The image Q of a point P (x1, y1, z1) w.r.t. a plane a x + b y + c z + d = 0 is x2 – x1 y______ z2 – z1 2 (ax1 + by1 + cz1 + d) 2 – y1 ______ = = ______ = – ____________________ a c b (a2 + b2 + c2) where Q = (x2, y2, z2).
x – x1 y_____ – y1 z_____ – z1 L1: _____ = m = n l1 1 1
and
x – x2 y_____ – y2 z_____ – z2 L2: _____ = m = n l2 2 2
3.29 Equation of the plane containing a given line and parallel to a given line is The equation of the plane containing a x – x1 ______ y – y1 L1: ______ = m = l1 1 and also parallel to another line x – x2 y_____ – y2 L2: _____ = m = l2 2
a l2 + b m1 + c n1 = 0
given line z – z1 _____ n 1
x – x1 l1 l2
y – y1 m1 m2
z – z1 n1 = 0 n2
x – x2 l1 l2
y – y2 m1 m2
z – z2 n1 = 0 . n2
or
z – z2 _____ n2
Exercises
(Problems based on Fundamentals)
ABC of 3D-Geometry 1. Find the distance between the points P (3, 4, 5) and Q (– 1, 2, – 3). 2. Find the shortest distance of the point P (a, b, c) from x-axis. 3. If the line passing through the points (5, 1, a) and 17 13 (3, b, 1) crosses the yz-plane at 0, ___ , – ___ , find 2 2 the value of a + b + 2. 4. Show that the points P (0, 7, 10), Q (– 1, 6, 6) and C (– 4, 9, 6) are the vertices of an isosceles rightangled triangle.
(
)
5. Find the locus of a point, the sum of distances from (1, 0, 0) and (– 1, 0, 0) is 10. 6. Find the ratio in which the plane x – 2y + 3z = 17 divides the line joining the points (– 2, 4, 7) and (3, – 4, 8). 7. Find the co-ordinates of points which trisects the line joining the points P (– 3, 2, 4) and Q (0, 4, 7). 8. If the centre of a tetrahedron OABC are (1, 2, –1), where A = (a, 2, 3), B = (1, b, 2), C = (2, 1, c) respectively, find the distance of P (a, b, c) from the origin. 9. Find the ratio in which the line joining the points P (– 2, 3, 7) and Q (3, – 5, 8) is divided by the yz-plane.
3D-Co-ordinate Geometry
10. The vertices of a triangle are A (5, 4, 6), B (1, – 1, 3) and C (4, 3, 2). The internal bisector of –BAC meets BC in D. Find AD. Direction cosines and direction ratios 11. If a line is equally inclined with the axes, find its direction cosines. 12. Find the angle at which the vector (4 i + 4 j + k) is inclined to the axes. 13. If a line makes an angle a, b, g with the co-ordinate axes, find the value of (i) S sin2 a (ii) S cos (2a). 14. A line OP makes with the x-axis and y-axis at an angle of 120°, 60°, respectively. Find the angle made by the line with the z-axis. 15. A vector r is inclined to x-axis and to y-axis at 60°. If |r| = 8, find the vector r. 16. Find the projection of the line joining (1, 2, 3) and (– 1, 4, 2) on the line having direction ratios (2, 3, – 6). 17. What is the angle between the lines whose direction
(
__
) (
__
)
÷3 3 ÷ 3 1 3 1 ___ cosines are – __ , __ , – ___ and – __ , __ , ? 4 4 2 4 4 2 18. Find the angle between any two diagonals of a cube. 19. Find the angle between one diagonal of a cube and a diagonal of one face. 20. A line makes angles a, b, g with the four diagonal of a cube, find the value of
S cos2 a (ii) S sin2 a (iii) S cos(2a) (i)
21. Find the angle between the lines whose direction cosines are satisfy the equations l + m + n = 0 and l 2 + m2 – n2 = 0. 22. Find the direction cosines of the two lines which are connected by the relation l – 5m + 3n = 0 and 7l 2 + 5m2 – 3n2 = 0. 23. Find the angle between the lines which are connected by the relation l + m + n = 0 and 2l m + 2l n + m n = 0. 24. Show that the straight lines whose direction cosines are given by the equations al + b m + cn = 0 and ul2 + v m2 + w n2 = 0 are perpendicular if a2 (u + v) + b2 (u + w) + c2 (u + v) =0 a2 b2 __ c2 and parallel if __ u + __ v + w = 0. 25. If a variable line in two adjacent positions has direction cosines (l, m, n) and (l + d l, m + d m, n + d n), show that the small angle dq between the two positions is given by (dq)2 = (d l)2 + (d m)2 + (d n)2.
6.11
Striaght Line 26. Find the equation of a line passing through (2, 3, 4) + 1 z____ –7 x – 3 y_____ and parallel to the line _____ = = . 3 4 5 27. Find the equation of a line passing through (– 3, 2, – 4) and parallel to the vector 2 i + 3 j + 7k. y – 2 x – 1 28. Find the point where the line _____ = _____ = 2 – 3 z_____ +3 meets the plane 2x – 2y – z = 7. 4 29. Find the equation of the line through 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to x = 2y = 3z.. 30. Prove that the lines 4x = 3y = – z and 3x = y = – 4z are perpendicular. 31. If the lines x = az + b, y = cz + d and x = a¢z + b¢, y = c¢z + d ¢ are perpendicular, prove that a ◊ a¢ + c ◊ c¢ = – 1. 32. Find the distance of the point (1, – 2, 3) from the plane x – y + z = 5 measured parallel to the y z x line __ = __ = ___ . 2 3 – 6 33. Find the foot of the perpendicular from the point – 1 z____ –7 x – 6 y_____ (1, 2, 3) to the line _____ = = and also 3 2 3 find the length of the perpendicular. 34. Find the image of a point (1, 6. 3) in the line –7 x y – 1 z____ __ = _____ = . 1 2 3 35. Find the equation of the line which can be drawn from the point (1, – 1, 0) to intersect the – 1 z____ y z+1 –3 x – 2 y_____ x – 4 __ lines _____ = = and _____ = = _____ . 2 3 4 4 2 5 orthogonally. 36. Find the equation of the line through (2, – 3, 1) parallel to the plane 2x – y + z = 6 so as to meet y z–2 x – 2 ___ the line _____ = = ____ at right angle. 2 – 3 – 1 Shortest distance between two lines 37. Find the shortest distance between the lines r = 3i + 8 j + 3k + l (3i – j + k) and r = – 3i – 7 j + 6k + m (– 3i + 2 j + 4k). 38. Find the shortest distance between the lines
– 3 z____ –4 x – 2 y_____ L1: _____ = = 3 4 5
and
+ 1 z_____ +2 x – 2 y_____ L2: _____ = = . 2 –1 3
39. Prove that the shortest distance between any two opposite edges of a tetrahedron formed by the planes __ y + z =__0, x + z = 0, x + y = 0 and x + y + z = ÷3 a is a ÷2 .
6.12 Integral Calculus, 3D Geometry & Vector Booster 40. If 2d be the shortest distance between the y z x __z 1 lines __ + __ c = 1, x = 0 and __ – c = __ y = 0, prove a b 1 1 1 1 that __ 2 + __ 2 + __ 2 = ___ 2 . a b c d Plane 41. Find the equation of the plane through the points P (1, 1, 1), Q (3, – 1, 2) and R (–3, 5, – 4). 42. Show that the four points (1, 2, 3), (2, 1, 4) (3, – 1, 3) and (2, 2, 6) are coplanar. 43. Find the intercepts of the plane 2x – 4y + 5z = 20 on the co-ordinate axes. 44. Find the equation of the plane passes through the points (3, 0, 0), (0, 4, 0) and (0, 0, 5). 45. Find the equations of the plane through the points (0, 4, – 3), (6, – 4, 3) other than the plane through the origin, which cut off from the axes intercepts, whose sum is zero, 46. A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point (p, q, r). Find the equation of the plane. 47. A variable plane cuts the co-ordinate axes in A, B, C and is of constant distance 3p from the origin. Find the locus of the centroid of the triangle ABC. x y __z 48. The plane __ a + __ + c = 1 meets the co-ordinate axes b at the points A, B, C respectively. Find the area of the D ABC. 49. A variable plane cuts the co-ordinate axes in A, B, C and is of constant distance p from the origin. Find the locus of the centroid of the tetrahedron OABC. 50. If a variable plane forms a tetrahedron of constant volume 64 k3 with the co-ordinate planes, find the locus of the centroid of the tetrahedron. 51. Find the angle between the planes x – 2y + 2z + 2014 = 0 and 2x + y + 2z + 2015 = 0. 52. Prove that the planes 3x – 2y + 3z + 10 = 0 and 2x – 3y – 4z + 7 = 0 are perpendicular. 53. Find the equation of the plane passing through (1, 2, 3) and the perpendicular to the planes 2x + 3y + 4z + 7 = 0, 3x + 4y + 5z + 10 = 0 x – 2 54. Find the angle between the line _____ = 3 y + 1 z – 3 _____ = ____ and the plane 3x + 4y + z + 5 = 0. – 1 – 2 – 4 z____ –5 x – 3 y_____ 55. Prove that the line _____ = = is parallel 2 3 4 to the plane 4x + 4y – 5z + 2 = 0. 56. Find the equation of a plane passing through (2, 3, – 1) and parallel to x – 2y + 3z + 10 = 0. 57. Find the equation of the plane passing through the points (2, 3, – 4), (1, – 1, 3) and parallel to x-axis. 58. Find the equation of the plane passing through the points (1, 2, 3), (4, 3, 1) and parallel to y-axis
59. Find the equation of the plane passing through the points (3, 4, 5), (– 2, 3, – 1) and parallel to z-axis. 60. Find the equation of the plane passing through (1, 2, 3) and parallel to xy-plane. 61. Find the equation of a plane parallel to yz-plane and passes through the point (2, 3, 4). 62. Find the equation of the plane passing through (3, 4, 5) and parallel to zx-plane. 63. Find the equation of the plane passing through the points (2, – 1, 0), (3, – 4, – 5) and parallel to the line 2x = 3y = 4z. 64. Find the equation of the plane passing through (2, 3, 4) and parallel to the lines x = 2y = 3z and 2x = 5y = z. 65. Find the equation of the plane passing through the intersection of the planes 2x + 5y – 5z = 6 and 2x + 7y – 8z = 7 and the point (– 1, 4, 3). 66. Find the equation of the plane passing through the point (2, – 1, 1) and the line 4x – 3y + 5 = 0 = y – 2z – 5. 67. Find the equation of the plane passing through the intersection of the planes 2x + 3y + 10z = 8 2x – 3y + 7z = 2 and the perpendicular to 3x – 2y + 4z = 5. 68. Find the equation of the plane passing through the line x + y + z + 3 = 0, 2x – y + 3z + 1 = 0 and x y __z parallel to __ = __ = . 1 2 3 69. Find the length of the perpendicular from a point P (1, 2, 3) to the plane 5x + 4y – 3z – 1 = 0. 70. Find the equation of the line through (– 1, 3, 2), perpendicular to the plane x + 2y + 2z = 3 and also find the co-ordinates of its foot. 71. Find the distance of the point (1, 2, 3) from the plane x + y + z = 11 measured parallel + 2 z____ –7 x + 1 y_____ to _____ = = . 1 – 2 2 72. Find the locus of a point, the sum of the squares of whose distances from the planes x + y + z = 0, x – z = 0, x – 2y + z = 0 is 9. 73. Two system of the co-ordinate axes have the same origin. If a plane cuts them at distances a, b, c and p, q, r respectively from the origin. Prove that 1 1 1 1 1 1 __2 + __ 2 + __ 2 = __ 2 + __ 2 + __ 2 . a b c p q r 74. Find the distance of the point (4, 1, 1) from the lines x + y + z – 4 = 0 = x – 2y – 2z – 4. 75. Find the distance between the planes x + 2y – 2z + 1 = 0 and 2x + 4y – 4z + 5 = 0 76. Find the co-ordinates of the point of intersection y – 2 z + 3 x – 1 of the line _____ = _____ = _____ with the plane 2 – 3 4 2x – 2y – z = 7.
3D-Co-ordinate Geometry
77. Find the equation of the plane passing through (1, 2, 0) – 1 z____ –2 x + 3 y_____ which contains the line _____ = = . 3 4 – 2 y – 2 z – 3 x – 1 78. Prove that the lines _____ = _____ = ____ and 2 3 4 y – 3 –4 x – 2 _____ z____ _____ = = are coplanar and also find the 3 4 5 equation of the plane where they lie. 79. Find the equation of the plane passing through the line of intersection of the planes 4x – 5y – 4z = 4 and 2x + y + 2z = 8 at the point (1, 2, 3). 80. Find the eqution of the plane bisecting the acute angle between the planes x + 2y + 2z – 3 = 0 and 3x + 4y + 12z + 1 = 0. 81. Find the image of the point (2, – 3, 4) with respect to the palne 4x + 2y – 4z + 3 = 0. 82. Find the equation of the image of the y – 2 z + 3 x – 1 line _____ = _____ = _____ in the plane 3x – 3y 9 –1 – 3 + 10z – 26. 83. Find the equation of the plane containning the y + 6 z + 1 x – 1 line _____ = _____ = _____ and is parallel to the 3 4 2 – 1 z_____ +3 x – 4 y_____ line _____ = = . 2 – 3 5 84. Find the equation of the plane containing the lines – 1 z_____ – 4 z____ +3 –6 x + 1 y_____ x – 2 y_____ _____ = = and _____ = = . 7 3 1 3 5 5
1. The shortest distance of the point (a, b, c) from the x-axis is ______
______
(a) ÷a 2 + b2
(b) ÷ a2 + c2
(c) ÷b 2 + c2
(d) ÷ a2 + b2 + c2
______
__________
2. The equation of the x-axis is x y __z x (a) __ = __ = (b) __ = 1 0 0 0 x y __z x (c) __ = __ = (d) __ = 0 0 1 1
y __ = 1 y __ = 0
(
z __ 0
5. The foot of perpendicular from (a, b, c) on the z-axis is (a) (a, 0, 0) (b) (0, b, 0) (c) (0, 0, 0) (d) (0, 0, c). 6. The angle between any two diagonals of a cube is __
(b) tan– 1 (÷2 ) __ 2 (c) cot– 1(÷2 ) (d) cos– 1 __ 3 7. The angle between a diagonal of a unit cube and a diagonal of a face is __ 2 2 – 1 __ –1 __ (a) cos (b) cos 3 3
( ) 1 ( __ ÷ 3 )
(c) cos– 1
3 ÷ (a) ___ 2 __
)
1 1 __ 1 3. If the direction cosines of a line are __ c , __ c , c , then c is (a) 1 (b) ± 2__ (c) ± 3 (d) ± ÷3 4. The number of lines, through the origin makes equal angles with the axes, is (a) 1 (b) 2 (c) 4 (d) 0
( )
( ÷ ) ( )
1 (d) cos– 1 __ 3
8. A line makes an angle a, b and g with the axes, the value of sin2a + sin2b + sin2g is (a) 1 (b) 2 (c) – 1 (d) 0. 9. A line makes an angle a, b and g with the axes, then the value of cos (2a) + cos (2b) + cos (2g) is (a) 1 (b) 2 (c) –1 (d) 0. 10. The shortest distance from the origin to the p l a n e 2x + 3y + 6z – 21 = 0 is (a) 2 (b) 1 (c) 3 (d) 4 11. A plane meets the co-ordinate axes at A, B and C, respectively. If the centroid of the triangle is (2, 2, 2), the equation of the plane is (a) x + y + z = 5 (b) x + y + z = 6 (c) x + y + z = 9 (d) x + y + z = 4 12. The locus of the first degree equation in x, y, z reprsents a (a) stline (b) plane (c) sphere (d) conicoid 13. If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1 (k) with x-axis, then k is __
z __ 1
__
(a) tan– 1 (2÷2 )
__
(Mixed Problems)
6.13
2 (b) __ 7
2 ÷ (c) ___ (d) 1 3 14. The distance between the parallel lines 2x – 3y + 6z + 5 = 0 and 6x – 9y + 18z + 21 = 0 is 3 1 (a) __ (b) __ 7 7 2 4 (c) __ (d) __ . 7 7 15. The equation of the plane passing through (1, 2, 3) and parallel to 2x + y + 2z + 10 = 0 is
6.14 Integral Calculus, 3D Geometry & Vector Booster (a) 2x + y + 2z – 10 = 0 (b) 2x + y + 2z – 12 = 0 (c) 2x + y + 2z + 12 = 0 (d) 2x + y + 2z + 10 = 0 16. The angle between the planes 2x – y + z = 6 and x + y + 2z = 7 is p p (a) __ (b) __ 4 3 p p __ __ (c) (d) 2 6 17. The line x = 1, y = 2 is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) none of these 18. The line y = 2 and z = 3 is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) none of these 19. The equation of a line passing through (2, 3, 4) and perpendicular to the plane 2x + 3y – z = 5 is – 3 z____ –4 x – 2 y_____ (a) _____ = = 2 1 3 + 3 z____ –4 x – 2 y_____ (b) _____ = = 2 1 3 + 3 z____ –4 x – 2 y_____ (c) _____ = = 2 1 3 – 3 z____ –4 x – 2 y_____ (d) _____ = = 2 4 5 y – 3 z – 4 x – 2 20. The line _____ = _____ = ____ is parallel to the 3 4 5 plane (a) 3x + 4y + 5z = 7 (b) 2x + 3y + 5z = 10 (c) 2x + y – 2z = 0 (d) 2x + 3y + 2z = 5 21. The angle between the plane 3x + 6y – 2z + 5 = 0 + 1 z____ –3 x – 2 y_____ and the line _____ = = is 2 – 1 2 8 4 (a) cos– 1 ___ (b) cos– 1 ___ 21 21 8 4 (c) sin– 1 ___ (d) sin– 1 ___ . 21 21 22. The co-ordinates of the point of intersection of the + 1 z_____ +3 x – 6 y_____ line _____ = = and the plane x + y – z = 3 –1 0 4 is (a) (2, 1, 0) (b) (1, 2, – 6) (c) (5, 1, 2) (d) (5, – 1, 1) x – 1 23. The shortest distance between the lines _____ = 2 y_____ – 2 z____ – 4 z____ –3 –5 x_____ – 2 y_____ = and = = is 3 4 3 4 5 1 1__ (a) __ (b) ___ 6 ÷6 1__ 1 (c) ___ (d) __ 3 3 ÷
( ) ( )
( ) ( )
24. The plane passing through (3, 2, 0) and the line – 6 z____ –4 x – 3 y_____ _____ = = is 1 4 5 (a) x – y + z = 1 (b) x + y + z = 5 (c) x + 2y – z = 1 (d) 2x – y + z = 5. ___›
25. If the projection of PQ on OX, OY and___ OZ are › respectively 12, 3 and 4, the magnitude of PQ is (a) 169 (b) 13 (c) 12 (d) 144 26. The distance of the plane passing through (1, 1, 1) and – 1 z____ –1 x – 1 y_____ perpendicular to the line _____ = = from 3 0 4 the origin is 3 4 (a) __ (b) __ 4 3 7 (c) __ (d) 1 5 27. A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the centroid D (x, y, z) of DABC satisfies the rela1 1 1 tion __ 2 + __ 2 + __ 2 = k, then k is x y z 28.
29.
30.
31.
32.
(a) 1 (c) 9
(b) 3 (d) 4 y – 3 x – 2 x – 1 4 – z The lines _____ = _____ = _____ and _____ = 1 1 k k y – 4 z____ –5 _____ = are coplanar if k is 2 1 (a) 1 (b) 2 (c) 0 (d) 4 The equation of the obtuse angle bisector between the planes 3x – 2y + 6z + 8 = 0 and 2x – y + 2z + 3 = 0 is (a) 5x – y – 4z = 3 (b) 5x + y – 4z = 5 (c) 4x + y – 5z = 5 (d) 7x + 3y – 9z The plane x – y – z = 4 is rotated through an angle 90° about its line of intersection with the plane x + y + 2z = 4. Then the new position of the plane is (a) 5x + 3y + 2z = 4 (b) 5x + y + 4z = 20 (c) 4x + y + 5z = 20 (d) 4x + 5y + z = 20 The value of m for which the straight lines 3x – 2y + z + 3 = 0 = 4x – 3y + 4z + 1 are parallel to the plane 2x – y + m z = 2 is (a) 6 (b) 8 (c) – 2 (d) – 11 A straight line is given by r = (1 + t) iˆ + 3t + (1 – t) , where t in R. If this line lies in the plane x + y + c z = d, the value of (c + d – 5) is
3D-Co-ordinate Geometry
(a) 8 (c) 4
(b) 9 (d) 2
– 1 z_____ +1 x – 2 y_____ 33. Let L1: _____ = = 1 0 2 – 1 ___ z x – 3 y_____ and L2: _____ = = 1 1 – 1 and let p be the plane which contains the line L1 and is parallel to L2. The distance of the plane p from the origin is __ 2 1 (a) __ (b) __ 7 7 __ (c) ÷6 (d) none of these 34. If the planes x = cy + bz, y = az + bx and z = bx + ay pass through the same line, the value of a2 + b2 + c2 + 2ab + 2 is (a) 2 (b) 1 (c) 3 (d) 4 35. The equation of the plane passing through (0, 2, 4) – 1 2____ –z x + 3 y_____ and containing the line _____ = = is 3 4 2 (a) x – 2y + 4z – 12 = 0 (b) 5x + y + 9z – 38 = 0 (c) 10x – 12y – 9z + 60 = 0 (d) 7x + 5y – 3z + 2 = 0 36. The two lines x = ay + b, z = cy + d and x = a¢y + b¢, z = c¢y + d¢ will be perpendicular if and only if (a) aa¢ + bb¢ + cc¢ + 1 = 0 (b) aa¢ + bb¢ + cc¢ = 0 (c) (a + a¢) + (b + b¢) + (c + c¢) = 0 (d) a a¢ + c c¢ + 1 = 0 1 – y _____ z + 2 x – 2 _____ 37. If the line _____ = = lies in the plane 3 2 6 x + 3y – a z + b = 0. Then the value of (a + b + 1) is (a) 2 (b) 1 (c) 3 (d) 4 y – y z z – 3 x x – 1 2_____ 38. If the lines __ = __ = __ ; _____ , = ____ and 1 2 3 3 1 4 – 1 z____ –2 x + k y_____ _____ = = are concurrent, then 3 2 h
÷
(a) h = – 2, k = – 6 (b) h = 6, k = 2 (c) h = 1/2, k = 2 (d) h = 2, k = 1/2 39. If the four points P (2 – x, 2, 2), Q(2, 2 – y, 2) R (2, 2, 2 – z) and S (1, 1, 1) are coplanar, then 1 __ 1 __ 1 (a) __ x + y + z = 1 (b) x + y + z = 1 1 1 1 (c) _____ + _____ + ____ = 1 1–x 1–y 1–z 1 1 1 (d) _____ + _____ + _____ = 2 1+x 1+y 1+z
6.15
40. The equation of the right bisector plane of the line segment joining the points (2, 3, 4) and (6, 7, 8) is (a) x + y + z = 15 (b) x + 2y + z = 15 (c) 3x + 2y + z = 15 (d) x + 2y + 5z = 15 41. The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2 + y2 + z2 + 4x – 2y – 6z = 155 is (a) 39 (b) 26 (c) 11 (d) 13 42. The intersection of the spheres x2 + y2 + z2 + 7z – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane (a) 2x – y – z = 1 (b) x – 2y – z = 1 (c) x – y – 2z = 1 (d) x – y – z = 1 43. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius (a) 3 (b) 1__ (c) 2 (d) ÷ 2 44. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the mid-point of the line joining the centres of the spheres x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8, then a is (a) – 1 (b) 1 (c) – 2 (d) 2 45. The intersection of the spheres x2 + y2 + z2 + 7x – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane. Then (a) 2x – y – z = 1 (b) x – 2y – z = 1 (c) x – y – 2z = 1 (d) x – y – z = 1.
(Problems for JEE-Advanced)
1. Show that the straight lines whose direction cosines are given by the equation al + bm + cn = 0 and ul2 + vm2 + wn2 = 0 are perpendicular if a2 (u + v) + b2 (u + w) + c2 (u + v) = 0 a2 b2 __2 and parallel if __ u + __ v + w = 0. 2. If P be any point on the plane l x + m y + n z = p and Q be a point on the line OP such that OP ◊ OQ, find the locus of the point Q. y x __ __z 3. A point P moves on a plane __ a + b + c = 1. A plane through P and perpendicular to OP meets the co-ordinate axes at A, B and C. If the plane through A, B and C and parallel to the planes x = 0, y = 0, z = 0 intersect at Q, find the locus of Q. 4. A plane passes through a fixed point (a, b, c). Show that the locus of the foot of the perpendicular to it from the origin is the sphere x2 + y2 + z2 – ax – by – cz = 0.
6.16 Integral Calculus, 3D Geometry & Vector Booster
5. The plane x – y – z = 4 is rotated through an angle of 90° about the line of intersection with the plane x + y + 2z = 4. Find its equation in the new position. 6. If the planes x – c y – b z = 0, c y – y + a z = 0 and b x + a y – z = 0 passing through a straight line, find the value of a2 + b2 + c2 + 2 abc. 7. Find the co-ordinates of those points on the + 2 z____ –3 x – 1 y_____ line _____ = = which are at a distance 2 3 6 of 3 units from the point (1, – 2, 3). 8. Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the + 2 6____ –z x – 2 y_____ line _____ = = . 2 3 6 9. Find the equation of the plane passing through (1, 2, 0) – 1 2____ –z x + 3 y_____ which contains the line _____ = = . 3 4 2 10. Find the equation of the projection of the + 1 z____ –3 x – 1 y_____ line _____ = = on the plane x + 2y + z 2 – 1 4 = 9. 11. Find the shortest distance between the lines 2x + y + z – 1 = 0 = 3x + y + 2z – 2 and x = y = z. 12. Find the distance of the point of intersection of the + 3 z_____ + 1 + 1 x – 4 y_____ x – 1 y_____ lines _____ = = and _____ = = 7 1 – 4 2 – 3 z + 10 ______ from (1, – 4, 7). 8
19. If the equation of the plane through the straight y + 2 z x – 3 line _____ = _____ = __ and perpendicular to 2 – 3 6 x – y + z + 2 = 0 is ax + by + cz + 4 = 0, find the value of a + b + c + 2. 20. Find the equation of the plane through (2, 3, 4) and also which is at the maximum distance from the origin. 21. A sphere of radius r passes through the origin and meets the axes in A, B, C. Find the locus of the centroid of DABC lies on the sphere. 22. A plane passes through a fixed point (f, g, h) and cuts the axes in A, B, C. Show that the locus of the centre of the sphere OABC. 23. A sphere of constant radius r passes through the origin and cuts the axes in A, B, C. Find the locus of the foot of the perpendicular from O to the plane ABC. 24. Find the equation of a sphere circumscribing the tetrahedron whose faces are x = 0, y = 0, z = 0 y __z x __ and __ a + b + c = 1 25. Find the equation of a sphere for which the circle x2 + y2 + z2 + 7y – 2z + 2 = 0 and 2x + 3y + 4z = 8 is a great circle.
(Tougher Problems for Jee-Advanced)
13. If the lines x = a y + b, z = c y + d and x = a1y + b1, z = c1y + d1 are perpendicular such that a a1 + b b1 + k = 0, find the value of (k2 + 4). 14. Find the distance of the point(1, 2, 3) from the plane x + y + z = 11 measured parallel – 12 z____ –7 x + 1 y______ to _____ = = . 1 – 2 2 15. If a variable plane forms a tetrahedron of constant volume 64k2 with the co-ordinate planes, find the locus of the centroid of the tetrahedron. y z 16. If 2d be the shortest distance between the lines __ + __ b c x __z 1 1 __ __ = 1, x = 0 and __ a – c = 1, y = c, prove that a2 + b2 1 1 + __ 2 = __ 2 . c d
3.
17. If the foot of the perpendicular from the origin to a plane is P (a, b, c), find the equation of the plane. 18. Find the equation of a plane which passes through y – 2 x – 1 the point of intersection of lines _____ = _____ = 3 1 y – 1 z–3 z–2 x–3 ____ and _____ = _____ = ____ and at the greatest 2 1 2 3 distance from the point (0, 0, 0).
4.
5.
6.
1. Show that the straight lines whose direction cosines are given by al + b m + c n = 0 and f m n + g n l + h l m = 0 are f g h (i) perpendicular if __ a + __ + __ = 0 b c __
2.
___
___
(ii) parallel if ÷ af + ÷ bg + ÷ ch = 0. If the edges of a rectangular parallelopiped are a, b and c, show that the angles between the four a2 b2 __ c2 diagonals are given by cos–1 __2 + __ 2 + 2 . a b c A variable plane passes through the fixed point (a, b, c) and meets the axes of reference in P, Q, R. Show that the locus of the point of intersection of the plane through P, Q, R parallel to the co-ordinate a b __c planes is __ x + __ y + z = 1. A variable plane makes intercepts on the co-ordinate axes, the sum of whose squares is constant and equal to k2. Find the locus of the foot of the perpendicular from the origin to the plane. Find the incentre of the tetrahedron formed by the planes x = 0, y = 0, z = 0 and x + y + z = a. If the image of the point (3, 5, 7) in the plane 2x + y + z = 16 is (a, b, c), find the value of a2 + b2 + c2 + 18.
(
)
3D-Co-ordinate Geometry
x – 1 7. If q be the acute angle between the lines _____ = a y_____ + 1 __z y – 3 –1 x + 1 _____ z____ = c and _____ = = c ,where a > b > c a b b and a, b and c are the zeroes of x3 + x2 – 4x – 4 = 0, find q. 8. If the planes x = cy + bz, y = az + cx and z = bx + ay meet in a line, show that the line of intersection y z x of these planes is _______ __ 2 = _______ __ 2 = ______ __ 2 1 – a 1 – b 1 – c ÷ ÷ ÷ 9. Let the plane x – y – z = 2 is rotated through 90° about its line, of intersection with the plane x + 2y + z = 2. Find the equation of the plane in the new position. y x __ __z 10. A point P moves on a plane __ a + b + c = 1. A plane through P and perpendicular to OP meets the co-ordinate axes at A, B and C. If the planes through A, B and C parallel to the planes x = 0, y = 0 and z = 0 intersect at Q, find the locus of Q.
Integer Type Questions
1. If the angle between the lines whose direction cosines are given by 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0 a –1 __ is cos , find the value of (a + b). b 2. A line makes angles a, b, g, d with the four diagonals of a cube, find the value of 3(cos2a + cos2b + cos2g + cos2d).
( )
3. If the image of the point P (1, 6, 3) in the –2 x y – 1 z____ line __ = _____ = is Q (a, b, g), find the vaue 1 2 3 of (a + b + g). 4. If P(a, b, c) be any point on the plane 3x + 2y + z = 7, find the least value of 2(a2 + b2 + c2). 5. Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the + 2 6____ –z x – 2 y_____ line _____ = = . 2 3 6 6. Find the shortest distance between the lines x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4 and the z-axis. 7. Find the number of planes that are equidistant from four non-coplanar points. 8. If the distance between the plane x – 2y + z = d – 2 x – 1 y_____ and the plane containing the lines _____ = = 2 3 __ y – 3 z – 3 z – 4 x – 2 ____ and _____ = _____ = ____ is ÷6 , find the 4 3 4 5 value of |d|.
6.17
9. If the angle between two faces of a regular tetrahedron be q, find the value of (3cosq + 2). 10. Find the number of lines, which are equally inclined to the axes. – 3 z____ –4 x – 2 y_____ 11. If the lines _____ = = and 1 1 – k – 4 z____ – 5 x – 1 y_____ _____ = = are coplanar, find the value 2 1 k of (k + 5). 9 __ – z y_____ – 3 x_____ – 2 2 12. If the straight lines = = _____ and 1 2 k y – 4 z x–1 _____ = _____ = __ intersect, find the value of k. 2 1 k 13. If the planes x = cy + bz, y = cx + az and z = bx + ay pass through a straight line, find the value of a2 + b2 + c + 2abc + 2. –2 z–k x – 4 y_____ 14. If the line _____ = = ____ lies in the plane 1 1 2 2x – 4y + z = 7, find the value of k. 15. If the plane ax – by + cz = 0 contains the – 2d z____ –c x – a y______ b line _____ = = c , find the value of __ . a b d
( )
Comprehensive Link Passages Passage I A line makes angles a, b, g with the co-ordinate axes, then cos2a + cos2b + cos2g = 1. 1. If a + b = 90°, then g is (a) 0 (b) 60° (c) 90° (d) 120° 2. If a – b = 90°, then g is (a) 30° (b) 60° (c) 120° (d) 90° 3. The direction cosines of a line equally inclined to the axes are 1 1 1 1 1 1 (a) __ , __ , __ (b) – __ , – __ , – __ , 3 3 3 3 3 3 1__ ___ 1 1__ 1__ ___ 1 1__ (c) ___ , __ , ___ (d) – ___ , – __ , ___ 3 ÷ 3 ÷ 3 3 ÷ 3 ÷ 3 ÷ ÷ Passage II A line makes angles a, b, g, d with the four diagonals of a unit cube. 1. cos2a + cos2b + cos2g + cos2d is (a) 2/3 (b) 4/3 (c) 8/3 (d) 10/3 2 2 2 2. sin a + sin b + sin g + cos2d is (a) 1/3 (b) 4/3 (c) 8/3 (d) 15/3
6.18 Integral Calculus, 3D Geometry & Vector Booster
3. cos (2a) + cos (2b) + cos (2g) + cos (2d) is (a) 1/3 (b) 4/3 (c) – 8/3 (d) – 4/3
Passage III
x1 1 x V = __ 2 6 x3 x4
y1 y2 y3 y4
z1 z2 z3 z4
1__ (a) ___ 3 ÷
1__ (b) ___ 2 ÷
2 (c) __ 3
1__ (d) ___ ÷6
÷
1. The volume of a tetrahedron with vertices (0, 0, 0), (2, 0, 0), (0, 3, 0) and (0, 0, 4) is (a) 1 (b) 4 (c) 6 (d) 2 2. The volume of the terahedron with vertices (1, 2, 3), (2, 0, 0), (0, 4, 0) and (0, 0, 3) is (a) 2 (b) 6 (c) 4 (d) 8 3. The volume of tetrahedron formed by the planes x + y = 0, y + z = 0, z + x = 0 and x + y + z = 1 is (a) 2/3 (b) 4/3 (c) 8/3 (d) 10/3 4. The volume of the tetrahedron formed by the y y y x x __ __z __ __ __ __z planes __ a + b + c = 1, a + b = 0, b + c = 0 x z and __ a + __ c = 0 is 1 (a) __ abc 3 2 __ (c) abc 3
4 (b) __ abc 3 10 ___ (d) abc 3
( )
__
( )
are perpendicular or, if a1a2 + b1b2 + c1c2 = 0 and parallel if a1 __ b1 __ c1 __ a2 = b2 = c2
1. The number of lines that are equally inclined with the axes is (a) 8 (b) 6 (c) 4 (d) 3 2. If the lines L1 : x = ay + b, z = cy + d and L2 : x = x¢y + b¢, z = c¢y + d ¢ are perpendicular, then (a) aa¢ + cc¢ = 0 (b) aa¢ + cc¢ = –1 (c) aa¢ – cc¢ = 0 (d) aa¢ – cc¢ = 1 – 14 z____ –3 1 – x 7y 3. Let L1: _____ = _______ = 3 2 25
If L1 meets L2 at right angles, the value of (11p – 64) is (a) 5 (b) 6 (c) 7 (d) 8. Passage VI x – x1 y_____ – y1 z_____ – z1 Two lines L1: _____ = = c a1 b1 1 x – x2 y_____ – y2 z_____ – z2 and L2: _____ = = c are coplanar if a2 b2 2
( ) __
(c) tan (2÷2 ) (d) cos ____ 2÷2 3 2. The angle between one diagonal of a unit cube and a diagonal of a face is __ (a) cot–1(÷2 ) (b) tan–1 ___ 1__ 3 ÷
(c) cos–1 __ 1 3
x – x2 y_____ – y2 z_____ – z2 and L2: _____ = y = c a2 2 2
– 5 6____ –z 7 – 7x y_____ and L2: ______ = = . 3p 1 5
1. The angle between two diagonals of a unit cube is 1 (a) tan–1(3) (b) tan–1 __ 3 –1
x – x1 y_____ – y1 z_____ – z1 Let L1: _____ = = c a1 b1 1
a1a2 + b1b2 + c1c2 __________ __________ . Then cos(q) = ______________________ 2 ÷a 1 + b21 + c 21 ÷a 22 + b 22 + c22
Passage V
1 1 1 1
Passage IV _› Let r 1 = a1 + b1 + c1 _› and r 2 = a2 + b2 + c2 .
3. If q be the angle between the vector a = 2i – j + k and the z-axis, then sinq is
__
The volume of the tetrahedron ABCD with vertices A(x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) is
–1
( ) ( ÷ )
__
2 (d) sin–1 __ 3
x2 – x1 a1 a2
y2 – y1 b1 b2
z2 – z1 c1 = 0 c2
y – 2 ____ y–1 z–3 x – 1 _____ x – 4 _____ 1. If two lines _____ = = and _____ = = z 2 3 4 2 5 are coplanar, the point of intersection is (a) (1, 1, 1) (c) (–1, –1, –1)
(b) (1, –1, –1) (d) (1, 1, –1)
6.19
3D-Co-ordinate Geometry
y + 1 ____ y – k __z z–1 x – 3 ____ x – 1 _____ 2. If the lines _____ = = and _____ = = 2 3 4 1 2 1 are coplanar, the value of k is (a) 3/2 (b) 9/2 (c) – 2/9 (d) – 3/2 y – 3 4 – z x – 2 x – 1 3. The lines _____ = _____ = ____ and _____ = 1 1 k k y_____ –4 = z – 5 are coplanar, if 2 (a) k = 0 (b) k = 1 (c) k = 2 (d) k = – 3
Passage VII – 2 z____ –3 x – 1 y_____ Let L1: _____ = = 2 3 4 – 4 z____ –5 x – 2 y_____ and L2: _____ = = 3 4 5
Let A = (2, 3, 1), P : 2x + y + z = 6 – 2 z____ –3 x – 1 y_____ and L: _____ = = 2 1 4 1. The foot of perpendicular from the point A to the line L is 37 43 11 43 ___ 11 37 ___ (a) ___ , ___ , (b) ___ , ___ , 21 21 21 21 21 21
( ) 43 11 37 ___ (c) ( – ___ , ___ , – 21 21 21 )
( ) 11 43 37 (d) ( ___ , – ___ , ___ 21 21 21 )
2. The image of the point A (2, 3, 1) w.r.t. the plane 2x + y + z is 2 7 __ 1 2 7 __ 1 (a) __ , __ , (b) – __ , __ , – 3 3 3 3 3 3 2 1 7 2 7 1 (c) __ , – __ , – __ (d) – __ , – __ , __ 3 3 3 3 3 3
( (
Matrix Match (For Jee-Advanced Examination Only)
Column I
)
)
( (
) )
Column II
(A)
S cos a
(P)
(B)
S sin2a
(Q) –1
(C)
S cos(2a)
(R) 2
(D)
S (cos2a + sin2a)
(S)
2
3
1
2. Match the following Columns: A line makes angles a, b, g, d with the four diagonals of a cube. Column I
( ) 7 4 (d) ( __ , __ , 11 ) 5 5
Passage VIII
3. The image of the line L with respect to the plane P is – 5 3z –8 3x – 1 3y (a) ______ = ______ = _____ 4 2 3 – 4 3z –5 3x – 1 3y (b) ______ = ______ = _____ 4 2 3 – 5 8_____ – 3z 3x – 1 3y (c) ______ = ______ = 4 2 1 – 3y 3z –8 3x – 1 5______ (d) ______ = = _____ 4 3 5
1. Match the following Columns: A line makes angles a, b, g with the co-ordinate axes.
and A = (2, 3, 5). 1. The shortest distance between the lines L1 and L2 is 1__ 1__ (a) ___ (b) ___ 3 ÷2 ÷ 1__ 1 (c) ___ (d) __ 2 ÷6 2. The equation of the plane passing through A and containing the line L1 is (a) 2x – z + 1 = 0 (b) 2x – y – z + 2 = 0 (c) 2x – y + 3z + 2 = 0 (d) 3x – y + 5z + 10 = 0 3. The point A = (2, 3, 5). is shifted to the line L2 by a distance 1. Then the co-ordinates of the new position of A is 7 11 __ 7 4 11 4 (a) __ , ___ , (b) __ , __ , ___ 5 5 5 5 5 5
( ) 7 11 (c) ( __ , ___ , 4 ) 5 5
Column II
(A)
S cos a
(P) 4
(B)
S sin2a
(Q) 4/3
(C)
S cos (2a)
(R) 8/3
(D)
S (cos2a + sin2a)
(S) – 4/3
2
3. Match the following columns: Column I
Column II
(A) The angle between any two (P) diagonals of a cube is
( ÷ ) __
2 cos–1 __ 3
( )
(B)
The angle between one (Q) sin–1 __ 1 diagonal of a unit cube and 3 a diagonal of a face is
(C)
The angle between a (R) diagonal of a cube and a diagonal of a face intersecting it
cot–1 ___ 1__ 3 ÷
(D) The angle between the (S) diagonal of the faces of the cube through the same (T) vertex is
cot–1( ÷2 ) 1 –1 __ cos 3
( ) __
( )
6.20 Integral Calculus, 3D Geometry & Vector Booster
4. Match the following Columns: Column I (A) The direction cosines of two lines are connected with l + m + n = 0 and l2 + m2 + n2 = 0, the angle between them is (B) The angle between the planes 3x – 4y + 5z = 0 and 2x – y – 2z = 5 is (C) The angle between the planes 2x – y + z = 6 and x + y + 2z = 7 is (D) The angle between the –2 +3 x – 1 y_____ line _____ = = _____ and 2 1 – 2 the plane x + y + 4 = 0 is
Column II (P) 60°
point (2, 6, 3) to the line –2 x y – 1 z____ __ = _____ = is 2 2 3 (C) The point of intersection of (R) (1, 3, 5) the lines –1 z–6 x – 2 y_____ _____ = = ____ 1 – 2 1 + 3 z_____ +3 x + 3 y_____ and _____ = = is 7 1 – 6
(Q) 45°
(R) 90°
(S)
(D) If the lines (S) (– 9, – 14, – 19) y + 2 z+3 x+1 _____ = _____ = _____ 2 3 4 – 2 z____ –1 x – 3 y_____ _____ and = = 3 4 5 are
120°
5. Match the following Columns: Column I y – 2 (A) x – 4 = _____ = If the line _____ 1 1 z____ –k lies on the plane 2 2x – 4y + z = 7, then k is y – 3 x – 2 (B) = _____ = The lines _____ 1 1 y – 4 4 – z x – 1 ____ and _____ = _____ = 2 k k z–5 ____ are coplanar, then k is 1 (C) If the lines + 1 z____ –1 x – 1 y_____ _____ = = 2 3 4 – k __z x – 3 y____ and _____ = = intersect, 1 2 1 then k is (D) The shortest distance between the lines – 3 z____ –6 x – 2 y_____ _____ = = 3 4 5 – 2 z____ –1 x – 5 y_____ and _____ = = is 1 2 3
(B) The co-ordinates of the foot (Q) (5, 10, of the perpendicular from the – 6)
Column II (P) – 3
(Q) 7
Column I
(R) 0 (S) 9/2 (T) 5/2
6. Match the following Columns: Column I
7. Match the following Columns: Let P (0, 3, – 2), Q (3, 7, –1) and R (1, – 3, –1) be three given points. Let L1 be the line passing through P and Q and L be the line through R and parallel to _› 2 the vector v = + .
Column II
(A) The co-ordinates of the point (P) (2, 3, 5) of intersection of the line + 2 z____ –2 x – 1 y_____ _____ = = 1 3 2 and the plane 3x + 4y + 5z = 15 is
Column II __
(A)
The perpendicular distance (P) of P from L2 is
7÷3
(B)
The shortest distance (Q) between L1 and L2 is
2
(C)
Area of the DPQR is
6
(D)
The distance from (0, 0, 0) (S) to the plane PQR is
(R)
19 _____ ____ 147 ÷
8. Match the following Columns: Column I
Column II
(A) The shortest distance from the (P) point (3, 4, 5) to the x-axis is
5
(B) The shortest distance from the (Q) point (3, 2, 4) to y-axis is
÷41
(C) If the plane 2x – 3y + 6z = 11 (R) makes an angle sin–1(k) with the x-axis, then 7k is
9
(D) A straight line (S) _ › r = (1 + t) + 5t + (1 – t) where t Œ R. If this line lies in (T) the plane x + y + cz = d, the value of (c + d) is
2
___
4
3D-Co-ordinate Geometry
9. Match the following Columns: Column I
Column II
(A) The volume of the tetrahedron, (P) whose vertices are (0, 0, 0), (2, 0, 0), (0, 3, 0) and (0, 0, 4) is
6
(B) The volume of the tetrahedron (Q) whose vertices are (0, 1, 2), (3, 0, 1), (4, 3, 6) and (2, 3, 2) is
4
(C) The volume of the tetrahedron (R) formed by the planes whose equations are x + y = 0, y + z = 0, z + x = 0 and x + y + z = 1 is
20
(D) The volume of the tetrahedron (S) formed by the planes whose equations are y z x y __ (T) + __ = 0, __ + __ = 0, 2 3 3 4 y z z x x __ + __ = 0 and __ + __ + __ = 1 4 2 2 3 4 is
16
2/3
Column II 2
2
2
(A)
If 2x + 3y + 4z + 2kxy (P) + 4yz = 0 represents two planes, the value of k2 is
2
(B)
If x2 + 5y2 + 5z2 + 2kxy (Q) + 4xz = 0 represents two planes, the value of 5k 2 is
5
(C)
If x2 + y2 + 4z2 – 2kyz = 0 (R) represents two planes, the value of (k2 + 2) is
10
(D)
If 9x2 + y2 + z2 – 6kxy = (S) 0 represents two planes, the (T) value of (k2 + 7) is
8 1
Questions asked in Previous Years’ JEE-Advanced Examinations
10. Match the following Columns: Column I
–2 x – 4 y_____ 1. The value of k, for which the line _____ = = 1 1 z____ –k lies in the plane 2x – 4y + z = 7, is 2 (a) 7 (b) 6 (c) no real value (d) – 7 [IIT-JEE, 2003] 2. Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1). [IIT-JEE, 2003]
6.21
3. If P be the point (2, 1, 6), find a point Q such that PQ is perpendicular to the plane in x + y – 2z = 3 and the mid-point of PQ lies on it. [IIT-JEE, 2003] y + 1 ____ y – k __z z–1 x – 3 ____ x – 1 _____ 4. If the lines _____ = = and _____ = = 2 3 4 1 2 1 intersect at a point, the value of k is (a) 3/2 (b) 9/2 (c) 2/9 (d) – 3/2 [IIT-JEE, 2004] 5. A parallelopiped T has one of its face as ABCD. The face opposite to ABCD is A¢B ¢C ¢D ¢. The parallelopiped T is compressed to another parallelopiped S with the face ABCD remaining the same but A¢B ¢C ¢D ¢ changes to A≤B≤C≤D≤. If the volume of S is 90% of T, show that the locus of A≤ is a plane. [IIT-JEE, 2004] 6. p1 and p2 are two planes passing through the origin. L1 and L2 are two lines passing through the origin such that L1 lies on p1 not on p2 and L2 lies on p2 not on p1. Show that there exist three points A, B, C whose permutation A¢, B¢, C ¢ can be choosen such that (a) A is on L1, B on p1 but not on L1 and C not on p1. (b) A¢ is on L2, B¢ on p2 but not on L2 and C ¢ not on p2. [IIT-JEE, 2004] 7. A plane p passes through the point (1, 1, 1) and is parallel to the vectors b = (1, 0, –1) and c = (–1, 1, 0). If p meets the axes in A, B and C, find the volume of the tetrahedron OABC. [IIT-JEE, 2004]
8. Find the equation of the planes passing through the lines 2x – y + z = 3, 3x + y + z = 5 and which at a 1 distance of ___ __ from the point (2, 1, –1). ÷6 [IIT-JEE, 2005]
9. A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If
the centroid G (x, y, z) of DABC satisfies the rela1 1 1 tion __ 2 + __ 2 + __ 2 = k, the value of k is x y z (a) 9 (b) 3 (c) 1 (d) 1/3 [IIT-JEE, 2005] 10. A plane p is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4 and passes through the point (1, – 2, 1). The distance of p from the point (1, 2, 2) is (a) 0 __ (b) 1 __ (c) ÷2 (d) 2÷2 [IIT-JEE, 2006]
6.22 Integral Calculus, 3D Geometry & Vector Booster 11. A point (a, b, g) lies on the plane x + y + z = 2 Let a = a i + b j + g k and k × (k × a) = 0, then g equals .......... [IIT-JEE, 2006] 12. A line L is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance L from the origin equals .......... [IIT-JEE, 2006] 13. A plane passes through (1, 2, 3) and is perpendicular to two planes x = 0 and y = 0. The distance of the plane from the point (0, –1, 0) equals .......... [IIT-JEE, 2006] 14. Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5. Statement I: The parametric equations of the line of intersection of given planes are x = 3 14t, y = 1 + 2t, and z = 15t. Statement II: The vector 14i + 2j + 15k is parallel to the given planes. [IIT-JEE, 2007] 15. Consider the following linear equations: ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0. Match the conditions/expressions in Column I with statements in Column II. Column I
Column II
(A) a + b + c π 0 (p) the equations represent planes meeting at a and a2 + b2 + c2 single point. = ab + bc + ca, (B) a + b + c = 0 (q) the equations represent the line x = y = z and a2 + b2 + c2 π ab + bc + ca, (C) a + b + c π 0 (r) the equations represent the identical planes and a2 + b2 + c2 π ab + bc + ca, (D) a + b + c = 0 (s) the equations represent the whole of the threeand a2 + b2 + c2 dimensional space. = ab + bc + ca, [IIT-JEE, 2007] 16.
Consider the planes P1 : x – y + z = 1, P2 : x + y – z = –1, P3 : x – 3y + 3z = 2. Let L1, L2 and L3 be the lines of intersection of the planes P2 and P3, P3 and P1 and P1 and P2, respectively. Statement I: At least two of the lines L1, L2 and L3 are non-parallel.
Statement II: The three planes do not have a common point. [IIT-JEE, 2008] 17. Consider the lines y + 2 z_____ +1 x + 1 _____ L1: _____ = = 3 1 2 + 2 _____ z–3 x – 2 y_____ L2: _____ = = 1 2 3 (i) The unit vector perpendicular to both L1 and L2 is i + 7j + 7k – i + 7j + 5k ___ __ (a) ___________ (b) ___________ ÷99 5÷3 – i + 7j + 3k – 7i + 7j + k __ ___ (c) ___________ (d) ___________ 5÷3 ÷99 (ii) The shortest distance between L1 and L2 is 17 __ (a) 0 (b) ___ 3 ÷ 17__ 41__ ____ ____ (c) (d) 5÷3 5÷3
(iii) The distance of the point (1, 1, 1) from the plane passing through the point (–1, – 2, –1) and whose normal is perpendicular to both the lines L1 and L2 is 7 2 ___ ___ (a) ____ (b) ____ ÷75 ÷75 13 23 ___ ___ (c) ____ (d) ____ ÷75 ÷75 [IIT-JEE, 2008] 18. A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the co-ordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals __ (a) 1 __ (b) ÷ 2 (c) ÷3 (d) 2 [IIT-JEE, 2009] 19. Let P (3, 2, 6) be a point in space and Q be a point on the line r = (i – j + 2k) + l (– 3i + j + 5k). Then the value of l for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is (a) 1/4 (b) –1/4 (c) 1/8 (d) –1/8 [IIT-JEE, 2009] 20. The equation of the plane containing the y z x straight line __ = __ = __ and perpendicular to 2 3 4 y z x the plane containing the straight lines __ = __ = __ 3 4 2 x y __z and __ = __ = is 4 2 3 (a) x + 2y – 2z = 0 (b) 3x + 2y – 2z = 0 (c) x – 2y + z = 0 (d) 5x + 2y – 4z = 0 [IIT-JEE, 2010]
3D-Co-ordinate Geometry
21. If the distance of the point P (1, – 2, 1) from the plane x + 2y – 2z = a, where a > 0, is 5, the foot of the perpendicular from P to the plane is
( ) 10 1 2 ___ (c) ( __ , __ , 3 3 3)
( ) 2 1 5 (d) ( __ , – __ , __ 3 3 2)
8 4 __ 7 (a) __ , __ , – 3 3 3
4 4 1 (b) __ , – __ , __ 3 3 3
[IIT-JEE, 2010] 22. If the distance between the plane x – 2y + z = d – 2 x – 1 y_____ and the plane containing the lines _____ = = 2 3 __ y – 3 z – 3 z – 4 x – 2 ____ and _____ = _____ = ____ is ÷ 6 , then |d| 4 3 4 1 is... [IIT-JEE, 2010] 24. Let a, b, c be three real numbers satisfying
[ ]
1 9 7 [a b c] 8 2 7 = [0 0 0] ... (E). 737 If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, the value of 7a + b + c is (a) 0 (b) 12 (c) 7 (d) 6. [IIT-JEE, 2011] 25. The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R(1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, the length of the line segment PS is __ 1__ (a) ___ (b) ÷ 2 2 ÷ __ (c) 2 (d) 2÷2 [IIT-JEE, 2012] 26. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and 2__ x – y + z = 3 and at a distance ___ from the point 3 ÷ (3, 1, –1) is
(a) 5x – 11y + z = 17
(c) x + y + z = ÷3
__
___ __ (b) ÷ 2x + y = 3÷2 – __ __ (d) x – ÷y = 1 – ÷ 2
1
[IIT-JEE, 2012] + 1 __z x – 1 y_____ x+1 27. If the straight lines _____ = = and _____ = 2 2 k 5 y + 1 z _____ = __ are coplanar, the planes containing these 2 k two lines is (are) (a) y + 2z = –1 (b) y + z = –1 (c) y – z = –1 (d) y – 2z = –1 [IIT-JEE, 2012]
6.23
28. The perpendiculars are drawn from points on the + 1 __z x + 2 y_____ line _____ = = to the plane x + y + z = 3. 2 –1 3 The feet of the perpendiculars lie on the line –2 –2 x y – 1 z____ x y – 1 z____ (a) __ = _____ = (b) __ = _____ = 8 –13 2 3 5 – 5 –2 –2 x y – 1 z____ x y – 1 z____ (c) __ = _____ = (d) __ = _____ = – 7 – 7 4 3 2 5 [IIT-JEE, 2013] 29. A line L passing through the origin is perpendicular to the lines L1 : (3 + t)i + (–1 + 2t) j + (4 + 2t) k and L2 : (3 + 2s) i + (3 + 2s) j + (2 + s) k, where – • < t, s 0 Hence, the obtuse angle bisector 3x – 2y + 6z + 8 2x – y + 2z + 3 _________ ________ _______________ = ______________ 4 + 1 + 4 ÷ ÷9 + 4 + 36 3x – 2y + 6z + 8 ______________ 2x – y + 2z + 3 fi _______________ = 7 3 fi
5x – y – 4z = 3
30. The equation of any plane passing through the intersection of the given planes
(x – y – z – 4) + l (x + y + 2z – 4) = 0
and
(l + 1)x + (l – 1) y + (2l – 1) x = 4l + 4
and
4a – 3b + 4c = 0
Solving, we get a b __ c fi __ = __ = 5 8 1 Since the line is parallel to the plane, so
10 – 8 + m = 0
fi m = – 2 32. Given striaght line can written as – 0 z____ –1 x – 1 y_____ _____ = = 1 3 –1 Since the lies in the plane, so 1 + 3 – c = 0 and 1 + 0 + c = d fi c = 4 and d = 5 Hence, the value of (c + d – 5) = 4 33. Let the equation of the plane be
a(x – 2) + b(y – 1) + c(z + 1) = 0
Since the plane contains the line L1 and parallel to L2, so a+b+c=0 and
a + 0 + 2c = 0
a b c fi __ = ___ = ___ 2 –1 –1 Hence, the equation of the plane is
2(x – 2) – (y – 1) + (z + 1) = 0
fi
2x – 4 – y + 1 + z + 1 = 0
fi
2x – y + z = 2
Hence, the distance from the origin
÷
__
2 2 2 = _________ ________ = ___ __ = __ 3 4 + 1 + 1 ÷6 ÷
34. Since the three planes pass through the same line, so 1 – c – b c a = 0 –1 b a –1 fi
|
2
|
(1 – a ) + c(– c – ab) – b(ac + b) = 0
which is perpendicular with x – y – z = 4. So,
fi
(1 – a2) – c2 – abc – abc – b2 = 0
fi
a2 + b2 + c2 + 2abc = 1
fi
a2 + b2 + c2 + 2abc + 2 = 1 + 2 = 3
fi
(l + 1) – (l – 1) – (2l – 1) = 0 3 l = __ 2
35. The equation of any plane through (0, 2, 4) is a(x – 0) + b(y – 2) + c(z – 4) = 0 which containing the line – 1 2____ –z x + 3 y_____ _____ = = 3 4 2 So,
3a + 4b – 2c = 0
and
– 3a – b – 2c = 0
Solving, we get a b c ____ = ___ = __ –10 12 9 Hence, the equation of the plane is
–10x + 12(y – 2) + 9(z – 4) = 0
fi
–10x + 12y – 24 + 9z – 36 = 0
fi
10x – 12y – 9z + 60 = 0
36. Ans. (d) x–2 37. Since the line _____ = 3 x + 3y – az + b = 0,
1 – y z_____ +2 _____ = lies in the plane 1 2 so
|
3D-Co-ordinate Geometry
6.41
|
– y x 0 x 0 – z = 0 x – 1 –1 –1 fi
x(0 – z) + y(– x + xz – z) = 0
fi
– xz – xy + xyz – yz = 0
fi
xy + yz + zx = xyz
1 __ 1 __ 1 fi __ x + y + z = 1 40. The equation of the line joining the points (2, 3, 4) and (6, 7, 8) is y–3 z–4 x–2 _____ = _____ = _____ 7–3 8–4 6–2 –3 x – 2 y_____ fi _____ = = 4 4 –3 x – 2 y_____ fi _____ = = 1 1
z–4 ____ 4 z–4 ____ 1
3 – 3 – 2a = 0
and
2 + 3 + 2a + b = 0
The mid-point of the line segment joining the points (2, 3, 4) and (6, 7, 8) is (4, 5, 6). Hence, the equation of the right bisector of the plane 1(x – 4) + 1(y – 5) + 1(z – 6) = 0
So,
a = 0, b = – 5
fi
Comprehensive Link Passages
Hence, the value of
(a + b + 7) = 2
x + y + z = 15
Clearly, l = 3m + 1, 2l = 2 – m
Passage I 1. Given,
cos2a + cos2b + cos2g = 1
Solving, we get
fi
cos2a + cos2 (90° – a) + cos2g = 1
fi
cos2a + sin2a + cos2g = 1
Hence, the point of intersection is (1, 2, 3).
fi
1 + cos2g = 1
–2 1 + k 2_____ – 1 3_____ Therefore, _____ = = 3 2 h 1 + k __ 1 1 fi _____ = = __ 3 2 h 3 1 fi h = 2, k = __ – 1 = __ 2 2
fi
cos2g = 0
fi
g = 90°
fi
cos2a + cos2(90° + a) + cos2g = 1
39. Given four points are
fi
cos2a + sin2a + cos2g = 1
P(2 – x, 2, 2), Q(2, 2 – y, 2)
fi
1 + cos2g = 1
R(2, 2, 2, – z) and S(1, 1, 1)
fi
g = 90°
Now,
PQ = (x, – y, 0)
PR = (x, 0, – z)
cos2a + cos2a + cos2a = 1
and
PS = (x – 1, –1, –1)
fi fi
3 cos2a = 1
fi
1 cos2a = __ 3
38. Let the point of intersection is (l, 2l, 3l).
l = 1, m = 0
It is given that, the vectors P, Q, R and S are coplanar, so the vectors PQ, PR, PS are coplanar.
2. cos2a + cos2b + cos2g = 1
3. cos2a + cos2b + cos2g = 1
6.42 Integral Calculus, 3D Geometry & Vector Booster 1 cos a = ___ __ 3 ÷
fi
(
)
1 1__ ___ 1 Hence, the direction cosines are ___ __ , ___ , __ 3 ÷3 ÷3 ÷ Passage II 2
0 0 0 1 1 1 –1 = __ 6 1 –1 1 –1 1 1
1 1 –1 1 __ = – 1 – 1 1 6 –1 1 1
= 2(cos2a + cos2b + cos2g + cos2d) – 4
4 = 2 ◊ __ 3 8 __ = – 4 3 4 = – __ 3
0 0 –1 1 2 0 1 __ = – 6 0 2 1
4 2 = __ = __ cu. units. 6 3
2
2
2
1. cos a + cos b + cos g + cos d = 4/3
2. sin2a + sin2b + sin2g + sin2d
= 4 – (cos2a + cos2b + cos2g + cos2d)
4 = 4 – __ 3 8 __ = 3 3. cos(2a) + cos(2b) + cos(2g) + cos(2d)
Passage III
0 12 = 60 0
0 0 3 0
0 0 0 4
1 1 1 1
2 0 0 1 0 3 0 – = 6 0 0 4
|
|
1 = – __ × 24 = |– 4| = 4 cu units. 6
(
) (
0 1 a 1 = __ 1 6 a 1 – a
0 1 b 1 – b 1 b
2. The volume of a tetrahedron
1 1 2 = __ 6 0 0
2 0 4 0
3 0 0 3
1 1 1 1
Ï0 0 1 2 1 Ô __ = Ì 4 0 1 – 2 4 6 Ô 0 Ó0 3 1 1 = __ [12 – 2{2(0 – 3) 6 1 = __ [12 + 12] 6 =4
1 1 1 1
Ê C1 Æ C1 + C3 ˆ ÁË C Æ C + C ˜¯ 2 2 3
4. On solving, we get the vertices of a tetrahedron as 1 1 __ 1 1 1 1 1 __ 1 __ 1 (0, 0, 0), __ a , __ , – c , __ a , – __ , __ and – __ a , b , c b b c respectively. Volume of a tetrahedron
1. Volume of a tetrahedron
3. On solving, we get the vertices of a tetrahedron as (0, 0, 0, 0), (1, 1, –1), (1, –1, 1), (–1, 1, 1). Volume of a tetrahedron
3 1¸ Ô 0 1˝ 4 1 Ô˛ – 4(0 – 0)}]
1 a 1 a 1 – a
1 b 1 – b 1 b
)
0 1 – c 1 c 1 c
1 1 1 1
1 c 1 c 1 c
–
1 = – __ 6
1 1 –1 1 __ = – abc 1 – 1 1 6 –1 1 1
4 = __ abc 6
2 = __ abc 3
(
)
3D-Co-ordinate Geometry
÷
_____
________
Passage IV
6.43
2 ___ 1 sin q = ÷ 1 – cos2q = 1 – __ = __ 3 ÷3 Passage V
1. Ans. (a)
x–b 2. L1: _____ = a
x – b¢ L2: _____ = a¢
y z–d __ = ____ c 1 y z – d¢ __ = _____ 1 c¢
Since L1 and L2 are perpendicular, so sum of the product of their sirection ratios is zero.
Thus,
a ◊ a¢ + 1 ◊ 1 + c ◊ c¢ = 0
fi
aa¢ + cc¢ = –1.
– 14 z____ –3 1 – x 7y L1: _____ = _______ = 3 2 25
1. OP = i + j + k
BN = i + j – k OP ◊ OL = 1 + 1 __
3. Given
x–1 _____ = – 3
__
÷ 3 ◊ ÷3 cos q = 1 + 1
y–2 z–3 _____ = ____ 2 25/7
– 5 6____ –z 7 – 7x y_____ L2: ______ = = . 3p 1 5
1 cos q = __ 3
and
1 q = cos–1 __ 3
y – 5 z____ –6 x–1 _____ = _____ = – 3p/7 1 – 5
Hence, the angle between any two diagonals is __ 1 q = cos–1 __ = tan–1 (2÷2 ) 3
Since L1 meets L2 at right angles, so
( ) ( )
2. OP = i + j + k
( )
3p 25 (– 3) – ___ + ___ – 10 = 0 7 7
9p 25 fi ___ = 10 – ___ 7 7
OL = i + j OP ◊ OL = 1 + 1 ÷ 3 ◊ ÷2 cos q = 1 + 1
9p fi ___ = 7
÷ 3 ◊ ÷2 cos q = 2
fi
÷ 3 cos q = ÷ 2
Passage VI
2 cos q = __ 3
__
__
__
__
__
__
÷
__
__
Hence, the angle between a diagonal and a diagonal of a face is
( ÷ ) __
__ 2 q = cos–1 __ = cot–1 (÷2 ) 3
3. a ◊ k = 2
|a| |i| cos q = 2 __
÷ 6 cos q = 2
÷
– 2 z____ –3 x – 1 y_____ _____ = = = l 2 3 4
Let two lines meet at P. So,
2l + 1 = 5m + 4, 3l + 2 = 2l + 1
and
4l + 3 = m
fi
3l – 3m = –1, 4l – m = – 3
l = –1 = m
Hence, the point of intersection is (– 1, – 1, – 1). 2. Since both the lines are coplaner, so
__
2__ 2 cos q = ___ = __ 3 ÷6
p=5
–1 x – 4 y_____ and _____ = = z = m 2 5
( ÷ )
2 q = cos–1 __ 3
1. Let
45 ___ 7
3–1 k +1 0 –1 2 3 4 =0 1 2 1
6.44 Integral Calculus, 3D Geometry & Vector Booster
fi
2 k + 1 –1 2 3 4 =0 1 2 1
x–2 1 2
y–3 z–5 1 2 =0 3 4
fi
2(3 – 8) – (k + 1)(2 – 4) – (4 – 3) = 0
fi
– 2(x – 2) + (z – 5) = 0
fi
–10 + 2(k + 1) – 1 = 0
fi
– 2x + 4 + z – 5 = 0
fi
2(k + 1) = 11
fi
– 2x + z – 1 = 0
fi
2x – z + 1 = 0
9 11 k = ___ – 1 = __ 2 2 3. Since the lines are coplaner, so
fi
3. Ans. (c)
Passage VIII Let
A = (2, 3, 1), P : 2x + y + z = 6
1– 2 4 – 3 5– 4 1 1 –k = 0 k 2 1
and
– 2 z____ –3 x – 1 y_____ L : _____ = = 2 1 4
fi
–1 1 1 1 1 –k = 0 k 2 1
1. Any point on the line is
M(2l + 1, l + 2, 4l + 3)
The direction ratios of AM are (2l – 1, l – 1, 4l + 2)
fi
(2 – k) – (1 + k2) – (1 + 2k) = 0
fi
k2 + 3k = 0
Clearly, AM is perpendicular to the line L. So,
fi
k(k + 3) = 0
2(2l – 1) + (l – 1) + 4(4l + 2) = 0
fi
k = 0, – 3
fi
21l + 5 = 0
fi
5 l = – ___ 21
Passage VII
1. The shortest distance between two lines
(r2 – r1) . (u × v) = ______________ |(u × v)|
Now,
(u × v)
i j k = 2 3 4 3 4 5
= – i + 2j – k
and
r2 – r1 = i + 2j + 2k
(
Let
and
–1 – 3 g_____ a – 2 b_____ 8 fi _____ = = = – __ 2 1 1 3 fi
16 8 8 a = 2 – ___ , b = 3 – __ , g = 1 – __ 2 3 3
fi
10 5 1 a = ___ , b = __ , g = – __ 3 3 3
(
)
z–3 ____ 4
10 1 __ 5 Hence the required image is ___ , __ , – . 3 3 3
z–5 ____ 3
Also, let A = (2, 3, 5).
2. Hence, the image of the point (2, 3, 1) w.r.t. the plane 2x + y + z = 6 is
–1 – 3 g_____ 2(4 + 3 + 1) a – 2 b_____ _____ = = = – __________ 2 1 1 (4 + 1 + 1)
(– 1 + 4 – 2) 1 ________ = __________ = ___ __ 1 + 4 + 1 ÷6 ÷ –2 x – 1 y_____ L1 : _____ = = 2 3 –4 x – 2 y_____ L2 : _____ = = 3 4
)
43 11 37 ___ – ___ , ___ , – 21 21 21
Hence, the shortest distance
Hence, the foot of the perpendicular is the
2. Hence, the equation of the plane passing through A and containing the line L1 is
3. Given plane is P : 2x + y + z = 6
– 2 z____ –3 x – 1 y_____ and the line is L : _____ = = 2 1 4 Let the image of the point A(2, 3, 1) w.r.t. to the given plane is Q(a, b, g ).
6.45
3D-Co-ordinate Geometry
S cos(2a) = cos(2a) + cos(2b) + cos(2g)
So, the image of the given line w.r.t. to the given –g – b z____ x – a y_____ plane is _____ = = 2 1 4
–1 – 3 g_____ a – 2 b_____ Let _____ = = = g 2 1 1
(D)
S (cos a + sin2a) = S 1 = 1 + 1 + 1 = 3
Let M is the mid-point of A and Q, i.e.
2. (A)
S cos2a = cos2a + cos2b + cos2g + cos2d
(
)
+1 + 3 g_____ a + 2 b_____ M = _____ , , 2 2 2
(
)
Clearly, the point M lies on the plane
(
) (
)
2l + 4 l+6 l+2 2 ______ + _____ + _____ = 0 2 2 2
fi
l l 2l + 4 + __ + 3 + __ + 1 = 0 2 2
fi
3l = – 8
fi
8 l = – __ 3
So,
a = 2l + 2, b = l + 3, g = l + 1
fi
16 8 8 a = 2 – ___ , b = 3 – __ , g = 1 – __ 3 3 3
fi
10 – 5 1 a = – ___ , b = __ , g = ___ 3 3 3
– 7 3z –1 3x + 2 3y fi ______ = ______ = _____ 2 1 4 10 5 7 x + ___ y – __ z + __ 3 3 3 fi ______ = _____ = _____ 2 1 4 – 7 3z +5 3x + 10 3y fi _______ = ______ = ______ 2 1 4
Match Matrix 1. (A)
(B)
2
(B)
= 4/3
S sin2a = sin2a + sin2b + sin2g + sin2d 4 8 = 4 – __ = __ 3 3
(C)
S cos(2a)
= cos(2a) + cos(2b) + cos(2g) + cos(2d) = 2(cos2a + cos2b + cos2g + cos2d) – 4
( )
8 4 4 = 2 __ – 4 = __ – 4 = – __ 3 3 3 (D)
S (cos2a + sin2a) = S 1
=1+1+1+1
=4
3. (A) Let two diagonals of a cube are ai + aj + ak and – ai + aj + ak Let q be the angle between them a ◊ – a + __a ◊ a + a ◊ a __ cos q = _______________ = a÷3 ◊ a÷3
a2 ___ 2 = 3a
1 __ 3
( )
1 fi q = cos–1 __ 3 (B) Let the diagonal of the cube be i + j + k and the diagonal of the face be i + j. Let q be the angle between them,
÷
__
1 __+ 1__ _____ 2 2 cos(q) = _____ = __ __ = __ 3 3 ÷ 2 ÷3 ÷ 2 ÷
( ÷ ) __
__ 2 fi (q) = cos–1 __ = cot–1 (÷2 ) 3
(C) Let the diagonal of the cube be i + j + k and the diagonal of the face intersecting it is j + k. Let q be the angle between them,
( ÷ ) __
S cos a = cos a + cos b + cos g = 1 2
= 2 – 3 = –1 2
10 5 7 x + ___ y – __ z + __ 3 3 3 ______ _____ _____ = = 2 1 4
Hence, the required image of the line w.r.t. to the plane is
2
= 2(cos2a + cos2b + cos2g) – 3
2l + 4 l________ + 3 + 3 l_____ +2 = ______ , , 2 2 2
(C)
2
S sin2a = sin2a + sin2b + sin2g
= 3 – (cos2a + cos2b + cos2g)
=3–1=2
__ 2 q = cos __ = cot–1 (÷2 ) 3 –1
(D) Let the diagonal of the faces of the same vertex are i + j, j + k. Let q be the angle between them. Then
6.46 Integral Calculus, 3D Geometry & Vector Booster
0 + __1 +__ 0 __ 1 cos q = ________ = 2 2 ÷ 2 ÷
1– 2 4 – 3 5– 4 1 1 –k = 0 k 2 1
fi
–1 1 1 1 1 –k = 0 k 2 1
fi
–1 0 0 1 2 1– k = 0 k k + 2 k +1
1 = ___ __ 3 ÷
fi
1__ q = cot–1 ___ 3 ÷
l + m + n = 0, l2 = m2 + n2
4. (A) Given
fi
– (m + n) = m2 + n2
fi
– 2m n = 0
fi
m=0=n
fi 2(k + 1) + (k – 1)(k + 2) = 0
When m = 0, then l = – n
fi 2k + 2 + k2 + k – 2 = 0
l m n 1 fi __ = __ = ___ = ___ __ 1 0 –1 ÷ 2 When n = 0, then l = – m l fi __ = 1
m ___ = –1
n __ = 0
1 ___ __ 2 ÷
fi k2 + 3k = 0 fi k = 0, – 3. (C) Since the given lines are intersect, so they are coplanar. Thus,
Let q be the angle between them. Then
cos (q) = l1l2 + m1m2 + n1n2
p q = __ 3
fi
(B) Let q be the angle between them. Then
6 + 4 – _________ 10 __________ cos q = _______________________ = 0 ÷9 + 16 + 25 ÷4 + 1 + 4
fi
p q = __ 2
(C) Let q be the angle between them. Then
3 1 2 – __1 +__ 2 __ cos (q) = ________ = = __ 6 2 ÷6 ÷ 6 p fi q = __ 3 (D) Let q be the angle between them. Then
fi
2 + __1 +__ 0 ___ 1 sin q = ________ = __ ÷9 ÷ 2 ÷2 p __ q = 4
5. (A) Since the line lies in the plane, so the point (4, 2, k) lies in the plane.
Thus,
8–8+k=7
k=7
fi
2 k + 1 –1 2 3 4 =0 1 2 1
1 1__ 1 = ___ __ ◊ ___ + 0 + 0 = __ 2 2 ÷2 ÷
3 – 1 k + 1 –1 2 3 4 =0 1 2 1
(B) Since the given lines are coplanar so
fi 2(3 – 8) – (k + 1)(2 – 4) – (4 – 3) = 0 fi –10 + 2(k + 1) – 1 = 0 fi 2k – 9 = 0 9 fi k = __ 2 (D) Given lines are – 3 z____ –6 x – 2 y_____ _____ = = 3 4 5 – 2 z____ –1 x – 5 y_____ and _____ = = 1 2 3 Hence, the shortest distance (r2 – r1) ◊ (u × v) = ______________ |(u × v)|
=0
where, (r2 – r1) =
(5 – 2, 2 – 3, 1 – 6) = (3, –1, – 5)
i j k (u × v) = 3 4 5 = 2i – 4j + 2k 1 2 3 _________
__
fi |u × v| = ÷ 4 + 16 + 4 = 2÷6
3D-Co-ordinate Geometry
6. (A) Let any point on the given line be
(l + 1, 3l – 2, 2 – 2l) Clearly, this point lies on the plane so, 3(l + 1) + 4(3l – 2) + 5(2 – 2l) = 15 fi 5l + 5 = 15 fi 5l = 10 fi l = 2 Hence, the point of intersection is (3, 4, – 2). y – 1 z____ – 2 x (B) Any point on the line __ = _____ = can 2 2 3 be considered as M(2l, 2l + 1, 3l + 2). Let the point P be P(2, 6, 3). The direction ratios of PM are (2l – 2, 2l – 5, 3l – 1). Clearly PM is perpendicular to the given line L. So, 2(2l – 2) + 2(2l – 5) + 3(3l – 1) = 0 fi 17l – 17 = 0 fi l – 1 = 0 fi l = 1 Hence, the foot of the perpendicular is (2, 3, 5). (C) Any point on the first line be (l + 2, 1 – 2l, l + 6). Any point on the second line be (7m – 3, – 6 – 3m, m – 3) Clearly, both the points are same. So, l + 2 = 7m – 3, l + 6 = m – 3 l – 7m = – 5, l – m = – 9 Solving we get 29 2 l = – ___ , m = – __ 3 3 Hence, the point of intersection is
(
)
23 61 ___ 11 ___ , ___ , – 3 3 3 (D) Clearly, 2l – 3m = 4, 3l – 4m = 4 Solving we get l = – 2, m = – 4 Hence, the point of intersection is (– 5, – 8, – 9) 7. Given P(0, 3, – 2), Q(3, 7, – 1) and R(1, – 3, – 1). +2 x y – 3 z_____ Clearly, L1 : __ = _____ = 3 4 1 + 3 z_____ +1 x – 1 y_____ and L2 : _____ = = 1 0 1 The equation of the plane PQR is
x 3 1
y–3 z+2 4 1 =0 –6 1
fi
10x – 2(y – 3) – 22(z + 2) = 0
fi
10x – 2y – 22z = 38
fi
5x – y – 11z = 19
(A)
6.47
Let the foot of the perpendicular on L2 is M. Thus, M = (l + 1, – 3, l – 1) The direction ratios of PM are (l + 1, – 6, l + 1) Since PM is perpendicular to L2, so
1(l + 1) + 0 + 1(l + 1) = 0
fi l = – 2 Hence, the foot of the perpendicular is (–1, – 3, – 3) Thus, the length of the perpendicular _________
___
=÷ 1 + 36 + 1 =÷ 38 (B) The shortest distance between L1 and L2
(r2 – r1) ◊ (u × v) = ______________ |(u × v)|
4 + 12 – 4 = _________ = 2, 6 where (r2 – r1) = (1, – 6, 1) = i – 6j + k i j k 3 4 1 = 4i – 2j – 4k fi u × v = 1 0 1 __________
fi |u × v| = ÷16 + 4 + 16 = 6
(C) Area of DPQR
1 = __ |PQ × PR| 2
i j k 1 __ = 3 4 1 2 1 –6 1
1 = __ |10i – 2j – 22k| 2 ____ __ ÷588 1 ____________ __ _____ = ÷ 100 + 4 + 484 = = 7÷3 2 2 (D) Hence, the distance from the origin to the plane 5x – y – 11z = 19
19 = ____________ ___________ = ÷25 + 1 + 121
19 _____ ____ 147 ÷
8. (A) The shortest distance from the given point to the x-axis ______
___
=÷ 4 2 + 52 = ÷41
6.48 Integral Calculus, 3D Geometry & Vector Booster
(B) The shortest distance from the given point to ______ the y-axis = ÷ 3 2 + 42 = 5
(C) Let q be the angle between them. Then 2 sin q = __ 7 2 –1 __ fi q = sin 7 2 __ Clearly, k = 7
( )
fi
fi
r = (1 + t)i + 5t j + (1 – t)k
= (i + k) + t(i + 5j – k)
g f =0 c
abc + 2fgh – af 2 – bg2 – ch2 = 0 2x2 + 3y2 + 4z2 + 2kxy + 4yz = 0
represents two planes if
h b f
(A) The given equation
7k = 2
(D) Given line is
a h g
2 k 0 k 3 2 =0 0 2 4
fi 2(12 – 4) – k(4k) = 0
Since the given line lies in the plane
fi 16 – 4k2 = 0
c = d so, 1 + 0 + 1 + 5 – c = 0
fi k2 – 4 = 0 fi k2 = 4
fi 1 + c = d c=6 Thus,
c = 6, d = 7
Hence, the value of c + d is 13 9. (A) Hence, the volume of the tetrahedron
1 1 = __ abc = __ × 2 × 3 × 4 = 4 6 6
(B) Let the points A, B, C and D are (0, 1, 2), (3, 0, 1), (4, 3, 6) and (2, 3, 2), respectively.
Now,
AB = (3, –1, –1)
AC = (4, 2, 4)
AD = (2, 2, 0)
fi 21 – k(5k) = 0 21 fi k2 = ___ 5
3 –1 –1 1 __ = 4 2 4 6 2 2 0
|
1 k 0 k 5 2 =0 0 2 5
fi 5k2 = 21
Hence, the volume of the tetrahedron 1 = __ [AB, AC, AD] 6
(B) x2 + 5y2 + 2kxy + 4xz = 0 represents a pair of planes if
(C) x2 + y2 + 4z2 – 2k yz = 0 represents a pair of planes if
1 0 –K
0 –K 1 0 =0 0 4
fi 4 – k2 = 0
|
fi k2 = 4
1 = __ (– 24 – 8 – 4) 6
fi k2 + 2 = 6
= 6 cu. units.
(C) Volume of a tetrahedron = 2/3 2 (D) Volume of a tetrahedron = __ abc 3 2 __ = × 2 × 3 × 4 3 = 16 10. As we know that the homogeneous equation of 2nd degree ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0 represents two planes if
(D) 9x2 + y2 + z2 – 6k xz = 0 represents a pair of straight lines if
9 0 0 1 0 – 3k
0 – 3k = 0 1
fi 1 – 9k2 = 0 fi 9k2 = 1
6.49
3D-Co-ordinate Geometry
1. We have, al + bm + cn = 0 (bm + cn) fi l = – ________ a Now, ul2 + vm2 + wn2 = 0. bm + cn 2 fi u _______ + vm2 + wn2 = 0 a
(
)
fi
u(bm + cn)2 + va2 m2 + wa2 n2 = 0
fi
u(b2 m2 + c2 n2 + 2bcmn) + va2 m2 + wa2 n2 = 0
(ub2 + va2) m2 + 2ubcmn + (uc2 + wa2) n2 = 0 m 2 m fi (ub2 + va2) __ n + 2ubc __ n + (uc2 + wa2) = 0 ...(i) m1 m2 ___ ___ Let its roots are n and n . 1 2 fi
( )
( )
(a) If two straight lines are parallel, so the Eq. (i) will provide us equal roots. Thus, D = 0 fi (2ubc)2 – 4(b2u + a2v)(c2u + a2w) = 0 fi (ubc)2 – (b2u + a2v)(c2u + a2w) = 0 fi (b2u + a2v)(c2u + a2w) – (ubc)2 = 0 fi a2b2 u w + a2c2uv + a4vw = 0 fi b2uw + c2uv + a2vw = 0 2
2
2. Let P = (a, b, g) and Q = (x1, y1, z1) Thus, the direction ratios of OP are (a, b, g) and the direction ratios of OQ are (x1, y1, z1). Since O, Q, P are collinear, we have a __ x1 =
Similarly, eliminating n, we get l1l2 m1m2 ________ 2 = ________ 2 2 wb + vc uc + wa2
g __ z 1
...(i)
As P(a, b, g) lies on the plane lx + my + nz = p so
la + mb + ng = p
fi
k(lx1 + my1 + nz1) = p
Given
...(ii)
2
OP ◊ OQ = p
____________
___________
fi ÷a 2 + b2 + g 2 ◊ ÷x 21 + y 21 + z21 = p2 ______________
___________
fi ÷k 2 (x21 + y 21 + z21) ◊ ÷x 21 + y 21 + z 21 = p2 _____________
fi ÷k 2(x21 + y 21 + z21) 2 = p2 fi
k(x21 + y 21 + z 21) = p2
...(iii)
On dividing Eq. (ii) and Eq. (iii), we get lx1 + my1 + nz1 __ 1 _____________ = p (x21 + y21 + z21) fi
p(lx1 + my1 + nz1) = (x21 + y 21 + z 21)
Hence, the locus of the point Q is p(lx + my + nz) = (x2 + y2 + z2) x y __z 3. Given plane is __ a + __ + c = 1 b
2
a b c __ __ fi __ u + v + w = 0 (b) When two straight lines are perpendicular. So, uc2 + wa2 the product of the roots = ________ ub2 + va2 2 m1 ___ m2 uc + wa2 ________ fi ___ ◊ = n1 n2 ub2 + va2 m1m2 n1n2 fi ________ 2 = ________ 2 2 2 uc + wa ub + va
b __ y = 1
Let
...(i)
P = (a, b, g)
b __g a __ Then __ a + b + c = 1
...(ii)
___________
So,
OP = ÷ a 2 + b 2 + g 2
Therefore, the direction ratios of OP are g b a ___________ ___________ ____________ , ____________ , ____________ ___________ 2 2 2 2 2 2 2 ÷a + b + g ÷a + b + g ÷a + b 2 + g 2
l1l2 m1m2 n1n2 Thus, ________ 2 2 = ________ 2 2 = ________ wb + vc uc + wa ub2 + va2
The equation of the normal through P and normal to OP is b a __________ ___________ x + ____________ ___________ y 2 2 2 2 ÷a + b + g ÷a + b2 + g 2
Two lines are perpendicular, if
___________ g + ____________ ___________ z = ÷a 2 + b 2 + g 2 2 2 2 ÷a + b + g
fi (wb2 + vc2) + (uc2 + wa2) + (ub2 + va2) = 0
fi
ax + by + g z = ÷ a 2 + b 2 + g 2
fi a2 (u + v) + b2 (u + w) + c2 (u + v) = 0
a 2 + b 2 + g 2 Clearly, A = ____________ , 0, 0 a
l1l2 + m1m2 + n1n2 = 0
Hence, the result.
___________
(
)
...(iii)
6.50 Integral Calculus, 3D Geometry & Vector Booster
(
) )
a 2 + b 2 + g 2 B = 0, ___________ , 0 g
(
2
2
2
a + b + g C = 0, 0, ____________ g
Let
Q = (p, q, r)
Then
a 2 + b 2 + g 2 a 2 + b 2 + g 2 p = ___________ , q = ___________ a b 2
and
2
2
a + b + g r = ___________ g
...(iv)
a2 + b2 + g 2 1 1 1 Now, __2 + __ 2 + __ 2 = ______________ p q r (a 2 + b 2 + g 2)2 1 1 1 1 fi __2 + __ 2 + __ 2 = _____________ 2 p q r (a + b2 + g 2)
...(v)
From Since From From
2 2 + b2 + g 2 __ + b2 + g 2 q a__________ p a___________ __ = , = a aa b bb 2 2 2 r a + b + g __c = ___________ cg
...(iv) ...(v) ...(vi)
ka 2 + kb2 + kg 2 – k(aa + bb + cg) = 0 fi a 2 + b 2 + g 2 – (aa + bb + cg ) = 0 Hence, the locus of the foot of the perpendicular P is x2 + y2 + z2 – (ax + by + cz) = 0. 5. Given planes are x – y – z = 4 ...(i) and
From Eq. (iv), we get
Eq. (ii) and (iii), we get ka a + kbb + kcg + D = 0 (a, b, g) lies in the plane (i), so Aa + Bb + Cg + D = 0 Eqs. (iii) and (v), we get ka 2 + kb 2 + kg 2 + D = 0 Eqs. (iv) and (vi), we get
x + y + 2z = 4
...(ii)
Since the required planes pass through the line of intersection of the planes (i) and (ii). Thus, its equation will be (x – y – z – 4) + l (x + y + 2z – 4) = 0 fi
(1 + l) x + (1 – l) y + (2 – l) z – 4 – 4l = 0 ...(iii) Therefore, Since Eq. (i) and (iii) are mutually perpendicular, 2 2 2 2 2 2 a + b + g a + b + g ____________ + ____________ so, aa bb (1 + l) – (1 – l) – (2 – l) = 0 2 2 2 a + b + g 2 + ___________ fi l = __ cg 3 p q __r 2 = __ a + __ + c = 1 Substituting the values of l = __ in Eq. (iii), we get 3 b 5x + y + 4z = 20. 1 1 1 1 ___ __ ___________ fi ___ 2 which is the required equation of the plane. ap + bq + cr = a 2 + b 2 + g 6. Given planes are 1 1 1 fi = __ 2 + __ 2 + __ 2 (from Eq. v) x – cy – bz = 0 ...(i) p q r cx – y + az = 0 ...(ii) Hence, the locus of Q(p, q, r) is bx + ay – z = 0 ...(iii) 1 1 1 1 1 1 ___ ___ __ __ __ __ ax + + cz = 2 + 2 + 2 The equation of any plane passing through the line by x y z of intersection of the planes (i) and (ii) may be taken 4. Let the equation of the plane be as (x – cy – bz) + l(cx – y + az) = 0 Ax + By + Cz + D = 0 ...(i) fi (1 + lc)x – (c + l)y + (al – b)z = 0
which is passing through (a, b, c). So,
Aa + Bb + Cc + D = 0
...(ii)
Let P(a, b, g) be the foot of the perpendicular from the origin to the plane (i). The direction ratios of OP are (a, b, g). Therefore, (a, b, g) be the direction ratios of the same line OP. So, g a b __ 1 __ = __ = = __ (say) ...(iii) A B C k
...(iv)
If planes (iii) and (iv) are the same, so the equations (iii) and (iv) are identical. (1 + lc) (c + l) (al – b) _______ = – ______ = _______ a –1 b (1 + lc) (c + l) Thus, _______ = – ______ a b fi
(a + bc) l = – _______ (ac + b)
...(v)
3D-Co-ordinate Geometry
(c + l) (al – b) – ______ = _______ a –1 (ab + c) fi l = – _______ (1 – a2) From Eqs (v) and (vi), we get
So,
and
3a + 4b – 2c = 0
(ac + b) (ab + c) – _______ = – _______ (a + bc) (1 – a2)
and – 4a – b + 2c = 0 Solving, we get a b c _____ = _____ = _______ 8 – 1 8 – 6 – 3 + 16 a b c fi __ = ___ = ___ 7 12 13 Hence, the equation of the plane is
fi
a – a3 + bc – a2bc = a2bc + ac2 + ab2 + bc
fi fi
3
2
2
...(vi)
2
a + ab + ac + 2a bc – a = 0 2
2
2
a + b + c + 2abc = 1
7. Any point on the given line is
P : (2l + 1, 3l – 2, 6l + 3)
Let
Q : (1, – 2, 3)
Given
PQ = 3
fi
PQ2 = 9
fi
(4l2 + 9l2 + 36l2) = 9
49l2 = 9 9 3 2 fi l2 = ___ = __ 7 49 3 fi l = ± __ . 7 Hence, the points are fi
( )
( ) ( ) ) ( ( ) 17 11 = ( ___ , 5, 17 ) and ( – ___ , – 9, – 11 ). 3 3
7 7 7 ± 2 __ + 1, ± 3 __ – 2, ± 6 __ + 3 3 3 3
8. The equation of any line through (1, 0, – 3) and + 2 6____ –z x – 2 y_____ parallel to the line _____ = = is 2 3 6 y z+3 x – 1 __ _____ = = _____ . 2 3 – 6 Any point on the line (i) is (2l + 1, 3l, – 6 – 3l) which is lies in the plane x – y – z = 9. So,
(2l + 1) – (3l) – (– 6 – 3l) = 9
fi
2l + 7 = 9
fi
2l = 2 fi l = 1
Thus, the point is (3, 3, – 9). Hence, the required distance
________________________
=÷ (3 – 1)2 + (3 – 0)2 + (– 9 + 3)2
=÷ 4 + 9 + 36
=÷ 49 = 7.
_________ ___
9. The equation of any plane passing through (1, 2, 0) is a(x – 1) + b(y – 2) + cz = 0 which contains the – 1 2____ –z x + 3 y_____ line _____ = = . 3 4 2
6.51
7(x – 1) + 2(y – 2) + 13z = 0
fi 7x + 2y + 13z = 11 10. Let the given line AB be + 1 z____ –2 x – 1 y_____ _____ = = ...(i) 2 –1 4 and the given plane is x + 2y + z = 9 ...(ii) Let DC be the projection of AB on plane (ii). Clearly the plane ABCD is perpendicular to the plane (ii). Equation of any plane through AB is a(x – 1) + b(y + 1) + c(z – 3) = 0 ...(iii) where 2a – b + 4c = 0 ...(iv) since the plane (iii) is perpendicular to the plane (ii), so a + 2b + c = 0 ...(v) Solving Eqs (iv) and (v), we get a b c ___ = __ = __ – 9 2 3 Putting these values of a, b and c in Eq. (iii), we get 9(x – 1) – 2(y + 1) – 5(z – 3) = 0 fi 9x – 2y – 5z + 4 = 0 ...(vi) Since the projection DC of AB on the plane (ii) is the line of intersection of the plane ABCD and the plane (ii). Thus, the equation of DC will be 9x – 2y – 5z + 4 = 0 fi x + 2y + z – 9 = 0 ...(vii) Let l, m, n be the direction ratios of the line of intersection of the planes (i) and (ii). So, 9l – 2m – 5n = 0 and l + 2m + n = 0 Solving, we get l m n __ = ____ = ___ 8 –14 20 l m n fi __ = ___ = ___ 4 – 7 10 Let the line of intersection of the planes (i) and (ii) meet xy-plane at P(a, b, 0). So, P lies on the planes (i) and (ii) Therefore, 9a – 2b + 4 = 0 and a + 2b – 9 = 0
6.52 Integral Calculus, 3D Geometry & Vector Booster 17 1 a = __ , b = ___ 2 4 Hence, the equation of DC is 17 1 x – __ y – ___ z 2 4 _____ ______ = = ___ – 7 4 10 – 17 ___ z 2x – 1 4y fi ______ = _______ = 8 – 28 10 – 17 __z 2x – 1 4y fi ______ = _______ = 4 –14 5 11. Any plane passing through the first line is (2x + y + z – 1) + l(3x + y + 2z – 2) = 0 ...(i) which is parallel to the 2nd line x = y = z. So, (2 + 3l) ◊ 1 + (1 + l) ◊ 1 + (1 + 2l) ◊ 1 = 0 fi
(2 + 3l) + (1 + l) + (1 + 2l) = 0
fi
6l + 4 = 0 2 fi l = – __ 3 2 __ Put l = – in Eq. (i), we get 3 2 (2x + y + z – 1) – __ (3x + y + 2z – 2) = 0 3 fi y – z + 1 = 0 ...(ii) 1__ Thus, distance from (0, 0, 0) to the plane (ii) = ___ . 2 ÷ 12. We have, 4 + l = 1 + 2m fi
l – 2m = – 3
Also,
– 3 – 4l = –1 – 3m
fi
4l – 3m = – 2
fi aa1 + cc1 + 1 = 0 fi k=1 Now, k2 + 4 = 1 + 4 = 5 The equation of any line through (1, 2, 3) parallel to the given line is – 2 z____ –3 x – 1 y_____ _____ = = . 1 – 2 2 Any point on the above line can be considered as (l + 1, 2 – 2l, 2l + 3) which is also a common point of the plane. Thus, l + 1 + 2 – 2l + 2l + 3 = 11 14.
So,
fi l=5 Therefore, the point is (6, – 8, 13). Hence, the required distance
_________________________
= ÷(6 – 1)2 + (– 8 – 2)2 + (13 – 3)2
= ÷25 + 100 + 100
=÷ 225
_____________ ____
= 15 15. Let the equation of the plane be y __z x __ __ a + b + c = 1
...(i) ...(ii)
From Eq. (i) and (ii), we get l = 2, m = 1 Thus, the point of intersection
= (4 + 1, – 3 – 4, –1 + 7)
= (5, – 7, 6)
Hence, the required distance from (5, – 7, 6) to (1, – 4, 7) _________ = ÷16 + 9 + 1
___
= ÷26 .
13. Let
L1 : x = ay + b, z = cy + d y z–d x – b __ fi _____ = = ____ a c 1 and L2 : x = a1y + b1, z = c1y + d1 x – b1 fi _____ = a1
y z – d1 __ = _____ c1 1 Since L1 and L 2 are perpendicular, so
aa1 + 1.1 + c ◊ c1 = 0
It is given that the = 64k3 0 0 1 a 0 fi __ 6 0 b 0 0
volume of the tetrahedron OABC 0 0 0 c
1 1 = 64k3 1 1
1 fi __ (abc) = 64k3 6 fi (abc) = 384 k3 Let the centroid be G(a, b, g). a b c Clearly, __ = a, __ = b, __ = g 4 4 4 fi a = 4a, b = 4b, c = 4g Therefore, 64(a b g) = 384k3 fi
(a b g) = 6k3
Hence, the locus of G(a, b, g) is xyz = 6k3
3D-Co-ordinate Geometry
y z x z 16. Given lines __ + __c = 1, x = 0 and __ a – __c = 1, y = 0 b can be written as z x y – b ___ L1 : __ = _____ = – c 0 b y z x – a __ and L2 : _____ = = __c a 0 Here, and Now,
r1 = bj r2 = ai u = bj – ck v = ai + ck (r2 – r1) = ai – bj i
and
k
u × v = 0 b –c a 0 c
fi
j
= bci – acj – abk _________________ (bc)2 + (ac)2 + (ab)2
|(u × v)| = ÷
Given shortest distance = 2d (r2 – r1) ◊ (u × v) fi ______________ = 2d |(u × v)| abc + abc _________________ fi __________________ = 2d ÷(bc) 2 + (ac)2 + (ab)2 2abc _________________ fi __________________ = 2d 2 + (ac)2 + (ab)2 ÷ (bc) abc _________________ fi __________________ = d 2 + (ac)2 + (ab)2 ÷ (bc) (abc)2 fi __________________ = d2 2 2 2 [(bc) + (ac) + (ab) ] ((bc)2 + (ac)2 + (ab)2) 1 fi ___2 = ___________________ d (abc)2 1 fi ___2 = d
1 __ 2 + a
1 __ 2 + b
1 __ 2 c
17. Let O be the origin and the direction ratios of OP are (a, b, c) Hence, the equation of the plane is
6.53
3l – 2 l_____ + 1 2l +1 fi ______ = = ______ 1 2 3 fi
l=1
Thus, the point of intersection of the two lines is P(4, 3, 5). Therefore, the equation of the plane perpendicular to OP, where O is (0, 0, 0) and passing through P is 4x + 3y + 5z = 50. 19. The equation of the plane through the line be A(x – 1) + B(y + 2) + C(z – 0) = 0, then
2A – 3B + 5C = 0
and
A–B+C=0
Solving, we get C A B __ __ = __ = 2 3 1 Hence, the equation of the plane is
2(x – 1) + 3(y + 2) + z = 0
fi
2x + 3y + z + 4 = 0
Thus, a = 2, b = – 3, c = 1 Hence, the value of a + b + c + 2.
=2–3+1+2
= 2.
20. The direction ratios of the normal to the plane is (2, 3, 4). The equation of any plane through (2, 3, 4) and the maximum distance from the origin is
2(x – 2) + 3(y – 3) + 4(z – 4) = 0
fi
2x + 3y + 4z = 29.
21. Let O be the origin (0, 0, 0). Let the co-ordinates of the points A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Thus, the equation of the sphere passing through O, A, B, and C is
x2 + y2 + z2 – ax – by – cz = 0
But the radius of the sphere is given by r.
÷
___________
So,
a2 + b2 + c2 r = ___________ 4
fi (a2 + b2 + c2) = 4r 2 ...(i) Let (a, b, g ) be the centroid of the DABC fi ax + by + cz = a2 + b2 + c2 a + 0 + 0 __ a a = ________ = 18. Let a point (3l + 1, l + 2, 2l + 3) of the first line Then 3 3 lies on the 2nd line. Thus, b c Similarly, b = __ = g = __ 3 3 3l + 1 – 3 l________ + 2 – 1 2l +3–2 _________ = = _________ Putting the values of a, b and c in Eq. (i), we get 1 2 3 9(a 2 + b 2 + g 2) = 4r 2
a(x – a) + b(y – b) + c(z – c) = 0
6.54 Integral Calculus, 3D Geometry & Vector Booster Hence, the locus of (a, b, g ) is
2
2
2
9(x + y + z ) = 4r
2
22. Let the co-ordinates of A, B and C be (a, 0, 0), (0, b, 0) and (0, 0, c), respectively Hence, the equation of the plane is y __z x __ __ a + b + c = 1 Since the given plane passes through (f, g, h), we get f __ g __ h __ ...(i) a + b + c = 1 The equation of the sphere passing through 0, A, B and C is x2 + y2 + z2 – ax – by – cz = 0 Let (a, b, g ) be its centre. a b c Thus, a = __ , b = __ , g = __ 2 2 2 fi a = 2a, b = 2b, c = 2g Putting the values of a, b and c in (i), we get f g h ___ + ___ + ___ = 1 2a 2b 2g f g h fi __ + __ + __ = 2 a b g Hence, the locus of the centre (a, b, g ) is f g __ h __x + __ y + z = 2 23. Let O be the origin (0, 0, 0). Let the co-ordinates of the points A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Then the equation of the plane ABC is y __z x __ __ ...(i) a + b + c = 1 Also, the equation of the sphere passing through O, A, B, and C is
x2 + y2 + z2 – ax – by – cz = 0
÷
2
2
2
So,
a +b +c r = __________ 4
fi
(a2 + b2 + c2) = 4r 2
...(ii)
Let (a, b, g ) be the foot of the perpendicular from the origin O to the plane (i) Now, the equation of the line through O and the perpendicular to the plane (i) is y z x ___ = ___ = ___ = r (say) 1/a 1/b 1/c Thus, the co-ordinates of any point on the line are r r __r __ a , __ , c ...(iii) b If it is the foot of the perpendicular from O to the plane (i), its co-ordinates will satisfy the equation of the plane.
(
)
(
)
Then from Eq. (iii), we get
and
and
1/a r a = __ a = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/b r b = __ = ____________ b 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/c r g = __ c = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
Now,
a 2 + b2 + g 2
1 1 1 __ 2 + __2 + __2 a b c = _____________ 2 = 1 1 1 __ 2 + __ 2 + __ 2 a b c
= aa = bb = cg
Thus,
a2 + b2 + g 2 a2 + b 2 + g 2 a = ___________ , b = ___________ a b
and
a 2 + b 2 + g 2 c = ____________ g
( (
) ) (
1 ____________ 1 1 1 __ __ 2 + 2 + __ 2 a b c
)
Putting the values of a, b and c in (ii), we get
(
)
1 1 1 (a 2 + b 2 + g 2) ___ 2 + ___ 2 + ___ 2 = 4r 2 a b g
Hence, the locus of the foot of the perpendicular (a, b, g ) is 1 1 1 (x2 + y2 + z2) __ 2 + __ 2 + __ 2 = 4r2 x y z 24. Given plane is y __z x __ __ a + b + c = 1 Let O be the origin. Clearly, the plane intersects the axes at (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Hence, the equation of the sphere passing through O, A, B and C is
(
But the radius of the sphere is given by r. __________
1 __r __ 1 __r __ 1 __r Hence, __ a ◊ a + b ◊ b + c ◊ c = 1 1 fi r = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
)
x2 + y2 + z2 – ax – by – cz = 0
25. The equation of any sphere through the given circle is x2 + y2 + z2 + 7y – 2z + 2 + l (2x + 3y + 4z – 8) =0 fi x2 + y2 + z2 + 2ll + (7 + 3l) y + (4l – 2) z + (2 – 8l) = 0
3D-Co-ordinate Geometry __ ___
The centre of the sphere is
(
)
7 + 3l 4l –2 – l, – ______ , – ______ 2 2 If the given circle is a great circle of the sphere, so the centre of the sphere lies on the plane
(
6.55
)
Thus,
7 + 3l 2(– l) + 3 – ______ + 4(1 – 2l) = 8 2
fi
– 4l – 21 – 9l + 8 – 16l = 16
fi
– 29l = 29
fi
l = –1
fi (af + bg ± 2 ÷af ÷bg ) = ch __
___
__
___
__
___
___
fi (÷af ± ÷ bg ) 2 = (÷ch ) 2 ___
fi (÷af ± ÷ bg ) = ± (÷ch ) ___
fi (÷af ± ÷ bg + ÷ ch ) = 0 which is the required condition. 2. Let OA = a, OB = b and OC = c
Hence, the equation of the sphere is x2 + y2 + z2 + 7y – 2z + 2 – 2x – 3y – 4z + 8 = 0 fi x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0
1. Given al + bm + cn = 0 and fmn + gnl + hlm = 0 Eliminating n, we get
(
)
(
)
al + bm al + bm = gl – _______ + hlm = 0 f m – _______ c c – afm – bfm2 – agl2 – bglm + chlm = 0 l l 2 m + bf = 0 fi ag __ m + (af + bg – ch) __ l1 l2 m . Let its roots are ___ m and ___ 1 2 Now, l1 ___ l2 bf ___ ___ m1 ◊ m2 = ag fi
( )
( )
l2l2 m1m2 ag fi ___ = _____ bf l2l2 m1m2 ____ n1n2 fi ___ = _____ ag = bf ch
Similarly, we can find the angle between other pairs of diagonals and we have six such pairs out of these four digonals and all these angles are given by
± a2 ± b2 + c2 cos q = ___________ (a2 + b2 + c2)
fi
a2 ± b2 ± c2 q = cos– 1 __________ a2 + b2 + c2
l1l2 = m1m2 + n1n2 = 0
f fi __ a +
g h __ + __ = 0 b c (ii) If the lines are parallel, so l1 m1 __ n1 __ = ___ m = n l2 2 2 So, the roots of the Eq. (i) have equal roots, i.e. D = 0 fi (af + bg – ch)2 = 4abfg __
(
___
fi (af + bg – ch) = ± 2÷af ÷bg
)
3. The equation of any plane passing through P, Q, R is
x __ + a
(i) If the lines are perpendicular, so
– a2 + b2 + c2 = ___________ (a2 + b2 + c2)
l2l2 m1m2 ____ n1n2 fi ____ = _____ = (f/a) (g/b) (h/c)
Now, OP, CN, AM and BL are four diagonals. Let q be the angle between the diagonals OP and AM. The direction ratios of OP and AM are (a, b, c) and (– a, b, c) – a◊a + b ◊ b__________ + c ◊ c __________ Thus, cos q = _______________________ 2 2 2 2 + c ÷ a + b2 + c2 ÷ a + b
where
y __ + b
z __ = 1 g
...(i)
P = (a, 0, 0), Q = (0, b, 0), R = (0, 0, g)
which is passing through (a, b, c) a b __c Thus, __ + __ + = 1 a b g The equation of any plane passing through Q (0, b, 0) and parallel to yz-plane is x = a The equation of any plane passing through R (0, 0, g) and parallel to -plane is y = b The Equation of any plane passing through P (a, 0, 0) and parallel to xy-plane is z = b Hence, the locus of the point (a, b, g) is x __ b __c __ a + y + z = 1
6.56 Integral Calculus, 3D Geometry & Vector Booster
4. Let the equation of the plane be y __z x __ __ a + b + c = 1 It is given that,
Putting the values of a, b and c in Eq. (ii), we get ...(i)
a2 + b2 + c2 = k2
...(ii)
Let P (a, b, g) be the foot of the perpendicular drawn from the origin O to the plane (i). Now the equation of the line through (0, 0, 0) and the perpendicular to the plane (i) is y z x ___ = ___ = ___ 1/a 1/b 1/c So, the co-ordinates of any point on the line are r r __r __ a , __ , c ...(iii) b
(
)
It is the foot of the perpendicular from O to the plane (i), its co-ordinates will satisfy the equation of the plane. 1 __r __ 1 __r __ 1 __r Hence, __ a ◊ a + b ◊ b + c ◊ c = 1 1 1 1 2 + __ 2 = 1 fi r __ 2 + __ a b c
(
)
and
1/a r a = __ a = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/b r b = __ = ____________ b 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/c r g = __ c = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
Now,
a2 + b2 + g2
1 1 1 __ 2 + __ 2 + __ 2 a b c = _____________ 2 1 1 1 __ 2 + __ 2 + __ 2 a b c
( ( (
)
Hence, the locus of P (a, b, g ) is
(
)
1 1 1 (x2 + y2 + z2) __ 2 + __ 2 + __ 2 = k2 x y z
5. Clearly, the planes x = 0, y = 0 and z = 0 meet in (0, 0, 0) Hence, the incentre lies on the perpendicular from (0, 0, 0) to the plane x + y + z = a and divides it in the ratio 3 : 1, i.e. 3 from the vertex (0, 0, 0) and 1 from the plane x + y + z = a. The equation of the perpendicular from (0, 0, 0) to the plane x + y + z x y __z = a is __ = __ = = l (say) 1 1 1 Any point on the perpendicular is (l, l, l) a If it lies on the plane x + y + z = a, so, l = __ 3 Thus, the perpendicular from (0, 0, 0) meets the plane a a __ a x + y + z = a in (l + l + l), i.e. __ , __ , . 3 3 3 Also, the incentre divides the join of (0, 0, 0) and a a __ a __ , __ , in the ratio 3 : 1. 3 3 3
(
1 r = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
Then from (iii)
)
(
)
(
(
1 1 1 (a2 + b2 + g 2 ) ___ 2 + __ 2 + __ 2 = k2 a b g
1 = ____________ = aa = bg = cg 1 1 1 __ __ 2 + 2 + __ 2 a b c
)
)
Let (a, b, g ) be its incentre. Then a 3 ◊ __ + 1.0 3 a a = ________ = __ 3+1 4 a Similarly, b = __ = g 4 a a __ a Thus, the required incentre is __ , __ , . 4 4 4 6. The equation of any line passing through (3, 5, 7) and parallel to the normal to the plane is
(
)
– 5 z____ –7 x – 3 y_____ _____ = = ...(i) 2 1 1 and the equation of any point on the line (i) is
)
)
M (2l + 3, l + 5, l + 7)
which lies on the plane 2x + y + z = 16. So, fi fi Thus,
(2l + 3) + (l + 5) + (l + 7) = 16 4l + 15 = 16 1 l = __ 4 1 1 1 M = __ + 3, __ + 5, __ + 7 2 2 2
( 15 7 11 ___ = ( __ , ___ , 2 2 2)
)
Thus,
a2 + b 2 + g 2 a2 + b2 + g2 a = __________ , b = __________ a b
and
a2 + b 2 + g2 c = __________ g
Let Q (a, b, c) is the image of P (3, 5, 7). Clearly, M is the mid-point of P and Q.
3D-Co-ordinate Geometry
a + 3 __ 15 7 b + 5 ___ 11 c + 7 ___ Thus, _____ = , _____ = , _____ = 2 2 2 2 2 2 fi a = 4, b = 6, c = 8 Hence, the value of a2 + b2 + c2 + 18 = 16 + 36 + 64 + 18 = 134. 7. Given lines are + 1 __z – 3 z____ –1 x – 2 y_____ x + 1 y_____ _____ = = c and _____ = c = a a b b ab + bc + ca Therefore, cos q = ____________ 2 ...(i) a + b2 + c2 Since a, b and c are the roots of x3 + x2 – 4x – 4 = 0, so a + b + c = – 1 ab + bc + ca = 4 and
abc = 4
Now,
a2 + b2 + c2
= (a + b + c)2 – 2 (ab + bc + ca)
=1+8=9
From Eq. (i), we get 4 cos q = – __ 9 Hence, the acute angle between the lines is
( )
Thus,
6.57
_____________
(ac + b) = ÷ (1 – a2) (1 – c2) _____________
Similarly, (bc + a) = ÷ (1 – b2) (1 – c2) l m n Hence, _______ _____ = _______ _____ = ______ _____ 2 2 ÷1 – b ÷ 1 – c2 ÷ 1 – a Since the three given planes pass through the origin, so the line of intersection of the planes also pass through the origin. Thefore, the equation of the line is y z x __ = __ m = __ n l y z x _____ fi _______ = _______ _____ = ______ _____ 2 2 ÷ 1 – b ÷ 1 – c2 ÷ 1 – a 9. Given planes are x – y – z = 2 ...(i) and x + 2y + z = 2 ...(ii) The equation of any plane passing through the line of intersection of planes (i) and (ii) can be written as (x – y – z – 2) + l (x + 2y + z – 2) = 0 ...(iii) The direction of the cosines of the normal to the plane (iii) are 1+l 2l – 1 l–1 ___________ ___________ ___________ _____________ , ____________ , _____________ 2 2 2 – 4l + 3 ÷6l – 4l + 3 ÷ 6l – 4l + 3 ÷ 6l
fi
(1 – a2) + (– c2 – abc) + (– abc – b2) = 0
The direction cosines of the normal to the plane (i) are 1__ ___ 1 1 ___ – __ , – ___ __ 3 ÷3 ÷3 ÷ Since the angle between the planes (i) and (ii) is 90°, so, (1 + l) (2l + 1) (l – 1) __ __ __ cos (90°) = ______ – _______ – ______ p÷3 p÷3 p÷3 1 where p = ____________ ___________ 2 – 4l + 3 ÷ 6l
fi
1 – a2 – b2 – c2 – 2abc = 0
fi
q = cos–1 __ 4 9 8. Since the planes x = cy + bz, y = az + cx and z = bx + ay meet in a line, so – c – b 1 c – 1 a b a – 1
fi
fi
| |
1(1 – a2) + c (– c – ab) – b (ca + b) = 0
2
2
2
a + b + c + 2abc = 1
Let l, m, n be the direction cosines of the line of intersection. Therefore, l – c m – b n = 0 and cl – m + an = 0 l m n Thus, ______ = ______ = _____ ac + b bc + a 1 – c2 Now, (ac + b)2 = a2c2 + b2 + 2ab
= a2c2 + b2 + 1 – a2 – b2 – c2 = 1 – a2 – c2 + a2c2 = (1 – a2) (1 – c2)
(1 + l) – (2l + 1) – (l – 1) = 0
3 l = __ 2 Hence, the required equation of the plane is 3 (x – y – z – 2) + __ (x + 2y + z – 2) = 0 2 fi 2 (x – y – z – 2) + 3 (x + 2y + z – 2) = 0 fi
fi
5x + 4y + z = 10
10. Given plane is y __z x __ __ a + b + c = 1 Let P (m, n, p) be a point on (i), so p m __ n __ __ a + b + c = 1
...(i)
...(ii)
6.58 Integral Calculus, 3D Geometry & Vector Booster __________
i.e
OP = ÷ m 2 + n2 + p2
1 1 1 1 fi ___ + ___ + __ = __________ aa bb cg m2 + n2 + p2
The direction cosines of OP are
(
1 = ___ 2 + a
1 ___ 2 + b
1 ___ 2 g
p m n ____________ __________ , ___________ __________ , ____________ __________ 2 + n2 + p2 ÷m 2 + n2 + p ÷ m 2 + n2 + p2 ÷ m
The equation of the plane through P and normal to OP is n y m x __________ ____________ + ____________ __________ 2 + n2 + p2 ÷m 2 + n2 + p2 ÷ m
1 1 1 1 1 1 ___ ___ __ __ __ ___ a x + b y + c z = x2 + y2 + z2
)
Hence, the required locus of Q is
Integer Type Questions
__________ 1. Let q be angle between the lines. p z 2 __________ + ____________ = ÷ m + n2 + p2 1 \ cos q = __ ÷m 2 + n2 + p2 3
fi
m x + n y + p z = (m2 + n2 + p2)
Thus,
(m2 + n2 + p2) A = ____________ 0, 0 m ,
(m2 + n2 + p2) B = 0, ____________ 0 n ,
and
(m2 + n2 + p2) C = 0, 0, ____________ p
( (
) )
(
)
( )
1 q = cos– 1 __ 3 Hence, (a + b) = 1 + 3 = 4 2. Four diagonals of a cube are OP, AM, CL, BN fi
Let the point Q be (a, b, g) Thus,
2 m2 + n2 + p2 m__________ + n2 + p2 a = __________ , b = m n
and
m2 + n2 + p2 g = ___________ p
...(iii)
Now, 1 1 1 ___2 + ___ 2 + ___ 2 = a b g
(m2 + n2 + p2) _____________ (m2 + n2 + p2)2
1 1__ ___ 1 Direction cosines of BN are ___ __ , – ___ , __ 3 ÷3 ÷3 ÷
1 = ____________ 2 (m + n2 + p2) 2
2
2
(m + n + p ) From Eqs. (iii), m = _____________ a 2 + n2 + p2) m (m _____________ fi __ = a aa 2 2 2 n (m + n + p ) Similarly, __ = _____________ b bb 2 p (m + n2 + p2) ____________ and __ = c cg 2 (m2 + n2 + p2) (m + n2 + p2) Now, ____________ + _____________ aa bb
1 1__ ___ 1 Direction cosines of OP are ___ __ , ___ , __ 3 ÷ 3 ÷3 ÷ 1 1 1__ __ , ___ __ , ___ Direction cosines of AM are – ___ 3 ÷ 3 ÷ 3 ÷ 1 1__ ___ 1 Direction cosines of CL are ___ __ , ___ , – __ 3 ÷ 3 ÷ 3 ÷
(m2 + n2 + p2) __ p m n __ + _____________ = a + __ + c = 1 cg b
Let l, m, n be the direction cosines of a line which is inclined at angles a, b, g, d. l + m__+ n cos a = ________ 3 ÷ – l + m__ + n l + m __– n Similarly, cos b = _________ , cos g = ________ 3 ÷3 ÷ l – m__+ n and cos d = ________ 3 ÷ Thus,
Thus,
cos2a + cos2b + cos2g + cos2d 1 = __ [(l + m + n)2 + (– 1 + m + n)2 3 + (l + m + n)2 + (l – m + n)2˚ 1 = __ [4 (l2 + m2 + n2)] 3 4 = __ 3
3D-Co-ordinate Geometry
Thus, 3(cos2a + cos2b + cos2g + cos2d) = 4 3. The image of P (1, 6, 3) w.r.t. the line –2 x y – 1 z____ __ = _____ = is Q (1, 0, 7) 1 2 3 It is given that Q (a, b, g) = Q (1, 0, 7) Thus,
a+b+g=8
4. Clearly, 3a + 2b + c = 7 Thus, the minimum value of (a2 + b2 + c2) is fi fi
|
|
3.0 +__________ 2.0 + 0 – 7 2 = ______________ 2 + 12 ÷ 3 + 22 49 7 = ___ = __ 14 2 7 (a2 + b2 + c2) ≥ __ 2 2 2 2 2 (a + b + c ) ≥ 7
Hence, the least value of (a2 + b2 + c2) is 7.
5. The equation of any line passing through (1, 0, – 3) + 2 6____ –z x – 2 y_____ and parallel to _____ = = 2 3 6 y z+3 x – 1 __ is _____ = = _____ ...(i) 2 3 – 6 Any point on (i) can be considered as P (2l + 1, 3l, – 34 – 6l) which is also a point of the plane 2l + 1 – 3l + 3 + 6l = 9 fi
5l + 4 = 9
fi
l=1
= ÷4 + 9 + 36
= ÷49
Shortest Distance = =
Distance of any point on z-axis 2 ____ __ = 2 ÷ 1 2 7. Let the points be A, B, C and D. The number of planes which have three points on one side and the fourth point on the othe side is 4. The number of planes which have two points on each side of the plane is 3. Thus, the number of planes is 7. 8. Let the equation of the any plane be ax + by + cz + d = 0 Since the plane (i) containing the lines, so 2a + 3b + 4c = 0 and 3a + 4b + 4c = 0. a b c Therefore, _______ = _______ = _____ 15 – 16 12 – 10 8 – 9 a b fi ___ = __ = – 1 2 a b fi __ = ___ = 1 – 2 Also,
a (x – 1) + b (y – 2) + c (z – 3) = 0
fi
(x – 1) – 2 (y – 2) + (z – 3) = 0
fi |d| = 6. 9. We shall find the angle between the faces OAB and OAC. The angle between the faces OAB and OAC is the angle between the normal _› to the faces. The _› normal to the_ face_ OAB is a × b and the normal to › › face OAC is a × c .
_________ ___
=7 6. The equation of any plane containing the given line is (x + y + 2z – 3) + l (2x + 3y + 4z – 4) = 0 fi (1 + l) x + (1 + 3l) y + (2 + 4l) z – (3 + 4l) = 0 ...(i) y z x If the plane is parallel to z-axis, i.e __ = __ = __ , 0 0 1 the normal to the plane (i) is perpendicular to z-axis (1 + l (0) + (1 + 3l) (0) + (2 + 4l) (1) = 0 fi
1 l = – __ 2
...(i)
c ___ – 1 c __ 1
| |
________________________
= ÷(3 – 1)2 + (3 – 0)2 + (– 9 + 3)2
From Eqs. (i) and (ii), we get 1 (x + y + 2z – 3) – __ (2x + 3y + 4z – 4) = 0 2 fi y + 2 = 0 ...(iii)
fi x – 2y + z = 0 __ Again distance between the plane is ÷ 6 . __ d –__0 fi _____ = ÷6 ÷6
Thus, the point P is (3, 3, – 9). Hence, the required distance
6.59
...(ii)
_›
Thus,
_›
_›
_›
(a × b ) ◊ |(a × c ) cos q = _______________ _› _› _› _› |(a × b )| |(a × c )| › _› _ a ◊ (b ×
_›
_›
(a × c )) = ______________ {1 ◊ 1 ◊ sin(60°)}2
6.60 Integral Calculus, 3D Geometry & Vector Booster _› _ _ _› › › a ◊ {(b ◊ c ) a –
› _ _› _ ›
(a ◊ b ) c } = ____________________ 3/4
(a ◊ a ) (b ◊ c ) – (a ◊ b ) (a ◊ c ) = _______________________ 3/4
› _ _ › _ › _ ›
› _ _ › _ › _ ›
1(1 ◊ 1 ◊ cos (60°)) – (1 ◊ 1 ◊ cos (60°))2 = _____________________________ 3/4 1 1 __ – __ 2 4 __ 1 = _____ = 3/4 3
Hence, the value of (3 cos q + 2) = 1 + 2 = 3. 10. Clearly, l + ± m = ± n l 2 + m2 + n2 = 1 1__ fi l = m = n = ± ___ 3 ÷ Hence, 8 possible direction cosines and 4 lines are possible. – 3 z____ –4 x – 2 y_____ Since the lines _____ = = 1 1 – k – 4 z____ –5 x – 1 y_____ and _____ = = are coplanar, so 2 1 k
fi 11.
|
|
y2 – y1 z2 – z1 x 2 – x1 a c1 b1 1 a2 c2 b2
| | | | | |
–4 – 3 5 1 – 2 4 fi = 0 – k 1 1 2 k 1 – 1 1 1 fi 1 1 – k = 0 k 2 1
0 0 –1 fi 1 1 = 0 – k 2 k 2+k 1+k fi fi fi
– {2 (1 + k) + (k – 1) (k + 2)} = 0 2
– {2 + 2k + k + k – 2} = 0 k2 + 3k = 0
fi k = 0, – 3 Hence, the value of (k + 5) is 5 or 2. 9 __ – z – 3 2_____ x_____ – 2 y_____ 12. Let = = = l 1 2 k –4 x – 1 y_____ and _____ = = 2 k Thus, and
z __ = m 1
9 x + l + 2, y = 2l + 3, z = __ – k l 2 x = m + 1, y = 2m + 4, z = m
Solving, we get 1 l = 1, m = __ and k = 4 2 Hence, the value of k is 4. 13. Given planes are x = cy + bz y = cx + az z = bx + ay fi x – cy – bz = 0 cx – y + az = 0 bx + ay – z = 0
| |
Since these three planes pass through a line, so – c – b 1 c – 1 a = 0 b a – 1 fi fi fi fi
1 (1 – a2) + c (– c – ab) – b (ca + b) = 0 (1 – a2) – (c2 + abc) – (abc + b2) = 0 1 – a2 – b2 – c2 – 2abc = 0 a2 + b2 + c2 + 2abc = 1
a2 + b2 + c2 + 2abc + 2 = 1 + 2 = 3 – 2 z____ –k x – 4 y_____ 14. Since the line _____ = = lies in the plane 1 1 2 2x – 4y + z = 7 So, the point (4, 2, k) lies in the plane Thus, 2(4) – 4(2) + = 7 fi 8–8+k=7 fi k=7 15. Given plane is a x – b y + c z = 0 ....(i) and the line is – 2d z____ –c x – a y______ _____ = = c ...(ii) a b fi
Since the plane (i), contains the line (ii), so,
a (a) – b (b) + c (c) = 0
a (a) – b (2d) + c (c) = 0
fi
a2 + c2 = b2
a2 + c2 = 2bd
Clearly, b2 = 2bd
b = __ = 2 d
Questions asked in Past Iit-JEE Examinations
1. Since the line lies on the plane, so, the point (4, 2, k) lies on the plane.
Thus,
8–8+k=7
fi
k = 7.
3D-Co-ordinate Geometry
2. The equation of the plane passing through (2, 1, 0), and (4, 1, 1) is
|
|
– 1 z x – 2 y 1 = 0 3 – 1 0 1 2 fi fi
(x – 2) (– 1 – 0) – (y – 1) (3 – 2) + z (0 + 2) = 0 x + y – 2z = 3
3. The equation of the line PQ is
– 1 z____ –6 x – 2 y_____ _____ = = 1 2 – 2 Any point on the above line is
Q (l + 2, l + 1, 6 – 2l)
6.61
Let V1, the volume of the new box = r ◊ (b × c) 90 9 It is given that V1 = ____ V2 = ___ V2. 100 10 Thus, V1 = (r – 0 ◊ 9a) ◊ (b × c) fi r lies on a plane passing through 0 ◊ 9 a, and perpendicular to (b × c). 6. We take A = A¢ as the origin and B = B¢ = any point other than A on the line intersection of p1 and p2. Now, consider C = C¢ = any point niether on p1 nor p2. Thus, in this case, both the conditions of (a) and (b) are fullfilled.
The mid-point of PQ is
(
)
l l __ + 2, __ + 1, 6 – l 2 2 Since the mid-point of PQ lies on the plane, so,
(
) ( __ l2 + 1 ) – 2 (6 – l) = 3
l __ + 2 + 2
fi
3l = 12
fi
l = 4.
Thus, the co-ordinates of Q are (6, 5, – 2). 4. Since the given lines intersect, so they are coplanar.
|
|
+ 1 – 1 2 k 2 4 = 0 3 1 2 1 fi
– 10 – (k + 1) (2 – 4) – (4 – 3) = 0
fi
– 10 + 2 (k + 1) – 1 = 0
fi
2k = 9
fi
9 k = __ . 2
Similarly if we take, A = non-origin point on L1 B = non-origin point on the line of intersection of p1 and p2 and C = non-origin point on L2. If we take A = C≤, B = B¢, C = A¢, both the conditions of (a) and (b) are fulfilled.
5. Let AA¢ = a, AB = b, AD = d Volume of the parallelopiped T,
V = a ◊ (b × c)
Let
AA≤ = r
7. As the plane p parallel to b = (1, 0, – 1) and c = (– 1, 1, 0) normal to the plane is given by
| |
i j k b × c 1 0 = i + j + k – 1 –1 1 1
\ The equation of the plane ABC is
1 ◊ (x – 1) + 1 ◊ (y – 1) + 1 ◊ (z – 1) = 0
fi
x+y+z–3=0
6.62 Integral Calculus, 3D Geometry & Vector Booster fi
x+y+z=3 x y __z fi __ + __ + = 1. 3 3 3 This planes meets the axes in A (3, 0, 0), B (0, 3, 0), C (0, 0, 3). Thus, volume of the tetrahedron OABC 1 = __ [OA OB OC] 6 1 = __ [3i 3j 2k] 6 27 = ___ [i j k] 6 9 = __ 2 8. The equation of the plane passing through the line of intersection 2x – y + z = 3, 3x + y + z = 5 is (2x – y + z – 3) + l (3x + y + z – 5) = 0 ...(i) fi (2 + 3l) x + (l – 1) y + (l + 1) z = (5l + 3) = 0 Also, it is given that, the distance from (2, 1, – 1) to 1 the above plane is ___ __ . ÷6
|
|
2 (2 + 3l) + (l – 1) – (l + 1) – (5l + 3) _________________________ ____________________________________ ÷(2 + 3l)2 + (l – 1)2 + (l + 1)2 1__ = ___ ÷6 l–1 1 _____________ fi _______________ = ___ __ 2 ÷6 + 12l + 6 ÷ 11l
|
|
fi 6 (l – 1)2 = 11l2 + 12l + 6 fi 5l2 + 24l = 0 24 fi l = 0, – ___ . 5 24 Putting the values of l = 0, – ___ in Eq. (i), we get 5 the required equations of the planes are 2x – y + z = 3 or 9.
62x + 29y + 19z – 105 = 0
Equation of any plane passing through A, B, C X is __ a + Thus, centroid
Y Z __ + __ = 1 b c = G (x, y, z)
a b __ c = __ , __ , . 3 3 3
| ÷
(
)
Also, it is given that OM = 1
|
0 + 0 + 0 – 1 fi ____________ __________ = 1 1 1 1 __ 2 + __ 2 + __ 2 a b c 1 1 1 fi __2 + __ 2 + __ 2 = 1 a b c 1 1 1 fi ____ + ____ + ___ = 1 9 x2 9 y2 9 z2 1 1 1 fi __2 + __ 2 + __ 2 = 9 x y z 10. The equation of any plane passing through (1, – 2, 1) is
a (x – 1) + b (y + 2) + c (z – 1) = 0 ...(i)
The plane (i) is perpendicular to the planes So, and
2x – 2y + z = 0 and x – y + 2z = 4 2a – 2b + c = 0 a – a + 2c = 0
a b c Thus, ______ = _____ = ______ – 4 + 1 1 + 2 – 2 + 2 a b c fi ___ = __ = __ – 3 3 0 a b c fi __ = ___ = __ 1 – 1 0 Therefore, the equation of the plane is (x – 1) – (y + 2) = 0 fi x – y – 3 = 0. Thus, the distance of p from the point (1, 2, 2)
|
|
__ 1 – _____ 2–3 4 is ________ = ___ __ = 2÷2 . 1 + 1 ÷2 ÷
11. Given
k × (k × a) = 0
fi
(k ◊ a) k – (k ◊ k) a = 0
fi
g k – a = 0
fi
g k – (a i + b j + g k) = 0
fi
– (a i + b j) = 0
fi
– (a i + b j) = 0 ◊ i + 0 ◊ j
fi
a = 0, b = 0
Since the point (a, b, g) lies on the plane x + y + z = 2, so a+b+g=2
3D-Co-ordinate Geometry
fi
0+0+g=2
fi
g=2
12. The equation of any line passing through (0, 1, 0) and perpendicular to x + 2y + 2z = 0 is x y–1 __ = _____ = 1 2
z __ . 2
6.63
So 3 x – 6 y = 15 and 2 x + y = 5. On solving, we get x = 3 and y = – 1 Hence, the equation of the line is + 1 ___ z x – 3 y_____ _____ = = = t (say) 14 2 15 The parametric equations of the line are x = 14t + 3, y = 2t – 1, z = 15t. 15. The given system of equations ax + by + cz = 0, bx + cy + az = 0, and cx + ay + bz = 0 can be written as
( ) ( ) ( )
a b c x b c a y = c ab z Clearly
OA = 1,
2 2 ________ AM = _________ = __ 1 + 4 + 4 3 ÷
Thus,
4 5 OM 2 = OA2 – AM2 = 1 – __ = __ 9 9
fi
| |
__
÷5 OM = ___ . 3
0 0 0
AX = O a b c Now, |A| = b c a c ab
= – (a3 + b3 + c3 – 3abc)
= – (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(a) Let a + b + c = 0
13. The equation of any plane passing through (1, 2, 3) is
Then (a2 + b2 + c2 – ab – bc – ca) = 0
1 fi __ ((a – b)2 + (a – b)2 + (a – b)2) = 0 2 fi (a – b) = 0 = (b – c) = (c – a)
a (x – 1) + b (y – 2) + c (z – 3) = 0
which is perpendicular to x = 0 and y = 0 So,
a = 0, b = 0.
Thus, the equation of the plane is
c (z – 3) = 0
fi
(z – 3) = 0
Thus, the required distance from (0, – 1, 0) to the plane (z – 3) = 0 is
| |
0 –__3 _____ = 3. 1 ÷ 14. Let a, b, c be the direction ratios of the line of intersection of 3x – 6y – 2z = 15 and 2x + y – 2z = 5. Then
3a – 6b – 2c = 0
and
2a + b – 2c = 0
a b c Thus, ______ = ______ = ______ 12 + 2 – 4 + 6 3 + 12 a b c fi ___ = __ = ___ 14 2 15 Clearly, the vector 14i + 2 j + 15k is parrallel to the line of intersections of the plane. Now, we shall find the equation of the line put z = 0 in the given planes.
fi a = b = c Thus, the three system of equations are identical planes. i.e x + y + z = 0. (b) Suppose a + b + c = 0 and a2 + b2 + c2 π ab + bc + ca In this case, at least one of the following is true. a2 π bc, b2 π ca, c2 = ab Consider b2 π ac We can write the first two equations are
a x + b y = (a + b) z
and b x + c y = (b + c) z Eliminating y, we get (ac – b2) x = (ac – b2) z fi x = z Putting this in the first two equations, we get ax + b y + c x = 0 fi by = – (a + c) x
6.64 Integral Calculus, 3D Geometry & Vector Booster by = b x fi x = y Thus, x = y = z (c) In this case |A| π 0. So, the system of equations provide us a trivial solution. Thus, x = 0 = y = z. (d) Given a + b + c = 0, a2 + b2 + c2 = ab + bc + ca 1 fi a + b + c = 0, __ [(a – b)2 + (b – c)2] 2 + (c – a)2) = 0 fi a + b + c = 0, a = b = c fi a = 0 = b = 0. Thus, the system of equations represent the whole three-dimensional space. 16. Let l, m, n be the direction ratios of the line L1. Then l+m–n=0 and
l – 3m + 3n = 0 l m n Thus, _____ = ______ = ______ 3 – 3 –1 – 3 – 3 – 1 l m n fi __ = __ = __ 0 1 1 Thus, the direction ratios of L1 are (0, 1, 1) Similarly, the direction ratios of L2 and L3 are (0, 1, 1) and (0, 1, 1), respectively. Therefore, the lines L1, L2 and L3 are parallel. So, Statement I is false. Put z = 0 in P2: x + y – z = – 1, P3: x – 3y + 3z = 2 3 1 we get, x = – __ , y = – __ 4 4
(
)
1 3 Thus, any point on L1 is – __ , – __ , 0 . 4 4 So, the equation of the line L1 is + 3/4 __z x + 1/4 y______ ______ = = 0 1 1 which is parallel to the plane P1. 1 3 Since – __ , – __ , 0 does not lie on P1 , so no point 4 4 of L1 lies on P1.
(
)
Therefore, the three planes do not have a common point. Thus, the statement II is true. 17. (i) The unit vector perpendicular to both L1 and L2 u×v is ______ |u × v|
| |
Now,
i j k u × v = 1 1 2 = – i – 7j + 5k 1 2 3
Thus,
– i – 7j + 5k u×v __ ______ = __________ . |u × v| 5÷3
(ii) The shortest distance between L1 and L2
(r2 ◊ r1) ◊ (u × v) = _____________ |(u × v)|
Here, and fi
r2 = – i – 2j – k r1 = 2i – 2j + 3k r2 – r1 = 3i + 4k
|
|
– 3 +__20 17 Shortest distance = _______ = ____ __ . 5÷3 5÷3 (iii) The equation of any plane passing through (– 1, – 2, – 1) is Thus,
a (x + 1) + b (y + 2) + c (z + 1) = 0
...(i)
which is perpendiculat to the lines + 2 z_____ +1 x + 1 y_____ L1: _____ = = 3 1 2 + 2 z____ –3 x – 2 y_____ and L2: _____ = = 1 2 3 So,
3a + b + 2c = 0
fi
a + 2b + 3c = 0
a b c Therefore, _____ = _____ = _____ 3–4 2–9 6–1 a b ___ c __ fi = __ = . 1 7 – 5 Hence, the equation of the plane is
(x + 1) + 7 (y + 2) – 5 (z + 1) = 0
fi
x + 7y – 5z + 10 = 0.
Therefore, the required distance
|
|
1 +__________ 7 – 5 + 10 13 = ____________ = ____ __ . 5÷3 ÷1 + 49 + 25
18. Let the equation of the line passing through P (2, – 1, 2) and makes equal angles with the co-ordinate axes is + 1 z____ –2 x – 2 y_____ _____ = = 1 1 1 ___ ___ ___ __ __ __ 3 3 3 ÷ ÷ ÷ + 1 z____ –2 x – 2 y_____ fi _____ = = . 1 1 1 Any point on the above line is
Q (l + 2, + l – 1, l + 2)
Since the point Q lies on the plane, so fi
2 (l + 2) + (l – 1) + (l + 2) = 9
3D-Co-ordinate Geometry
fi
4l = 9 – 5 = 4
21. Given
fi
l = 1.
1 – ________ 4–2–a fi ___________ = 5 1 + 4 + 4 ÷
Thus, the point Q is (3, 0, 3). Therefore, the length of PQ
______________________ __ (3 – 2)2 (0 + 1)2 + (3 – 2)2 = ÷3
= ÷
19. The cartesian form of the given line is
Let
PQ = 5
|
|
| |
a+5 fi _____ = 5 3 fi fi
+ 1 x_____ x – 1 y_____ –2 _____ = = – 3 1 5
6.65
a + 5 = ± 15 a = – 5 ± 15 = 10, – 15
Since a is positive, so a = 10. The equation of the line PQ is
Q = (1 – 3l, l – 1, 5l + 2)
The direction ratios of PQ is ( –2 – 3l, – l – 3, 5l – 4) Since PQ is parallel to the given plane. So,
(– 2 – 3l) – 4 (l – 3) + 3 (5l – 4) = 0.
fi
1 l = __ 4
+ 2 z____ –1 x – 1 y_____ _____ = = . 1 2 – 2
1 Hence, the value of l is __ . 4 20. Let plane 1: ax + by + cz = 0 contains the line x y __z __ = __ = 2 3 4 Thus, 2a + 3b + 4c = 0 ...(i) Let plane 2: a; x + b¢y + c¢z = 0 is perpendicular to the plane containing the lines x __ = 3
y __ = 4
z x __ and __ = 2 4
y __ = 2
Let
Q = (l + 1, 2l – 2, 1 – 2l)
Also,
PQ = 5
fi
l + 1 + 2 (2l – 2) – 2 (1 – 2l) = a = 10
fi
9l = 10 + 5 15 5 fi l = ___ = __ 9 3 8 4 __ 7 Thus the point Q is __ , __ , – 3 3 3
(
z __ 3
)
Thus,
3a¢ + 4b¢ + 2c¢ = 0
fi
22. Let the equation of the any plane be
4a¢ + 2b¢ + 2c¢ = 0
ax + by + cz + d = 0 Since Eqs. (i) containing the lines, so
a¢ b¢ c¢ Hence, ______ = _____ = ______ 12 – 4 8 – 9 6 – 16
a¢ b¢ c¢ fi __ = ___ = ____ 8 – 1 – 10 Also, Plane 1 is perpendicular to Plane 2 So,
a ◊ a¢ + bb¢ c ◊ c¢ = 0
fi
8a – b – 10c = 0
From Eqs. (i) and (ii), we get a b c _______ = _______ = _______ – 30 + 4 32 + 20 –2 – 24 a b c fi ____ = ___ = ____ – 26 52 – 26 a b c fi __ = ___ = __ 1 – 2 1 Hence, the euation of the required plane is x – 2y + z = 0
...(i)
2a + 3b + 4c = 0
and ...(ii)
3a + 4b + 5c = 0. a b c Therefore, _______ = _______ = _____ 15 – 16 12 – 10 8 – 9 a b c fi ___ = __ = ___ – 1 2 – 1 a b c fi __ = ___ = __ 1 – 2 1 Also, a (x – 1) + b (y – 2) + c (z – 3) = 0
fi
(x – 1) – 2 (y – 2) + (z – 3) = 0
fi
x – 2y + z = 0
__
Again distance between the plane is ÷ 6 .
| |
__ d –__0 fi _____ = ÷ 6 ÷6
fi
|d| = 6.
6.66 Integral Calculus, 3D Geometry & Vector Booster 23. Let
and
– 1 z_____ +1 x – 2 y_____ L1: _____ = = 1 – 2 1 8 x – __ y + 3 –1 3 _____ z____ _____ L2: = = 2 – 1 1
Let the line be y z __ = __ ...(i) b c intersects the lines L1 and L2. So, the shortest distances between Eq. (i) and L1, and Eq. (i) and L2 are zero. For PQ and L1, x __ a =
| |
24. Given
1 9 7 [a b c] 8 2 7 = [0 0 0] 737
fi and
a + 8b + 7c = 0 9a + 2b + 3c = 0 a+b+c=0
Solving, we get b = 6 a, c = – 7 a Since the point P lies on the plane, so 2a + b + c = 1 fi 2a + 6a – 7a = 1 fi a=1 Thus, b = 6 and c = – 7 Now the value of 7a + b + c =7+6–7 = 6. 25.
| |
j i k a b c = (b + 2c) i + (c – a) j – (2a + b) k 1 – 2 1 Since PQ is perpendicular to L1 So,
2 (b + 2c) + (c – a) + (2a + b) = 0
fi
a + 3b + 5c = 2
| |
For PQ and L2
j i k a b c = (b + c) i – (a – 2c) j – (a + 2b) k 2 – 1 1 Since PQ is perpendicular to L2, 8 __ So, (b + c) + 3 (a – 2c) – (a + 2b) k 3 fi 3a + b – 5c = 0 ...(ii) Solving Eq. (i) and (ii), we get a b c _______ = ______ = _____ – 15 – 5 15 + 5 1 – 9
Hence, the required equation of the line be y z x __ = ___ = __ 5 – 5 2 Thus the point of intersections P and Q are
(
(
is (1, 4, 1). 1 13 __ , ___ 3 3
)
The direction ratios of PT is (2, 2, – 1) The angle between OR and PT is 9 2 +___ 8 –__1 ____ 1 cos q = ________ = __ = ___ __ 18 ÷ 9 9÷2 ÷2 ÷ p fi q = __ 4 and the length of PT
÷( ) ( ) ( ) _______________
a b c fi __ = ___ = __ 5 – 5 2
22 22 12 = __ + __ + __ = 1 3 3 3
\
p cos __ = 4
)
10 10 8 P = (5, – 5, 2), Q = ___ , – ___ , __ 3 3 3 Therefore, d 2 = PQ2 = 6.
The direction ratios of QR 4 The co-rdinates of P = __ , 3
PS 1 fi PS fi ___ ( ) ___ 1 ÷2 p TS 1 sin ( __ ) = ___ fi TS = ___ 4 1 ÷2 __
__
1 PS = TS = ___ __ . 2 ÷ 26. The equation of the required plane is P: (x + 2y + 3z – 2) + l (x – y + z – 3) = 0 Thus,
3D-Co-ordinate Geometry
fi (1+ l) x + (2 – l) y + (3 + l) z – (2 + 3l) = 0
Thus,
(2l – 2) + (– l – 1) + 3l = 3
2 Its distance from (3, 1, – 1) is ___ __ . 3 ÷
fi
4l = 6 3 l = __ 2
|
|
3(1 + l) + (2 – l) – (3 + l) – (2 + 3l) ________________________ fi _________________________________ + 1)2 + (2 – l)2 + (3 + l)2 ÷ (l
fi
(
)
5 9 So, the point P is 1, – __ , __ . 2 2 The foot of the perpendicular from the point (– 2, – 1, 2 __ = ___ 0) on the plane is the point Q (0, 1, 2). 3 ÷ 7 5 The direction ratio of PQ = 1, – __ , __ = (2, – 7, 5) 2 2
(
(– 2l)2 4 fi ____________ = __ 3l2 + 4l + 14 3
)
Hence, the equation of the line is
fi 4l + 14 = 0
–2 x y – 1 z____ __ = _____ = . – 7 2 5
7 fi l = – __ . 2 Hence, the required equation of the plane is
| |
29. The common perpendicular is along
z 17 5 11 – __ x + ___ y – __ + ___ = 0 2 2 2 2
i j k 1 2 2 = 2i + 3j – 2k 22 1
fi
– 5x + 11y – z + 17 = 0
Let
fi
5x – 11y + z = 17
2l – 3 – 3l + 1 2l –4 So, ______ = _______ = ______ 1 2 2
| |
27. Since given lines are coplanar, so we get 2 k 2 5 2 k = 0 2 0 0 fi fi
6.67
M = (2l, – 3l, 2l)
fi
l=1
Thus,
M = (2, – 3, 2)
Let the required point be P.
k2 – 4 = 0
Given,
k = ± 2
___
PM = ÷17
The equation of the plane containing these two lines is
fi (3 + 2s – 2)2 + (3 + 2s + 3)2 + (2 + s – 2)2 = 17
+ 1 z x – 1 y 2 = 0 k 2 k 2 5
fi (s + 2) (9s + 10) = 0 10 fi s = –2, – ___ 9 8 7 7 __ Thus, P = (–1, –1, 0) or __ , __ , 9 9 9 30. Given lines are – y z x – 5 _____ L1: _____ = = ___ 0 a – 3 – 2
|
|
fi (x – 1) (k2 – 4) – (y + 1) (2k – 10) + z (4 – 5k) = 0 fi – (y + 1) (2k – 10) + z (4 – 5k) = 0 fi – (y + 1) (± 4 – 10) + (4 – ± 10) = 0 Taking positive sign, we get
( + 1) – z = 0
fi y – z = – 1 Taking negative sign, we get
– (y + 1) (– 14) + 14z = 0
(y + 1) + z = 0
fi
fi
y + z = – 1
28. Any point P on the given line is (2l – 2, – l – 1, 3l) The point P lies on the given plane for some l
fi 9s2 + 28s + 20 = 0
(
and
x–a L2: _____ = 0
|
)
y z ___ = _____ – 1 2 – a
Since L1 and L2 are coplanar, so
|
5 – a 0 0 3 = 0 – a 0 – 2 2–a 0 –1 fi
(5 – a) (3 – a) (2 – a) – 2) = 0
fi
(a – 5) (a 2 – 5a + 4) = 0
fi
(a – 5) (a – 1) (a – 4) = 0
fi
a = 1, 4, 5.
6.68 Integral Calculus, 3D Geometry & Vector Booster 31. Ans. (A) y z+ 3 ___ = _____ – 1 1
x– 1 L1: _____ = 2
y + 3 z_____ +3 x – 4 _____ L2: _____ = = 1 1 2
| |
i j k Normal of plane P : n = 7 1 2 3 5 – 6
= – 16i + 48j + 32k
fi
n = i – 3 j – 2 k
The point of intersection of L1 and L2 are
2l + 1 = m + 4, – l = m – 3
1 = 3m – 2
m=1 Thus, the point of intersection is (5, – 2, – 1). The equation of the plane passing through (5, – 2, – 1) is a (x – 5) + b (y + 2) + c (z + 1) = 0 ...(i) Also, ax + by + cz = d is perpendicular with P1: 7x + y + 2z = 3, P2: 3x + 5y – 6z = 4 Thus,
7a + b + 2c = 0
and
3a + 5b – 6c = 0 a b c Therefore, _______ = ______ = ______ – 6 – 10 6 + 42 35 – 3 a b c fi ____ = ___ = ___ – 16 48 32 a b c fi __ = ___ = ___ 1 – 3 – 2 Now, from Eq. (i), we get
(x – 5) – 3 (y + 2) – 2 (z + 1) = 0.
fi
x – 3y – 2z = 13
Thus, a = 1, b = – 3, c = – 2, d = 13 32.
x L1: __ = 1 x L2: __ = 1
y z–1 __ = ____ = r, Q (r, r, 1) 1 0 y z+1 ___ = _____ = s, Q (s, – s, – 1,) – 1 0
fi
(l – 1) (l + 1) = 0 ( l = r)
fi
l = – 1, 1
For l = 1, the point P and Q are concides Hence, the value of l = – 1. 33. Let the required plane be x + z + ly – 1 = 0
| |
l______ –1 Now, _______ = 1 2 ÷ l + 2 fi
(l – 1)2 = l2 + 2
fi
– 2l + 1 = 2 1 l = – __ 2
fi Thus,
P3: 2x – y + 2z – 2 = 0
Distance of P3 from (a, b, g ) is 2.
|
|
2a – b + 2g – 2 ________ fi ____________ = 2 4 + 1 + 4 ÷ fi
2a – b + 2g – 2 = ± 6
fi
2a – b + 2g + 4 = 0 and 2a – b + 2g – 8 = 0
34. Line L will be parallel to the line of intersection of P1 and P2. Let a, b and c be the direction ratios of line L. Thus,
a + 2b – c = 0 and 2a – b + c = 0
fi a : b : c :: 1 : – 3, : – 5 The equation of the line L is – 0 z____ –0 x – 0 y_____ fi _____ = = 1 – 3 – 5 Again foot of the perpendicular from origin to plane 1 1 1 P1 is – __ , – __ , __ 6 3 6
(
)
The equation of the projection of line L on plane P1 is
1 1 1 x + __ y + __ z – __ 6 3 6 _____ = _____ = _____ = k 1 – 3 – 5
(
) (
)
5 2 1 1 1 Clearly points 0, __ , – __ and – __ , – __ , __ 6 3 6 3 6
PQ = (l + r) i + (l – r) j + (l – 1) k
satisfy the line of projection, i.e. M.
and
l – r + l – r = 0, as PQ is ^ t0 L1
fi
2l = 2r
fi
l=r
PR = (l – s) i + (l + s) j + (l + 1) k
35. Points O, P, Q, R, S are (0, 0, 0), (3, 0, 0), (3, 3, 0), 3 3 (0, 3, 0), __ , __ , 0 respectively. 2 2 1 The angle between OQ nad OS is cos– 1 ___ __ . 3 ÷
and l–s–l–s=0fis=0 Since PQ ^ PR, so (l – r) (l – s) + (l – r) (l + s) + (l – 1) (l + 1) = 0
(
)
( )
The equation of the plane containing the points O, Q and S is x–y=0
3D-Co-ordinate Geometry
The perpendicular distance from P (3, 0, 0) to the plane x – y = 0 is
| |
3 –__0 fi _____ = 2 ÷
3 ___ __ 2 ÷
line RS x y–3 __ = _____ = 1 – 1
x – y + z = 3 is obtained by – 1 z____ – 1 + 7 – 3) – 7 – 2 (3 x – 3 y_____ _____ = = = _______________ 1 – 1 1 3 i.e.
and the perpendicular distance from O (0, 0, 0) to the
÷
___
z 15 __ is ___ . 2 2
36. The mirrior image of (3, 1, 7) w.r.t. the plane
6.69
P = (– 1, 5, 3)
The equation of the plane passing through line and
| |
P = (– 1, 5, 3) is x y z _› n = – 1 5 3 = 0] 1 21 fi
x – 4y + 7z = 0
A
y
a
b a
b
yz-plane
y
z
f(x) dx
x xy-plane
first octant
a
b a
Integral Calculus, 3D Geometry & Vector Booster 11 March 2017 11:32:07 AM
f(x)
A
y
b
f(x) dx
Integral Calculus, 3D Geometry & Vector Booster with Problems & Solutions
JEE
Main and Advanced
About the Author REJAUL MAKSHUD (RM) Post Graduated from Calcutta University in PURE MATHEMATICS. Presently, he trains IIT Aspirants at RACE IIT Academy, Jamshedpur.
Integral Calculus, 3D Geometry & Vector Booster with Problems & Solutions
JEE
Main and Advanced
Rejaul Makshud M. Sc. (Calcutta University, Kolkata)
McGraw Hill Education (India) Private Limited chennai
McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto
McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai - 600 116, Tamil Nadu, India Integral Calculus, 3D Geometry & Vector Booster Copyright © 2017, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-5260-576-7 ISBN (10): 93-5260-576-4 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Bharati Composers, D-6/159, Sector-VI, Rohini, Delhi 110 085, and text and cover printed at Cover Designer: Creative Designer Visit us at: www.mheducation.co.in
Dedicated to My Beloved Mom and Dad
Preface
INTEGRAL CALCULUS, 3D GEOMETRY & VECTOR BOOSTER with Problems & Solutions for JEE Main and Advanced is meant for aspirants preparing for the entrance examinations of different technical institutions, especially NIT/IIT/BITSAT/ IISc. In writing this book, I have drawn heavily from my long teaching experience at National Level Institutes. After many years of teaching I have realised the need of designing a book that will help the readers to build their base, improve their level of mathematical concepts and enjoy the subject. This book is designed keeping in view the new pattern of questions asked in JEE Main and Advanced Exams. It has six chapters. Each chapter has the concept booster followed by a large number of exercises with the exact solutions to the problems as given below: Level - I
: Problems based on Fundamentals
Level - II
: Mixed Problems (Objective Type Questions)
Level - III
: Problems for JEE Advanced
Level - IV
: Tougher problems for JEE Advanced
(0.......9)
: Integer type Questions
Passages
: Comprehensive Link Passages
Matching
: Matrix Match
Reasoning
: Assertion and Reason
Previous years’ papers : Questions asked in Previous Years’ IIT-JEE Exams Remember friends, no problem in mathematics is difficult. Once you understand the concept, they will become easy. So please don’t jump to exercise problems before you go through the Concept Booster and the objectives. Once you are confident in the theory part, attempt the exercises. The exercise problems are arranged in a manner that they gradually require advanced thinking. I hope this book will help you to build your base, enjoy the subject and improve your confidence to tackle any type of problem easily and skillfully. My special thanks goes to Mr. M.P. Singh (IISc. Bangalore), Mr. Yogesh Sindhwani (Head of School, Lancers International School, Gurugram), Mr. Manoj Kumar (IIT, Delhi), Mr. Nazre Hussain (B. Tech.), Dr. Syed Kashan Ali (MBBS) and Mr. Shahid Iyqbal, who have helped, inspired and motivated me to accomplish this task. As a matter of fact, teaching being the best learning process, I must thank all my students who inspired me most for writing this book. I would like to convey my affectionate thanks to my wife, who helped me immensely and my children who bore with patience my neglect during the period I remained devoted to this book. I also convey my sincere thanks to Mr Biswajit Das of McGraw Hill Education for publishing this book in such a beautiful format.
viii Preface I owe a special debt of gratitude to my father and elder brother, who taught me the first lesson of Mathematics and to all my learned teachers— Mr. Swapan Halder, Mr. Jadunandan Mishra, Mr. Mahadev Roy and Mr. Dilip Bhattacharya, who instilled the value of quality teaching in me. I have tried my best to keep this book error-free. I shall be grateful to the readers for their constructive suggestions toward the improvement of the book. Rejaul Makshud M. Sc. (Calcutta University, Kolkata)
Contents
Preface
1. Indefinite Integrals
1.1–1.161
Definition 1.1 Geometrical interpretation of Integration 1.1 Basic formulae on integration 1.1 Standard Methods of Integration 1.2 Integration by Parts 1.5 Choice of the first function and the second function 1.5 Partial Fractions 1.6 Integration of Irrational Functions 1.8 Euler’s Substitution 1.9 Integration by Reduction Formula 1.9 Inexpressible integrals 1.12 Exercises 1.12 Answers 1.39 Hints and Solutions 1.41
2. Definite Integrals
vii
What is definite Integral? 2.1 Evaluation of Definite Integrals 2.1 Evaluation of definite integrals by substitution 2.1 Geometrical interpretation of definite integral 2.1 Definite integral as the limit of sum 2.2 Evaluation of the limit of the sum using Newton-Leibnitz Formula 2.2 Properties of Definite Integrals 2.2 Mean Value of a Function over an Interval 2.7 Improper Integrals 2.7 Gamma Function 2.7 Beta Function 2.7 Walli’s Formula 2.7 Exercises 2.8 Answers 2.37 Hints and Solutions 2.38
2.1–2.131
x Contents
3. Area Bounded by the Curves
4. Differential Equation
4.1–4.103
Introduction 4.1 Definition 4.1 Ordinary Differential Equation 4.1 Partial differential equation 4.2 Order of a differential equation 4.2 Degree of a differential equation 4.2 Linear and non-linear differential equation 4.3 Formation of a differential equation 4.3 Differential equation of first order and first degree 4.3 Orthogonal Tragectories 4.6 First Order Higher Degree Differential Equation 4.6 Higher Order Differential Equation 4.7 Applications of Differential Equation 4.8 Exercises 4.9 Answers 4.26 Hints and Solutions 4.28
5. Vectors
3.1–3.62
Rules to Draw Different Types of Curves 3.1 Area of the Cartesian Curve 3.2 area of the region bounded by a single curve and the co-ordinate axes 3.2 Area between two curves 3.3 Area of the region bounded by the several curves 3.4 Least value of a variable area 3.4 Exercises 3.4 Answers 3.17 Hints and Solutions 3.18
Introduction 5.1 Physical quantities 5.1 Representation of vectors 5.1 types of vectors 5.2 Algebra of vectors 5.3 Left and right handed orientation 5.4 Position vector of a point in a space 5.4 Linear Combination 5.4 Linearly dependent vectors 5.4 Linearly independent vectors 5.4 Section formulae 5.4 Bisector of angle between vectors A and b 5.6 straight line 5.6 Plane 5.6 Definition 5.7 Geometrical interpretation of a ◊ b 5.7 Properties of dot product of two vectors 5.7 __› __› Component of a vector B along and perpendicular to vector A 5.8
5.1–5.80
Contents
Physical significance of the dot prodcut of two vectors 5.8 Introduction 5.9 _› _ › Geometrical Interpretation of a × b 5.9 Properties of vector product of two vectors 5.9 Introduction 5.10 Properties of scalar triple product of vectors 5.10 Introduction 5.11 Geometrical significance of a × (b × c) 5.11 Explanation of Vector Triple Product 5.11 Properties of vector triple product 5.11 Scaler product of four vectors 5.12 Vector product of four vectors 5.13 Geometrical interpretation of (a × b) × (c × d) 5.13 Reciprocal system of vectors 5.13 Exercises 5.15 Answers 5.35 Hints and Solutions 5.36
6. 3D-Co-ordinate Geometry
xi
Introduction 6.1 Rectangular co-ordinate system 6.1 Position vector of a point in a space 6.1 Distance between two points in space 6.2 Section formulae 6.2 Direction cosines and direction ratios of a vector or a line 6.3 Definition 6.4 Equation of a line passing thorugh a point and parallel to a vector 6.4 Equation of a Line Passing Through Two Points A (r1) and B (r2) is r = r1 + l(r2 – r1) 6.5 Angle between two straight lines 6.5 Skew lines 6.5 Definition 6.5 General Form 6.5 Equation of a plane passing through a point (x, y, z,) 6.6 Equation of a plane passing through three non-collinear points 6.6 Coplanarity of four points 6.6 Intercept form of a plane 6.6 Normal to a plane 6.6 Vector form 6.6 Equation of a plane in normal form 6.6 Normal form of a Plane 6.7 Theorem 6.7 Angle between two planes 6.7 Angle between a line and a plane 6.7 Equation of a plane parallel to a given plane 6.7 Equation of a plane parallel to the axes 6.7 Equation of a plane passing thorugh (x1, y1, z1), (x2, y2, z2) and parallel to the line having direction ratios a, b, c 6.8
6.1–6.69
xii Contents
Equation of the plane passing through a point (x 1, y1, z1) and parallel to two lines having direction ratios (a1, b1, c1) and (a2, b2, c2) is 6.8 Equation of a plane passing through the line of intersection of planes 6.8 Distance of a point from a plane 6.9 Distance between two parallel planes 6.9 Sides of a plane 6.9 Intersection of a line and a plane 6.9 Condition for a line to lie in a plane 6.9 Condition of coplanarity of two lines 6.9 Equation of the planes bisecting the angle between the planes 6.9 Bisector of the acute and obtuse angles between two planes 6.9 Foot of perpendicular of a point w.r.t. a plane 6.10 Image of a point w.r.t. a plane 6.10 Equation of the plane containing a given line and parallel to a given line 6.10 Equation of the plane containing two given lines 6.10 Exercises 6.10 Answers 6.24 Hints and Solutions 6.25
Chapter
1
Indefinite Integrals
Concept Booster
1. Definition The inverse process of differentiation is called integration. d Let g (x) be a differentiable function of x such that ___ dx (g(x) + c) = f (x). Then (x) dx = g(x) + c. Thus g(x) is called a primitive or anti-derivative or an indefinite integral or simply integral of f (x) with respect to x, where f (x) is called the integrand, c is called the constant of integration.
2. Geometrical interpretation
of Integration
Let f (x) be a given continuous function and g(x) one of its anti-derivatives such that y = If y =
(x) dx = g(x) + c, then y = g(x) + c represents
3. Basic formulae
on integration
Formula 1
xn + 1 1. Ú xndx = _____ + c n+1 2. Ú 0 . dx = c
3. Ú k . dx = kx + c 1 4. Ú __ x . dx = log |x| + c 5. Ú ex = ex + c ax 6. Ú ax dx = ____ + c log a Formula 2 1 1 1. Ú __2 dx = – __ x + c x __ 2 3/2 __ 2. Ú ÷x dx = x + c 3
(x) dx = g(x) + c.
__ 1__ 3. Ú ___ dx = 2 ÷x + c x ÷ Formula 3
1. Ú sin x dx = – cos x + c 2. Ú cos x dx = sin x + c 3. Ú sec2 x dx = tan x + c 4. Ú cosec2 x dx = – cot x + c 5. Ú sec x ◊ tan x dx = sec x + c 6. Ú cosec x ◊ cot dx = – cosec x + c Formula 4
dx 1. Ú ______ _____ = sin– 1 x + c 2 ÷ 1 – x
dx 2. Ú _____ = tan– 1 x + c 1 + x2
dx 3. Ú ________ _____ = sec– 1 x + c 2 |x|÷x – 1
Type 1: Integrals of the form dx Ú _______ 1 ± sin x sin x Ú _______ dx, 1 ± sin x dx Ú ________ , 1 ± cos x
(
)
(
)
cos x Ú ________ dx, 1 ± cos x sec x dx Ú ___________ , sec x ± tan x cosec x Ú _____________ dx cosec x ± cot x
1.2 Integral Calculus, 3D Geometry & Vector Booster Rule: Simply rationalize the denominator. f (x) Type 2: Integrals of the form Ú ____ dx, g (x)
( )
where f (x) and g(x) be two polynomials such that deg of f(x) ≥ deg of g(x). Rule: Simply divide the numerator by the denominator. Type 3: An integral is related to inverse trigonometric functions. Rule: Simply write the simplest form of the expression under the inverse term.
4. Standard Methods
cos x 2. Ú cot x dx = Ú ____ dx sin x = log |sin x| + c sec x(sec x + tan x) 3. Ú sec x dx = Ú ________________ dx (sec x + tan x) = log |sec x + tan x| + c
of Integration
There is no general method to find the integral of a function. If the integral is not a derivative of some known functions, the corresponding integrals cannot be determined. In general, we use the following three types of integration: (i) Integration by substitution (ii) Integration by parts (iii) Integration by partial fractions. Now we shall discuss about the integration by the substitution method. Type 4: Integrals of the form Ú f (ax + b) dx Rule: Simply put a x + b = t. Type 5: Integrals of the form dx ______ ______ . Ú _________________ ax + b ± ÷ ax – d ÷ Rule: Simply rationalize the denominator. Type 6: Integrals of the form ____
f (x)
____ dx, Ú _____ dx. Ú f (x) ÷g(x) g(x)
Rule: Put
g(x) = t2
÷
f ¢(x) Type 7: Integrals of the form Ú ____ dx f(x) Rule: Put f (x) = t Formula 5
|
| | |
|
1 + sin x = log ________ + c cos x p 1 + cos __ – x 2 = log ______________ + c p __ sin – x 2 p x 2cos2 __ – __ 4 2 = log _____________________ + c p x p x 2 sin __ – __ cos __ – __ 4 2 4 2
(
(
|
| |
Proof sin x 1. Ú tan x dx = Ú ____ dx cos x = – log |cos x| + c = log |sec x| + c
|
( ) ( ) ( ) ( ) |
|
Type 8: Integrals of the form
Ú (f (x))n ◊ f ¢(x) dx
( ) |
|
1 – cos x = log _______ + c sin x x 2sin2 __ 2 = log ______________ x x + c 2 sin __ cos __ 2 2 x = log tan __ + c 2
2. Ú cot x dx = log |sec x| + c
)|
)
= log |cosec x – cot x| + c
(
) (
)
cosec x(cosec x – cot x) 4. Ú cosec x dx = Ú ____________________ dx (cosec x – cot x)
Rule: Put f (x) = t Type 9: Integrals of the form
| |
|
( ) | = log tan ( __ | p 2 – ( __ p 4 – __ 2x ) ) | + c = log tan ( __ | p 4 + __ 2x ) | + c
1. Ú tan x dx = log |sec x| + c
3. Ú sec x dx = log |sec x + tan x| + c p x = log tan __ + __ + c 4 2 4. Ú cosec x dx = log |cosec x – cot x| + c x = log tan __ + c. 2
(
)
)
p x = log cot __ – __ + c 4 2
(
Ú sinm x cos nx dx, where m and n are real numbers.
Case I: When m is odd and n is even Rule: Put cos x = t Case II: When m is even and n is odd Rule: put sin x = t. Case III: When m and n both are odd Rule: Put either sin x = t or cos x = t Case IV: When m and n both are even Rule: In this case, we shall use the following formulae:
Indefinite Integrals
1 + cos 2x = 2 cos2x 2
or 1 – cos 2x = 2 sin x. Case V: When m is odd and n is zero Rule: put cos x = t Case VI: When m is zero and n is odd Rule: Put sin x = t Case VII: When m is even and n is zero. Rule: In this case we shall use the following formulae: 1 + cos 2x = 2 cos2x
1 – cos 2x = 2sin2x.
Case VIII: When m is zero and n is even Rule: In this case, we shall use the following formulae:
1 + cos 2x = 2 cos2x
= Р2k (say), where k ΠN
Rules 1. Divide the numerator and denominator by cos2k x 2. Put tan x = t. Formula 6 dx x 1 1. Ú ______ = __ tan– 1 __ a + c x2 + a2 a
|
x–a 1 ___ log _____ + c x+a 2a
dx 2. Ú ______ = 2 x – a2
dx ______ 3. Ú _______ = log x + ÷ x 2 + a2 + c 2 2 ÷ x + a
dx ______ 4. Ú ________ = log x + ÷x 2 – a2 + c. 2 2 ÷ x – a
|
_______
|
______
|
|
( )
dx x ______ 5. Ú ________ = sin– 1 __ a + C. 2 2 ÷ a – x Note: You should keep in mind that, the coefficient of x2 must be unity every time. If not, first we make the coefficient of x2 be unity.
dx Type 10: Integrals of the form Ú ___________ ax2 + bx + c Rules 1. Make the co-efficient of 2 unity 2. Express the denominator as a sum or difference of two perfect squares. Type 11: Integrals is of the form
Type 12: Integrals of the form
dx ___________ . Ú ____________ 2 ÷ax + bx + c
Rules 1. Make the coefficient of x2 unity. 2. Express the term under the square root as a sum or difference of two perfect squares. dx dx ______ 3. Use Ú ______ 2 or Ú _______ 2 2 x ±a ÷a – x2
|
_______
( )
|
x = log x + ÷ x 2 ± a2 + c or sin–1 __ a + c
Type 13: Integrals of the form
f ¢(x)dx
___________________ , Ú ____________________ 2
÷a{f (x)} ± b{f (x)} ± c
dx ___________ i.e. reducible to Ú ____________ 2 + bx + c ÷ ax
( )
|
d (Denominator) = Numerator + k 2. Then separate the Numerator into two terms and use dx . Ú x______ 2 ± a2
1 – cos 2x = 2 sin2x. Case IX: When m + n = – ve even integer
1.3
px + q dx. Ú ___________ 2 ax + bx + c
Rules 1. Reduce the denominator in such a way that
Rules 1. Put f (x) = t 2. Apply (Type 12) Type 14: Integrals of the form
px + q
___________ dx Ú _____________ 2
÷ax + bx + c
Rules 1. Reduce the term under the square root in such a way that
d (ax2 + bx + c) = numerator + k
2. Separate the numerator into two terms and use
|
_______
|
dx ______ = log x + ÷x 2 ± a2 + c Ú ________ 2 2 ÷ x ± a
Type 15: Integrals of the form
dx , Ú ______________ 2 a sin x + b cos2x
dx dx , Ú _________ , Ú a_________ ± b sin2x a ± b cos2x
dx 2 , Ú __________________ 2 a sin x ± b cos x ± c
1.4 Integral Calculus, 3D Geometry & Vector Booster
dx _______________ , Ú (a sin x ± b cos x)2
dx , Ú ____________________________ (a sin x ± b cos x)(c sin x ± d cos x)
dx . Ú __________ a + b sin2x
Rules
÷ a2 + b2
sin x cos x , ________ Ú _______ sin 3x dx Ú cos 3x dx
Rules 1. First we cancel the common factors from the numerator and the denominator. 2. Divide the numerator and the denominator by cos2x. 3. Put tan x = t. Type 17: Integrals of the form
dx dx , _________ , Ú _____________ a cos x + b sin x Ú a + b sin x
dx dx , ________________ . Ú a_________ + b cos x Ú a cos x + b sin x + c
Rules
|
Type 19: Integrals of the form
|
(
( ) |
)|
a sin x + b cos x dx. Ú _____________ c sin x + d cos x
Rules 1. Write the numerator = l (denominator) + m (denominator). 2. Compare the coefficients of sin x and cos x 3. Solve for l and m. Type 20: Integrals of the form
a sin x + b cos x + c dx. Ú ________________ p sin x + q cos x + r
Rules 1. Write numerator = l (denominator) + m (denominator) + n 2. Compare the coefficient of sin x, cos x and the contant term. 3. Find the values of l, m and n. Type 21: Integrals of the form
( ) ( )
x 2 tan __ 2 1. Replace sin x by ___________ , x 2 __ 1 + tan 2
Rules
( ) ( )
x 1 – tan __ 2 cos x by ___________ x . 1 + tan2 __ 2 2
2. Reduce the denominator as sin (A ± B) or cos (A ± B) 3. use p x + __ + c Ú sec x dx = log tan __ 4 2 or Ú cosec x dx = log tan __ 2x + c
Rules 1. Divide the numerator and denominator by the highest power of cos x. 2. Put tan x = t. Type 16: Integrals of the form
1. Divide the numerator and the denominator by ______
(cos x ± sin x)
dx Ú ___________ f (sin2x)
1. Put sin x ± cos x = t
2. Use (sin x ± cos x)2 = 1 ± sin2 x
Type 22: Integrals of the form 2
x x 2. Replace 1 + tan __ in the numerator by sec2 __ 2 2 x 3. Put tan __ = t 2
x +1 dx, Ú ___________ x4 ± l x2 + 1
dx x –1 dx, Ú ___________ , Ú ___________ 4 2 4 x ± lx + 1 x ± lx2 + 1
dx dx 4. Use Ú _______ or Ú ______ . x2 + a2 x2 – a2
where
l ΠR
2
( )
( )
Type 18: Integrals of the form
dx , Ú _____________ a cos x + b sin x _______
where
0 < ÷a 2 + b2 £2
( )
2
Rules 1. Divide the numerator and the denominator by x2. 2. Reduce the denominator in such way that 1 d x ± __ x = numerator ± k dx 3. Use Ú _______ . 2 x ± a2
(
)
1.5
Indefinite Integrals
Type 23: Integrals of the form
Type 24: Integrals of the form
Ú tan x dx, Ú cot x dx
Ú (tan x + cot x) dx
Type 25: Integrals of the form
Type 26: Integrals of the form
Rules 1. Put tan x or cot x = t2 2. Apply the form 24. by
Ú ex {( f + g) + ( f ¢ + g¢ )} dx = ex ( f + g) + c
Ú (tan x – cot x) dx.
5. Integration
Ú ex {f (x) + f ¢(x)} dx = ex f (x) + c
Ú ekx {k f (x) + f ¢(x)} dx = ekx f (x) + c
Type 27: Integrals of the form
Parts
If u and v be two functions of x, then
{
}
du Ú (u v) dx = u Ú v dx – Ú ___ Ú v. dx dx dx
Ú eax sin bx dx, Ú eax cos bx dx.
Rule 1
Ú eax sin bx dx eax = _______ 2 2 (a sin bx – b cos bx) + c a + b
Proof: We know that, d ___ ( f (x).g(x)) = f (x) . g¢(x) + g(x) . f ¢(x) dx fi
Ú [f (x) . g¢(x) + g(x) . f ¢(x)] dx = f (x) . g(x)
fi
Ú ( f (x) . g¢(x)dx = f (x) g(x) – Ú g(x) . f ¢(x)dx
Put
d f (x) = u, and ___ [g(x)] = v dx
fi Thus,
of the
eax = _______ (a cos bx – b sin bx) + c a2 + b2
Proof: Let
u = Ú eax cos bx dx
and
v = Ú eax sin bx dx
Then
u + i v
= Ú eax cos bx dx + i Ú eax sin bx dx
Ú v dx = g(x)
= Ú eax (cos bx + i sin bx) dx
du Ú v . dv } dx. Ú (u v) dx = u Ú v dx – Ú { ___ dv
= Ú eax◊ eibxdx
= Ú eax + ibxdx
= Ú e(a + ib)xdx
e(a + ib)x = ______ + c a + ib
(a – ib)e(a + ib)x = _____________ + c (a2 + b2)
i.e. the integral of the product of two function = (First function) × (Integral of the second function) – Integral of [(derivative of the first function) × (Integral of the second function)]
6. Choice
Ú eax cos bx dx.
Rule 2
first function
and the
second function
1. We can choose the first fnction as the function which comes first in the word ILATE, where I stands for inverse trigonometric functions L stands for logarithmic functions A stands for algebraic functions T stands for trigonometric functions E stands for exponential functions. 2. If the integrand be logarithmic functions or inverse trigonometric functions alone, take the second function as unity. 3. If both the functions are trigonometric, consider the second function, whose integral being simpler and the other as the first function.
Comparing the real and imaginary part, we get the required result. Formula 7 ______
______
x a2 x 1. Ú ÷a 2 – x2 dx = __ ÷a 2 – x2 + __ sin– 1 __ a + c 2 2
2. Ú ÷x 2 + a2 dx
( )
_______
_______
______
x a2 = __ ÷x 2 + a2 + __ log x + ÷x 2 – a2 + c 2 2 ______ 3. Ú ÷x 2 – a2 dx _______ ______ x a2 = __ ÷x 2 – a2 – __ log x + ÷ x 2 – a2 + c 2 2
| |
|
|
1.6 Integral Calculus, 3D Geometry & Vector Booster Proof:
1. Let
fi fi fi fi fi fi fi
2. Let
______
I = Ú ÷a 2 – x2 dx ______
(
)
1 ______ × – 2x I = ÷a 2 – x2 ◊ x dx Ú dx – Ú ________ 2 ÷a 2 – x2 ______
(
)
2 _______ I = x ÷a 2 – x2 + Ú x ________ dx 2 ÷ a – x2
______
(
______
)
2
2
2
fi 3. Let
______
______
|
______
|
______
|
______
|
|
fi 2I = x ÷x 2 – a2 – a2 log x + ÷ x 2 – a2 + c
|
___________
( ) ( ) ( )
I = Ú ÷x 2 + a2 dx
___________
_______
(
_______
(
)
)
(
2 + bx + c dx Ú ( px + q) ÷ ax
Rules 1. Reduce (px + q) as a derivative of (ax2 + bx + c) 2. Use ______
)
_______ (x2 + a2) – a2 ______ = x ÷x 2 + a2 – Ú ____________ dx ÷ x 2 + a2
_______
dx Ú ÷x 2 ± a2 dx. Ú ÷a 2 – x2
7. Partial Fractions
A special type of Aational (proper) function is known as the partial fraction, where the degree of the numerator < degree a dx ______ = dx + Ú _______ of denominator. Ú ÷x 2 + a2 ÷ x 2 + a2 ______ ______ f (x) h (x) = ____ . = x ÷x 2 + a2 – I + a2 log x + ÷ x 2 + a2 + c Let g(x) _______ x ÷x 2 + a2 –
_______
2
|
2I =
_______ x ÷x 2 + a2 + a2 log x
|
|
______
|
+ ÷x 2 + a2 + c
_______
_______
x a2 I = __ ÷x 2 + a2 + __ log |x + ÷ x 2 + a2 |+c 2 2 ______
= x ÷x 2 – a2 – I – a2 log x + ÷ x 2 – a2 + c
2
x2 = x ÷x 2 + a2 – Ú _______ ______ dx ÷ x 2 – a2
fi
dx 2 + bx + c dx Ú ÷ax ______ I = x ÷a – x – Ú ÷a – x dx + a Ú _______ 2 2 ÷ a – x Rules ______ x 1. First we make the coefficient of x2 unity I = x ÷a 2 – x2 – I + a2 sin–1 __ a + C 2. Express the term under the square root as a sum or ______ x difference of two perfect squares. 2 2 2 – 1 __ 2I = x ÷a – x + a sin a + c _______ ______ 2 2 ______ 2 3. Use x ± a d x or a 2 – x2 dx ÷ ÷ Ú Ú x a x 2 2 –1 I = __ ÷ a – x + __ sin __ + c a 2 2 Type 29: Integrals of the form ______ 2
Type 28: Integrals of the form
1_______ × 2x = ÷x 2 + a2 ◊ x dx Ú dx – Ú _________ 2 ÷x 2 + a2
a2 ______ = x ÷x 2 – a2 – Ú ÷x 2 – a2 dx – Ú ________ dx ÷ x 2 – a2
______
______
x a2 fi I = __ ÷ x2 – a2 – __ log x + ÷ x 2 – a2 + c 2 2
(a2 – x2) – a2 ______ I = x ÷a 2 – x2 – Ú ____________ dx ÷ a 2 – x2 ______
______
I = Ú ÷x – a dx 2
2
______
(
)
1______ × 2x = ÷x 2 – a2 ◊ x dx Ú dx – Ú ________ 2÷x 2 – a2 ______
(
)
x2 = x ÷x – a – Ú _______ ______ dx ÷ x 2 – a2
______ (x2 – a2) + a2 ______ = x ÷x 2 – a2 – Ú ____________ dx ÷ x 2 – a2
2
2
(
)
Type I: When the denominator is expressible as the product of non-repeating linear factors. Let g(x) = (x – a1) (x – a2) (x – a3) ... (x – an) f (x) Then ____ g(x) A3 An A1 A2 = _______ + _______ + _______ + ... + _______ (x – a1) (x – a2) (x – a3) (x – an) where A1, A2, ..., An are constant and can be determined by equating the numerator on RHS to the numerator on LHS and then substituting x = a1, a2, a3, ..., an Type II: When the denominator is expressible as the product of linear factors such that some of them are repeating. Let
g(x) = (x – a1)k(x – a2)(x – a3)...(x – an)
Indefinite Integrals
A1 A1 A1 f (x) _______ Then ____ = + _______ + ... + _______ g(x) (x – a1) (x – a1)2 (x – a1)k Bn B1 B2 + _______ + _______ + ... + _______ (x – a2) (x – a3) (x – an) Type III: When the denominator is expressible as the product of linear and quadratic factors but non-repeating. Let
g(x) = (x – a1)(x – a2)(x – a3) ... (ax2 – bx + c)
A3 A1 A2 f (x) _______ Then ____ = + _______ + _______ g(x) (x – a1) (x – a2) (x – a3) Type IV: Integrals of the form
Ú cosec2r + 1x dx = Ú cosec2r – 1x . cosec2x dx and then integrate it by parts, where consider cosec2x as the first function. consider sec2x as the second function.
2
Rules 1. Put x2 = t 2. Donot find its derivative 3. Use the concept of partial fractions. Type V: Integrals of the form
Rules 1. If m is even or odd integer and n is even positive integer, put cot x = t. 2. If m is odd positive integer and n Πeven positive integer, put cosec x = t. 3. If m = 0 and n = 2 r + 1, " r ΠN, write
Bx + C + ... + ___________ Type 3: Integrals of the form ax2 + bx + c dx dx , Ú ____________ , Ú ________ (n – 1) ______ x(xn + 1) 2 n n x (x + 1)
x dx Ú ______________ 2 (x + a)(x2 + b)
(x2 + a)(x2 + b) dx Ú ______________ (x2 + c)(x2 + d)
Rules 1. Put x2 = t 2. Do not find its derivative 3. Reduce it into a partial fraction (degree of the numerator < Degree of denominator) 4. Use the concept of partial fractions.
dx , where n Œ N Ú ___________ n x (1 + xn)1/n
Rules 1. Take common xn from the denominator.
1 + x– n = t.
2. Put
Type 4: Integrals of the form m
x dx, where m, n ŒN Ú ________ (ax + b)n
Rule put
ax + b = t.
Type 5: Integrals of the form dx Ú __________ , where m, n ŒN xm(ax + b)n
(
)
ax + b Rule put ______ x = t
Advanced Level
Type 6: Integrals of the form
Type 1: Integrals of the form
Ú tanmx . secnx dx
Rules 1. If is even or odd integer and n is even positive integer, put tan x = t 2. If m is odd positive integer and n Πeven positive integer, put sec x = t. 3. If m = 0 and n = 2 r + 1, " r ΠN, write
Ú sec2r + 1x dx = Ú sec2r – 1x . sec2x dx
and then integrate it by parts, where Type 2: Integrals of the form
Ú cotmx . cosecnx dx
1.7
Rules
( (
dx , where m, n ŒN Ú (x_____________ – a)n(x – b)n
) )
x–a 1. _____ = t, when m < n x–b x – b 2. _____ x – a = t, when m > n
Type 7: Integrals of the form dx , where n ŒN Ú _________ x(a + bx n) 1 xn = __ t Type 8: Integrals of the form Rule: put
2m + 1
x ________ dx Ú (ax 2 + b)n
1.8 Integral Calculus, 3D Geometry & Vector Booster Rule Put
(a x2 + b) = t
Rule Put (ax + b) = t p, where p is the LCM of m and n.
Type 9: Integrals of the form
Type 3: Integrals of the form
(a sin x + b)
dx Ú ___________ (a + b sin x)2
1
Rules 1. Divide the numerator and the denominator by cos2x 2. Put a sec x + b tan x = t. Type 10: Integrals of the form
Rule Put
dx ___________ Ú (a + b sin x)2
Rule Put
L(x) = t2
dx_____ Ú __________ L (x) Q (x)
Rule Put
1 L (x) = __ t
(
)
)
______ n
) } dx Ú { f ( x ± ÷x 2 + a2
Type 8: Integrals of the form
Functions
( (
dx , where m + p = N Ú __________ xm(a + bx)p
and
m+p>1
Rule Put
a + bx = t x.
Type 9: Integrals of the form
Type 1: Integrals of the form
))
a n ax + b __ ______ f x , dx, Ú cx + d
where
a, b, c, d, a ,
ΠR
)
ax + b Rule Put ______ = t n cx + d Type 2: Integrals of the form Ú f ( x, (ax + b)
Q2(x) 1 x = __ or t2 = _____ t Q1(x)
______
Rules 1. Express the numerator = l (Denominator) + m × derivative of (Denominator) 2. Compare the coefficients of ex and e– x 3. Find l and m.
(
2
Rule Put ( x ± ÷x 2 + a2 ) = t
a e + b e x dx Ú __________ p e + q e– x
of Irrational
÷
Type 7: Integrals of the form
Type 13: Integrals of the form
8. Integration
dx_____ Ú ___________ Q (x) Q (x) 1
a cos x + b t = __________ a + b cos x – x
÷
Type 6: Integrals of the form
Rule Put
x
÷
Type 5: Integrals of the form
Type 12: Integrals of the form dx ___________ Ú (a + b cos x)2
(
L2(x) = t 2 dx_____ Ú _________ Q(x) L (x)
2
a sin x + b Rule Put _________ =t a + b sin x
Rule put
÷
Type 4: Integrals of the form
(a cos x + b)
dx Ú ____________ (a + b cos x)2
Rules 1. Divide the numerator and the denominator by sin2x 2. Put a cosec x + b cot x = t. Type 11: Integrals of the form
dx_____ Ú __________ L (x) L (x)
a /n
, (ax + b)
where m, n are positive integers.
) dx,
b /m
dx , where m, n ŒR Ú _____________ L (x))m(L (x))n 1
2
Rules
L1(x) 1. If n > m, Put _____ = t L2(x)
L 2 (x) 2. If m > n, Put _____ =t L 1(x)
dx ______ Type 10: Integrals of the form Ú ________ x ÷ax n + b Rule Put
axn + b = t2
Indefinite Integrals
Type 11: Integrals of the form dx where a, b Œ R – {0} Ú (a__________ + bx2)3/2 Rule Put
1 x = __ t
dx ___________ , Ú __________________ r (x – k) ÷ax 2 + bx + c
where
r ΠN and k ΠR Р{0}
1 Rule Put x – k = __ t Type 13: Integrals of the form
___________
2 + bx + c ) dx Ú R ( x, ÷ax
Rules
___________
__
1. Put ÷ax 2 + bx + c = t ± x ÷a , if a > 0
2. Put ÷ax 2 + bx + c = t x ± ÷c , if c > 0,
3. Put (x – a) t, or (x – b) t, where a and b are the real roots of ax 2 + bx + c.
___________
Type 12: Integrals of the form
(ax + b)
___________ dx Ú ___________________ 2
(cx + d) ÷px + qx + r
Rules 1. Put (ax + b) = A(cx + d) + B 2. Find the values of A and B 3. Reduce the given integral into two separate integrals. Type 14: Integrals of the form (ax2 + bx + c) ___________ dx Ú ___________________ (dx + e) ÷px 2 + qx + r
1.9
__
___________ ÷a x 2 + bx + c =
10. Integration
by
Reduction Formula
A reduction formula is defined as a formula or a connection by means of which the power of the integrand is reduced, therefore, making the integration easier. The basic technique of obtaining a reduction formula is the integration by parts. In some cases the method of differentiation or other special devices is adopted. Type I: Reduction formula for Ú sinnx dx Soln. Let In = Ú sinnx dx
= Ú sinn – 1x ◊ sin x dx = sinn – 1x Ú sin x dx – Ú (n – 1)sinn – 2x (– cos2x) dx = sinn – 1x(– cos x) + (n – 1) Ú sinn – 2x (1 – sin2x) dx = sinn – 1x(– cos x) + (n – 1) I n – 2 – In
Rules 1. Put (ax2 + bx + c) = L (dx + e) (2px + q) + M(dx + e) + N 2. Compare the co-efficents of the like terms of both the sides and find L, M and N. 3. Integrate the given integral.
Thus, (1 + n – 1) In = – cos x ◊ sinn – 1x + (n – 1)In – 2
Type 15: Integrals of the form
Type 2: Reduction formula for Ú cosnx dx
Ú xb (a + bx g )a dx
Rules 1. If a Œ I +, expand the integral by the concept of binomial expansion. 2. If a Œ I –, we put x = t p, where p is the LCM of the denominator of b and g.
b+1 3. If _____ ΠI and a is a fraction, put (a + bxg ) = t p, where p is the denominator of a . b + 1 4. If _____ + a ΠI, put (a + bxg ) = t pxg, where p is the denominator of a .
9. Euler’s Substitution Type 16: Integrals of the form
fi
nIn = – cos x ◊ sinn – 1x + (n – 1) In – 2
cos x ◊ sinn – 1x n–1 In = ___________ + _____ n In – 2 n
( )
which is the required reduction formula.
Soln. Let In = Ú cosnx dx = Ú cosn – 1x ◊ cos x dx = cosn – 1x Ú cos x dx + (n + 1) Ú cosn – 2x ◊ sinnx dx = cosn – 1x ◊ sin x + (n + 1) Ú cosn – 2x ◊ (1 – cos2x) dx = cosn – 1x ◊ sin x + (n + 1) I n – 2 – In Thus, (1 + n + 1)In = cosn – 1x ◊ sin x + (n + 1) I n – 2 + C
(
)
cosn – 1x ◊ sin x n+1 fi In = ___________ + _____ I + C (n + 2) n + 2 n – 2 which is the required reduction formula. Type 3: Reduction formula for Ú tann x dx Soln. Let
In = Ú tann x dx
1.10 Integral Calculus, 3D Geometry & Vector Booster 2 sin(n – 1) x = ___________ + In – 2 (n – 1)
= Ú tann – 2x ◊ tan2 x dx
= Ú tann – 2x ◊ (sec2 x – 1)dx
which is the required reduction formula.
= Ú tann – 2x ◊ sec2 x dx – Ú tann – 2x dx
dx Type 7: Reduction formula for Ú ________ 2 (x + k)n
tann – 1x = ______ – In – 2 + C n–1
dx Soln. Let In = Ú ________ (x2 + k)n
(
)
which is the required reduction formula.
dx Then In – 2 = Ú __________ 2 (x + k)n – 1
Type 4: Reduction formula for Ú cot x dx n
Soln. Let
(
In = Ú cotnx dx
)
1 = Ú ___________ 2 dx (x + k)n – 1 ◊1
= Ú cotn – 2x ◊ cot2 x dx
= Ú cotn – 2x ◊ (cosec2 x – 1) dx
– (n – 1) 1 = _________ 2 n – 1 Ú 1 ◊ dx – Ú ________ ◊ 2x ◊ x dx (x + k) (x2 + k)n
= Ú cotn – 2x ◊ cosec2 x dx – Ú cotn – 2x dx
x x2 = __________ 2 + 2(n – 1) Ú _______ 2 n dx n – 1 (x + k) (x + k)
(
(
)
(
n – 1
cot x = – ______ – I n – 2 + C n–1
)
)
(x2 + k) – k x ___________ = _________ + 2 (n – 1) dx Ú (x2 + k)n – 1 (x2 + k)n
which is the required reduction formula.
(
)
x k 1 = _________ 2 n – 1 + 2 (n – 1) Ú __________ 2 n – 1 – ________ dx 2 (x + k) (x + k) (x + k)n
Type 5: Reduction formula for Ú secnx dx Soln. Let In = Ú sec x dx n
= Ú secn – 2x ◊ sec2 x dx
= secn – 2x Ú sec2 x dx – Ú (n – 2)secn – 2x ◊ tan2 x dx
= secn – 2x tan x – (n – 2) Ú secn – 2x ◊ (sec2 x – 1)dx
= secn – 2x tan x – (n – 2) Ú (secnx – secn – 2 x)dx
= secn – 2x tan x – (n – 2) In + (n – 2) In – 2
x = _________ + 2 (n – 1) (In – 1 – k In) (x2 + k)n – 1 x = _________ + 2(n – 1)In – 1 – 2(n – 1)k In (x2 + k)n – 1 fi 2 (n – 1) k In
x = _________ + 2(n – 1) In – 1 – In – 1 (x2 + k)n – 1
secn – 2x tan x n–2 fi In = __________ + _____ I +C n–1 n – 1 n – 2
x = _________ 2 n – 1 + (2n – 3) In – 1 (x + k)
which is the required reduction formula.
x fi In = __________________ 2 + 2(n – 1) k (x + k)n – 1
fi (n – 2) In = secn – 2x tan x – (n – 2) In – 2
(
)
sin nx Type 6: Reduction formula for Ú _____ dx sin x sin nx Soln. Let In = Ú _____ dx sin x
(
]
dx x = _________ 2 n – 1 + 2 (n – 1) In – 1 – In – 1 – k Ú _______ 2 (x + k) (x + k)n
(
)
1 2n – 3 __ ______ I k 2n – 2 n – 1
which is the required reduction formula. Type 8: Reduction formula for Ú x m(log x)n dx
)
sin(n – 2)x = Ú 2cos(n – 1)x + __________ dx sin x
Soln. Let Im, n = Ú x m(log x)n dx
[ sin nx – sin (n – 2)x = 2 cos (n – 1) x sin x
]
(
( )
)
1 x m + 1 = (log x)n Ú x m dx – Ú n(log x)n – 1 __ x _____ dx m+1
sin(n – 2)x sin n x fi _____ = 2 cos (n – 1)x + __________ = (log x)n, sin x sin x
x m + 1 n _____ – _____ ( x m(log x)n – 1 ) dx m+1 m+1 Ú
Indefinite Integrals
1.11
x m + 1 n = (log x)n, _____ – _____ I m + 1 m + 1 m, n – 1
m m – 1 – __ n Ú – cos x(sin nx cos x – sin(n – 1)x) dx
x m + 1 n fi Im, n – 1 = (log x)n, _____ – _____ I m + 1 m + 1 m, n – 1 which is the required reduction formula.
cosmx ◊ cos nx __ m = – ___________ – n Ú cos mx sin nx dx n m m – 1 + __ n Ú cos x sin (n – 1) x dx
Type 9: Reduction formula for
Ú x m(1 – x)n
cosmx ◊ cos nx __ m m = – ___________ – n Im, n + __ n Im – 1, n – 1 n
Im, n = Ú x m(1 – x)n dx
Soln. Let = (1 – x)
n
(
Ú x dx + Ú n(1 – x) m
m + 1
(
m fi 1 + __ n Im, n
)
x ◊ _____ dx m+1
n – 1
x m + 1 n = (1 – x)n _____ + _____ [x m + 1 ◊ (1 – x)n – 1] dx m + 1 m + 1Ú x m + 1 n = (1 – x)n _____ + _____ x m + 1 . (1 – x)n – 1dx m + 1 m + 1Ú
cosmx ◊ cos nx __ m = – ___________ + n Im – 1, n – 1 n
fi
cosmx cos nx _______ m Im, n = – ___________ + I (m + n) (m + n) m – 1, n – 1
Note: Similarly, we can easily formulate the reduction formula for m x sin nx m __________ + _______ I Ú cosmx cos nx dx = cos m+n (m + n) m – 1, n – 1
x m + 1 _____ n = (1 – x)n _____ + m+1 m+1
Type 11: Reduction formula for
Ú [x m ◊ (1 – x)n – 1 ◊ {– 1(1 – x)}] dx
x m + 1 n = (1 – x) _____ + _____ [I – Im, n] m + 1 m + 1 m, n – 1 n
m + 1
x n n = (1 – x)n _____ + _____ I – _____ – Im, n m + 1 m + 1 m, n – 1 m + 1 n fi 1 + _____ I m + 1 m, n
(
which is the required reduction formula.
x m + 1 n = (1 – x)n _____ + _____ [x m . (1 – x)n – 1 ◊ x] dx m + 1 m + 1Ú
)
)
m + 1
x n _____
n = (1 – x) + _____ I m + 1 m + 1 m, n – 1
Ú sinmx sin nx dx
Soln. Let Im, n = Ú sinmx sin nx dx
= sinmx Ú sin nx dx
(
)
cos n x – Ú m sinm – 1x ◊ cos x ◊ – ______ dx n
sinmx ◊ cos nx = – ___________ n
x m + 1 (1 – x)n n fi Im, n = ____________ + _________ I (m + n + 1) m + n + 1 m, n – 1
m m – 1 + __ n Ú sin x cos x cos nx dx
hich is the required reduction formula. w Type 10: Reduction formula for
Ú cosmx sin nx dx Soln. Let Im, n = Ú cosmx sin nx dx
m m – 1 + __ n Ú sin x (cos(n – 1)x – sin nx sin x)dx
= cosmx Ú sin nx dx
(
)
cos nx – Ú – _____ ◊ m ◊ cosm – 1x◊ – sinx dx n cos nx = cosmx – _____ n
(
)
m m – 1 – __ n Ú – cos x (cos nx sin x) dx
cosmx ◊ cos nx = – ___________ n
sinmx ◊ cos nx = – ___________ n
sinmx ◊ cos nx = – ___________ n
m m – 1 + __ n Ú sin x (cos(n – 1) x dx m m – __ n Ú sin x sin nx dx sinmx ◊ cos nx m fi 1 + __ n Im, n = – ___________ n
(
)
m + __ n Im – 1, n – 1
1.12 Integral Calculus, 3D Geometry & Vector Booster fi
sinmx ◊ cos nx _______ m Im, n = – ___________ + I (m + n) (m + n) m – 1, n – 1
Note: Similarly, we can easily formulate the reduction formula for sin nx m + _______ I Ú sinmx cos nx dx = sinmx _______ (m + n) (m + n) m – 1, n – 1 Type 12: Reduction formula for
which is the required reduction formula.
11. Inexpressible integrals functions, the integral is known as computable. But if an
n + 1
Soln. Let P = sin x cos x dp fi ___ = (m – 1) sinm – 2 x cos n + 2x dx – (n + 1) sinm x cos nx
sinm – 1x cosn + 1x ______ m–1 = – Ú _____________ + I m+n m + n m – 2, n
If an integral Ú f (x) dx as expressible in terms of elementary
Ú sinmx cos nx dx m – 1
Ú sinmx cosnx dx
fi
integral Ú f (x) dx is not expressible in terms of elementary functions, the integral is known as inexpressible or ‘cannot be found’. Some inexpressible integrals are
= (m – 1) sinm – 2x cosnx ◊ cos 2x
(i) Ú ex dx 2
dx – (n + 1) sinmx cos nx (ii) Ú e– x 2
____
(iii) Ú ÷sin x dx
= (m – 1) sinm – 2 x cosnx ◊ (1 – sin 2x )
– (n + 1) sinmx cos nx (iv) ÷_____ dx Ú cos x m – 2 n 2 = (m – 1) sin x cos x ◊ (1 – sin x ) (v) Ú x tan x dx
– (n + 1) sinmx cos nx
sin x (vi) Ú ____ dx x = (m – 1) sin x cos x cos x (vii) Ú ____ dx x – (m – 1) sinmx cosnx – (n + 1) sinmx cos nx dx (viii) Ú ____ = (m – 1) sinm – 2 x cosnx log x _____ – (m – 1 + n + 1) sinmx cosnx (ix) Ú ÷1 + x3 m – 2 n = (m – 1) sin x cos x ______ 3 1 + x3 (x) Ú ÷ dx – (m + n) sinmx cosnx (xi) Ú sin (x2) dx On integration, we get m – 2
(xii) Ú cos (x2) dx
P = (m – 1) Ú sinm – 2x cosnx dx – (m – n) Ú sinmx cosnx dx
fi
n
(m – n) Ú sinmx cosnx dx
__________
= – P + (m – 1) Ú sinm – 2x cosnx dx
x2 (xiii) Ú _____ dx 1 + x5
2 (xiv) Ú ÷1 – k2sin x dx
__
__
(xv) Ú ÷x cos ÷x dx.
Exercises
(Problems based on Fundamentals) ABC of Integration
1. Evaluate: Ú logx x dx
2. Evaluate: Ú ( 3log x2 – 2 log x3 )
(
m 3. Evaluate: Ú xm + mx + mm + __ x dx
4. Evaluate: Ú 2 ◊ 3 ◊ dx
5. Evaluate: Ú tan2x dx
6. Evaluate: Ú cot2x dx
dx 7. Evaluate: Ú _________ 2 sin x cos2x
x
)
x
Indefinite Integrals
(
(
) ) (
(
))
3p p 8. Evaluate: Ú 1+ tan x + ___ 1 + tan __ – x dx 8 8 9. Evaluate: Ú (tan x + cot x)2dx
dx 10. Evaluate: Ú _________ 1 + cos2x
dx 31. Evaluate: Ú ________________________ (tan x + cot x + sec x + cos x)
11. Evaluate: Ú ( 3log 5x – 2 log 5x ) dx
(
) ) (
(
(
Type 2
))
p p 12. Evaluate: Ú 1+ tan __ – x 1+ tan __ + x dx 8 8
(
)
81 + x + 41 + x 13. Evaluate: Ú __________ dx 22x x m 14. Evaluate: Ú __ m + __ x + xm + mx dx
(
)
(ax + b x)2 15. Evaluate: Ú ________ dx axbx (2x + 3 x)2 16. Evaluate: Ú ________ 2x ◊ 3xdx p 1 1 17. If f ¢(x) = __ x + ______ ______ and f (1) = __ , find f (x). 2 2 ÷ 1 – x 18. If f ¢(x) = a cos x + b sin x and
( )
p f ¢(0) = 4, f ¢(0) = 3, f __ = 5, find f (x). 2 dx 19. Evaluate: Ú _______ 1 – sin x
) )
sin4x + cos4x 20. Evaluate: Ú ____________ dx sin2x cos2x 6
6
sin x + cos x 21. Evaluate: ____________ 2 dx sin x cos2x
(
)
cos 2x – cosa 22. Evaluate: Ú ____________ dx cosx – cosa
( Ú (
)
cos4x – sin4x 23. Evaluate: Ú ___________ _________ dx 1 + cos4x ÷
)
1 + tan2x 24. Evaluate: ________ dx 1 + cot2x
(
(1 + x)2 33. Evaluate: Ú ________ dx x (1 + x2) x2 – 2 34. Evaluate: Ú ______ dx x2 + 1 x–1 35. Evaluate: Ú _____________ dx (x2/3 + x1/3 + 1)
( Ú ( Ú ( Ú ( Ú ( Ú ( Ú (
) ) )
x4 + 2 36. Evaluate: Ú ______ 2 dx x +2 x4 – 3 37. Evaluate: ______ 2 dx x +1 x6 – 1 38. Evaluate: ______ 2 dx x +1
)
___________
)
x4 40. Evaluate: _____ 2 dx x +1
)
x4 + x2 + 1 41. Evaluate: __________ dx x2 + x + 1
)
x6 + 1 42. Evaluate: ______ 2 dx x + 1 Type 3 43. Evaluate: Ú sin– 1(sin x) dx 44. Evaluate: Ú sin– 1(cos x) dx
( ÷
_________
)
cos x – cos 2x 25. Evaluate: Ú ___________ dx 1 – cos x
(
x 32. Evaluate: Ú _____ dx x+1
x8 + x4 + 1 39. Evaluate: __________ 4 dx x + x2 + 1
Type 1
( Ú (
( ) cos x – cos2x 30. Evaluate: Ú ( ____________ dx 1 – cosx )
cos 5x + cos 4x 29. Evaluate: Ú _____________ dx 1 + 2 cos3x
)
)
1 – cos 2x 45. Evaluate: Ú tan _________ dx 1 + cos 2x – 1
(
)
+ 2 ÷ x 4 + x– 4 26. Evaluate: Ú ____________ dx 3 x
sin 2x 46. Evaluate: Ú tan– 1 _________ dx 1 + cos 2x
5cos3x + 3sin3x 27. Evaluate: Ú ______________ dx cos2x sin2x
1 – cos 2x 47. Evaluate: Ú tan– 1 _________ dx 1 + cos 2x
(
(
)
)
cos x – sin x 28. Evaluate: Ú ___________ (1 + sin2x) dx cos x + sin x
( ÷
_________
(
)
)
sin x 48. Evaluate: Ú tan– 1 ________ dx 1 – cos x
1.13
1.14 Integral Calculus, 3D Geometry & Vector Booster
( ÷
________
)
dx _________ _________ 71. Evaluate: Ú ______________________ 2x + 2014 + ÷2x + 3013 ÷ Type 6
1 – sin x 49. Evaluate: Ú tan ________ dx 1 + sin x –1
(
)
sin x 50. Evaluate: Ú tan–1 ________ dx 1 + cos x
x _____ 72. Evaluate: Ú ______ dx x – 1 ÷
( ) 1 – sin x ( ________ cos x ) dx
cos x 51. Evaluate: Ú tan–1 ________ dx 1 – sin x 52. Evaluate: Ú tan– 1 53. Evaluate: Ú
(
__
x ÷ 73. Evaluate: Ú _____ dx x+1
_________ _________ ÷(1 + sin x) + ÷(1 – sin x) – 1 ______________________ _________ dx tan _________ ÷(1 + sin x) – ÷(1 + sin x)
)
54. Evaluate: Ú tan– 1 (sec x + tan x) dx
(
)
sin 2 x 55. Evaluate: Ú tan– 1 __________ dx 1 + cos 2 x Type 4 56. Evaluate: Ú (3x + 2) dx dx 57. Evaluate: Ú ______ 2x – 3 dx 58. Evaluate: Ú ______ 5 – 2x 59. Evaluate: Ú eax + b dx 60. Evaluate: Ú 34x + 5 dx 61. Evaluate: Ú cos (5x + 3) dx 62. Evaluate: Ú sin 2x dx ______
63. Evaluate: Ú ÷3x + 2 dx dx ______ 64. Evaluate: Ú _______ 3x + 4 ÷
x ______ 74. Evaluate: Ú _______ dx 3x + 1 ÷ x______ +1 75. Evaluate: Ú _______ dx 2x – 1 ÷ x_____ –1 76. Evaluate: Ú ______ dx x + 4 ÷ x 77. Evaluate: Ú _____ dx x2 + 1 Type 7 cos x – sin x 78. Evaluate: Ú ___________ dx sin x + cos x 3 cos x 79. Evaluate: Ú _________ dx 2 sin x + 5 cos x – sin x 80. Evaluate: Ú __________ dx 2 + sin 2x xe x + ex 81. Evaluate: Ú ________ dx cos2(xex) dx 82. Evaluate: Ú _________ x(1 + ln x)2 cos x – sin x + 1 – x 83. Evaluate: Ú _________________ dx ex + sin x + x dx 84. Evaluate: Ú _____ 1 + ex
Type 5
dx 85. Evaluate: Ú ________ 3 x(x + 1)
dx _____ _____ 65. Evaluate: Ú ______________ x + 2 – ÷x + 1 ÷
dx 86. Evaluate: Ú ________ 4 x(x + 1)
dx ______ 66. Evaluate: Ú _________________ ______ (÷2x + 5 – ÷2x + 3 ) dx ______ ______ 67. Evaluate: Ú ________________ 3x + 4 – ÷3x + 1 ÷ dx ______ ______ 68. Evaluate: Ú ________________ 2x + 3 + ÷2x – 3 ÷ dx _____ 69. Evaluate: Ú ___________ __ x + 1 + ÷x ÷ dx _____ 70. Evaluate: Ú _______________ _____ x + a + ÷x + b ÷
dx 87. Evaluate: Ú _______ 5 x(x – 1) sin 2x 88. Evaluate: Ú __________ dx sin 5 x sin 3x dx 89. Evaluate: Ú _________________ sin(x – a) sin (x – b) e x – e– x 90. Evaluate: Ú ________ dx e x + e– x
Indefinite Integrals
dx 91. Evaluate: Ú ______ 1 + e– x
dx 112. Evaluate: Ú ____________ __ __ 2 x (4 + 3÷x ) ÷
dx 92. Evaluate: Ú _____ 1 + e x
113. Evaluate: Ú 33 ◊ 33 dx
sin 2 x 93. Evaluate: Ú ______________ dx 2 a sin x + b cos2x
114. Evaluate: Ú tan3 x ◊ sec2xdx
sin(x – a) 94. Evaluate: Ú _________ dx sin x sin x 95. Evaluate: Ú ________ dx sin (x – a) sin(x + a) 96. Evaluate: Ú _________ dx sin(x + b)
3x
x
115. Evaluate: Ú sin3 x ◊ cos xdx (log x)3 116. Evaluate: Ú ______ dx x sin x _________ 117. Evaluate: Ú ___________ dx 3 + 2 cos x ÷ ________
dx 97. Evaluate: Ú _________ __ __ x (÷ x + 1) ÷
÷2 + log x 118. Evaluate: Ú _________ dx x dx 119. Evaluate: Ú ______ __ 1 + ÷x
1 + tan x 98. Evaluate: Ú ___________ dx x + log sec x
120. Evaluate: Ú x3sin x4 dx
sin 2x 99. Evaluate: Ú ___________ dx sin 5x . sin 3x
121. Evaluate: Ú 55 ◊ 55 ◊ 5x dx
cos x – sin x 100. Evaluate: Ú ___________ dx 1 + sin 2x dx 101. Evaluate: Ú _________ sin x . cos2x dx 102. Evaluate: Ú __________ . sin2x . cos2x dx 103. Evaluate: Ú _________________ sin (x – a) sin (x – b) dx 104. Evaluate: Ú __________________ cos (x – a) cos (x – b) dx 105. Evaluate: Ú _________________ sin (x – a) sin (x – b) x x(1 + ln x) 106. Evaluate: Ú __________ dx xx + 1 cos x – sin x + 1 – x 107. Evaluate: Ú _________________ dx ex + sin x + x 108. Evaluate: sin3x Ú _________________________________ dx 4 2 (cos x + 3cos x + 1)tan– 1(sec x + cos x) Type 8 109. Evaluate: Ú 3x2 sin (x3) dx (1 + ln x)3 110. Evaluate: Ú _________ dx x dx 111. Evaluate: Ú __________ 2 x (1 + x4)3/4
5x
x
sin x – cos x 122. Evaluate: Ú __________ dx ex + sin x dx 123. Evaluate: Ú _________ x (1 + x 3) Type 9 case I: 124. Evaluate: Ú sin3 x ◊ cos4x dx 125. Evaluate: Ú sin x ◊ cos6x dx 126. Evaluate: Ú sin5x ◊ cos9x dx case II: 127. Evaluate: Ú sin2x ◊ cos3x dx 128. Evaluate: Ú sin4x ◊ cos3x dx 129. Evaluate: Ú sin6x ◊ cos5x dx case III: 130. Evaluate: Ú sin3x ◊ cos3x dx 131. Evaluate: Ú sin5x ◊ cos5x dx 132. Evaluate: Ú sin5x ◊ cos7x dx case IV: 133. Evaluate: Ú sin2x ◊ cos2x dx 134. Evaluate: Ú sin2x ◊ cos4x dx
1.15
1.16 Integral Calculus, 3D Geometry & Vector Booster 135. Evaluate: Ú sin4x ◊ cos2x dx case V: 136. Evaluate: Ú sin3x dx 137. Evaluate: Ú sin5x dx 138. Evaluate: Ú sin7x dx case VI: 139. Evaluate: Ú cos5x dx 140. Evaluate: Ú cos3x dx 141. Evaluate: Ú cos7x dx case VII: 142. Evaluate: Ú sin4x dx 143. Evaluate: Ú sin2x dx 144. Evaluate: Ú sin6x dx case VIII: 145. Evaluate: Ú cos x dx 6
146. Evaluate: Ú cos2x dx 147. Evaluate: Ú cos4x dx case IX: dx 148. Evaluate: Ú ____________ 1/2 sin x cos3/2x dx 149. Evaluate: Ú ____________ 3/2 sin x cos5/2x ____ ÷tan x ________
150. Evaluate: Ú dx sin x cos x sin x 151. Evaluate: Ú _____ dx. cos5x dx 152. Evaluate: Ú _________ 3 sin x cos5x
dx 156. Evaluate: Ú ___________ 7 1 __ __ 2 2 sin x ◊ cos x dx __________ 157. Evaluate: Ú ___________ 3 5 x ◊ cos x ÷ sin ABC of Formula 6 dx 158. Evaluate: Ú ______ x 2 + 4 dx 159. Evaluate: Ú _______ 9x 2 + 1 dx 160. Evaluate: Ú ______ 2 x – 4 x 4 – 1 161. Evaluate: Ú _______ x2 + 5dx dx _____ 162. Evaluate: Ú _______ 4 ÷ x + 4 dx ______ 163. Evaluate: Ú ________ 2 + 1 ÷ 4x
(
)
x+4 164. Evaluate: Ú _______ dx x3 + 4x
( )
x4 + 1 165. Evaluate: Ú ______ 2 dx x +1 dx __________ 166. Evaluate: Ú ___________ – x)2 + 1 ÷ (2 x+9 167. Evaluate: Ú _______ dx x3 + 9x 1 + x 168. Evaluate: Ú _____2 dx 1+x 1+x 169. Evaluate: Ú _____ dx x3 + x dx 170. Evaluate: Ú _____ x4 + 1 dx 171. Evaluate: Ú _____ 3 x +x Type 10
sin2x dx 153. Evaluate: Ú _______ cos6x
dx 172. Evaluate: Ú __________ 3 x + 4x + 4
dx 154. Evaluate: Ú _________ sin x cos3x
dx 173. Evaluate: Ú ___________ 2 x + 6x + 10
dx 155. Evaluate: Ú _________ 2 sin x cos4x
dx 174. Evaluate: Ú ___________ 2 2x + 5x + 6
Indefinite Integrals
dx 175. Evaluate: Ú _________ x2 + x + 1 dx 176. Evaluate: Ú _________ 1 + x + x2 dx 177. Evaluate: Ú __________ 2 x + 4x + 3
3x + 2 195. Evaluate: Ú __________ dx x2 – 3x + 4 x 195. Evaluate: Ú _________ dx 2 x + x + 1 4x + 1 196. Evaluate: Ú __________ dx x2 + 3x + 2
dx 178. Evaluate: Ú ____________ 2 4x + 7x + 10
dx 197. Evaluate: Ú ____________ 2e2x + 3ex + 1
dx 179. Evaluate: Ú _______ x2 – 2ax
(3 sin x – 2) cos x dx 198. Evaluate: Ú _________________ (5 – cos2x – 4sin x)
dx 180. Evaluate: Ú _______ 2 x + 2ax
ax3 + bx 199. Evaluate: Ú ________ dx. x4 + c2
dx 181. Evaluate: Ú ________ 2 a + 2a x
cos x – sin x 200. Evaluate: Ú ___________ × (2 + 2 sin 2 x) dx cos x + sin x
dx 182. Evaluate: Ú _______ 2a x – x2
sin x + cos x 201. Evaluate: Ú ___________ dx 5 + 3sin2x
dx 183. Evaluate: Ú ______________ 2 (x + 1) (x2 + 4)
sin x – cos x 202. Evaluate: Ú ___________ dx 3 + 5sin 2x
x2 184. Evaluate: Ú ________ 6 x + 1dx
Type 12
cos x dx 185. Evaluate: Ú _______________ 2 sin x + 3sin x + 2 x
x (1 + log x) 186. Evaluate: Ú ___________ dx x2x + x x + 1 dx 187. Evaluate: Ú ________ 4 x(x + 1) x dx 188. Evaluate: Ú __________ 4 x + x2 + 1 ex dx 189. Evaluate: Ú ___________ e2x + 6ex + 5 5
x 190. Evaluate: Ú _______ dx 1 + x12 dx 191. Evaluate: Ú ________ 4 x(x + 1) dx 192. Evaluate: Ú ________ x(x3 + 1)
(
dx _________ 203. Evaluate: Ú ___________ 2 ÷ x + x + 1 dx ________ 204. Evaluate: Ú _________ 2 ÷ x – 2ax dx _______ 205. Evaluate: Ú ________ – x2 ÷ 4x dx _________ 206. Evaluate: Ú __________ ÷ 6 – x – x2 dx _________ 207. Evaluate: Ú __________ ÷ 1 + x + x2 dx _________ 208. Evaluate: Ú __________ ÷ 1 + x – x2 dx ________ 209. Evaluate: Ú _________ 2 ÷ x + 2ax dx ________ 210. Evaluate: Ú _________ – x2 ÷ 2ax Type 13
dx 193. Evaluate: Ú ________ x(xn + 1)
x dx __________ 211. Evaluate: Ú ___________ 4 + 1 ÷ x – x2
Type 11
212. Evaluate: Ú ÷sec x – 1 dx
2x + 3 194. Evaluate: Ú __________ dx 2 x + 4x + 5
)
________
dx _______ 213. Evaluate: Ú ___________ 3/4 1/2 x ÷ x – 1
1.17
1.18 Integral Calculus, 3D Geometry & Vector Booster
÷
_________
sin (x – a) 214. Evaluate: Ú _________ dx sin (x + a)
dx 233. Evaluate: Ú ______________ . 3 sin2x + 4 cos2x
ex ______ 215. Evaluate: Ú _______ dx ÷ 4 – e2x
dx 234. Evaluate: Ú ______________ (2sin x + 3 cos x)2
sec2x _________ 216. Evaluate: Ú __________ dx ÷16 + tan x ________
217. Evaluate: Ú ÷sec x + 1 dx _________
218. Evaluate: Ú ÷cosec x – 1 dx dx ______ 219. Evaluate: Ú ________ ÷ 1 – e2x
÷
_________
sin (x – a) 220. Evaluate: Ú _________ dx sin(x + a) dx ______ 221. Evaluate: Ú ___________ 2/3 x ÷x 2/3 – 4
÷
______
x 222. Evaluate: Ú ______ dx x 3 – x3
sin x cos x 235. Evaluate: Ú ____________ dx sin4x + cos4x dx 236. Evaluate: Ú _____________ (sin x + 2 cos x)2 dx 237. Evaluate: Ú _____________ (sin x + 2sec x)2 sin 2 x dx 238. Evaluate: Ú ____________ sin4x + cos4x dx 239. Evaluate: Ú _______________ (2 sin x + 3 cos x)2 dx 240. Evaluate: Ú _________ 2 + cos2 x Type 16
sinq – cosq _________ 224. Evaluate: Ú ___________ dq ÷2 – sin 2q
sin x 241. Evaluate: Ú _____ dx sin 3 x cosec 3x 242. Evaluate: Ú _______ dx cosec x sec 3x 243. Evaluate: Ú _____ dx sec x
Type 14
Type 17
x–1 __________ 225. Evaluate: Ú ___________ dx 2 + 2 ÷ x – 3x
dx 244. Evaluate: Ú _________ 1 + 2 sin x
(
)
cosq + sinq 223. Evaluate: Ú ___________ _________ dq ÷5 + sin 2q
3x + 4 __________ 226. Evaluate: Ú ___________ dx 2 + 2 ÷ x + 5x x+2 __________ 227. Evaluate: Ú ___________ dx 2 + 6 ÷ x + 5x 6x – 5 ___________ 228. Evaluate: Ú ____________ dx 2 – 5x + 1 ÷ 3x _____ a – x 229. Evaluate: Ú _____ dx a+x
÷
dx 245. Evaluate: Ú _________ 3 cos x + 4 dx 246. Evaluate: Ú ______________ 1 + sin x + cos x 1 + sin x 247. Evaluate: Ú _____________ dx sin x (1 + cos x) dx 248. Evaluate: Ú _________ 1 + 2 sin x
a2 – x2 230. Evaluate: Ú x2 _______ 2 dx a + x2
dx 249. Evaluate: Ú ________________ 3 sin x + 4 cos x + 5 dx 250. Evaluate: Ú ____________ cos x + cos a
4 – x3 231. Evaluate: Ú x2 ______ dx 4 + x3
dx 251. Evaluate: Ú _______________ tan x + 4 cot x + 4
Type 15
Type 18
dx 232. Evaluate: Ú _________ 3 + 4 sin2x
dx 252. Evaluate: Ú __________ sin x + cos x
÷ ÷
_______
______
Indefinite Integrals
dx __ 253. Evaluate: Ú _____________ 3 sin x + cos x ÷ dx__ 254. Evaluate: Ú _____________ sin x + ÷3 cos x dx __ 255. Evaluate: Ú _____________ 3 sin x + cos x ÷ dx __ 256. Evaluate: Ú _____________ . 3 sin x – cos x ÷ Type 19 2 sin x + cos x 257. Evaluate: Ú _____________ dx 3 sin x + 2 cos x sin x 258. Evaluate: Ú ___________ dx sin x + cos x 2 sin x + 3 cos x 259. Evaluate: Ú _____________ dx 3 sin x + 4 cos x
(
)
sin x 260. Evaluate: Ú __________ dx sin x + cos x
( ) 2 cos x – sin x 276. Evaluate: Ú ( ____________ dx 9 + 16 sin 2x ) cos x – sin x 275. Evaluate: Ú ___________ dx 5 – 7sin 2x
Type 22 x2 + 1 277. Evaluate: Ú ______ dx x4 + 1 x2 – 1 278. Evaluate: Ú ______ dx x4 + 1 x4 + 1 279. Evaluate: Ú ______ dx x6 + 1 x4 + 3x + 1 280. Evaluate: Ú __________ dx x4 + x2 + 1
( )
dx 1 – x2 ____________ 281. Evaluate: Ú ______ _________ 2 1+x ÷ 1 + x2 + x4
( )
cos x 261. Evaluate: Ú __________ dx sin x + cos x 1 262. Evaluate: Ú _______ dx 1 + tan x
dx x–1 282. Evaluate: Ú _____ × ___________ __________ 3 x+1 + x ÷ x + x2 2 dx x – 1 _______ 283. Evaluate: Ú ______ × _____ x2 + 1 ÷x 4 + 1
1 263. Evaluate: Ú _______ dx 1 – tan x
x2 – 1 ___________ 284. Evaluate: Ú ______________ dx x3 ÷ x 4 – 2x2 + 1
1 264. Evaluate: Ú _______ dx 1 + tan x
x4 – 1 __________ 285. Evaluate: Ú _____________ dx x2 ÷ x 4 + x2 + 1
(
Type 20 3 sin x + 2 cos x + 4 265. Evaluate: Ú _________________ dx 3 cos x + 4 sin x + 5 3 cos x + 2 266. Evaluate: Ú _______________ dx sin x + 2 cos x + 3 2 cos x + 3 sin x 267. Evaluate: Ú ________________ dx 2sin x + 3 cos x + 5 Type 21
(
)
sin x + 2 cos x 268. Evaluate: Ú ____________ dx 9 + 16 sin 2x 269. Evaluate: Ú (cos x – sin x) (2 + 3 sin 2x) dx sin x + cos x 270. Evaluate: Ú ___________ dx 9 + 16 sin 2x dx 271. Evaluate: Ú ____________ cos x + cosec x 272. Evaluate: Ú (sin x + cos x) (2 + 3 sin 2x) dx 273. Evaluate: Ú (sin x – cos x) (3 – 4 sin 2x) dx
(
)
cos x – sin x 274. Evaluate: Ú ___________ dx 3 + 2 sin 2x
)
dx______ 286. Evaluate: Ú _____________ 2 x (x + ÷ 1 + x2 ) x2 – 3x – 1 287. Evaluate: Ú __________ dx x4 + x2 + 1 dx 288. Evaluate: Ú ___________ 4 sin x + cos4x dx 289. Evaluate: Ú _____ x4 + 1 x2dx 290. Evaluate: Ú _____ x4 + 1 2dx 291. Evaluate: Ú _____ . 4 x +1
( )
x4 + 1 292. Evaluate: Ú ______ 6 dx x +1 dx 293. Evaluate: Ú __________ x2 (1 + x4)3/4
(
)
x4 – 1 __________ 294. Evaluate: Ú ______________ dx x2 ÷x 4 + x2 + 1
1.19
1.20 Integral Calculus, 3D Geometry & Vector Booster dx 295. Evaluate: Ú __________ dx x2(1 + x5)4/5
(
315. Evaluate: Ú x2 sin x dx 316. Evaluate: Ú log x dx
)
x2_____ –1 296. Evaluate: Ú ________ dx x ÷1 + x4
317. Evaluate: Ú (log x)2dx
dx ___________ 297. Evaluate: Ú _____________ 4 x ÷x + 3x2 + 1
318. Evaluate: Ú log(x2 + 1) dx
( )
dx x2 + 1 298. Evaluate: Ú ______ × ___________ __________ 4 1 – x2 + 1 ÷ x + x2
(
)
dx x2 + 1 299. Evaluate: Ú ______ x × ____________ ___________ 4 + 1 ÷ x + 3x2
319. Evaluate: Ú tan– 1x dx
( )
1 – x2 320. Evaluate: Ú cos– 1 ______ dx 1 + x2 __
321. Evaluate: Ú e÷ x dx
(
)
xx(x2x + 1) (ln x + 1) 300. Evaluate: Ú _________________ dx (x4x + 1)
x + sin x 322. Evaluate: Ú ________ dx 1 + cos x
x2 – 1 ___________ 301. Evaluate: Ú _____________ dx x3 ÷x 4 – 2x2 + 1
1 – x 323. Evaluate: Ú tan– 1 _____ dx 1+x
(
)
x2009 302. Evaluate: Ú __________ dx (1 + x2)1006
x tan– 1x 326. Evaluate: Ú _________ dx (1 + x2)3/2
Type 23
(
____
)
÷cot x – ÷tan x 304. Evaluate: Ú _____________ dx 1 + 3 sin 2x ____
305. Evaluate: Ú ÷tan x dx ____ ____ (÷ tan x – ÷cot x ) dx
306. Evaluate: Ú
______
307. Evaluate: Ú ÷cot x dx
( ) 1 309. Evaluate: Ú ( cot x + _____ dx cot x ) 1 310. Evaluate: Ú ( tan x + _____ dx tan x ) ____ 1 308. Evaluate: Ú ÷cot x – _____ ____ dx cot x ÷ ____ ÷
____ ÷
____ ÷
____ ÷ 4 ____ 4 ____ 2 311. Evaluate: ( ÷tan x + ÷cot x ) dx
Ú
dx 312. Evaluate: Ú _______________ 4 dx ____ _____ (÷sin x + ÷ cos x )
Integration by Parts ABC of integration by parts 313. Evaluate: Ú x ex dx 314. Evaluate: Ú x sin x dx
__
__
– 1 sin– 1 ÷x – cos ÷x 324. Evaluate: Ú ________________ dx __ – 1 – 1 __ sin ÷x + cos ÷x
x2 325. Evaluate: Ú _____________ dx (x sin x + cos x)2
(2x + 1) dx 303. Evaluate: Ú _____________ 2 (x + 4x + 1)3/2 ____
( ÷ ) _____
_____
( ÷
)
x 327. Evaluate: Ú sin– 1 _____ dx a+x 328. Evaluate: Ú log(1 + x) dx x – sin x 329. Evaluate: Ú ________ dx 1 – cos x ____
330. Evaluate: Ú sin ÷x dx 331. Evaluate: Ú x log (1 + x) dx 332. Evaluate: Ú (sin– 1x)2 dx x – sin x 333. Evaluate: Ú ________ dx 1 – cos x 334. Evaluate: Ú x sin3x dx
(
)
sec 2 x – 1 335. Evaluate: Ú x _________ dx sec 2 x + 1
( ) ( )
2x 336. Evaluate: Ú sin–1 _____ 2 dx 1+x 2x 337. Evaluate: Ú tan–1 _____ 2 dx 1–x x3sin– 1(x2) _____ dx 338. Evaluate: Ú _________ ÷ 1 – x4
Indefinite Integrals
sec x (2 + sec x) 339. Evaluate: Ú ______________ dx (1 + 2sec x)2
359. Evaluate: Ú ex(2sec2x – 1) tan x dx 360. Evaluate: Ú ex (log (sec x + tan x) + sec x) dx
1–x 340. Evaluate: Ú ______ dx. ex + x
(
Type 24
Type 25
341. Evaluate: Ú ex(sin x + cos x) dx
(
{
)
( (
)
)
(
} )
{ } 2 + sin 2x 349. Evaluate: Ú e { _________ dx 1 + cos 2x }
367. Evaluate: Ú e3x(3sin x + cos x) dx 368. Evaluate: Ú e2x(sec2x + 2 tan x) dx
x
(
)
370. Evaluate: Ú e2x (– sin x + 2 cos x) dx
(
log x 351. Evaluate: Ú __________ 2 dx (1 + log x)
)
)
373. Evaluate: Ú e2x(2 × log (sec x + tan x) + sec x) dx Type 27
}
( )
374. Evaluate: Ú ex sin 3x dx 375. Evaluate: Ú e4x cos 3x dx
1 – x 2 355. Evaluate: Ú e _____ dx 1 + x2
(
376. Evaluate: Ú e2x sin 3x dx
)
377. Evaluate: Ú e–x cos x dx
1 – x 2 356. Evaluate: Ú ex _____ dx 1+x
(
)
378. Evaluate: Ú e2xcos (3x + 4) dx
x2 + 1 357. Evaluate: Ú e _______ dx (x + 1)2 x
(
)
1 – sin x 372. Evaluate: Ú e– x/2 ________ dx 1 + cosx
1 1 354. Evaluate: Ú ____ – ______ dx log x (log x)2 x
( ÷
________
x–1 353. Evaluate: Ú ex _______ dx (x + 1)3
{
)
1 + sin 2x 371. Evaluate: Ú e2x _________ dx 1 + cos 2x
1 352. Evaluate: Ú ex log x + __ x dx
(
)
2 sin 4x – 4 369. Evaluate: Ú e2x __________ dx 1 – cos 4x
x2 + 1 350. Evaluate: Ú e _______ dx (x + 1)2
(
)
Type 26
1 – sin x 348. Evaluate: Ú ex ________ dx 1 – cos x
(
)
x3 – x + 2 366. Evaluate: Ú ex _________ 2 dx (x – 1)2
1 1 346. Evaluate: Ú ex __ x – __ 2 dx x x 347. Evaluate: Ú ex _______ dx (x + 1)2
x
(
esin x(x cos3x – sin x) 365. Evaluate: Ú _________________ dx cos2x
x2 + 1 345. Evaluate: Ú e _______ dx (x + 1)2
{ (
}
x4 + 2 364. Evaluate: Ú ex _________ dx (1 + x2)5/2
1 + x + x dx 344. Evaluate: Ú e _________ (1 + x2)3/2 x
)
1 363. Evaluate: Ú log (log x) + ______ 2 dx (log x)
2 + sin 2x 343. Evaluate: Ú ex _________ dx 1 + cos 2x 3
(
1 362. Evaluate: Ú ex log x + __2 dx x
x ex 342. Evaluate: Ú _______ dx (x + 1)2
x
)
x cos3x – sin x 361. Evaluate: Ú esin x ____________ dx cos2x
_____
)
(x + 1) + ÷1 – x2 _____ 358. Evaluate: Ú ex _______________ dx 2 2 (x + 1) ÷1 – x
379. Evaluate: Ú excos2x dx 1 380. Evaluate: Ú __3 sin(log x) dx. x
1.21
1.22 Integral Calculus, 3D Geometry & Vector Booster ______
2x 404. Evaluate: Ú _____________ dx (x2 + 1)(x2 + 2)
_______
cosq 405. Evaluate: Ú __________________ dq (2 + cosq )(3 + cosq )
Type 28 381. Evaluate: Ú ÷4 – x2 dx 382. Evaluate: Ú ÷1 – 9x2 dx _____
383. Evaluate: Ú ÷x 2 + 1 dx ______
384. Evaluate: Ú ÷3x 2 + 1 dx ______
dx 407. Evaluate: Ú ___________ sin x – sin 2 x
_______
Type 2
385. Evaluate: Ú ÷x 2 – 9 dx 386. Evaluate: Ú ÷4x 2 – 1 dx
__________ ÷x 2 + 2x + 3 dx __________ ÷3 – 4x – x2 dx ________ ÷2ax – x2 dx ________ ÷2ax + x2 dx
387. Evaluate: Ú 388. Evaluate: Ú 389. Evaluate: Ú 390. Evaluate: Ú
(1 – cos x) 406. Evaluate: Ú _____________ dx cos x (1 + cos x)
_______
391. Evaluate: Ú ÷x – 4x2 dx Type 29 392. Evaluate: Ú (2x +
__________ 1) ÷x 2 + 3x + 4 dx ______
393. Evaluate: Ú (x – 5) ÷x 2 + x dx
_________
394. Evaluate: Ú (3x – 2) ÷x 2 + x + 1 dx _________ (4x + 1) ÷x 2 – x – 2 dx _________ x ÷1 + x – x2 dx.
395. Evaluate: Ú 396. Evaluate: Ú
Partial Fractions Type 1 (2x + 1) dx 397. Evaluate: Ú ____________ (x + 2)(x + 3) dx 398. Evaluate: Ú ____________ (x – 1)(x – 2) dx 399. Evaluate: Ú __________________ (x + 1)(x + 2)(x + 3) dx 400. Evaluate: Ú ________________________ (x + 1) (x + 2) (x + 3) (x + 4) x–1 401. Evaluate: Ú ____________ dx (x + 1)(x – 2)
dx 408. Evaluate: Ú _____________ (x + 1)(x + 1)2 2x + 1 409. Evaluate: Ú ____________ dx (x + 2)(x – 3)2 3x + 1 410. Evaluate: Ú ____________ dx (x – 2)2(x + 2) x2 + 1 411. Evaluate: Ú _____________ dx (x – 1)2 (x + 3) x2 412. Evaluate: Ú ____________ dx (x – 1)3 (x + 1) (x – 1) 413. Evaluate: Ú ________ dx x2(x + 4) (2x – 1) 414. Evaluate: Ú _______ dx x3(x – 2) Type 3 (2x + 3) dx 415. Evaluate: Ú _____________ (x + 1)(x2 + 4) (3x – 2) dx 416. Evaluate: Ú ____________ (x – 1)(x2 + 9) 2x – 1 417. Evaluate: Ú _____________ dx (x + 1)(x2 + 2) x 418. Evaluate: Ú _____________ dx (x + 1) (x2 + 4) 8 419. Evaluate: Ú _____________ dx (x + 2) (x2 + 9) x 420. Evaluate: Ú ____________ dx (x + 1)(x2 + 1) Type 4
2x – 1 402. Evaluate: Ú ___________________ dx (x + 1) (x + 2) (x + 3)
x2 421. Evaluate: Ú _____________ dx (x2 – 1) (x2 + 1)
x3 403. Evaluate: Ú ____________ dx (x – 1)(x – 2)
x2 422. Evaluate: Ú _____________ dx (x2 – 3) (x2 + 4)
Indefinite Integrals
(x2 + 4) 423. Evaluate: Ú _____________ dx (x2 + 5)(x2 + 7) x2 424. Evaluate: Ú _____________ dx (x2 – 1)(x2 – 2) x2 425. Evaluate: Ú ___________________ dx (x2 – 1)(x2 – 2)(x2 – 3) Type 5
(
)
(x2 + 3)(x2 + 1) 426. Evaluate: Ú _____________ dx (x2 – 1)(x2 + 2) 2
2
(x + 1)(x + 2) 427. Evaluate: Ú _____________ dx (x2 + 3)(x2 + 4) (x2 – 1)(x2 + 3) 428. Evaluate: Ú _____________ dx. (x2 + 2)(x2 + 1)
Type 3 dx 17. Evaluate: Ú ________ 7 x(x + 1) dx 18. Evaluate: Ú __________ 2 4 x (x + 1)3/4 dx 19. Evaluate: Ú ________ 5 x (x + 1) dx 20. Evaluate: Ú ________ 4 x (x + 1) dx 21. Evaluate: Ú __________ 2 7 x (x + 1)6/7 dx 22. Evaluate: Ú __________ 3 x (1 + x3)1/3 Type 4 x2 23. Evaluate: Ú _______ dx (x + 3)2 x3 dx 24. Evaluate: Ú ________ (2x + 3)2
Type 1
x2 25. Evaluate: Ú _______ dx (x + 2)3
1. Evaluate: Ú tan x ◊ sec x dx
2. Evaluate: Ú sec3x dx
3. Evaluate: Ú sec5x dx
4. Evaluate: Ú tan2x ◊ sec4x dx
dx 27. Evaluate: Ú _________ x (1 + x3)2
5. Evaluate: Ú tan3x ◊ sec6x dx
6. Evaluate: Ú tan3x ◊ sec5x dx
x4 28. Evaluate: Ú ________ dx (3x – 2)3
7. Evaluate: Ú sec7x dx
4
2
8. Evaluate: Ú sec9 x dx
Type 2
9. Evaluate: Ú cosec2 x ◊ cot2 x dx
10. Evaluate: Ú cot3 x ◊ cosec3 x dx 11. Evaluate: Ú cosec3 x dx 12. Evaluate: Ú cot2 x ◊ cosec4 x dx 13. Evaluate: Ú tan–5x ◊ sec6 x dx 14. Evaluate: Ú cot3 x ◊ cosec–8 x dx
x2 26. Evaluate: Ú ________ dx (ax + b)2
Type 5 dx 29. Evaluate: Ú _________ 2 x (3x + 2)3 dx 30. Evaluate: Ú _________ 3 x (b + ax)2 dx 31. Evaluate: Ú ________ 2 3 x (x + 2) dx 32. Evaluate: Ú _________ 3 x (a + bx)2 dx 33. Evaluate: Ú _________ 2 x (1 + x2)3
15. Evaluate: Ú cosec x dx
dx 34. Evaluate: Ú _________ 2 x (a – bx)2
16. Evaluate: Ú cosec7 x dx
dx 35. Evaluate: Ú _________ 4 x (2x + 1)3
5
1.23
1.24 Integral Calculus, 3D Geometry & Vector Booster Type 6 dx 36. Evaluate: Ú _____________ (x – 1)3 (x – 2)4 dx 37. Evaluate: Ú _______________ 4 _____________ – 1)3 (x + 2)5 ÷ (x dx 38. Evaluate: Ú _____________ (x – 1)3 (x – 2)2 dx 39. Evaluate: Ú ______________ (x – 3)4 (x – 2)5 dx 40. Evaluate: Ú ________________ 3/2 (x – 3) (x – 2)7/2 dx ______________ 41. Evaluate: Ú _______________ 5 + 1)4 (x + 3)6 ÷ (x dx ______________ 42. Evaluate: Ú _______________ 5 ÷(x + 1)4 (x + 3)7 Type 7 dx 43. Evaluate: Ú _________ x (2 + 3x3) dx 44. Evaluate: Ú _________ x (3 + 5x5) dx 45. Evaluate: Ú _________ x (2 + 3x2) dx 46. Evaluate: Ú _________ x (3 + 4x3) dx 47. Evaluate: Ú _________ x (2 – 5x3) dx 48. Evaluate: Ú _________ x (1 – 4x4) dx 49. Evaluate: Ú _________ x (3x4 + 1) Type 8 x5 50. Evaluate: Ú _______ 2 4 dx (x – 1) x9 51. Evaluate: Ú _________ 5 dx 2 (2x + 3) x3 52. Evaluate: Ú _______ 2 4 dx (x + 1) 5
8x9 55. Evaluate: Ú ________ 2 5 dx (3x – 2) 10x11 56. Evaluate: Ú ________ 2 dx (3x + 5)4 Type 9 2 sin x + 3 57. Evaluate: Ú __________ dx (3sin x + 2)2 (2 sin x + 5) 58. Evaluate: Ú ___________ dx (2 + 5 sin x)2 (3 sin x – 2) 59. Evaluate: Ú ___________ dx (2 – 3 sin x)2 Type 10
(
)
4 cos x + 3 60. Evaluate: Ú ___________ dx (3 cos x + 4)2 (cos x + 2) 61. Evaluate: Ú ___________ dx (1 + 2 cos x)2 3 cos x + 4 62. Evaluate: Ú ___________ dx (3 + 4 cos x)2 Type 11 dx 63. Evaluate: Ú ___________ (3 + 4 sin x)2 dx 64. Evaluate: Ú ___________ (5 + 4 sin x)2 dx 65. Evaluate: Ú __________ (1 – 2 sin x)2 Type 12 dx 66. Evaluate: Ú ___________ (2 + 3 cos x)2 dx 67. Evaluate: Ú _____________ (12 + 13 cos x)2 dx 68. Evaluate: Ú ___________ (3 – 4 cos x)2 Type 13
(
)
4ex + 6e–x 69. Evaluate: Ú _________ x dx. 9e – 4e–x
x 53. Evaluate: Ú _______ 2 4 dx (x – 3)
3ex – 2e–x 70. Evaluate: Ú _________ dx 2ex + 5e–x
x7 54. Evaluate: Ú ________ 2 4 dx (3x – 2)
4ex + 3e–x 71. Evaluate: Ú _________ dx 3ex + 7e–x
Indefinite Integrals
Integration of Irrational Functions Type 1 dx ______________ 72. Evaluate: Ú _______________ + 2)5 (x + 1)3 ÷ (x
(
)
dx x + 2 1/2 ___ ◊ x 73. Evaluate: Ú ______ 2x + 3 (x + 1) 74. Evaluate: Ú ______________ dx (x + 2) (x + 3)3/2
÷
_____
32 – x 2 75. Evaluate: Ú _______ 2 _____ dx (2 – x) 2 + x
Type 2 dx 76. Evaluate: Ú ______________ _____ _____ 3 x + 1 x + 1 + ÷ ÷ dx 77. Evaluate: Ú _________________ _____ 1/4 x + 1 – (x + 1) ÷ 1 + x1/2 – x1/3 78. Evaluate: Ú ____________ dx 1 + x1/3
dx 90. Evaluate: Ú _________ 2 __ (x + 1) ÷x x dx 91. Evaluate: Ú _________________ _____ 2 (x + 2x + 2) ÷x + 1 dx 92. Evaluate: Ú _________ 2 __ (x – 1) ÷x x dx 93. Evaluate: Ú _________________ _____ (x2 – 2x + 2) ÷x – 1 Type 5 dx _____ 94. Evaluate: Ú ____________ (x + 1) ÷x 2 + 1 dx __________ 95. Evaluate: Ú _________________ (x + 1) ÷x 2 + 2x + 2 dx 96. Evaluate: Ú ________ _____ x ÷x 2 + 4 dx _____ 97. Evaluate: Ú ____________ (x – 1) ÷x 2 + 4
dx 79. Evaluate: Ú ________ 1/2 1/3 x +x
dx _____ 98. Evaluate: Ú _____________ (2x – 1)÷x 2 + 1
x ÷ 80. Evaluate: Ú 4_______ _____ dx 3 ÷x + 1
dx _____ 99. Evaluate: Ú _____________ (3x + 2) ÷x 2 – 4
x ÷ 81. Evaluate: Ú _______ __ dx __ x + ÷ 3 x ÷
Type 6
__
__
Type 3
dx 100. Evaluate: Ú ________ _____ 2 2 x ÷ x – 1
dx _____ 82. Evaluate: Ú ___________ (x – 1) ÷x + 3
dx _____ 101. Evaluate: Ú ______________ 2 (x + 1) ÷x 2 + 2
dx _____ 83. Evaluate: Ú ___________ (x + 3) ÷x + 2
dx _____ 102. Evaluate: Ú _____________ 2 (x – 1) ÷x 2 + 2
__
x ÷ 84. Evaluate: Ú _____ dx x+1 dx 85 .Evaluate: Ú _______ _____ x÷x – 2 dx 86. Evaluate: Ú ________ __ (x + 3) ÷x
dx ______ 87. Evaluate: Ú ____________ (x + 3) ÷2x + 1 Type 4 dx 88. Evaluate: Ú ________ 2 _____ x ÷ x – 1 dx _____ 89. Evaluate: Ú ____________ 2 (x – 4) ÷x + 1
dx _____ 103. Evaluate: Ú _____________ 2 (1 + x ) ÷1 – x2 x dx_____ 104. Evaluate: Ú _____________ 4 (x – 1) ÷x 4 + 3 dx __________ 105. Evaluate: Ú __________________ 2 (x – 1) ÷x 2 + 4x + 5 Type 7 _____
106. Evaluate: Ú (x + ÷ x 2 + 1 )10 dx _____
107. Evaluate: Ú (x – ÷x 2 + 4 )5 dx _____
108. Evaluate: Ú (x + ÷ 1 + x2 )n dx
1.25
1.26 Integral Calculus, 3D Geometry & Vector Booster dx _____ 109. Evaluate: Ú _____________ (x + ÷x 2 – 4 )5/3
dx 127. Evaluate: Ú ________ ______ x ÷3x – 2
dx _____ 110. Evaluate: Ú _____________ (x – ÷x 2 + 9 )10
dx 128. Evaluate: Ú _________ ______ x ÷3x 9 – 2
dx_____ 111. Evaluate: Ú ____________ 2 x (x – ÷ x 2 + 9 )
dx 129. Evaluate: Ú __________ _______ x ÷2 x 10 – 3
Type 8
Type 11
dx 112. Evaluate: Ú ____________ x1/2 (2 + 3x)3/2
dx 130. Evaluate: Ú __________ (2 + 3x2)3/2
dx 113. Evaluate: Ú ____________ x2/3 (2 + 3x)4/3
dx 131. Evaluate: Ú _________ (c + dx2)3/2
dx 114. Evaluate: Ú ____________ 3/4 x (3x – 1)5/4
dx 132. Evaluate: Ú __________ (3 + 5x2)3/2
dx 115. Evaluate: Ú ____________ 1/3 x (2x + 1)5/3
dx 133. Evaluate: Ú _________ (3 – 4x2)3/2
dx 116. Evaluate: Ú ____________ 1/2 x (2 + 3x)5/2
x dx 134. Evaluate: Ú _________ (2 – 5x4)3/2
dx 117. Evaluate: Ú ___________ x2(2 + 3x2)5/2
x2 dx 135. Evaluate: Ú _________ (1 – 4x6)3/2
Type 9
Type 12
dx 118. Evaluate: Ú _____________ (x – 1)3 (x + 2)4
dx _________ 136. Evaluate: Ú _________________ 2 2 (x – 2) ÷ x – 4x + 7
dx 119. Evaluate: Ú _____________ (x – 1)3 (x – 2)2
dx _________ 137. Evaluate: Ú _________________ 3 2 (x + 1) ÷ x + 2x + 4
dx 120. Evaluate: Ú _____________ (x – 1)2 (x – 2)3
dx ____________ 138. Evaluate: Ú ____________________ 3 (x – 2) ÷4x 2 – 16x + 20
dx ______________ 121. Evaluate: Ú 4_______________ ÷ (x – 1)3 (x + 2)5
dx _____________ 139. Evaluate: Ú _________________________ 2 (x – 6x + 9) ÷4x 2 – 24 x + 20
dx 122. Evaluate: Ú ________ 2 x (x + 5)4
dx ___________ 140. Evaluate: Ú ________________________ 2 (4x + 4x + 1) ÷4x 2 + 4x + 7
dx 123. Evaluate: Ú ________________ 3/2 (x – 1) (x + 1)5/2
dx __________ 141. Evaluate: Ú __________________ 3 (x + 1) ÷x 2 + 2x – 4
Type 10
Type 13
dx 124. Evaluate: Ú _________ ______ x ÷3x 3 + 4
(2x + 3) __________ 142. Evaluate: Ú __________________ dx (3x + 4)÷x 2 + 2x + 4
dx 125. Evaluate: Ú _________ ______ x ÷5x 4 + 3
(2x + 3) __________ 143. Evaluate: Ú _________________ dx (x + 1) ÷x 2 + 2x + 9
dx 126. Evaluate: Ú _________ ______ x ÷2 – 5x6
(4x + 7) __________ 144. Evaluate: Ú _________________ dx (x + 2) ÷x 2 + 4x + 7
Indefinite Integrals
Type 14 2
x + __________ 4x + 2 145. Evaluate: Ú _________________ dx 2 (x + 1) ÷x + 2x + 3 x2 + __________ 5x + 6 146. Evaluate: Ú _________________ dx 2 (x + 2) ÷x + 5x + 4 x2 + 10x +6 __________ 147. Evaluate: Ú _________________ dx 2 (x + 2) ÷x + 4x + 9
dx __________ 164. Evaluate: Ú ____________ 2 x ÷x – 3x + 2 dx ______ 165. Evaluate: Ú __________ x+÷ x 2 – 1 x dx _____ 166. Evaluate: Ú _________ x+÷ x 2 – 1
Reduction Formulae Type 1
Type 15 __
167. Evaluate: Ú sin5 x dx
__
3 3 148. Evaluate: Ú ÷ x (1 + ÷ x ) dx
__
149. Evaluate: Ú ÷x 2 (3 + x–2/3)–2 dx 3
(
______ __
)
x ÷ 1 +__÷4 150. Evaluate: Ú _______ dx 3 4 ÷ x dx 151. Evaluate: Ú ________ 2 _____ 7 x ÷ 1 + x4
170. Evaluate: Ú cos8 x dx Type 3 171. Evaluate: Ú tan5 x dx
153. Evaluate: Ú x–1/2 (2 + 3x1/3)–2 dx 154. Evaluate: Ú
________ __ 3 1 + ÷x 4
÷( 7
)
155. Evaluate: Ú x–6 (1 + 2x3)2/3 dx dx 156. Evaluate: Ú ________ 3 _____ x ÷ 1 + x5 dx 157. Evaluate: Ú _________ 2 _____ 11 x ÷ 1 + x4
(
)
________ __ _________ (1 + 4÷x ) 4 __ dx ÷x 3
158. Evaluate: Ú
÷
3
Type 2 169. Evaluate: Ú cos7 x dx
dx 152. Evaluate: Ú ___________ __ __ 4 10 x ( ÷ x + 1 ) ÷
__ 3÷x ×
168. Evaluate: Ú sin6 x dx
Type 16 dx _________ 159. Evaluate: Ú _______________ 2 (1 + ÷ x + x + 1 ) dx _________ 160. Evaluate: Ú _____________ x+÷ x 2 – x + 1 x dx ___________ 161. Evaluate: Ú ____________ – 10 – x2 ÷ 7x dx _________ 162. Evaluate: Ú _____________ 2 x – ÷x – x + 2 dx __________ 163. Evaluate: Ú ______________ 2 x–÷ x – 2x + 4
172. Evaluate: Ú tan6 x dx Type 4 173. Evaluate: Ú cot7 x dx 174. Evaluate: Ú cot6 x dx Type 5 175. Evaluate: Ú sec3 x dx 176. Evaluate: Ú sec4 x dx 177. Evaluate: Ú sec5 x dx 178. Evaluate: Ú sec7 x dx Type 6 sin 3 x 179. Evaluate: Ú _____ dx sin x sin 5 x 180. Evaluate: Ú _____ dx sin x sin 6 x 181. Evaluate: Ú _____ dx sin x sin 8 x 182. Evaluate: Ú _____ dx sin x Type 7 dx 183. Evaluate: Ú _______ 2 2 (x + 2)
1.27
1.28 Integral Calculus, 3D Geometry & Vector Booster dx 184. Evaluate: Ú _______ (x2 + 3)3
x + 1 185. Evaluate: Ú ____________ dx 2 2 (x + 3x + 2)
Type 8 186. Evaluate: Ú x2 log x dx 187. Evaluate: Ú x2 (log x)2 dx
188. Evaluate: Ú x3 (log x)2 dx 189. Evaluate: Ú x2 (1– x)3 dx
Mixed Problems
190. Evaluate: Ú sin8 x dx
sin x 3. Ú _____ dx = sin 3 x
| |
|
__
3 + tan x ÷ 1__ __ (a) ___ log _________ + c 3 ÷3 – tan x ÷ __
|
3 + tan x ÷ 1__ __ (b) ____ log _________ + c 2÷3 ÷3 – tan x
|
__
|
3 + tan x ÷ __ (c) log _________ + c 3 – tan x ÷
|
__
|
3 + tan x ÷ 1__ __ (d) ___ log _________ + c 3 ÷3 – tan x ÷ dx 4. Ú __________________ = sin (x – a) cos (x – b)
191. Evaluate: Ú cos10 x dx
1 (a) _________ (log |sin (x – a)| + log |sec (x – b)|) + c cos (a – b)
192. Evaluate: Ú tan8 x dx
(b) ( log | sin (x – a) | + log | sec (x – b) | ) + c
(c) (log |sin (x – a)| + c)
(d) (log |sec (x – b )| + c)
193. Evaluate: Ú cot8 x dx 194. Evaluate: Ú cot x dx
195. Evaluate: Ú sec9 x dx
196. Evaluate: Ú xn ex dx
9
dx 197. Evaluate: Ú _______ (x2 + 2)2
1.
Ú x2 cos x dx =
5. Ú tan3 2 x ◊ sec 2x dx = 1 1 (a) __ sec3 2 x – __ sec 2 x + c 3 2 1 1 3 (b) – __ sec 2 x – __ sec 2 x + c 2 6 1 1 (c) __ sec3 2 x – __ sec 2 x + c 2 6 1 1 3 (d) __ sec 2 x + __ sec 2 x + c 3 2 1 6. If Ú f (x) sin x ◊ cos x dx = _______ 2 2 log ( f (x)) + k, then (b – a ) f (x) is
1 (a) ________________ + c 2 2 a sin x + b2 cos2 x
1 (b) _______________ + c a2 sin2 x – b2 cos2 x
Ú ex cos2 x dx =
1 (a) ex + ___ ex (cos 2 x + 2 sin 2 x) + c 10
1 (c) _______________ + c a2 cos2 x + b2 sin2 x
1 (b) __ ex + ex (cos 2 x + 2 sin 2 x) + c 2
1 (d) _______________ + c a2 cos2 x – b2 sin2 x
(a) x2 sin x (b) x2 sin x (c) x2 sin x (d) x2 sin x
2.
+ + + +
2 x cos x – 2 sin x + c 2 x cos x + 2 sin x + c 2 x cos x + sin x + c 2 cos x + sin x + c.
1 (c) __ ex + (cos 2 x + 2 sin 2 x) + c 2
1 1 (d) __ ex + ___ ex (cos 2 x + 2 sin 2 x) + c 2 10
7. If
cos 4 x + 1 __________ dx = k cos 4 x + c, then Ú ( cot x – tan x )
1 (a) k = – __ 2 1 (c) k = – __ 4
1 (b) k = – __ 8 1 (d) k = __ 6
dx 8. Ú ______ = 6 x + x4
1 1 (a) – ___ 2 + __ x + cosec–1x + c 3x 1 1 (b) – ____ 2 + __ x + cot–1 x + c 3 x 1 1 (c) – ___ 3 + __ + tan–1 x + c x 3x 1 1 (d) ____ 2 – __ x + sin–1 x + c 3 x
|
|
sin x – (x + 1) cos x – 1 1 (a) __ log _____________________ + c 2 sin x – (x + 1)cos x + 1 1 (b) __ tan–1 {sin x – (x + 1) cos x} + c 2 1 (c) __ sin–1 {sin x – (x + 1) cos x} + c 2 1 (d) __ sin–1 (sin x + cos x) + c 2
{
cos x (a) _________ + c 2 + 5 sin x
– cos x (b) _________ + c 2 + 5 sin x
1 (c) _________ + c 2 + 5 sin x
sin x (d) _________ + c 2 + 5 sin x
}
(log x – 1) 10. Ú __________ dx = 1 + (log x)2
x ex (a) _____ + c 1 + x2
x ex (b) __________ + c 1 + (log x)2
log x (c) __________ + c 1 + (log x)2
x (d) _____ + c 1 + x2
}
( ( ( (
) ) )
x 1 (a) ex _______ _____ + _________ _______ + c 2 ÷(1 + x2)3 ÷ 1 + x
x 1 (b) ex _______ _____ – _________ _______ + c 2 ÷(1 + x2)3 ÷ 1 + x
x 1 (c) ex _______ _____ + _________ _______ + c 2 ÷ ÷ 1 + x (1 + x2)5
x 1 (d) ex ______ _____ – ________ _______ + c 2 ÷(1 + x2) ÷ 1 – x
)
x cos x + 1 ___________ 15. Ú ____________ dx = 3 e sin x + x2 ÷ 2x
( ( ( (
_________
) ) ) )
2x e sin x + 1 – 1 ÷ __________ (a) log ______________ + c sin x + 1 + 1 ÷ 2 xe
2
cosec x – 2005 11. Ú _____________ dx = cos2005 x
_________
cot x (a) _______ + c cos2005 x
tan x (b) _______ + c cos2005 x
sin x – 1 + 1 ÷ 2x e (b) log ______________ __________ + c sin x + 1 + 1 ÷ 2 xe
– tan x (c) _______ + c cos2005 x
– cot x (d) _______ + c cos2005 x
2x e sin x + 1 +1 ÷ _________ (c) log ______________ + c sin x – 1 + 1 ÷ 2 xe
sin x _________ + 1 + 1 ÷ 2x e (d) log ______________ + c sin x – 1 – 1 ÷ 2 xe
sin 2 x + 2 tan x 12. Ú __________________ dx = ( cos6x + 6 cos2 x + 4 )
÷
1.29
2x2 1 14. Ú ex _______ _____ + 1 – _________ ________ dx = ÷ 1 + x2 ÷ (1 + x2)5
2 sin x + 5 9. Ú ___________ dx = (2 + 5 sin x)2
{
Indefinite Integrals
________
_________
_________
1 + cos2 x (a) 2 ________ +c cos7x
16. Let f (x) be a function such that f (0) = f ¢(0) = 0, f ¢¢(x)
1 + cos2 x 1 (b) tan ___ __ × ________ +c 2 ÷ cos7x
1 + cos2 x 1 (c) ___ log ________ + c 12 cos7x
6 1 4 (d) ___ log 1 + _____ 4 + _____ 6 + c 12 cos x cos x
–1
( ) ÷
= sec4 x + 4, the function is
________
(
(
)
)
(1 + x) sin x 13. Ú __________________________ dx = 2 (x + 2x) cos2 x – (1 + x) sin 2 x
1 (a) log (sin x) + __ tan3 x + cx 3 2 1 (b) __ log (sec) + __ tan2 x + 2x2 3 6 x2 1 (c) log (cos x) + __ cos2 x + __ 6 5 (d) none.
17. The value of the integral I = where p x Π0, __ 2
( )
____
____
+ ÷cot x ) dx, Ú (÷tan x
1.30 Integral Calculus, 3D Geometry & Vector Booster
__
(a) ÷2 sin–1 (cos x – sin x) + c
(b) ÷ 2 sin (sin x – cos x) + c
ln |x| _______ 23. Ú ____________ dx = x × ÷1 + ln |x|
(c) ÷2 sin–1 (sin x + cos x) + c
(d) – ÷2 sin–1 (sin x + cos x) + c
__
–1
__
__
____
____
18. The value of the integral I = Ú ( ÷tan x + ÷cot x ) dx,
3p where x Πp, ___ 2
( )
(a) ÷2 sin–1 (cos x – sin x) + c
(b) ÷2 sin–1 (sin x – cos x) + c
__ __ __
(c) ÷2 sin (sin x + cos x) + c (d) – ÷2 sin–1 (sin x + cos x) + c ____
____
)
p 3p where x Œ 0, __ » p, ___ 2 2
____ ____ __ tan x – ÷cot x –1 ÷ ____________ __ (a) ÷2 tan + c ÷2 ____ ____ __ ÷tan x + ÷cot x –1 _____________ __ (b) ÷2 tan + c ÷2 ____ ____ __ ÷tan x – ÷cot x –1 _____________ __ (c) – ÷2 tan + c ÷2 ____ ____ __ ÷tan x + ÷cot x __ (d) – ÷2 tan–1 _____________ + c ÷2
) )
( (
+ 2) + c
(a) – 1/9 (b) – 1/3 (c) 1/3 (d) Non existent. 25. The number of values of x satisfying the equation 3 __ x x + 1 28 2 2 ___ _______________ ______ Ú 8t + t + 4 dt = , 3 log(x + 1) ( ÷(x + 1) ) –1
(
)
is (a) 0 (c) 2
(b) 1 (d) 3.
________________________
26. Ú ÷(1 + 2 cot x (cot x + cosec x)) dx =
) )
(
– 2) + c
0
__
( (
+ 2) + c
24. If Ú t2 dt = x cos (p x), the value of f ¢(9) is
19. The value of the integral I = Ú (÷tan x + ÷ cot x )dx,
– 2) + c
f (x)
–1
( ) (
2 _______ (a) __ ÷1 + ln |x| × (ln |x| 3 2 _______ (b) __ ÷ 1 + ln |x| × (ln |x| 3 1 _______ (c) __ ÷ 1 + ln |x| × (ln |x| 3 1 _______ (d) __ ÷ 1 + ln |x| × (ln |x| 3
)
x4 cos3 x – x sin x + cos x 20. Ú e(x sin x + cos x) ____________________ dx x2 cos2 x
( ) 1 ( ______ x cos x ) + c 1 ( ______ x cos x – x ) + c 1 ( ______ x cos x + x ) + c
( ( (
( ) ) ( ) ) ( ) )
x (a) 2 ln cos __ + c 2 x (b) 2 ln sin __ + c 2 x 1 (c) __ ln sin __ + c 2 2 (d) None.
1 (a) e(x sin x + cos x) x – ______ x cos x + c
27. A differentiable function satisfies 3f 2(x) f ¢(x) = 2x. Given f (2) = 1, the value of f (3) is
(b) e(x sin x + cos x)
(a) ÷24
(b) ÷6
(c) e(x sin x + cos x)
(c) 6
(d) 2.
(d) e(x sin x + cos x)
1 x 28. Ú __ x log __ x dx = e
x
dt _____ 21. Let f (x) = Ú _______ and g be the inverse of f. Then 2 ÷ 1 + t4
the value of g¢(0) (a) 1 ___ (c) ÷17
(b) 17 (d) None.
3 ___
3
( ( ) )
1 (a) __ ex – ln x + c 2 1 (c) __ ln 2 x – ex + c 2
29.
Ú x × 2 ln(x + 1) dx =
t dt 22. If f (x) = eg(x) and g (x) = Ú _____ 4 , then f ¢(2) is 21 + t
2 ln (x + !) (a) ________ + c 2 2(x + 1)
(x2 + 1) 2 ln (x + !) (b) _____________ + c 2 (x2 + 1)
x
(a) 2/17 (c) 1
(b) 0 (d) cannot be determined.
2
2
2
__
1 (b) __ ln x – ex + c 2 (d) None.
Indefinite Integrals
(
)
(x2 + 1) 2 ln2 + 1 (c) ____________ + c 2 (ln 2 + 1)
(d) None.
(
)
(
)
x (a) __________ _________ + c 2 ÷ x + rx + 1 x (b) __________ _________ + c ÷ x 2 + rx + 1
(
)
(
)
1 (d) cos–1 __ (sin x + cos x) + c. 3
}
(b) B = cos a
(c) A = cos a
(d) B = sin a
36.
Ú x log (x2 + 1) dx = f (x) log (x2 + 1) + g (x) + c, then
(a) ex tan x + c
(b) ex cot x + c
(c) ex cosec2 x + c
(d) ex sec2 x + c.
{
(
_____
)}
2 etan x ( –1 1 – x2 32. Ú _______2 sec ÷1 + x2 ) + cos–1 _____2 dx (1 + x ) 1+x
(a) etan–1 x ◊ tan–1 x + c
etan–1 x ◊ (tan–1 x)2 (b) ______________ +c 2
( ( ) x ◊ ( cosec ( ÷1 + x ) ) + c 2
(c) etan–1 x ◊ sec–1 ÷1 + x2 ) + c
(d) etan–1
_____
2
( ( (
__
)
(1 + ÷x ) (b) ________ + c (1 – x)2
__
__
) )
(1 – ÷x ) (c) ________2 + c (1 – x)
2 (1 – ÷ x ) _____ (d) ________ + c – ÷1 – x
(
(
__
)
(a) A = 1/2 (c) f (x) = tan x
(b) A = 1/3 (d) f (x) = tan 2 x.
dx 38. If Ú __________________ cos (x – a) cos (x – b)
2
(a) A = sin (a – b) (b) f (x) = cos (x – a) (c) g (x) = cos (x – b) (d) A = sin (b – a).
Ú tan4 x dx = k tan3 x + L tan x + f (x), then
(a) k = 1/3
(b) L = –1
(c) f (x) = x + c
(d) k = 2/3.
(More than one options are correct)
(
)
x + sin x 40. If Ú ________ dx = f (x) tan (g (x)) + c, then 1 + cos x
)
(a) f (x) = x2
(b) f (x) = x
x2 (c) g (x) = __ 2
x (d) g (x) = __ 2
(
)
x–1 ex 41. If Ú ex _______ dx = ______ + c, then 3 (g (x))m (x + 1)
)
cos x – sin x ________ 34. Ú __________ dx = 8 – sin 2 x ÷
(a) sin–1 (sin x + cos x) + c
1 (b) sin–1 __ (sin x + cos x) + c 3
(
(
39. If
dx _____ 33. Ú _____________ = __ (1 + ÷ x )÷x – x2 2 (1 + ÷x ) (a) _________ + c 2 (1 – x)
1+ x2 (a) f (x) = _____ 2 1 + x2 (b) g (x) = _____ 2 1 + x2 (c) g (x) = – _____ 2 x2 – 1 (d) f (x) = _____ . 2 1 + sin 2 x 37. If Ú e2x _________ dx = Ae2x ◊ f (x) + c, then 1 – sin 2 x
1 = __ (log |f (x)| + log |g (x)| + c, then A
_____
–1
)
(a) A = sin a
cos x (2 sin x + 1) 2 sin2 x – 1 ______________ 31. Ú ex _________ + dx = cos x 1 + sin x
–1
(
)
{
x2 (c) __________ _________ + c ÷ x 2 + rx + 1 1 _________ (d) __________ + c 2 ÷ x + rx + 1
(
(c) cos–1 ((sin x + cos x)) + c
sin x 35. If Ú _________ dx = Ax + B log |sin (x – a)| + c, then sin (x – a)
2x + 1 30. Ú _____________ dx = (x2 + 4x + 1)3/2 3
1.31
)
(a) g (x) = x (c) m = 1
(b) g (x) = x + 1 (d) m = 2.
1.32 Integral Calculus, 3D Geometry & Vector Booster
(
)
dx 1 B __ 42. If Ú __________ = A 1 + + c, then x2 (x4 + 1)3/4 x4
cos4 x 50. If Ú ___________________ 3 dx sin x {sin5 x + cos5 x}3/5 m __ 1 = – __ (B + cot5 x) n + c, then A (a) A + B = 3 (b) m + n = 7
(a) A = –1 (c) A = 1/2
(b) B = 1/4 (d) B = 1/2.
(
)
logx e + logex e + log 2 e ex 43. If Ú _____________________ dx x = A (log (loge x)) + B loge (1 + loge x) + C log (2 + loge x) + K, then
(a) A + B = 2 (c) A – B = 0
(
____
(
(b) L – M = 6 (d) L + M – N = 14.
)
(c) m + n = 8
(d) A + B = 4.
(b) A – C = 0 (d) A + B + C = 3
)
tan x ÷ 44. If Ú ________ dx = A (f (x))1/m + c, then sin x cos x
(a) A = 2
(b) m = 2
(c) f (x) = tan x
(d) A + m = 5
(
(a) L + M = 18 (c) L + M + N = 22
cos 2 x – cos x 1. Evaluate: Ú ____________ dx 1 – cos x cos 5 x + cos 4 x 2. Evaluate: Ú _____________ dx 1 – 2 cos 3 x
____ cot x ÷ 45. If Ú ________ dx = A ( – ÷f (x) ) + c, then sin x cos x
sin6 x + cos6 x 3. Evaluate: Ú ___________ dx sin2 x ◊ cos2 x
(a) A = 2
(b) f (x) = cot x
(c) f (x) = tan x
(d) A = 4.
cos 2 x 4. Evaluate: Ú __________ dx cos2 x ◊ sin2 x
sin 2 x 5. Evaluate: Ú ___________ dx sin 5 x ◊ sin 3 x
sin x 6. Evaluate: Ú __________ dx sin x + cos x
dx 7. Evaluate: Ú ____________ sec x + cosec x
8. Evaluate: Ú tan 3 x ◊ tan 2 x ◊ tan x dx
b 9. Evaluate: Ú ______ dx a + c ex
____
(
)
)
( )
______
x9/2 L 46. If Ú _______ ______ dx = __ log xN/P + ÷1 + x11 + c, 11 M ÷ 1 + x then
(a) L (b) L (c) L (d) L
(
+ – + +
M M M M
= = + +
)
÷
_________
L P __ N + __ 4 + c , then M x
dx 10. Evaluate: Ú _____ 1 + ex
(a) L = 1 (b) M = 2 (c) L + M + N + P = 5 (d) L + M + N – P = 3.
( )
tan–1 x 48. If Ú ______ 4 dx = x
(a) A = 3 (c) C = 6
(
x+9 11. Evaluate: Ú _______ dx x3 + 9 x sin x – cos x 12. Evaluate: Ú __________ dx ex + sin x
| |
tan–1 x __ x2 + 1 1 1 – ______ + log _____ + ____ 2 + k, 3 B A x x2 C x
then
|
13 9 N + P = 26 N = 15.
dx 47. If Ú ________ _____ = 3 x ÷ 1 + x4
|
(b) B = 6 (d) A + B + C = 15.
)
2 (cos x + sec x) sin x 49. If Ú _________________ 6 dx cos x + 6 cos2 x + 4 N 1 M = __ log 1 + _____ 4 + _____ 6 + c, L cos x cos x then
( (
))
cos x – sin x + 1 – x 13. Evaluate: Ú ________________ dx ex + sin x + x sin (x + a) 14. Evaluate: Ú _________ dx sin (x + b) dx 15. Evaluate: Ú _________________ sin (x – a) sin (x – b) dx 16. Evaluate: Ú _________________ sin (x – a) cos (x – b) ____
tan x ÷ 17. Evaluate: Ú ___________ sin x ◊ cos x dx
Indefinite Integrals 3 ____
dx _____ 18. Evaluate: Ú ________ x ÷x 4 – 1
40. Evaluate: Ú ÷tan x dx dx 41. Evaluate: Ú _______ (ex – 1)2
dx 19. Evaluate: Ú _______ (x2 + 1)2
tan–1 x 42. Evaluate: Ú ______ dx x4 dx ____ 43. Evaluate: Ú _______________ 4 _____ ( ÷ cos x + ÷ sin x )
÷
_________
sin (x – a) 20. Evaluate: Ú _________ dx sin (x + a) x 21. Evaluate: Ú _________ dx x4 + x2 + 1
1 + x– 2/3 44. Evaluate: Ú _______ dx 1+x
22. Evaluate: Ú sec3 x dx 23. Evaluate: Ú cosec3 x dx
dx 45. Evaluate: Ú ___________ 2 sin x + sec x
x2 – 1 __________ 24. Evaluate: Ú ____________ dx x ÷x 4 + 3x2 + 1
x4 + 1 46. Evaluate: Ú ______ dx x6 + 1
dx 25. Evaluate: Ú ____________ x4 + 18x2 + 81
dx 47. Evaluate: Ú _______ (ex – 1)2
(
)
x + sin x 26. Evaluate: Ú ________ dx 1 + cos x
dx _____ 48. Evaluate: Ú ____________ (x – 1) ÷x + 2
27. Evaluate: Ú sin 4 x etan2 x dx
dx _____ 49. Evaluate: Ú ________ x2 ÷ x + 1
(
)
2x + 2 ____________ 28. Evaluate: Ú sin–1 _____________ dx 2 + 8x + 13 ÷ 4x
x4 – 1 _________ 50. Evaluate: Ú ____________ dx x2÷ x 4 + x2 + 1
x2 29. Evaluate: Ú _____________ dx (x sin x + cos x)2
x2 – 1 ___________ 51. Evaluate: Ú ______________ dx x3 ÷2x 4 – 2x2 + 1
x2 (x sec2 x + tan x 30. Evaluate: Ú _______________ dx (x tan x + 1)2
______
cos 2 x ÷ 52. Evaluate: Ú _______ dx sin x
secx (2 + sec x) 31. Evaluate: Ú _____________ dx (1 + 2 sec x)2
(
)
cos q + sin q 32. Evaluate: Ú cos (2 q) × log ___________ dq cos q – sin q
(
)
x4 + 2 33. Evaluate: Ú ex _________ dx (1 + 2)5/2
(
sec2 x 54. Evaluate: Ú _____________ dx (sec x + tan x)9/2 tan 2 q ____________ 55. Evaluate: Ú ______________ dq 6 6 q + cos q ÷ sin
)
x cos3 x – sin x 34. Evaluate: Ú esinx _____________ dx cos2 x dx 35. Evaluate: Ú ____________ cos x + cosec x
Q. Evaluate the following integrals.
36. Evaluate: Ú cot (1 + x + x ) dx –1
cos q + sin q ___________ 53. Evaluate: Ú ____________ dq ÷ 5 + sin q (2q)
2
37. Evaluate: Ú tan–1 (1 + x + x2) dx sin x 38. Evaluate: Ú ________ sin 4 x dx
x2 + 6 1. Ú _______________ dx (x sin x + 3 cos x)6
log (1 + sin2 x) 2. Ú _____________ dx cos2 x
3. Ú x 3 1 + x3 dx
39. For any natural number m, evaluate Ú (x
3m
2m
+x
m
2m
+ x ) (2x
m
1/m
+ 3x + 6)
dx.
2 – __
(
)
2 –1 __
1.33
1.34 Integral Calculus, 3D Geometry & Vector Booster
(
(
) ) ( (
(
)))
5p p 4. Ú 1 + tan ___ – x 1 + tan – ___ + x dx 16 16
5. Ú tan (x – a) ◊ tan (x + a) ◊ tan2 x dx
( ( ÷ ) ) _____
1–x 6. Ú cos 2 cot _____ dx 1+x
(x – 1) ÷x 4 + 2x3 – x2 + 2x + 1 7. Ú __________________________ dx x2 (x + 1)
p tan __ – 4 4 __________________ 8. Ú ________________________ dx 2 3 cos x ÷tan x + tan2 x + tan x
tan–1 x 9. Ú ______ 4 dx x
–1
(
)
x + 2 2 25. Ú ex _____ dx x+4 ______ ÷1 + x8 _______
___________________
(
dx 24. Ú ____________________ __________________ 3 + a) cos (x – b ) ÷ sin (x
)
26. Ú
dx x13 dx 27. Ú ___________ 3 sin x + cos3 x _____
(x + ÷1 + x2 )3 _____ 28. Ú ____________ dx ÷ 1 + x2 (x + 1) 29. Ú __________ dx x (1 + x ex)2
÷
_____________
(1 – x sin x) 10. Ú ____________ dx x (1 – x3 e3cosx)
sin x 30. Ú _____________ dx 2 sin x + 3 cos x
(1 + x cos x) 11. Ú ____________ dx x (1 – x2 e2sinx)
cos 4 x + 1 31. Ú __________ dx cot x – tan x
dx 12. Ú ____________ 2 sin x + sec x
cos3 x + cos5 x 32. Ú ____________ dx sin2 x + sin4 x
13.
dx
dx 33. Ú ____________ 6 sin x + cos6 x
_____________________ ________________ Ú cos x sin (2x + a) + sin a
÷
(
( )
( ) )
f (x) 34. Evaluate Ú _____ 3 dx, where f (x) is a polynomial of x – 1 degree 2 in x such that f (0) = f (1) = – 3 = 3 f (2).
1 1 14. Ú 3 2 tan __ x – x sec2 __ x dx __________
3 cos 2 x – 1 ÷ 15. Ú ____________ dx cos x
(
) ( (
)
( ) )
Integer Type Questions
1 2x 2x 16. Ú _____ 8 cos–1 _____ 2 + tan–1 _____ 2 dx 1–x 1+x 1–x x2 – x3 17. Ú ___________________ dx (x + 1) (x3 + x2 + x)3/2
cos x + cos 2 x 1. If of Ú ____________ dx = A sin x + Bx + c, find the 2 cos x – 1 value of A + B + 3.
4
cos x 18. Ú ___________________ sin3x (sin5 x + cos5 x)3/5
sin3 x dx 19. Ú _________________________________ 4 (cos x + 3 cos2 x + 1) tan–1 (sec x + cos x)
2 ÷ p x = ___ log tan __ + __ L M N
____ ____ ÷cot x – ÷tan x _____________
20. Ú
dx 4 + 3 sin2 x
x2 (x sec2 x + tan x 21. Ú _______________ dx (x tan x + 1)2 ___________
3 – sin x ÷ sin x 22. Ú ____________ dx 3 1 – sin x x
__
|
(
)|
P + __ tan–1 (sin x – cos x) + c, 3 then find the value of L + M – N – P.
dx 3. If Ú ____________ sin6 x + cos6 x
= tan–1 (L tan x + M cot x) + c, find the value of L + M + N + 4.
2
e (2 – x ) _____ 23. Ú ____________ dx (1 – x) ÷1 – x2
dx 2. If Ú ___________ 3 sin x + cos3 x
x+1 4. Let Ú ______ dx x3 + x
( )
| |
xM 1 = tan–1 x + __ log ______ N + c. L x +1
)
L + M + N Find the value of __________ 3 sin x 5. Let Ú __________ dx sin x + cos x = Ax + B log |sin x + cos x| + c Find the value of A + B + 1
(
(
)
x2 6. Let Ú ________ dx (2x + 3)2
(3 + 2x) __ 9 1 = _______ + log |3 + 2x| – ________ + C. L M 8(2x + 3) Find the value of L + M + 4
(
2
)
(
)
(b) x (ln x) + c
x (c) ___ + c ln x
(d) x + ln x + c
x ex dx 2. Ú _______ = (1 + x)2
(a) x ex + c
ex (b) __________ (1 + x)2 + c
1 (c) ex – _____ + c 1+x
ex (d) _____ + c 1+x
find the value of L + M + N + P + Q.
(
)
cos4 x 8. If Ú ___________________ 3 dx sin x (sin5 x + cos5 x)3/5
=
(
)
M __ 1 1 + tan x N – __ ________ + 5 L 5
tan x
c,
9. If Ú x13/2 (1 + x5/2)1/2 dx
8M L = ___ (1 + x5/2)7/2 – ___ (1 + x5/2)5/2 35 25 N + ___ (1 + x5/2)3/2 + c, 15 find the value of L + M + N. 10. If Ú cos4 x dx = L x + M sin 2 x + N sin 4 x + c, find the value of 8L + 4M + 32 N.
Comprehensive Link Passages Passage I In some of the cases we can split the integrand into the sum of the two functions such that the integration for which one of them integrate by parts and other one be fixed.
( (
) )
) ( ) (
) )
x4 – 2 _________ 1. Ú ____________ dx = x2 ÷x 4 + x2 + 2
÷ 2 (b) x + __ + 1 + c ÷ x 1 (c) x + __ + c ÷ x 2 (d) x + __ ÷ x + c. 1 (a) x2 + __ 2 + 1 + c x __________
2
2
_______
= Ú e f (x) dx + Ú e f ¢(x) dx
= ex f (x) – Ú ex f ¢(x) dx + Ú ex f ¢(x) dx + c = e f (x) + c
)
_________
)
) ( ) (
( (
Ú ex {f (x) + f ¢(x)} dx
x
( (
) )
a a a Ú f x – __ x 1 + __ 2 dx, put x – __ x = t, x a a a __ __ Ú f x + x 1 – 2 dx, put x + __ x = t, x a a a Ú f x2 – __ 2 x + __ 3 dx, put x2 – __ 2 = t x x x a a a Ú f x2 + __ 2 x – __ 3 dx, put x2 + __ 2 = t x x x The following integrands can be brought into above forms by suitable transformations.
Consider
x
)
( (
For integrals
x
(
1 3. Ú ex log x – __ 2 dx x 1 x (a) e log x – __ x + c 1 (b) ex log x + __ x + c 1 (c) ex log x + __ 2 + c x 1 (d) ex log x + ____ + c log x
Passage II
find the value of L + M + N.
)
(a) ln (ln x) + c
xM + xN + P 1 = __ tan–1 ___________ + k, L Q
(
1.35
1 1 1. Ú ___ – _____ dx = ln x (ln x)2
3x + 2x 7. If Ú _________________________ dx 6 5 x + 2x + x4 + 2x3 + 2x2 + 5
Indefinite Integrals
2
2
______
2
2
(x – 1) dx __________ 2. Ú _________________ = (x + 1) ÷x 3 + x2 + x
1.36 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 (a) tan–1 x + __ x + 1 + c __________
÷(
)
1 (b) tan–1 x + __ x + 1 + c
1 (c) 2 tan x + __ x + 1 + c
(d) tan–1
__________
) ÷( 1 ÷( x – __ x + 1 ) + c –1
__________
(
4
5
)
5x + 4x 3. Ú __________ dx = (x5 + x + 1)2
(a) x5 + x + 1 + c
x5 (b) ________ 5 + c x +x+1
( ( (
__
__
)
Matrix Match (For JEE-Advanced Examination only)
) ) )
4
(
3
x + ÷x 2 + ÷4 x __ 3. Ú ___________ dx 3 4 x+÷ x 3 6 __ (a) __ x2/3 + 6 tan–1 (÷ x ) + C 2 __ 3 (b) __ x2/3 – 6 tan–1 (÷ 6 x ) + C 2 __ 3 (c) – __ x2/3 – 6 tan–1 (÷ 6 x ) + C 2 __ 3 6 (d) – __ x2/3 + 6 tan–1 ( ÷6 ) + C 2
Q1. Match the following columns: Column I
Column II x 1 (P) ___ __ tan–1 ___ __ + c 2 ÷2 ÷
dx
x (c) ________ 5 + c x +x+1
(A)
__ Ú ________ (x + 2)÷x
x4 (d) ________ 5 + c x +x–1
(B)
dx __ __ Ú _________ (÷x + 2)÷x
(C)
Ú _______ x (x + 2)
(D)
2 Ú _______ (x + 2)
Passage III Let us consider the integrals of the form
(
Ú R x,
b __ xq ,
a __ xp ,
)
__g x r dx
In this case, we shall put x = t m, where m is the LCM of (p, q, r). __
÷x __ 1. Ú _______ dx = 4 3 ÷ x + 1 __
___
(
__
__
)
__
2 2 12 (b) ___ 4Rx9 – ___ ÷x 13 + C 27 13
13 2 12 (c) ___ 4Rx9 – ___ ÷x 13 + C 27 13
___
(
___ 2 ___ ÷x +
2 (d) – ___ 4Rx9 – 27 13
12 13
)
x 1 (S) __ + c log _____ 2 x+2
(
)
(B)
Ú sin4 x dx
3 3 (Q) – __ (tanx)– 8/3 + __ (tan x)4/3 8 4 3 – __ (tan x)–2/3 + c 2
(C)
Ú sin7 x cos3 x dx
sin5 x _____ sin7 x (R) _____ – + c 7 5
(D)
Ú cosec11/3 x ◊
sin8 x _____ sin10x (S) _____ – + c 8 10
Q.3 Match the following columns: (Write the suitable substitutions)
___
(R) 2 log (2 + ÷ x ) + c
dx
sec7/3 x dx
2 4 2 12 (a) ___ ÷x 9 + ___ ÷x 13 + C 27 13
Column II sin 4 x 1 _____ (A) sin4 x ◊ cos3 x dx (P) __ 3 x + – 2 sin 2 x Ú 8 4 + c,
___
__
Q.2 Match the following columns:
÷ x 3 – ÷3 x 2. Ú ________ dx = __ 6 – 4÷x
__
(Q) ÷2 tan–1
__
dx
Column I
4 3 4 4 (a) __ ( ÷x 3 – ln ( ÷ x + 1 ) ) + C 3 __ ___ 2 3 4 2 3 (b) __ ( ÷ x – ln ( ÷ x + 1 ) ) + C 3 __ ___ 2 3 5 2 3 (c) __ ( ÷ x – ln ( ÷ x + 1 ) ) + C 3 __ ___ 4 3 8 4 (d) __ ( ÷x 3 – ln ( ÷ x + 1 ) ) + C 3
( ) x ( __ ÷ 2 ) + c
C
Column I (A)
sin x cos x
dx Ú ____________ sin2 x + cos2 x
Column II (P) put sin x = t
Indefinite Integrals
(B)
sin x◊cos x dx Ú ___________________ 2 2
(Q) put sin x = t
dx
(C)
Ú 3___________ + sin x cos x
(D)
cos x – sin x dx Ú _____________ 2 2
2
(R) put cos x = t
2
(S) put sin x cos x=t
1 – sin x cos x
Q.4 Match the following columns: Column I
(
Column II p (P) __ 2
)
x + sin x (A) If f (x) = ________ dx and Ú 1 + cos x p f (0) = 0, then f __ 2
( )
(B) Let
(
Q.6 Match the following columns: Column I
sin x + 2 cos x + cos x
Column II x 1 __ _____ dx (A) _____ Ú x2 + x (P) – x – log x + 1 + c x _____ dx (B) ______ + c (Q) log x + 1 Ú x3 + x2 x 1 1 __ 1 ___ ___ _____ dx (C) ______ + c (R) – 3 + 2 – x – log x + 1 Ú x4 + x3 3x 2x (D)
)
x _____ dx f (x) = Ú esin–1 x 1 – ______ ÷ 1 – x__2 k ÷3 ep/6 1 and f (0) = 1, if f __ = _______ , p 2
( )
p dx (C) Let f (x) = Ú _____________ (R) __ and 4 (x2 + 1) (x2 + 9) __ 5 f (0) = 0, if f (÷3 ) = ___ k, then k 56 is
x 1 1 – ____ 2 + __ x + log _____ + c x +1 2 x
dx
(S) Ú ______ x5 + x4
| | | | | | | |
(S) p. ) dx and ( sin x cos x
Let f (x) = Ú
( )
p 2k f (0) = 0, if f __ = ___ p , then k is 4
Column II 1 dx _____ (P) – ______ _____ (A) ___________ Ú (x – 1)÷ x – 1 ÷ _____ x + 4 –1 1__ –1 x_____ ___ – tan + c 2 2 ÷
| (
( ÷ ) )
1 1 dx ___ __ (B) ____________ _____ (Q) – __ log t – 3 Ú (x2 – 1)÷ 3 ÷ x – 1 __________ 2 1 __ t2 – __ t + + + c, 3 3 1 _____ t = x+2
÷
_____
(
_____
)
dx
x3 1 __ _____ + c (Q) log 3 3 x +1
Ú ________ x (x5 + 1)
dx
x2013 1 log _______ 2013 + c (R) _____ 2013 x +1
Ú dx/x (x2013 + 1)
x2 1 log _____ 2 + c (S) __ 2 x +1
Ú ________ x (x3 + 1)
(C) (D)
Q.8 Match the following columns: Column I 3
Column II
2
x – 3x
(A)
dx Ú _______ ex + x3
(B)
dx Ú _____ ex + x
(C)
dx Ú __________ ex + sin x
(D)
dx Ú ___________ ex + tan x
(P) x – log |x + ex| + c
x–1
x (Q) log |sin x + e | – x + c
cos x – sin x
tan x – sec2 x
x (R) x – log |tan x + e | + c
3 x (S) x – log |x + e | + c
Q.9 Match the following columns:
(A)
dx _____ (R) __ ÷x + 4 – 3 1 (C) ____________ _____ log _________ + c Ú 3 ÷x + 4 + 3 (x + 2) ÷x 2 – 1 dx ______ (S) ___ (D) _____________ ÷x 2 + 4 1 __ Ú 2 – __ tan–1 _______ + c 2 ÷3 x ÷3 (x + 1) ÷ x + 4
x5 1 log _____ 5 + c (P) __ 5 x +1
(B)
|
|
dx
Ú ________ x (x2 + 1)
Q.5 Match the following columns: Column I
Column II
(A)
____ ÷ tan x ________
|
| | | |
Q.7 Match the following columns:
then k is
(D)
| | | |
Column I p (Q) __ 3
1.37
(B)
(
Column I
5p – x ) Ú 1 – tan ( ___ 4
(
(
(
(
)
))
(P)
2x + c
))
(Q)
3x + c
3p 1 + tan ___ – x dx 2
Ú ( 1 + tan ( __ p8 – x ) )
Column II
p 1 + tan __ + x dx 8
1.38 Integral Calculus, 3D Geometry & Vector Booster
(C)
(
5p – x ) Ú 1 + tan ( ___ 4
16. The value of the integral
)
(R)
(
(
))
(
(
(S)
))
5x + c
Questions asked in Previous Years’ JEE-Advanced Examinations
dx 1. Ú _______ 1 – cot x x dx 2. Ú _____ 1 + x4
[IIT-JEE, 1978] [IIT-JEE, 1979]
÷
( )
[IIT-JEE, 1980]
6. Ú (elogx + sin x) cos x dx
[IIT-JEE, 1981]
x–1 dx 7. Ú ex _______ (x + 1)3
[IIT-JEE, 1983]
dx 8. Ú __________ 2 4 x (x + 1)3/4
[IIT-JEE, 1984]
(
÷
1 – ÷x dx 9. Ú ______ __ 1+÷ x
(
__
dx ____________ 18. Ú ______________ is equal to – p)3 (x – q) ÷ (x
__
)
–1 sin–1 ÷x – cos ÷x 10. Ú _______________ dx __ –1 –1 __ sin ÷x + cos ÷x
______
____
12. Ú (÷tan x + ÷ cot x ) dx
(
1 ___________ (c) _____________ + c – p) (x – q) ÷ (x
(d) none. [IIT-JEE, 1996]
dx 19. If Ú _________________ (sin x + 4) (sin x – 1)
( ( ) )
4 tan x + 1 1 2___ (a) A = __ , B = – _____ , f (x) = _________ 5 5 5÷15
[IIT-JEE, 1986]
4 tan (x/2) + 1 1 2___ ___ (b) A = – __ , B = – _____ , f (x) = ____________ 5 5÷15 ÷15
[IIT-JEE, 1987]
4 tan x + 1 2 2___ (c) A = __ , B = – _____ , f (x) = _________ 5 5 5÷15
[IIT-JEE, 1988]
4 tan (x/2) + 1 2 2___ ___ (d) A = __ , B = – _____ , f (x) = ____________ 5 5÷15 ÷15
)
then A = ..., B = ... and C = ...
[IIT-JEE, 1990]
log (1 + x1/6) 1 14. Ú ________ 1/3 1/4 + ___________ dx __ 1/3 x +x x + x ÷
[IIT-JEE, 1992]
(
)
)
cosq + sinq 15. Ú cos 2q log __________ dq cosq – sinq
A = ___________ + B tan–1 (f (x)) + C, x __ tan – 1 2
[IIT-JEE, 1985]
4 ex + 6 e–x 13. If I = __________ x dx = Ax + log |(9 e2x – 4)| + C 9e – 4 e–x
(
______
cos 2 x ÷ 11. Ú _______ sin x dx ____
÷( ) x–p 2 (b) – ______ ( _____ + c (p – q) ÷ x – q ) ______
x–p 2 (a) ______ _____ + c (p – q) x – q
then
______ __
[IIT-JEE, 1996]
[IIT-JEE, 1980]
)
6 tan–1 (sin x) + c 2 (sin x)–1 + c 2 (sin x)–1 – 6 tan–1 (sin x) + c 2 (sin x)–1 – 5 tan–1 (sin x) + c
x+1 17. Ú __________ dx x 2 x (1 + x e )
x dx 4. Ú 1 + sin __ 2 x2 _____ 5. Ú ______ dx 1 – x ÷
– – – –
[IIT-JEE, 1995]
[IIT-JEE, 1979]
___________
(a) sin x (b) sin x (c) sin x (d) sin x
x2 dx 3. Ú ________ (a + bx)2
)
cos3 x + cos5 x Ú ____________ dx is sin2 x + sin4 x
(1 + tan x) dx
7p – x Ú 1 + tan ___ 8 (D) 3p 1 + tan ___ + x dx 8
(
4x + c
[IIT-JEE, 1994]
[IIT-JEE, 1997] cos x – sin x 20. Ú ___________ (2 + 2 sin 2 x) is equal to cos x + sin x
(
(a) sin 2 x + c (c) tan 2 x + c
÷
21. Ú
)
_______ __ 1 – ÷x dx ______ × ___ x __ 1+÷ x
(b) cos 2 x + c (d) None [IIT-JEE, 1997]
dx ___________ 22. Ú ___________________ is equal to (2x – 7) ÷x 2 – 7x + 12
[IIT-JEE, 1997]
Indefinite Integrals
(a) 2 sec–1 (2x – 7) + c (b) sec–1 (2x – 7) + c (c) 1/2 sec–1 (2x – 7) + c (d) None of these.
[IIT-JEE, 1997]
2
x + 3x + 2 23. Ú _____________ dx (x2 + 1)2 (x + 1)
(
[IIT-JEE, 1999]
)
2x + 2 ____________ 24. Ú in–1 _____________ dx 2 + 8x + 12 ÷ 4x 3m
25. (x
2m
+x
2m
m
+ x ) (2 x
[IIT-JEE, 2001] m
+ 3x + 6)
2
____________
2 4 – 2x + 1 ÷ 2x (a) _____________ +c x2
(b)
4 – 2x 2 + 1 ÷ 2x (c) _____________ +c x
( ( ( (
e4x – e2x + 1 1 (a) __ log ___________ + c 2 e4x + ex + 1
e2x + ex + 1 1 (b) __ log __________ + c 2 e2x – ex + 1
e2x – ex + 1 1 (c) __ log __________ + c 2 e2x + ex + 1
e4x + e2x + 1 1 (d) __ log ___________ + c 2 e4x – e2x + 1
+c
x3
[IIT, 2008]
sec2 x 29. The integral Ú _____________ dx equals (sec x + tan x)9/2
____________
____________
2 4 – 2x + 1 ÷ 2x (d) _____________ + c [IIT, 2006] 2 2x x 27. Let f (x) = _________ for n ≥ 2 and g (x) = ( f O fO (1 + xn)1/n
{ { { {
}
1 1 (a) – ______________ ___ 1 – __ (sec x + tan x)2 + K (sec x + tan x)11/2 11 7
1 1 (b) ______________ ___ 1 – __ (sec x + tan x)2 + K (sec x + tan x)11/2 11 7
1 1 (c) – ______________ ___ 1 + __ (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x) 1 1 (d) ______________ ___ 1 + __ (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x)
n–2
... O f ) (x), then x
) ) ) )
____________ ÷ 2x 4 – 2x2 + 1 _____________
For an arbitrary constant C, the value of J – I is
dx where m > 0
x – 1 dx ____________ 26. Ú _______________ 3 4 x÷ 2x – 2x2 + 1
[IIT, 2007]
e–x J = ____________ –4x dx. e + e–2x + 1
1/m
[IIT-JEE, 2002]
1 __ 1 (d) ______ (1 + n xn)1 + n + c (n + 1) ex 28. Let I = Ú __________ dx, e4x + e2x + 1
1.39
g (x) dx is
1 __ 1 (a) _______ (1 + n xn )1 – x + c n (n – 1)
1 __ 1 (b) ______ (1 + n xn)1 – x + c (n – 1)
1 __ 1 (c) _______ (1 + n xn)1 + n + c n (n – 1)
} } }
[IIT, 2012]
Answers Level IIB 1. (a) 6. (b) 11. (a) 16. (d) 21. (c)
2. (b) 7. (a) 12. (b) 17 (a) 22. (d)
3. (c) 8. (b) 13. (a) 18. (b) 23. (a)
4. (d) 9. (c) 14. (b) 19. (a) 24. (b)
5. (a) 10. (d) 15. (c) 20. (b) 25. (a)
26. (b) 27. (c) 28. (d) 29. (a) 31. (a) 32. (b) 33. (c) 34. (d) 36. (b) 37. (a) 38. (b) 39. (c) 41. (b, d) 42. (a, b) 43. (a, b, c, d) 45. (a, b) 46. (a, b, c) 47. (a, b, c, d) 49. (a, b, c, d) 50. (a, c)
30. (b) 35. (a) 40. (b, d) 44. (a, b, c) 48. (a, b)
1.40 Integral Calculus, 3D Geometry & Vector Booster 1 18. – __ (1 + cot5 x)2/5 + c 2 19. log |tan–1 (cos x + sec x)| + c
Level IV
– x sec x 1. ____________ + tan x + c x sin x + 3 cos x __
2
2. tan x log (1 + sin x) – 2 ÷2 tan (t÷2 ) + c, where t = tan x –1 1/3
3. 3 tan (x ) + c 1 4. – __ log |cos (2x)| + log |cos (x – a)| 2
( ÷ ) _____
_____
6. – ÷1 – x + c
(t + 1) – ÷2 1 1 7. __ log |t2 + 2t – 3| – ____ __ log ___________ __ 2 2÷2 (t + 1) + ÷ 2
|
( ) 1 ( t + __ ÷ t + 1 ) + c
__
(
(
|
)
+1 1__ –1 3t + – ___ sin _____ + c 2 3 ÷
(
( )
| | | |
(x ex)3 – 1 1 10. __ log ________ + c 3 (x ex)3
_________
where t = ÷ 2 + 3 cot x
(x esin x)2 1 11. __ log __________ sin x 2 + c 2 (x e ) – 1
( ) ( )
1 31. – __ cos (4x) + c 8
)|
x p 1__ 1 12. ____ log tan __ + __ – ___________ + c 2 8 (sin x + cos x) 2÷2
32. sin x – 2 cosec x – 6 tan–1 (sin x) + c 33. tan–1 (tan x – cot x) + c
|
tan x cos a + sin a __ 13. _______________ + c 2 cos a ÷
( )
1 14. x3 tan __ x + c
( ÷
__________
)
]
)
_________
(
)
Integer type Questions
2 (x2 + 1) (x + 1) p 1 1 _______ 16. __ Ú _____ 2 + _____ 2 + Ú _______ – Ú dx 4 4 8 1–x x +1 x +1 1+x
( ÷
|
(x2 + x + 1) 2x +__ 1 2 34. log __________ + ___ __ tan–1 ______ + c (x – 1) 3 ÷3 ÷
__ __ __ cosec2 x – 3 15. ÷ 6 sin–1 (÷3 s in x) + 2÷2 tan–1 __________ + c 2
( ÷
| |
__ 1 t + __ – ÷2 t 1__ –1 1 1 30. – ____ tan t – __ + ____ __ log ___________ __ + c t 1 3÷2 6÷2 t + __ + ÷ 2 t
where tanq = x
and then you do it.
)|
| |
)
[ (
(
(x + ÷1 + x2 )3 28. ____________ +c 3 t 1 29. log ____ + ______ + c, where t = x ex t+1 (t + 1)
q 1 1 9. – ______ – __ ______ + log |sin q| + c 3tan3q 3 2 sin2 q
|
|
_____
(
( )
1 1 + x8 3/2 26. – ___ _____ + c 12 x8 x p 2__ 2 27. ____ log tan __ + __ – __ tan–1 (sin x – cos x) + c 2 8 3 3÷2
1 where t = x + __ x
)
x 25. ex ______ + c (x + 4)
________
8. – 2 tan–1
____________________________
24. 2 sec (a + b) ÷ cos (a + b) tan (x – b) + sin (a + b) + c
1 where t = x + __ x
2 22. __ sin–1 (sin3/2 x) + c 3 1+x 23. ex _____ + c 1–x
+ log |cos (x + a) + c
|
x 21. – __________ + 2 log |x sin x + cos x| + c (x tan x + 1)
1 5. – __ log |cos (2x)| + log| cos (x – a)| 2
2
|
__
(tan x + cot x) ÷2 – 1 1__ __ 20. ____ log __________________ + c 2÷2 (tan x + cot x) ÷2 + 1 2
+ log |cos (x + a)| + c
__
–1
)
)
1 1 17. 2 _________ ________ + tan–1 x + __ x + 1 + c 1 __ x + x + 1
1. 5 6. 4
2. 7 7. 9
3. 5 8. 9
4. 2 9. 7
Comprehensive Link Passages Passage I: Passage II: Passage III:
1. (b) 1. (b) 1. (b)
2. (d) 2. (c) 2. (b)
3. (a) 3. (b) 3. (a)
5. 1 10. 5
Indefinite Integrals
Matrix Match 1. (A)Æ(Q), (B)Æ(R), 2. (A)Æ(R), (B)Æ(P), 3. (A)Æ(P, Q, R), (C)Æ(R),
(C)Æ(S), (D)Æ(P) (C)Æ(S), (D)Æ(Q) (B)Æ(Q), (D)Æ(R, S)
Hints
Ú logx x dx = Ú 1 ◊ dx = x + c 2 – 2log 3) dx = Ú (2log 3 – 2log 3)dx Ú (3log x
x
x
3. we have, m x ) dx Ú ( xm + mx + mm + __
x m + 1 = _____ + m+1
mx _____ + mmx + m log |x| + c log m
4. we have,
Ú 2 ◊ 3 dx = Ú (2 ◊ 3) dx = Ú 6 dx
6x = ____ + c log 6
x
x
dx
dx
1 = __ tan x + c 2 11. The given integral is
x – xlog 3) dx = Ú (3log x – 3log x) dx Ú (3log 5
5
dx sin2x + cos2x 2 2 = Ú ____________ dx Ú _________ sin x cos x sin2x cos2x
(
)
Ú ( 1 + tan ( __ p8 – x ) ( 1 + tan ( __ p8 + x ) ) ) dx
= Ú (2) ◊ dx
13. The given integral is
= 2x + C
= – cot x – x + c
(
1 + x
1 + x
= 8 Ú 2xdx + Ú 4 dx 8 ◊ 2x = ____ + 4x + c log 2
= Ú (sec2x + cosec2x) dx
14. The given integral is
= tan x – cot x + c
) ) (
(
))
2 + 2x
= Ú (23 + x + 22) dx
(
3 + 3x
3p p 8. Ú 1 + tan x + ___ 1 + tan __ – x dx 8 8
)
(
)
+4 2 +2 dx = Ú ___________ dx Ú 8__________ 2x 2 22x
1 1 = Ú _____ + _____ dx cos2x sin2x
(
6
= Ú (0) ◊ dx = c
6
12. The given integral is
= tan x – x + c
1 = ______ = __ sec2x dx Ú 1_________ + cos 2x Ú 2cos2x Ú 2
Ú cot2x dx = Ú (cosec2x – 1) dx
= tan x – cot x + c
6. we have,
7. we have,
10. We have,
Ú tan2x dx = Ú (sec2x – 1) dx
= Ú (sec2x + cosec2x) dx
x
5. we have,
x
Ú (tan x + cot x)2dx = Ú (tan2x + cot2x + 2) dx
x
= Ú 0 ◊ dx = c
(D)Æ(S) (D)Æ(Q) (D)Æ(R) (D)Æ(R) (D)Æ(R) (D)Æ(P)
9. we have,
(C)Æ(R), (C)Æ(S), (C)Æ(S), (C)Æ(P), (C)Æ(Q), (C)Æ(P),
solutions
2. we have,
(B)Æ(Q), (B)Æ(P), (B)Æ(P), (B)Æ(Q), (B)Æ(P), (B)Æ(P),
p = Ú 2dx [ if A + B = __ , 4 then (1 + tan A) (1 + tan B) = 2] = 2x + c.
1. We have,
and
4. (A)Æ(P), 5. (A)Æ(R), 6. (A)Æ(Q), 7. (A)Æ(S), 8. (A)Æ(S), 9. (A)Æ(P),
1.41
m m + __ x + xm + mx ) dx Ú ( __ x
x2 xm + 1 = ___ + m log x + _____ + 2m m+1
mx _____ + c log m
1.42 Integral Calculus, 3D Geometry & Vector Booster
)
((sin2x + cos2x)2 – 2sin2x cos2x) = Ú ___________________________ dx sin2x cos2x
a x b x = Ú __ + __ a + 2 dx b
(1 – 2 sin2x cos2x) = Ú _______________ dx sin2x cos2x
(a/b)x (b/a)x = _______ + _______ + 2x + c log(a/b) log(b/a)
(sin2x + cos2x – 2sin2x cos2x) = Ú ________________________ dx sin2x cos2x
= Ú (sec2x + cosec2x – 2) dx
= tan x – cot x – 2x + c
15. The given integral is
(
(ax + bx)2
2x
2x
x x
a + b + 2a b dx = Ú _______________ dx Ú ________ x x ab a xbx
( ( ) ( ) )
16. The Given integral is (2x + 3x)2 dx = Ú _________ 2x ◊ 3x
(
Ú ( __ 23 ) + ( __ 32 ) + 2 dx x
x
)
(2/3)x (3/2)x = _______ + _______ + 2x + c log (2/3) log (3/2)
21. We have,
17. It is given that 1 1 _____ f ¢(x) = __ x + ______ ÷ 1 – x2 On integration, we get
f (x) = log |x| + sin–1x + c
p x = 1 and f (1) = __ , then c = 0 2 Hence, the function is f (x) = log |x| + sin–1x 18. It is given that f ¢(x) = a cos x + b sinx On integration, we get f (x) = a sin x – b cos x + c When x = 0 and f ¢(0) = 4, then a = 4 p When x = 0, f (0) = 3 and f __ = 5 2 then, we get b = – 2 and c = 1 Hence, the function is
(
sin6x + cos6x
f (x) 4 cos x + 2sin x + 1
19. We have, (1 + sin x) dx = Ú _________________ dx Ú 1_______ – sin x (1 – sin x) (1 + sin x)
(1 + sin x) 1 + sin x = Ú _________ dx = Ú _______ dx (1 – sin2x) cos2 x
= Ú (sec2x + sec x ◊ tan x) dx
= tan x + sec x + c
sin4x + cos4x
dx Ú ____________ sin2x cos2x
)
(sin2x)3 + (cos2x)3 = Ú _______________ dx sin2x ◊ cos2x
(sin4x + cos4x – sin2x ◊ cos2x) = ________________________ dx sin2x cos2x
(1 – 3sin2x ◊ cos2x) = _______________ dx sin2x cos2x
(sin2x + cos2x – 3sin2x cos2x) = ________________________ dx sin2x cos2x
= Ú (sec2 x + cosec2 x – 3)dx
= tan x – cot x – 3x + c
)
)
)
22. We have,
– cosa ____________ dx Ú ( cos 2x cos x – cosa )
(
)
2cos2x – 1 – 2cos2a + 1 = Ú ____________________ cos x – cosa dx
cos2x – cos2a = Ú 2 ____________ dx cos x – cosa
= Ú 2 (cos x + cosa) dx
= 2 (sin x + x cosa) + c
)
(
23. We have,
20. We have,
( Ú ( Ú ( Ú (
When
( )
)
dx Ú ___________ sin2x ◊ cos2x
(
cos4x – sin4x
)
________ dx Ú ____________ ÷1 + cos4x
(
)
(cos2x – sin2x) (cos2x + sin2x) _________ = Ú _________________________ dx 1 + cos 4x ÷
Indefinite Integrals
( (
cos 2x = Ú ________ __ dx 2 cos 2x ÷
x = ___ __ + c 2 ÷
( )
24. We have,
(
27. We have,
)
)
(cos2x – sin2x) __ = Ú ____________ dx 2 cos 2x ÷
)
+ tan2x 1 + tan2x dx = Ú ________ dx Ú 1________ 2 1 1 + cot x 1 + _____ 2 tan x
(
(
5cos3x + 3sin3x
)
dx Ú ______________ cos2x sin2x
(
)
5cos3x 3sin3x _________ = Ú _________ 2 + dx cos x sin2x cos2x sin2x
= 5 Ú cosec x cot x dx + 3 Ú sec x tan x dx
= – 5 cosec x + 3 sec x + c
28. We have,
)
cosx – sin x __________ (1 + sin 2x) dx Ú ( cos x + sin x )
tan2x (1 + tan2x) = Ú ______________ dx (1 + tan2x)
= Ú (tan x) dx
cos x – sin x = Ú __________ (cos x + sin x)2dx cos x + sin x
= Ú (sec x – 1) dx
= Ú (cos x – sin x) (cos x + sin x) dx
= Ú (cos2x – sin2x) dx
= tan x – x + c
2
2
– cos 2x ___________ dx Ú ( cos x 1 – cos x )
( Ú ( Ú ( Ú (
)
cos x – 2cos2x + 1 = Ú ________________ dx 1 – cos x
2cos2x – cos x – 1 = – _______________ dx 1 – cos x
)
(cos x – 1) (2cos x + 1) = – ___________________ dx 1 – cos x
(1 – cos x) (2cos x + 1) = ___________________ dx 1 – cos x
=
= 2 sin x + x + c.
)
)
Ú ((2 cos x + 1))dx
26. We have,
(
___________
(
)
= Ú cos2x dx sin 2x = _____ + c 2 29. The given integral is
25. We have,
) ( ÷ Ú ( ÷
__________
) )
+ 2 ÷ x 4 + x– 4 x4 + x– 4 + 2 dx = Ú __________ dx Ú ____________ 3 x x6 ___________
dx Ú ( _____________ 1 – 2cos 3x ) cos 5x + cos 4x
sin 3x (cos 5x + cos 4x) = Ú ___________________ dx sin 3x – sin 6x
( ) ( ) ( ) 3x x = – Ú 2cos ( ___ ) cos ( __ ) dx 2 2 =
Ú
( ) ( ) ( )
9x 3x 3x x 2sin ___ cos ___ 2cos ___ cos __ 2 2 2 2 _____________________________ dx 9x 3x ___ ___ – 2cos sin 2 2
= – Ú (cos 2x + cos x) dx
sin 2x = – _____ + sin x + c 2
(
)
30. The given integral is
– cos 2x ___________ dx Ú ( cos x 1 – cos x )
x8 + 2x4 + 1 = __________ dx x10
x4 + 1 = Ú _____ dx x5
2cos2x – cosx – 1 = – Ú _______________ dx 1 – cos x
1 = Ú x– 5 + __ x dx
(cos x – 1) (2cos x + 1) = – Ú ___________________ dx 1 – cos x
1 = log |x| – ___ 4 + c 4x
= Ú (2cos x + 1) dx
= (2sin x + x) + C
( ) (
1.43
)
1.44 Integral Calculus, 3D Geometry & Vector Booster 31. The given integral is
dx
Ú __________________________ (tan x + cot x + sec x + cosec x) sin x cos x dx = Ú ______________ 1 + sin x + cos x
=
_____________ Ú sec x + tan x + 1
=
Ú _____________________ (1 + tan x)2 – sec2x
=
= Ú (x1/3 – 1) dx
3 = – __ x– 2/3 – x + c 2
36. We have,
sin x(1 + tan x – sec x)dx
(
(
)
4
– 1 + 3 3 x –1 dx = Ú _____ + _____ 2 dx Ú x_________ 2 2 x +1 x +1 x +1
Ú ____________________ 2tan x
1 = __ Ú cos x (1 + tan x – sec x) dx 2
(x2 – 1) (x 2 + 1) _____ 3 = Ú _____________ + 2 dx x2 + 1 x +1
3 = Ú (x2 – 1) + _____ 2 dx x +1
x3 = __ – x + 3 tan–1x + c 3
(
sin x(1 + tan x – sec x)dx
32. We have,
(x + 1 – 1) x dx = Ú __________ dx Ú x_____ +1 (x + 1)
1 = Ú 1 – _____ dx x+1
= x – log |x + 1| + C
(
)
(1 + x)2 ________ dx = Ú x (1 + x2)
Ú
(1 + x2 + 2x) ___________ dx x (1 + x2)
)
37. We have,
)
(
(
x4 – 1 – 2
[ ( ) ] ) ]
)
4
x –1 2 dx = Ú _____ – _____ dx Ú _________ x2 + 1 x2 + 1 x2 + 1
[ (
(x2 + 1) (x2 – 1) 2 = Ú ______________ – _____ dx 2 2 x +1 x +1
2 = Ú (x2 – x) – _____ 2 dx x +1
x3 = __ – x – 2 tan–1x + c 3
(
33. We have,
)
4
1 = __ Ú (cos x + sin x – 1) dx 2 1 = __ (sin x + cos x – x) + C 2
=
sin x dx
(x1/3 – 1) (x2/3 + x1/3 + 1)
dx Ú _____________________ (x2/3 + x1/3 + 1)
38. we have,
(
)
) [ ( ) Ú [ ( )
]
(1 + x2) + 2x = ____________ dx x (1 + x2)
dx dx = Ú ___ x + 2 Ú _____ 1 + x2
= log |x| + 2 tan–1x + c
(x2 + 1) (x4 – x2 + 1) 2 = _________________ – _____ dx 2 2 x +1 x +1
=
x5 = __ – 5
34. We have,
Ú
)
(
)
3 = Ú 1 – _____ dx x2 + 1
–1
= x – 3 tan x + c
35. We have,
(
x2 – 2 x2 + 1 – 3 ______ 2 dx = Ú ________ dx x +1 x2 + 1
(x1/3)3 – 1 x–1 _____________ d x = dx Ú _____________ Ú (x 2/3 + x1/3 + 1) (x2/3 + x1/3 + 1)
x6 + 1 – 2
(
(
]
)
2 2 dx Ú (x4 – x2 + 1) – _____ x +1 x3 __ + x – 2 tan–1x + c 3
39. We have,
6
x +1 2 dx = Ú _____ – _____ dx Ú _________ 2 2 2 x +1 x +1 x +1
x8 + x4 + 1
)
dx Ú __________ x4 + x2 + 1
(
)
(x4 – x2 + 1) (x4 + x2 + 1) = Ú ______________________ dx (x4 + x2 + 1)
= Ú ((x4 + x2 + 1)) dx
Indefinite Integrals
x5 x3 = __ + __ + x + c 3 5
( )
(x4 – 1) + 1
4
x 2 dx = Ú __________ dx Ú _____ x2 + 1 x +1
(
)
1 = Ú x2 – 1 + _____ 2 dx x +1
x3 = __ – x + tan–1x + c 3
(
)
41. The given integral is
(
4
)
2
(x2 – x + 1) (x2 + x + 1) _____________________ dx (x2 + x + 1)
=
= Ú (x2 – x + 1)dx
x3 x2 = __ – __ + x + C 3 2
(
)
( )
(x2 + 1) (x4 – x2 + 1)
x6 + 1
dx = Ú __________________ dx Ú ______ x2 + 1 (x2 + 1) = Ú (x4 – x2 + 1) dx
(
5
3
)
43. We have,
x2 + c Ú sin–1(sin x) dx = Ú x dx = __ 2
44. We have,
Ú sin–1(cos x) dx = Ú sin–1 [ sin ( __ p2 – x ) dx ]
(
p = Ú __ – x dx 2 p x2 = __ x – __ + c 2 2
)
45. We have,
(
= Ú tan–1(tan x) dx
= Ú x dx
x2 = __ + c 2
( ÷
– cos 2x dx Ú tan–1 ÷ 11_________ + cos 2x _________
( )
= Ú tan–1(tan x) dx
= Ú x dx
x2 = __ + c 2
)
( ÷ )
sin x dx Ú tan–1 ( _______ 1 – cos x )
(
)
2sin (x/2) cos (x/2) = Ú tan–1 _______________ dx 2sin2(x/2)
[ ( ) ]
x = Ú tan–1 tan __ dx 2 x = Ú __ dx 2
( )
x2 = __ + c 4
49. The given integral is
Ú tan
( ÷
________
)
1 – sin x ________ dx 1 + sin x
–1
( (
)
cos (x/2) – sin (x/2) = Ú tan–1 _________________ dx cos (x/2) + sin (x/2)
1 – tan(x/2) = Ú tan–1 __________ dx 1 + tan(x/2)
_______
[ (
)
)]
p x = Ú tan–1 tan __ – __ dx 4 2 p x = Ú __ – __ dx 4 2
(
2 sin2x = Ú tan–1 ______ dx 2 cos2x
= Ú tan (tan x) dx = Ú x dx
x2 = __ + c 2
50. The given integral is
–1
)
– cos 2x sin x dx = Ú tan–1 ____ cos x dx Ú tan–1 11_________ + cos 2x
x x = __ – __ + x + c 3 5
)
2sin x cos x = Ú tan–1 _________ dx 2 cos2x
48. The given integral is
42. The given integral is
(
_________
sin 2x dx Ú tan–1 ( _________ 1 + cos 2x )
47. The given integral is
x + x + 1 dx Ú __________ x2 + + 1
Ú
46 We have,
40. The given integral is
1.45
(
)
)
p x x2 = ___ – __ + c 4 4
1.46 Integral Calculus, 3D Geometry & Vector Booster
sin x dx Ú tan–1 ( 1________ + cos x )
53. The given integral is
(
)
2 sin (x/2) cos (x/2) = Ú tan–1 ________________ dx 2cos2(x/2)
= Ú tan–1 [tan (x/2)] dx
x = Ú __ dx 2
( )
x2 = __ + c 4 51. The given integral is
( (
(
(
)
)
((( (( )) (( )))) (( (( )) (( ))))) ( (( )) )
)
= Ú tan–1
)
p x p x 2 sin __ – __ cos __ – __ 4 2 4 2 _____________________ dx p x 2sin2 __ – __ 4 2 p x cot __ – __ dx 4 2
–1
x cos __ 2 _______ dx x sin __ 2
[ ( ) ] p x [ tan ( __ – __ dx 2 2 )]
(
p x = Ú __ – __ dx 2 2
)
p x x2 = ___ – __ + c 2 4 54. The given integral is
Ú tan–1 (sec x + tan x)dx
(
) p 1 + cos ( __ – x ) 2 _____________ dx p __ sin ( – x ) 2 p x 2cos ( __ – __ 4 2) ____________________ dx p x p x 2sin ( __ – __ ) cos ( __ – __ ) 4 2 4 2 p x [ cot ( __ – __ dx 4 2 )] p p x [ tan { __ – ( __ – __ dx 2 4 2 )} ]
1 + sin x = Ú tan–1 _______ dx cos x
= Ú tan–1
–1
(
( (
)
= Ú tan–1
– sin x ) dx Ú tan–1 ( 1_______ cos x
= Ú tan–1
= Ú tan–1
p = Ú __ + 4
p x x2 = ___ + __ + c 4 4
p x x2 = ___ + __ + c 4 4
52. The given integral is
= Ú tan–1
= Ú tan–1
( (
(
)
p 1 – cos __ – x 2 _____________ dx p sin __ – x 2 p x 2sin2 __ – __ 4 2 _____________________ dx p x p x 2 sin __ – __ cos __ – __ 4 2 4 2
( ( ( )
(
(
(
))
)
)
) ) (
p x = Ú tan–1 tan __ – __ dx 4 2 p x = Ú __ – __ dx 4 2 p x x2 = ___ – __ + c 4 4
)
x = Ú tan–1 cot __ dx 2
) ( ) = Ú tan ( ) = Ú tan ( ( )) p p x = Ú tan ( tan ( __ – ( __ – __ dx 2 4 2 ))) p x = Ú tan ( tan ( __ + __ dx 4 2 )) p x = Ú ( __ + __ dx 4 2) –1
(
_________
x x x x cos __ + sin __ + cos __ – sin __ 2 2 2 2 __________________________________ dx x x x x cos __ + sin __ – cos __ – sin __ 2 2 2 2
= Ú tan
p sin __ – x 2 _____________ = Ú tan dx p 1 – cos __ – x 2 –1
= Ú tan
–1
(
_________
+ sin x) + ÷(1 – sin x) ÷ (1 _________ dx ______________________ _________ + sin x) –÷ (1 + sin x) ÷ (1
–1
cos x dx Ú tan–1 ( _______ 1 – sin x ) –1
Ú tan
–1
)
)
(
2
)
x __ dx 2
55. The given integral is
sin 2x dx Ú tan–1 ( _________ 1 + cos 2x )
(
)
2sin x cos x = Ú tan–1 _________ dx 2cos2x
)
)
1.47
Indefinite Integrals
= Ú tan–1 (tan x) dx
= Ú x dx 2
x = __ + c 2
56. We have,
+c Ú (3x + 2) dx = Ú ________ 2.3
(3x + 2)2 = Ú ________ + c 6 1 __ log |2x – 3| + c 2
= Ú (÷x + 2 + ÷ x + 1 ) dx
2 2 = __ (x + 2)3/2 + __ (x + 1)3/2 + c 3 3
ax + b
e a + c Ú eax + bdx = _____ 4x + 5
3 +c Ú 34x + 5dx = _____ 4log 3
sin (5x + 3) +c Ú cos (5x + 3) dx = __________ 5 62. We have, cos 2x + c Ú sin 2x dx = – _____ 2
(÷2x + 5 + ÷2x + 3 ) dx = Ú ____________________ (2x + 5) – (2x + 3)
______ ______ 1 = __ Ú (÷2x + 5 +÷ 2x + 3 ) dx 2
1 2 __ 1 2 1 = __ __ × (2x + 5)3/2 + __ × __ (2x + 3)3/2 + c 2 3 2 3 2
1 = __ ((2x + 5)3/2 + (2x + 3)3/2) + c 6
(3x + 2)3/2
+ 2 dx = _________ +c Ú ÷3x 3 __
◊ 3 2
64. We have,
Ú (3x +
1 – __ 4) 2 dx
(3x + = ________ +c 1 __ ◊ 3 2 1 __ 4)2 +
______
______ ______ 1 = __ Ú (÷3x + 4 +÷ 3x + 1 ) dx 3
+ 4)3/2 (3x + 1)3/2 1 (3x = __ _________ + _________ + c 3 9/2 9/2
(
)
dx
______ ______ Ú _________________ ÷2x + 3 + ÷2x – 3 ______
(÷2x + 3 – ÷2x – 3 ) = Ú _________________ dx (2x + 3 – 2x + 3)
______ ______ 1 = __ Ú (÷2x + 3 –÷ 2x – 3 ) dx 6
+ 3)3/2 (2x – 3)3/2 1 (2x = __ _________ – _________ + c 3 3 6
c
(
)
69. The given integral is
dx
_____ _____ Ú ______________ ÷x + 2 – ÷x + 1 _____
dx
______ ______ Ú ÷ _______________ 3x + 4 – ÷3x + 1
______
65. We have,
)
______
1 __ 4)2
2 = __ (3x + 3
(
68. The given integral is
2 = __ (3x + 2)3/2 + c 9
______
(÷3x + 2 + ÷3x + 1 ) = Ú __________________ dx (3x + 4 – 3x– 1)
63. We have, ______
______
(÷2x + 5 +÷ 2x + 3 )dx _____ ______ ______ ______ = Ú __________________________________ (÷x + 5 – ÷ 2x + 3 ) (÷2x + 5 +÷ 2x + 3 )
67. The given integral is
61. We have,
dx
______ ______ Ú _________________ (÷2x + 5 – ÷2x + 3 )
______
60. We have,
_____
______
dx 1 = – __ log |5 – 2x| + c Ú ______ 5 – 2x 2 59. We have,
_____
57. We have, dx = Ú ______ 2x – 3 58. We have,
_____
66. We have,
(3x + 2)2
_____
(÷x + 2 + ÷x + 1 ) dx = Ú _________________ (x + 2) – (x + 1)
_____
(÷x + 2 + ÷ x + 1 ) dx _____ _____ _____ _____ = Ú _______________________________ (÷x + 2 – ÷x + 1 ) (÷x + 2 + ÷ x + 1 )
dx
_____
__
_____ __ = Ú (÷x + 1 – ÷ x )dx Ú ÷__________ x + 1 + ÷ x 2 = __ ((x + 1)3/2 – (x)3/2 ) + c 3
1.48 Integral Calculus, 3D Geometry & Vector Booster 70. The given integral is
74. We have,
dx
_____ _____ Ú ÷______________ x + a + ÷x + b
(
_____
_____
)
x + a – ÷x + b ÷ = Ú ______________ dx x+a–x–b
_____ _____ 1 = ______ Ú (÷x + a – ÷ x + b ) dx (a – b)
2 = _______ ((x + a)3/2 – (x + b)3/2) + c 3(a – b)
_________
_________
_________
_________
÷2x + 2014 – ÷2x + 2013 = Ú _______________________ dx (2x + 2014 – 2x – 2013)
2 t3 = __ __ – t + c 3 3
3/2 ______ 2 (3x + 1) = __ _________ –÷ 3x + 1 + c 3 3
( )
(
)
1 = __ [(2x + 2014)3/2 – (2x + 2013)3/2] + c 3
fi 2 dx = 2t dt fi dx = t dt
= 2 Ú ( + 1) dt
( )
t3 = 2 __ + t + c 3
(
2/3
_____
)
(x – 1) = 2 ________ + ÷ x – 1 + c 3
__
x ÷ t dx = Ú _____ 2 ◊ 2t dt, Ú _____ x+1 t + 1
Let x = t2
fi dx = 2t dt
t2 = 2 Ú _____ dt t2 + 1 2
(t + 1 – 1) = 2 Ú __________ dt t2 + 1
(
)
1 = 2 Ú 1 – _____ 2 dt t +1
= 2 (t – tan–1t) + c
–1 = 2 (÷x – tan (÷x ) ) + c
__
__
(t2 + 1)
t dt Ú ______ 2t
\
=
1 = __ Ú (t2 + 1) dt 2
1 t3 = __ __ + t + c 2 3
3/2 1 (2x – 1) = __ _________ + (2x – 1)1/2 + c 2 3
( )
2
Let 2x – 1 = t2
2 = __ Ú (t2 – 1) dt 3
= Ú (÷2x + 2014 – ÷2x + 2013 ) dx
2 (t______ + 1) x ______ _____ d x = × 2t dt. Ú ÷ x – 1 Ú t Let x – 1 = t2 fi dx = 2t dt
73. We have,
Let 3x + 1 = t2 2t dt fi dx = ____ 3
75. We have, x______ +1 dx Ú _______ 2x + 1 ÷
72. We have,
(Ú )
t2 – 1 _____ 3 _______ × 2t dt, t
dx
_________ _________ Ú ÷______________________ 2x + 2014 + ÷2x + 3013
=
71. The given integral is
x
______ dx Ú _______ ÷3x + 1
(
)
76. We have,
x–1
_____ dx Ú ÷______ x + 4
Let x + 4 = t 2
fi dx = 2t dt
( ) 2
–5 × t dt Ú t_____ t
=
= Ú (t2 – 5) dt
t3 = __ – 5t + c 3
(x – 1)3/2 = ________ – 5(x – 1)1/2 + c 3
(
)
77. We have,
x
2 dx Ú _____ x +1
Let x2 + 1 = t
Indefinite Integrals
1.49
fi 2x dx = dt
Let xex = t
1 fi x dx = __ dt 2
fi (xe x + ex) dx = dt
fi (x + 1) ex dx = dt
1 dt = __ Ú __ t 2 1 = __ log |t| + c 2 1 = __ log |x2 + 1| + c 2
78. We have,
cos x – sin x __________ dx Ú sin x + cos x
Let sin x + cos x = t
fi (cos x – sin x)dx = dt
dt = Ú _____ 2 cos t
= Ú sec2t dt
= tan t + c
= tan(xex) + c
82. We have,
dx
Ú _________ x (l + ln x)2
dt = Ú __ t = log |t| + c
= log |sin x + cos x| + c
dt = __ 2 t
1 = – __ + c t
1 = – __ + c t
1 = – _______ +c 1 + ln x
79. We have,
Let 1 + ln x = t 1 fi __ x dx = dt
3cos x
________ dx Ú 2sin x +5
Let 2 sin x + 5 = t fi 2 cos x dx = dt 1 fi cos x dx = __ dt 2
83. We have, cos x – sin x + 1 – x
3 dt = __ Ú __ t 2
dx Ú _________________ ex + sin x + x
3 = __ log |t| + c 2
(e x + cos x + 1) – (ex + sin x + x) = Ú ____________________________ dx ex + sin x + x
3 = __ log |2sin x + 5| + c 2
(e x + cos x + 1) (ex + sin x + x) = Ú _____________ – _____________ dx x e + sin x + x (e x + cos x + 1)
(ex + cos x + 1) (ex + sin x + x) = _____________ – _____________ dx x (e + sin x + x) (ex + sinx + x)
80. We have,
cos x – sin x
cos x – sin x
dx = Ú _____________ dx Ú __________ 2 + sin 2x 1 + (1 + sin 2x)
( Ú ( Ú (
)
cosx – sin x = Ú ________________ dx 1 + (sin x + cos x)2
(ex + cos x + 1) = _____________ – 1 dx (e x + sin x + x)
Let sin x + cos x = t
= log |ex + sin x + x| – x + c
fi (cos x – sin x) dx = dt
84. We have,
dt = Ú _____ 1 + t2
= tan–1(t) + c –1
= tan (t) + c
81. We have,
xe x + ex
(x + 1)ex
dx = Ú _________ dx Ú ________ cos2(xe x) cos2 (x e x)
dx
) )
e– x
x = Ú ______ – x dx Ú _____ 1+e e +1
Let e– x + 1 = t
fi – e– xdx = dt
dt = – Ú ____ t+1
= – log |t + 1| + c
= – log |e– x + 1| + c
1.50 Integral Calculus, 3D Geometry & Vector Booster 85. We have,
dx x2dx 3 = Ú ________ 3 3 Ú ________ x (x + 1) x (x + 1) dt 1 = __ Ú ______ 3 t(t + 1)
(
Put x3 = t
1 1 ____ 1 = __ Ú __ – dt t t+1 3
t 1 = __ log ____ + c 3 t+1
x3 1 = __ log _____ 3 + c 3 x +1
| |
dt 1 = __ Ú _______ Let x4 = t 4 t (t + 1) dt fi x3dx = __ 4
)
1 1 1 = __ Ú __ – ____ dt t t+1 4
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
)
| |
sin (5x – 3x)
dx = Ú ___________ dx Ú __________ sin 5x sin 3x sin 5x sin 3x
Let ex + e– x = t
fi (ex – e– x) dx = dt
dt = Ú __ t = log |t| + c
= log |e x + e–x| + c
91. The given integral is
ex
dx
= Ú ______ x dx = log |ex+ 1| + c Ú 1______ + e– x e + 1 e– xdx
dx
x = Ú ______ Ú 1_____ +e e– x + 1
= – log |e– x + 1| + c
ex = log _____ x + c e +1
| |
93. The given integral is
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 5x sin 3x sin 5x cos 3x __________ cos 5x sin 3x = Ú __________ – dx sin 5x sin 3x sin 5x sin 3x
(
ex – e– x
dx Ú _______ ex + e–x
x –1 1 = __ log _____ + c 5 x5 sin 2x
|)
92. The given integral is
5
88. We have,
1 = _________ [log |sin (x – b)| – log |sin (x – a)|] + c sin (b – a)
t–1 1 = __ log ____ + c t 5
1 = _________ [cot (x – b) – cot (x – a)] dx sin (b – a) Ú
dt 1 = __ Ú ______ Let x5 = t 5 t (t – 1) dx __ dt ___ x = 5t 1 1 1 = __ Ú ____ – __ dt 5 t–1 t
| |
)
90. The given integral is
sin (x – a) cos (x – b) – cos (x – a)sin (x – b)
( |
dx x4dx ________ 5 = Ú ________ 5 5 Ú x (x – 1) x (x – 1)
(
(
dx Ú ______________________________________ sin (x – a) sin (x – b)
sin (x – b) 1 = _________ log _________ + c sin (b – a) sin (x – a)
| |
87. We have,
dx
Ú __________________ sin (x – a) sin (x – b)
1 = _________ sin (b – a)
| |
89. We have,
4 = Ú ________ 4 4 Ú ________ x (x + 1) x (x + 1)
(
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
sin ((x – a) – (x – b)) dx 1 = _________ Ú ____________________ sin (x – a) sin (x – b) sin (b – a)
x3dx
dx
sin (b – a)dx 1 = _________ Ú __________________ sin (b – a) sin (x – a) sin (x – b)
| |
86. We have,
= Ú (cot 3x – cot 5x) dx
)
)
sin 2x
_____________ 2 dx Ú a sin x + b cos2x
Let a sin2x + b cos2x = t
Indefinite Integrals
(a – b) sin 2x dx = dt
dt sin 2x dx = ______ (a – b) dt 1 = ______ __ (a – b) Ú t 1 = ______ log |t| + c (a – b)
1 = ______ log |a cos2x + b sin2x| + c (a – b)
94. The given integral is
sin (x – a)
sin x cos a – cos x sin a
dx = Ú __________________ dx Ú _________ sin x sin x = Ú (cos a – sin a cot x) dx
1.51
98. The given integral is 1 + tan x dx Ú __________ x + log sec x
Let x + log sec x = t
fi (1 + tan x) dx = dt
dt = Ú __ t = log |t| + c
= log |x + log (sec x)| + c 99. The given integral is
sin(5x – 3x)
sin 2x
dx = Ú ___________ dx Ú __________ sin 3x sin 5x sin 3x sin 5x
= x cos a = sin a log |sin x| + c 95. The given integral is sin (t + a) sin x ________ dx = Ú _________ dt, Ú sin (x – a) sin t
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 3x sin 5x
= Ú (cot 3x – cot 5x) dx
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
100. The given integral is
Let t = x – a fi dt = dx
cos x – sin x
(cos x – sin x)
dx = Ú ________________ dx Ú ___________ 1 + sin 2x 1 + (sin x + cos x)2
= t cos a + sina log |sin t| + c = (x – a) cos a + sin a log |sin(x – a)| + c 96. The given integral is
Put (sin x + cos x) = t
sin (x + a) Ú _________ dx sin (x + b)
fi (cos x – sin x) dx = dt
sin [t + (a – b)] = Ú _____________ dt, sin t
dt = Ú _____ 1 + t2
Let t = (x + b)
= tan–1(t) + c
fi dt = dx
= tan–1(sin x + cos x) + c
sin (t) cos (a – b) + cos t sin (a – b) = Ú _____________________________ dt sin t
101. The given integral is
= Ú (cos (a – b) + sin (a – b) cot t) dt
= t cos (a – b) + sin (a – b) log |sin t| + c
= Ú (sec x tan x + cosec x)dx
x = sec x + log tan __ + c 2
= (x + b) cos (a – b) + sin (a – b) log |sin (x + b)| + c
|
97. The given integral is
dx
__ __ Ú ÷__________ x (÷x + 1)
dt
Ú __ t
=2
= 2 log |t| + c
= 2 log |÷x + 1| + c
__
(sin2x + cos2x) dx
dx
= Ú ______________ Ú _________ sin x cos2x sin x cos2x
( ) |
102. The given integral is __
Let (÷x + 1) = t dx fi ____ = dt __ 2÷x dx __ fi ___ = 2dt x ÷
(sin2x + cos2x)
dx
2 2 = Ú ____________ dx Ú _________ sin x cos x sin2x ◊ cos2x = Ú (sec2x + cosec2x) dx
= tan x – cot x + c. 103. The given integral is
dx
__________________ Ú sin x (x – a) sin (x – b)
1.52 Integral Calculus, 3D Geometry & Vector Booster sin (a – b) 1 = _________ Ú _________________ dx sin (x – a) sin (x – b) sin (a – b) sin [(x – b) – (x – a)] 1 = _________ ___________________ dx sin (a – b) Ú sin (x – a) sin (x – b)
= log |t| + c
107. The given integral is
= log |xx + 1| + c
1 = _________ sin (a – b)
cos x – sin x + 1 – x
dx Ú _________________ ex + sin x + x (cos x + 1) – (sin x + x) = Ú ____________________ dx (e x + sin + x)
sin (x – b) cos (x – a) – cos (x – b)sin (x – a) Ú ______________________________________ sin (x – a) sin (x – b) dx 1 = _________ [cot (x – a) – cot (x – b)] dx sin (a – b) Ú
=
1 = _________ [log |sin (x – a)| – log |sin (x – b)|] + c sin (a – b)
( |
|)
sin (x – a) 1 = _________ log ________ + c sin (a – b) sin (x – b)
(ex + cos x + 1) – (ex + sin x + x)
dx Ú _____________________________ (ex + sin x + x)
(
)
(ex + cos x + 1) = Ú ________________ dx (ex + sin x + x) – 1
= log |ex + sin x + x| – x + c
108. The given integral is
104. The given integral is
sin3x
4 dx Ú __________________________________ (cos x + 3cos2x + 1) tan–1(sec x + cos x)
dx Ú _________________ cos (x – a) cos (x – b)
sin (a – b) 1 = _________ Ú _________________ dx sin (a – b) cos (x – a) cos (x – b)
Let tan–1(sec x + cos x) = t
sec x tan x – sin x fi ________________ dx = dt 1 + (sec x + cos x)2
sin3x fi ______________________ dx = dt cos2x (1 + (sec x + cos x)2)
sin3x fi ____________________ dx = dt 2 2 cos x (sec x + cos2x + 3)
sin3x fi _________________ dx = dt 4 (cos x + 3cos2x + 1)
sin [(x – b) – (x – a)] 1 = _________ ___________________ dx sin (a – b) Ú cos (x – a) cos (x – b) 1 = _________ [tan(x – b) – tan (x – a)] dx sin (a – b) Ú 1 = _________ [– log |cos (x – b)| + log |cos (x – a)|] + c sin (a – b) cos (x – a) 1 = _________ log _________ + c sin (a – b) cos (x – b)
( |
|)
dt = Ú __ t
= log |t| + c
= log |tan–1(sec x + cos x)| + c
cos (a – b) 1 = _________ _________________ dx cos (a – b) Ú sin (x – a)cos (x – b)
cos [(x – b) – (x – a)] 1 = _________ Ú ___________________ dx cos (a – b) sin (x – a) cos (x – b)
= – cos (t) + c
1 = _________ [tan (x – b) tan (x – b)]dx cos (a – b) Ú
= – cos (x3) + c
cos (x – a) 1 = _________ log _________ + c cos (a – b) cos (x – b)
105. The given integral is dx Ú _________________ sin (x – a) cos (x – b)
( |
|)
106. The given integral is x
x (1 + ln x)
dt
x dx = Ú __ , Let t = xx Ú __________ t x +1
109. We have,
Ú 3x3sin (x3) dx = Ú sin(t) dt
where x3 = t
110. We have,
(1 + ln x)3
dx = Ú t3dt Ú _________ x
Let (1 + ln x) = t dx fi ___ x = dt
t4 = __ + c 4
(1 + ln x)4 = _________ +c 4
Indefinite Integrals
111. We have,
114. The given integral is dx
dx
x x 4 + 1 x
Let tan x = t
fi sec2x dx = dt
2 = Ú _______________ Ú __________ 3/4 1 2 4 __ x (1 + x4)3/4
=
( ( ) ) Ú ( ) Ú ( ) dx ___________ 3/4 1 5 __ x 4 + 1 x
4t3 1 = – ___ 3 dt, Let __ 4 + 1 = t4 t x – 4 dx fi _____ 5 = t3dt x
= – 4 Ú dt
= – 4t + c
1/4 1 = – 4 __ 4 + 1 + c x
(
)
__
Let 4 + 3÷x = t
fi
3 ____ __ dx = dt 2÷x
fi
dx ___ __ = x ÷
2 __ dt 3
dt __ 2 t 2 1 = – __ × __ + c t 3 2 = __ Ú 3
Ú 33 ◊ 33 ◊ 3xdx 3x
x
Let 3 x = t
1 fi 3xdx = _____ dt loge 3
t4 = __ + c 4
tan4x = _____ + c 4 115. The given integral is
Ú sin3x ◊ cos x dx
= Ú t3dt,
(log x)3 ______ dx Ú x
= Ú t3dt,
1 = ______ 2 Ú 33 ◊ 3tdt, (log3) t
3v = ______ +c (log3)3
33 = ______ +c (log3)3
3x
Let t = log x dx fi dt = ___ x
sin x
_________ dx Ú __________ ÷3 + 2cos x
Let 3 + 2 cos x = t2
fi – 2 sin x dx = 2t dt
fi sin x dx = – t dt
t dt = – Ú ___ t
= – Ú dt
Let 3t = v
= – t + c
fi logs 3dt = dv
= – ÷3 + 2 cos x + c
1 = ____ Ú 3 3 ◊ 3tdt log3
t4 = __ + c 4
t
Let t = sin x fi dt = cos x dx
sin4x = _____ + c 4 116. The given integral is
113. We have,
t4 = __ + c 4 (log x)4 = ______ + c 4 117. The given integral is
2 1 = __ × ________ __ + c 3 (4 + 3÷x )
= Ú t3dt
dx Ú ÷ __________ __ __ 2 x (4 + 3 ÷x )
Ú tan3x ◊ sec2x dx
112. We have,
1.53
_________
118. The given integral is ________
÷ 2 + log x dx Ú _________ x
Let 2 + log x = t2 dx fi ___ x = 2t dt
1.54 Integral Calculus, 3D Geometry & Vector Booster
= Ú (t ◊ 2t) dt
122. The given integral is
= 2 Ú t2 dt
t3 = 2 __ + c 3
( )
2 = __ [(2 + log x)3/2] + c 3 119. The given integral is
sin x – cos x
dx Ú __________ ex + sin x
(ex + sin x) – (ex + cos x) = Ú _____________________ dx (ex + sin x)
(ex + cos x) = Ú 1 – _________ dx (ex + sin x)
= (x – log |ex + sin x| + c)
(
dx
__ Ú 1______ +÷ x Let x = t2 fi dx = 2t dt
2t = Ú ____ dt 1+t
1 = 2 Ú 1 – ____ dt 1+t
= 2 (t – log |1 + t|) + c
(
__
Let x4 = t
dt fi x dx = __ 4
1 = __ Ú sin (t)dt 4 1 = – __ cos (t) + c 4 1 = – __ cos (x4) + c 4
Ú 5
x
5x
x
◊ 5 ◊ 5 dx
Put 5x = t
dt fi 5xdx = ____ log 5
t 1 = ____ 55 ◊ 5tdt log5 Ú
1 = ______ 2 Ú 5z ◊ dz, z = 5t (log5)
1 = ______ 2 5z ◊ dz Ú (log5)
5z = ______ +c (log5)3
= Ú (1 – cos2x)cos4x sin x dx
= – Ú (1 – t2) t4dt
55 = ______ +c (log5)3
Let cos x = t fi sin xdx = dt
= Ú (t6 – t4) dt
t7 t5 = __ – __ dt 7 5
cos7x _____ cos5x = _____ – + c 7 5
( (
)
)
127. We have,
Ú sin2x cos3x dx = Ú sin2x(1 – sin2x) cos x dx = Ú t2(1 – t2) dt, Let sin x = t
fi cos xdx = dt
= Ú (t2 – t4) dt
t3 t5 = __ – __ + c 3 5
sin3x _____ sin5x = _____ – + c 3 5
5x
Ú sin3 x cos4 x dx
121. The given integral is 5 5
| |
| |
124. We have,
3
)
x3 1 = __ log _____ 3 + C 3 x +1
3
fi 4x dx = dt
(
1 1 1 = __ Ú __ – ____ dt t t+1 3 t 1 = __ log ____ + C 3 t+1
x2dx
d
________ = Ú ________ 3 3 Ú x (1 + x3) x (x + 1)
Ú x3sin x4 dx
dt 1 = __ Ú _______ , Let t = x3 3 t (t + 1) fi dt = 3x2
= 2 (÷x – log |1 + ÷ x | ) + c 120. The given integral is
123. The given integral is
)
__
)
( (
)
)
1.55
Indefinite Integrals
130. We have,
Ú sin3x cos3x dx
= Ú sin x (1 – sin )cos x dx
= Ú t (1 – t ) dt
3
3
2
Let sin x = t fi cos xdx = dt
= Ú (t3 – t5) dt
sin3x _____ sin5x = _____ – + c 3 5
(
)
134. we have,
Ú sin2x cos2x dx = __ 14 Ú (2 sin2x) (2 cos2x)dx
1 = __ Ú (1 – cos 2x) (1 – cos 2x)dx 4
1 = __ Ú (1 – cos22x)dx 4
1 = __ Ú sin22x dx 4
1 = __ Ú 2 sin22x dx 8 1 = __ Ú (1 – cos 4x) dx 8 sin 4x 1 = __ x – _____ + c 8 4
(
)
135. We have,
Ú sin3x dx = Ú sin2x ◊ sin x dx = Ú (1 – cos2x) ◊ sin x dx = – Ú (1 – t2)dt, Let cos x = t
139. We have,
fi –sin xdx = dt
= Ú (t2 – 1)dt
( ) (
t3 = __ – t + c 3 3 cos x = _____ – cos x + c 3
)
Ú cos5x dx = Ú cos4x ◊ cos dx
= Ú (1 – sin2x)2 ◊ cos dx
= Ú (1 – t ) dt,
t5 = __ – 5
sin5x __ 2 = _____ – sin3x + sin x + c 3 5
( (
2 2
Let sin x = t fi cos xdx = dt
)
2 __ t3 + t + c 3
)
142. We have,
= Ú (t4 – 2t2 + 1)dt
2
Ú sin4 x dx = __ 14 Ú (2 sin2x)2dx
1 = __ Ú (1 – cos 2x)2dx 4 1 = __ Ú (1 – 2 cos 2x + cos22x) dx 4 1 1 = __ Ú (1 – 2 cos 2x)dx + __ Ú (2cos22x) dx 4 8 1 1 = __ Ú (1 – 2 cos 2x) dx + __ Ú (1 + cos 4x) dx 4 8 2 sin 2x sin 4x 1 1 = __ Ú x – ______ + __ x + _____ + c 4 2 8 4 3 1 1 = __ x – __ sin 2x + ___ sin 4 x + c 8 4 32
(
) (
)
145. We have,
Ú cos6x dx = __ 18 Ú (2 cos2x)3dx
1 = __ Ú (1 + cos 2x)3dx 8 1 = __ Ú (1 + 3 cos 2x + 3 cos22x + cos32x) dx 8 3 1 = __ Ú (1 + 3 cos 2x) dx + ___ Ú (2 cos22x) dx 8 16 1 + ___ Ú (4 cos32x) dx 32
3 1 = __ Ú (1 + 3 cos 2x)dx + ___ Ú (1 + cos 4x) dx 8 16 1 + ___ Ú (cos 6x + 3 cos 2x) dx 32
(
(
)
)
3 3 sin 4x 1 = __ x + __ sin 2x + ___ x + _____ 8 2 4 16
(
)
3 1 sin 6x __ + ___ _____ + sin 2x + c 32 2 6
5 15 3 1 = ___ x + ___ sin 2x + ___ sin 4x + ____ sin 6x + c 192 16 64 64 148. We have,
dx
1/2 Ú ___________ sin x cos3/2x
Divide the numerator and the denominator by cos2x, we get
1.56 Integral Calculus, 3D Geometry & Vector Booster 2t dt sec2x ____ dx = Ú ____ , Let tan x = t2 Ú _____ t tan x ÷ fi sec2 xdx = dt
= 2 Ú dt
= 2t + c
=
t2 t4 = __ + __ + C 2 4
tan2x _____ tan4x = _____ + + C 2 4
152. The given integral is
____ 2÷tan x +
c
sec8x
dx
3 5 = Ú _____ 3 dx Ú _________ tan x sin x cos x
149. We have, dx 3/2 Ú ___________ sin x cos5/2x Divide the numerator and the denominator by cos4x, we get
Ú
(1 + tan2x) sec2x sec4x ______ 3/2 dx = Ú ______________ dx tan x tan3/2x (1 + t2)dt = Ú ________ Let tan x = t2 t3/2 fi sec2 xdx = dt
=
Ú (t– 3/2 + t1/2)dt
_ 2 = 2 ÷t + __ t3/2 + c 3 ____ 2 = 2 ÷tan x + __ (tan x)3/2 + c 3
150. The given integral is
____ ÷tan x ________
dx = Ú dx Ú sin x cos x sin x cos x ____ sec x ÷tan x __________ 2
= Ú
=
2t dt = Ú ____ , Let tan x = t2 t
dx tan x
sec2x
____ dx Ú _____ ÷tan x
fi sec2 xdx = 2t dt
= Ú 2 dt
= 2t + c
=
_____ 2÷ tan x +
c
sin x
5 dx = Ú sec4x tan x dx Ú _____ cos x
= Ú (1 + tan2x) tan x sec2x dx
= Ú (1 + t ) t dt Let tan x = t fi sec2 xdx = dt 2
= Ú (t + t3) dt
(
)
where t = tan x 153. The given integral is
sin2x dx
= Ú tan2x sec4x dx Ú _______ 6 cos x =
=
Ú tan2x (1 + tan2x) sec2x dx Ú t2(1 + t2) dt Let tan 2 x = t
fi sec x dx = dt
= Ú (t4 + t2) dt
t 5 t3 = __ + __ + c 3 5
tan5x ____ tan3 = _____ + + c 3 5
154. The given integral is sec4x
dx
__________ = Ú _____ dx Ú sin x cos x 3 tan x x
Ú
(1 + tan2x)sec2 x _____________ dx tan x
Ú
(1 + t2) ______ dt Let tan x = t t
=
=
=
t2 = __ + log |t| + c 2
tan2x = _____ + log |tan x| + c 2
Let t = tan x fi dt = sec2 xdx
(1 + 3t2 + 3t4 + t6) = Ú _________________ dt t3 1 3 = Ú __ 3 + __ + 3t + t3 dt t t 3t2 t4 1 = – ___2 + 3log |t| + ___ + __ + c 2 4 2t
151. The given integral is
(1 + t2)3 = Ú _______ dt, t3
____ ÷tan x ________
(1 + tan2x)3sec2 = Ú _____________ dx tan3x
Ú ( 1 + __ 1t ) dt
Indefinite Integrals
155. The given integral is
dx sec x 2 4 = Ú _____ 2 dx Ú _________ sin x cos x tan x (1 + tan2x)2sec2x = Ú ______________ dx tan2x
(1 + t2)2 = Ú _______ dt Let tan x = t t2 fi sec2 xdx = dt
(
)
158. We have, x 1 2 = Ú ______ = __ tan–1 ( __ ) + c Ú _____ 2 x +4 x2 + 22 2 dx
dx
159. We have, dx
dx
1 1 2 = __ Ú ________ 2 = __ × 3tan–1 (3x) + c Ú _______ 9 1 __ 9 x + 1 9 2
( )
x + 3
(t4 + 2t2 + 1) = Ú ___________ dt t2
(
tan2x = 2 _____ + log |tan x| + C 2
6
160. We have,
)
1 = Ú t2 + __ 2 + 2 dt t
t3 1 = __ – __ + 2t + c t 3
x–2 1 = ___ log _____ + c 2.2 x+2
x–2 1 = __ log _____ + c 4 x+2
tan3x = _____ + 2tan x – cot x + c 3 156. The given integral is
dx 1 = Ú __________ 7 __ __ sin2 x cos2 x
sec4x ____ dx Ú ÷_____ tan x 2
=
2
(1 + tan x)sec x
____ dx Ú _____________ ÷tan x
(1 + t2)2t = Ú ________ dt t
= 2 Ú (1 + t ) dt
t3 = 2 t + __ + c 3
Let tan x = t fi sec2 xdx = dt
dx
=
dx Ú ____________ x2 + 4
=
dx Ú ____________________ x2 + 4
5 = Ú x2 – 4 + ______ dx 2 x + 22
5 x3 x = __ – 4x + __ tan–1 __ + c 3 2 2
sec x
dx
(1 + tan2x) sec2x (1 + t )2t
dt, Ú ________ 3 t
(1 + t2) = 2 Ú ______ dt t
1 = 2 Ú t + __ dt t
t2 = 2 __ + log |t| + C 2
(
dx
÷x + 4 ÷ x +2
)
= log (x + ÷x 2 + 4 )+c
163. We have, dx
Let tan x = t 2
÷ ( ) 1 = __ log x + x + __ 2 | ÷ 14 | + c
÷4x + 1
x2 + __ 2 ______ 2
2
fi sec x dx = 2t dt
dx
1 ______ = __ Ú _________ ________ Ú ________ 2 2 1 2
dx Ú ______________ tan3/2x
( )
( )
÷sin x cos x
)
_____
dx
=
((x2 + 4) (x2 – 4)) + 15
_____ = Ú _______ ______ Ú _______ 2 2 2
4
2
(x4 – 16) + 15
(
162. We have,
_________ = Ú ______ 3/2 dx Ú __________ 3 5 tan x =
x4 – 1
dx Ú _____ x2 + 4
157. The given integral is
| | | |
161. We have
3
tan x = 2 tan x + _____ + c 3
dx
2 = Ú _______ 2 2 Ú _____ x –4 x – (2)
2
( ) ( )
1.57
164. We have,
(
)
(
)
x+4 x 4 Ú ______ dx = Ú ______ 3 + ______ dx x3 + 4x x + 4x x3 + 4x
(
)
1 4 = Ú _____ 2 + ________ dx x + 4 x (x2 + 4)
1.58 Integral Calculus, 3D Geometry & Vector Booster
169. The given integral is
dx 4x = Ú _____ 2 + Ú ________ 2 2 dx x +4 x (x + 4)
( )
x 1 = __ tan–1 __ + 2 2 2
dt
, Let t = x2 Ú _______ t (t + 4) fi dt = 2x2dx
(1 + x) dx
1+x
dx = Ú _________ 2 Ú ______ x3 + x x (x + 1) (1 + x2) + x – x2
=
dx Ú _______________ x (x2 + 1)
=
x + Ú _____ 2 – Ú _____ dx Ú ___ x +1 x2 + 1
x 1 2 1 ____ 1 = __ tan–1 __ + __ Ú __ – dt t t+4 2 2 4
1 = __ tan–1 2
) ( ) ( x 1 ( __ ) + __ log ____ 2 2 | t +t 4 | + c
1 = __ tan–1 2
x __ + 2
1 = log |x| + tan–1x – __ log |x2 + 1| + c 2 170. The given integral is
( )
| |
x2 1 __ log _____ 2 + c 2 x +4
165. The given integral is
4 +1 dx = Ú xx_____ 2 +1
(x4 – 1) + 2 ___________ dx Ú x2 + 1
( (
) )
2 = Ú x2 – 1 + _____ dx 2 x +1
x3 = __ – x + 2 tan–1x + c 3
=
dt
_____ , Ú ______ 2
Let t = (x – 2)
÷t + 1
fi dt = dx
_____ ÷t 2 + 1 |+
= log |t +
= log |(x – 2) + ÷ (x – 2)2 + 1 | + c
c
dx
(x – 1) (x + 1)
(
=
Ú
x + (9 + x2) – x2 _______________ dx x (x2 + 9)
dx = Ú _______ 2 + (x + 9)
1 = __ tan–1 3
dx
x dx
x – Ú _______ 2 Ú ___ (x + 9)
( )
x 1 __ + log |x| – __ log| x2 + 9| + c 3 2
1 1 1 = __ Ú _____ – _____ dx 2 x2 – 1 x2 + 1
x–1 1 __ 1 = __ log _____ – tan–1x + c 2 2 x+1
( |
1+x
dx
x
dx = Ú _____ 2 + Ú _____ dx Ú ______ 2 1 + x 1+x 1 + x2
1 = tan–1(x) + __ log |1 + x2| + c 2
|
)
171. The given integral is
dx
dx
3 = Ú ________ 2 Ú _____ x +x x(x + 1) x dx = Ú ________ 2 2 x (x + 1)
dt 1 = __ Ú _______ , Let x2 = t2 2 t (t + 1) 2x dx = 2t dt
(
)
1 1 ____ 1 = __ Ú __ – dt t t+1 2
t 1 = __ log ____ + c 2 t+1
x2 1 = __ log _____ 2 + c 2 x +1
| |
| |
172. We have,
dx
dx = Ú _______ 2 2 Ú __________ 2 (x + 2)
1 = – ______ + c (x + 2)
x + 4x + 4
173. We have,
168. The given integral is
)
167. The given integral is
x
dx 4 = Ú ______________ Ú x_____ 2 2 –1
__________
x+9 x+9 dx = Ú ________ dx Ú _______ 3 x + 9x x (x2 + 9)
dx
166. The given integral is dx dx __________ = Ú ____________ __________ Ú ____________ 2 ÷(2 – x) + 1 ÷(x – 2)2 + 1
dx
dx dx = Ú ________________ Ú ___________ 2 2 x + 6x + 10
(x + 3) + (10 – 9)
dx = Ú __________ (x + 3)2 + 1
= tan–1(x + 3) + c
1.59
Indefinite Integrals
174. We have,
178. The given integral is
1 dx dx = __ Ú ___________ Ú ___________ 2 2 2 __ 5 2x + 5x + 6
x + x + 3 2 1 dx = __ Ú _________________ 2 2 5 25 x + __ + 3 – ___ 4 16
( ) (
)
dx 1 ___ = __ Ú _______________ 2 ÷23 2 5 2 __ ____ x + + 4 4
( ) ( ) 5 ( x + __ ) 4 1 ____ 4 __ _______ = × tan + c
23 ÷ ____ 4
)
( ) ( ) 1 ( x + __ ) 2 2 = ___ tan _______ + c –1
__
÷3
( ) __
(
3 ÷ ___ 2
)
dx
dx
( ) ( ) ÷5 1 ( x – __ ) – ___ 2 2 1 = – ___ log ___________ + c ÷5 ÷5 1 __ ( x – ) + ___ 2 2
|
| |
dx Ú (x__________ + 2)2 – 1
| | | |
|
(x – a) – a 1 = ___ log __________ + c 2a (x – a) + a
(x – 2a) 1 = ___ log _______ + c x 2a
|
|
180. The given integral is
dx
dx
2 = Ú __________ Ú x_______ + 2ax (x + a)2 – a2
|
|
x 1 = ___ log ______ + c 2a x + 2a
dx
1 2 = ___ log |a2 + 2ax| + c Ú a_______ 2a + 2ax
182. The given integral is
dx
dx = – Ú _______ Ú ________ 2 x2 – 2ax 2ax – x
dx = – Ú ___________ (x – a)2 – a2
x – a – a 1 = – ___ log ________ + c x–a+a 2a
x 1 = ___ log ______ + c 2a x – 2a
__
dx 2 = Ú _________ x + 4x + 3
(x – a) – a
dx __ = – Ú ______________ ÷5 2 1 2 __ ___ x – – 2 2
177. The given integral is
dx
dx 2 = Ú ____________ Ú x_______ 2 2 2 – 2ax
181. The given integral is
= – Ú ________ Ú 1________ + x – x2 x2 – x – 1
__
)
179. The given integral is
2x +__ 1 2 = ___ __ tan–1 ______ + c 3 3 ÷ ÷
)
(4x + 5) 2 ___ = ____ ___ tan–1 _______ + c 23 ÷23 ÷
(
___
__
(
( )
–1
___ ÷32
176. The given integral is
( ) ( )
7 2 = _____ ____ tan–1 8x + _____ ____ + c 111 111 ÷ ÷
dx __ dx = Ú ______________ Ú _________ 2 2 ÷3 2 1 x +x+1 x + __ + ___ 2 2
dx 1 ____ = __ Ú ________________ 4 2 111 2 ÷ 7 __ _____ x + + 8 8
(
)
x + x + 4 2
8 7 1 = __ × _____ ____ tan–1 8x + _____ ____ + c 4 ÷ 111 111 ÷
175. The given integral is
(
4x + 7x + 10
2
dx
1 dx = __ Ú ____________ Ú ____________ 2 4 5 7 2 __ __
|
183. The given integral is
(x + 2) – 1 1 = __ log __________ + c 2 (x + 2) + 1
(x + 1) 1 = __ log ______ + c 2 (x + 3)
|
|
|
(
)
1 1 1 2 = ___ Ú _____ 2 – _____ 2 dx Ú (x_____________ 2 3a + 1) (x + 4) x +1 x +4 dx
(
1 1 = __ tan–1x – __ tan–1 3 2
( ) )
x __ + c 2
1.60 Integral Calculus, 3D Geometry & Vector Booster 184. We have, x2 1 3x2 6 dx = __ _____ dx Ú _____ 3 x6 + 1 x +x 1 dt = __ _____ , Let t = x3 3 t2 + 1 fi dt = 3x3 dx
1 = __ tan–1 (t) + c 3
1 = __ tan–1(x3) + c 3
cos x dx _______________ 2 Ú sin x + 3 sin x + 2
dt = Ú _________ 2 t + 3t + 2
Let t = sin x fi dt = cos x dx
)
1 1 ____ 1 = __ Ú __ – t t+1 4
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
| |
| |
194. We have,
185. We have,
(
(2x + 4) – 1
2x + 3 dx = Ú ___________ dx Ú __________ 2 2
(2x + 4) = Ú __________ dx – 2 x + 4x + 5
(2x + 4) dx = Ú __________ dx – Ú __________ 2 (x + 2)2 + 1 x + 4x + 5
= log |x2 + 4x + 5| – tan–1(x + 2) + c
x + 4x + 5
x + 4x + 5
dx
2 Ú __________ x + 4x + 5
dt = Ú ___________ (t + 1) (t + 2)
1 1 = Ú ____ – ____ dt t+1 t+2
t+1 = log ____ + c t+2
sin x + 1 = log _______ + c sin x + 2
9 3 (2x – 3) + 2 + __ __ 2 2 = Ú __________________ dx 2 x – 3x + 4
(2x – 3) 3 13 dx = __ Ú __________ dx + ___ Ú __________ 2 x2 – 3x + 4 2 x2 – 3x + 4
(
195. We have,
)
| | | |
186. We have,
xx(1 + log x)
dx = Ú ________ dt Ú ___________ 2x x 2 x
+x +1
t + t + 1
dt __ = Ú ______________ 3 2 ÷ 1 2 __ ___ t + + 2 2
( ) ( ) 1 ( t + __ ) 2 2 = ___ tan ______ + c
__
–1
÷3
( ) (
)
2x x __ +1 2 = ___ __ tan–1 ______ + c 3 3 ÷ ÷
187. We have,
dx
x3dx
________ 4 = Ú ________ 4 4 Ú x (x + 1) x (x + 1) 3
4x dx 1 = __ Ú ________ 4 x4(x4 + 1)
dt 1 = __ Ú _______ , Let t = x4 4 t (t + 1) fi dt = 4 x3 dx
x – 3x + 4
(
)
(2x – 3) dx 3 13 __ = __ Ú __________ dx + ___ Ú ______________ 2 2 x – 3x + 4 2 ÷7 2 3 2 – __ + ___ 2 2
( ) ( ) 3 ( x – __ ) 2 13 ______ ___ – 3x + 4| + tan + c
( )
3 = __ log |x2 2
3 13 2x –__ 3 = __ log |x2 – 3x + 4| + ___ __ tan–1 ______ + c 2 7 ÷7 ÷
__
3 ÷ ___ 2
3x + 2 dx Ú __________ 2
–1
__
÷7
__
7 ÷ ___ 2
(
)
195. The given integral is 1 (2x + 1) – 1 x Ú _________ dx = __ Ú ___________ dx 2 2 (x2 + x + 1) x + x + 1 (2x + 1) dx 1 1 __ = __ Ú __________ dx – __ Ú ______________ 2 (x2 + x + 1) 2 ÷3 2 1 2 __ ___ x + + 2 2
( ) ( )
(
)
2x +__ 1 1 1 = __ log |x2 + x + 1| – ___ __ tan–1 ______ + C 2 3 3 ÷ ÷
Indefinite Integrals
196. The given integral is
200. The given integral is.
1 2x + __ 4x + 1 2 _________ ___________ dx = 2 Ú dx Ú x2 + 3x +2 (x2 + 3x + 2)
(2x + 3) – (5/2) = 2 Ú ______________ dx (x2 + 3x + 2)
(2x + 3) dx = 2 Ú ___________ dx – 5 Ú ___________ 2 (x + 1) (x+ 2) (x + 3x + 2)
| |
____________ 2x = Ú ______________ x Ú 2e (2e + 1) (ex + 1) + 3ex + 1
x+1 = 2 log |x2 + 3x + 2| – 5 log _____ + c x+ 2
dx
dx
( Ú (
) )
× (2 + 2 sin2 x)dx Ú ( __________ cos x + sin x ) cos x – sin x
( ) 1+t = Ú ( _____ dt, Let cos x + sin x = t t )
cos x – sin x = Ú __________ (1 + (sin x + cos x)2) dx cos x + sin x 2
197. The given integral is
fi (– sin x + cos x) dx = dt
( )
1 = Ú t + __ dt t t2 = __ + log |t| + c 2 (sinx + cos x)2 = ____________ + log |sin x + cos x| + c 2
201. The given integral is (sin x + cos x) dx
2 1 = Ú ________ x – _______ dx (2e + 1) (ex + 1)
2e–x e– x = _______ – ________ – x dx –x (2 + e ) (e + 1)
(sin x + cos x)dx = Ú ______________ 8 – 3 (1 – sin 2x)
= log |e– x + 1| – 2 log |e– x + 2| + c
(sin x + cos x)dx = Ú _________________ 8 – 3 (sin x – cos x)2
198. The given integral is
(3sin x – 2) cos x = Ú _______________ dx sin2x – 4sin x + 4
(3t – 2) = Ú _________ dt, 2 t – 4t + 4
sin x + cos x dx = Ú __________________ Ú ___________ 5 + 3 sin 2x 5 + 3 (1 – 1 + sin 2x)
dt = Ú ______ 2 , Let t = (sin x – cos x) 8 – 3t dt = (cos x + sin x)dx
(3sin x – 2) cos x dx Ú _________________ (5 – cos2x – 4sin x)
Let t = sin x fi dt = cos x dx
dt = Ú ______ 2 8 – 3t dt 1 __ = – __ Ú _________ 3 2 8 2 __ t – 3
( ÷ )
|
(3t – 2) = Ú _______ dt (t – 2)2
(3t – 6 + 4) = Ú __________ dt (t – 2)2
dt dt = 3 Ú ______ + 4 Ú ______ (t – 2) (t – 2)2
4 = 3 log |t – 2| – ______ + c (t – 2) 4 = 3 log |sin x – 2| – ________ + c (sin x – 2)
sin x – cos x = Ú ________________ dx 5 (sin x + cos x)2 – 2
cos x – sin x = Ú ________________ dx 2 – 5 (sin x + cos x)2
=
ax3 + bx
x3
x dx
dx = a Ú ______ 4 2 dx + b Ú ______ 4 2 Ú _______ x4 + c2 x +c x +c
( )
a b x2 = __ log |x4 + c2| + ___ tan–1 __ c + c 4 2c
___
|
t – (÷8/3 ) 1 ___ = – _____ ___ log ________ + c 2÷24 t + (÷8/3 )
where t = sin x + cos x 202. The given integral is
199. The given integral is
1.61
sin x – cos x
sin x – cos x
dx = Ú _______________ dx Ú ___________ 3 + 5 sin 2x 5 (1 + sin 2x) – 2
dt
2 , Let t = (sin x + cos x) Ú 2______ – 5t
fi dt = (cos x – sin x)dx
1.62 Integral Calculus, 3D Geometry & Vector Booster dt
2 Ú ______ 2 – 5t
=
dt 1 ___ = – __ Ú __________ 2 5 t2 – (÷2/5 )
t – (÷2/5 ) 1 ___ = – ____ ___ log _________ + c 10 t + (÷2/5 ) ÷
=–
| |
Ú
|
| (
)
1 = log x + __ + 2 dx
dx = Ú _________________ ________________ 25 1 2 ___ – x + __ – 2 4
÷ { ( )
÷( ) ( ) 1 ( x + __ ) 2 _______ = sin + c –1
(
|
= log (x – a) +
dx
dx
dx
dx
______ = Ú _______________ ______________ Ú _______ 2 2 ÷4x – x
c
÷( ) ( )
÷1 + x + x
3 ÷ 1 ___ x + __ + 2 2
| (
dx = Ú ____________ ___________ 2 ÷2 – (x – 2)2
x – 2 = sin–1 _____ + c 2
( )
dx = Ú ________________ _______________ 5 1 2 __ – x – __ – 2 4
÷ { ( ) }
dx = Ú ______________ _____________ – 2)2 – 4} ÷ – {(x
=
dx ____________ __________ ÷4 – (x – 2)2
x – 2 = sin _____ + c 2
Ú
206. The given integral is
–1
( )
dx
_______________ __ Ú ________________ 2 2
÷( ) ( )
=
1 x – __ 2 __ = sin–1 ______ + c ÷5 ___ 2
2x –__ 1 = sin–1 ______ + c ÷5
÷5 1 __ ___ – x – 2 2
(( )) (
205. The given integral is dx dx ______ = Ú __________ _________ Ú _______ 2 2 – x ÷– (x – 4x) ÷ 4x
|
208. The given integral is dx dx _________ = Ú ______________ ____________ Ú __________ 2 2 + x – x ÷– {x – x – 1} ÷ 1
÷4 – 4 + 4x – x
dx = Ú _________________ ________________ 2 2 ÷2 – (x – 4x + 4)
)
_________
1 = log x + __ + ÷x 2 + x + 1 + c 2
|
)
_________ = Ú ________________ _______________ __ Ú ___________ 2 2 2
÷x – 2ax ÷x – 2ax + a – a
_______ + ÷ x 2 – 2ax
5 __ 2
207. The given integral is
dx
dx = Ú _____________ ___________ ÷(x – a)2 – a2
( )
2x + 1 = sin–1 ______ + c 5
|
}
dx = Ú _______________ ______________ 2 5 1 2 __ __ – x + 2 2
_________ ÷x 2 + x + 1 + c
205. We have,
÷6 – x – x ÷– (x + x – 6)
c
_______ = Ú _________________ ________________ Ú ________ 2 2 2 2
dx
|
÷( ) ( )
204. We have,
___
dx dx __________ ________ = Ú ________________ _______________ __ 2 2 3 2 ÷ ÷ x + x + 1 1 x + __ + ___ 2 2
dx
_________ = Ú _____________ ___________ Ú __________ 2 2
___ (sin x + cos x) – (÷2/5 ) 1 ____ ___________________ ___ + ___ log ÷10 (sin x + cos x) + (÷2/5 )
203. We have,
)
209. The given integral is
dx
dx
2 = Ú ____________ __________ Ú x_______ 2 2 + 2ax ÷(x + a) – a
________
|
210. The given integral is
|
= log (x + a) + ÷ x 2 + 2ax + c
dx
dx
________ = Ú ____________ ___________ Ú _________ 2 2 ÷2ax – x ÷ – {x – 2ax}
Indefinite Integrals
dx = Ú _______________ ______________ – a)2 – a2} ÷ – {(x
dx = Ú ____________ ___________ 2 a)2 ÷ a – (x –
x–a = sin–1 _____ + c a
(
dt 1 _______ = __ Ú ________ , 2 2 ÷ t + t +1
dt fi x dx = __ 2
(sin xcos a – cos x sin a) ___________ = Ú ___________________ dx 2 2x – sin a ÷ sin
sin x = cos a Ú ____________ ___________ dx 2 2 x – sin a ÷ sin
|
1 1 = __ log t + __ + ÷ t 2 + t + 1 + c 2 2
1 1 = __ log x2 + __ + ÷x 4 + x2 + 1 + c 2 2
_________
|
_______
– 1 dx Ú ÷sec x
cosx – sin a Ú ____________ ___________ dx 2 2 x – sin a ÷ sin
212. We have,
÷
____________________
sin (x – a) ___________ = Ú ____________ dx 2 2x – sin a ÷ sin
)
sin x = cos a Ú ____________ ___________ dx 2 2 a – cos x ÷ cos
– sin a
÷ Ú ÷
___________________
(1 + cos x) 1 – cos x __________ = Ú _______ × dx cos x (1 + cos x) ______________
1 – cos2x = _____________ dx cos x (1 + cos x)
sin x = Ú ______________ _____________ dx + cos x) ÷ cos x (1
dt = – Ú ________ _______ Let cos x = t + 1) ÷ t (t fi – sin x dx = dt
_________
sin (x – a) sin (x – a) _________ dx = Ú _________ dx Ú sin (x + a) sin (x + a)
_________
)
÷
_________
Let x2 = t
÷( ) ( )
| ( | (
÷
c
sin (x – a) _________ sin (x – a) = Ú _________ × dx sin (x + a) sin (x – a)
dt 1 _______________ = __ Ú ________________ __ 2 ÷3 2 1 2 __ ___ t + + 2 2
|
x dx _________ Ú __________ 4 ÷ x + x2 + 1
|
214. We have,
211. We have,
______ ÷x 1/2 – 1 +
= 4 × log x1/4 +
|
= 4 log t + ÷t 2 – 1 + c
)
_____
|
dt = – Ú _______________ _____________ 2 1 1 2 __ t + __ – 2 2
÷( ) ( ) = – log ( t + __ | 12 ) + ÷t + t | + c = – log ( cos x + __ | 12 ) + ÷ cos x + cos x | + c _____ 2
___________ 2
213. We have, dx dt ______ = 4 Ú ______ _____ , Let x1/4 = t Ú __________ 3/4 1/2 2 x ÷ x – 1 ÷ t – 1 1 fi __ x–3/4 = dt 4
1.63
cos x
___________ dx Ú ____________ 2 2 ÷sin x – sin a
( ) ___________ – sin a log | sin x + ÷sin x – sin 2a | + c
cos x = – cos a sin–1 _____ cos a
2
215. The given integral is
exdx
ex
______ dx = Ú _________ ________ Ú _______ 2x x 2
÷4 – (e ) dz = Ú ______ _____ , Let z = ex 2 ÷ 4 – z fi dx = ex dt
÷4 – e
( )
z = sin–1 __ + c 2 ex = sin–1 __ + c 2
( )
216. The given integral is sec2x ________ dx Ú ÷__________ 16 + tan x dt = Ú ______ _____ , 2 ÷ t + 4
Let t = tan x fi dt = sec2 x dt
_____
| | ________ = log | tan x + ÷ tan x + 4 | + c = log t + ÷ t 2 + 4 + c 2
1.64 Integral Calculus, 3D Geometry & Vector Booster 217. The given integral is
_______ ÷sec x – 1 dx
Ú
________
÷
1 – cos x = Ú _______ dx cos x
÷
__________________
(1 – cos x) (1 + cos x) = Ú __________________ dx cos x (1 + cos x)
sin x = Ú ______________ _____________ dx + cos x) ÷ cos x (1
dt = – Ú ______ _____ , Let t = cos x 2 ÷ t + t fi dx = – sin x dx dt = – Ú _______________ _____________ 2 1 1 2 __ __ t + – 2 2
÷( ) ( ) = – log ( t + __ | 12 ) + ÷t + t | + c = – log ( cos x + __ | 12 ) + ÷ cos x + cos x | + c _____ 2
___________ 2
218. The given integral is
÷
________
_________
1 – sin x – 1 dx = Ú _______ dx Ú ÷cosecx sin x
÷
__________________
220. The given integral is
÷
_________
sin (x – a) Ú _________ dx sin (x + a)
÷
__________________
sin (x – a) sin (x – a) = Ú __________________ dx sin (x – a) sin (x + a) sin(x – a) ___________ = Ú ____________ dx 2 2x – sin a ÷ sin =
sin x cosa – cos x sina
___________ dx Ú ___________________ 2 2
÷sin x – sin a
sin x dx cos x dx = cosa Ú __________________ _________________ Ú _____________ ____________ 2 2 2 a – sin x – sina ÷sin 2x – sin a ÷ sin sin x dx cos x dx = cosa Ú _____________ ____________ – sina Ú _____________ ____________ 2 2 2 a – cos x ÷sin 2x – sin a ÷ cos
(
)
cos x = – cos a sin–1 _____ cosa ___________
|
(1 – sin x) (1 + sin x) = _________________ dx sin x (1 + sin x)
221. The given integral is
cos x = Ú ______________ ____________ dx + sin x) ÷ sin x (1
dt = Ú _______ _____ , 2 ÷ t + t
Let t = sin x fi dt = cos x dt
dt = Ú ______________ _____________ 1 1 2 t + __ – __ 2 2
÷( ) ( ) = log ( t + __ | 12 ) + ÷ t + t | + c = log ( sin x + __ | 12 ) + ÷ sin x + sin x | + c _____ 2
__________ 2
219. The given integral is
e– x
dx
_______ = Ú ___________ __________ dx Ú ________ 2x 2 ÷1 – e ÷(e – x) – 1
dt = – Ú ______ _____ , 2 ÷ t – 1
Let t = e– x
| = – log | e
fi dt = – e –x dt
_____
|
= – log t + ÷ t 2 – 1 + c – x
_______
|
+ ÷e – 2x – 1 + c
|
2 – sin a log sin x + ÷sin 2x – sin a + c
dx
dx
______ ________ = Ú ____________ Ú __________ 2/3 2/3 2/3 1/3 2
x ÷x
– 4 x ÷(x ) – 4
dt 1 _____ = __ Ú ______ , 3 ÷t 2 – 4
dt 1 _____ = __ Ú ______ 3 ÷t 2 – 4
Let t = x1/3 1 fi dt = __ x–2/3 dx 3
_____
| |
|
1 = __ log t + ÷ t 2 – 4 + c 3 ______ 1 = __ log x1/3 + ÷x 2/3 – 4 + c 3 222. The given integral is ______ x 3 3 dx Ú ______ a –x Let x = a sin2/3q
|
÷
fi dx = a ◊ sin–1/3 q dq
__
1/3 –1/3 ÷a sin q sin q cosq = Ú ___________________ 3/2 a cosq 1 = __ a Ú dq
q = __ a + c
x 3/2 sin–1 __ a = ____________ + c a
{ ( ) }
1.65
Indefinite Integrals
223. The given integral is
(
cosq + sinq
)
_________ dq Ú ___________ ÷ 5 + sin 2q cosq + sinq _____________ = Ú _______________ dq ÷6 – (1 – sin 2q)
cosq + sinq ________________ = Ú _________________ dq 2 ÷ 6 – (sinq – cosq)
Put (sinq – cosq) = t fi (cosq + sinq)dq = dt
dt _____ = Ú _______ – t2 ÷ 6 t = sin–1 ___ __ + c ÷6
( )
(
)
sinq – __cos q = sin–1 __________ + c ÷6
Let
(x2 – 3x + 2) = t2
fi
(2x + 3) dx = 2t dt
| (
_________
sinq – cosq ______________ = Ú _______________ dq ÷ 3 – (1 + sin 2q)
sinq – cosq ________________ = Ú _________________ dq 2 ÷3 – (sinq + cosq)
dt = – Ú ______ _____ , Let t = (sinq + cosq) 3 – t2 ÷ dt = (cosq – sinq) dq
( )
t = cos–1 ___ __ + c 3 ÷
sinq + cosq __ = cos–1 __________ + c 3 ÷
(
3x + 4
÷x + 5x + 2
(
)
3 15 __ (2x + 5) + 4 – ___ 2 2 _________ = Ú ___________________ dx 2 + 2 ÷ x + 5x
(2x + 5)dx 3 dx 7 __________ __________ = __ Ú ___________ – __ Ú ___________ 2 2 ÷x 2 + 5x 2 + 2 + 2 ÷ x + 5x dx 3 2t dt __ 7 = __ Ú ____ – Ú ___________________ __________________ t 2 2 2 5 25 __ x + + 2 – ___ 2 4
÷( ) (
)
Let (x2 + 5x + 2) = t2
fi (2x + 5) dx = 2t dt dx 7 ________________ = 3 Ú dt + __ Ú _________________ ___ 2 ÷17 2 5 2 __ ____ x + – 2 2
÷( ) ( )
| ( | (
_________
)
|
5 7 = 3t + __ log x + __ + ÷ x 2 + 5x + 2 + c 2 2
)
)
_________
|
x–1 __________ dx Ú ___________ 2 + 2 ÷ x – 3x
227. We have,
÷x + 5x + 6
÷x + 5x + 6
(2x + 5) 1 1 1 __________ __________ = __ Ú ___________ dx – __ Ú ___________ dx 2 2 ÷x 2 + 5x 2 + 6 ÷x + 5x + 6
– 3) + 1 1 (2x __________ = __ Ú ___________ dx 2 ÷x 2 – 3x + 2
(2x – 3) dx __________ 1 1 __________ __________ = __ Ú ___________ dx + __ Ú ___________ = x 2 + 5x + 6 ÷ 2 2 ÷x 2 – 3x 2 + 2 ÷x – 3x + 2
(2x + 5) 1 1 1 __________ = __ Ú ___________ dx – __ Ú _______________ ______________ dx 2 ÷x 2 + 5x 2 5 2 1 2 + 6 __ x + __ – 2 2
dx 1 2t dt 1 ________________ = __ Ú ____ dx + __ Ú _________________ t 2 2 9 3 2 x – __ + 2 – __ 2 2
÷( ) ( )
(2x + 5) – 1
x+2
1 __________ __________ dx = __ Ú ___________ dx Ú ___________ 2 2 2
1 2 x – 2 __________ = __ Ú ___________ dx 2 ÷x 2 – 3x + 2
)
5 7 = 3 ÷x 2 + 5x + 2 + __ log x + __ + ÷ x 2 + 5x + 2 + c 2 2
225. We have,
|
_________
__________ dx Ú ___________ 2
_________
| (
c
226. We have,
1.65
|
__________ + 2 + ÷ x 2 – 3x
1 1 = ÷x 2 – 3x + 2 + __ log x – __ + ÷x 2 – 3x + 2 + c 2 2
sinq – cosq
________ dq Ú ___________ ÷2 – sin 2q
)
1 1 = t + __ log x – __ + 2 2
224. The given integral is
÷( ) ( )
| (
)
_________
|
5 1 – __ log x + __ + ÷ x 2 + 6x + 6 + c 2 2
1.66 Integral Calculus, 3D Geometry & Vector Booster 231. We have,
6x – 5 __________ 228. Ú ___________ dx 2 – 5x + 1 ÷ 3x
Let 3x 2 – 5x + 1
fi (6x – 5) dx = 2t dt
2t dt = Ú ____ t
= 2 Ú dt
= 2t + c
= 2÷3x 2 – 5x + 1 + c
_____
____________
÷
÷
a______ – x = Ú _______ dx 2 ÷ a – x2
÷ 4 1–t = __ Ú ____ dt 3 ÷1 + t
–1
= a sin
÷
______
( )
x __ a + ÷ a 2 – x2 + c
______
a2 – x2 230. Ú x ______ dx a2 + x2
4 – 4t 4 = __ Ú ______ dt 3 4 + 4t
Let x2 = a2t 2
fi
a2(1 – t) a2 fi __ Ú ________ dt 2 a2(1 + t)
2x dx = a dt
÷ Ú ÷
– t) 4 (1_____ = __ Ú ______ dt 3 ÷ 1 – t2
dt t 4 = __ Ú _______ _____ – Ú ______ _____ dt 3 ÷1 – t2 ÷1 – t2
4 = __ ( sin–1(t) + ÷ 1 – t2 ) + c 3
x3 x3 2 4 = __ sin–1 __ + 1 – __ + c 3 4 4
(
______
(1 – t) a2 = __ Ú ______ dt 2 (1 + t)
(1 – t) 1____ a2 –t = __ ______ × dt 2 (1 + t) 1 – t
÷
( ( ) ÷ ( ) )
232. We have, sec2x dx
dx
= Ú ______________ Ú _________ 3 + 4sin2x 3 sec2x + 4 tan2x
=
________
_____________
__
÷
– t) a2 (1_____ = __ Ú ______ dt 2 ÷ 1 – t2
__
7 1 ___ = ____ × tan–1 tan x __ + c 3 21 ÷
233. We have,
(
)
dt t a2 = __ Ú ______ _____ – Ú ______ _____ dt 2 2 ÷1 – t ÷1 – t2
_______
÷ ( )
)
a2 x2 x 2 = __ sin–1 __ 2 + 1 – __ a + c 2 a
sec2x dx
dx
2 = Ú _________ Ú _____________ 3sin x + 4cos2x 3tan2x + 4 =
a2 = __ ( sin–1(t) + ÷ 1 – t2 ) + c 2
( ( )
( ÷ ) ( ÷ )
7 7 1 ÷ = __ × ___ tan–1 t __ + c 7 3 3 __
_____
( ÷ )
______
sec2x dx
Ú __________ 3 + 7 tan2x
sec2x dx 1 = __ Ú ________ 7 3 tan2x + __ 7 dt 1 __ = __ Ú _________ 7 2 3 2 t + __ 2
(1 – t)2 a2 = __ Ú ______ dt 2 (1 – t2)
)
________
4 x2dx = __ dt 3
_____
dx x = a Ú _______ ______ – Ú ________ _______ dx 2 2 2 ÷a – x2 ÷ a – x
fi
–x a – x a_____ –x dx = Ú _____ × dx Ú aa_____ +x a+x a–x
Let x3 = 4t
_____
229. We have,
3
–x dx Ú x 44_____ + x3
______
__________
÷
______
2
dt
2 Ú ______ 3t + 4
Let tan x = t fi sec2 x dx = dt
dt 1 = __ Ú _________ 2 3 2 2 t + ___ __ 3 ÷
( )
Indefinite Integrals
1 = __ × 3
__
( ) ( ) __
3 3 ÷ ÷ ___ tan–1 ___ t + c 2 2
__
3 ÷ 1 = ___ __ tan–1 ___ tanx + c 2 3 ÷
234. We have,
2
dx sec x dx 2 = Ú __________ Ú _____________ (2sinx + cosx) (2tan x + 3)2
dt = Ú _______ (2t + 3)2
1 = – ________ + c 2(2t + 3)
1 = – __________ + c 2(2tanx + 3)
sec2x dx 1 = __ Ú _____________________ 4 1 2 1 2 __ ___ tan x + + 1 – 4 16
[ (
) (
1 = __ tan–1 (tan2x) + c 2 2
dx sec x = Ú _________ dx Ú _____________ (sin x + 2 cos x)2 (tan x + 2)2 =
dt Ú ______ (t + 2)2
sec2x dx 1 = __ Ú _________________ 4 15 2 tan x + 1 2 ___ _______ + 4 16
[ (
) ] ÷15 1 Let ( tan x + __ ) = ____ tanq 4 4
15 ___ dq 16 2 1 ÷ = __ × ____ × Ú _____ 4 4 15 sec2q
1 = – _________ + c (tan x + 2)
( )
8 = ______ 3/2 Ú (2cos2q)dq (15) 8 = ______ 3/2 Ú (1 + cos2q)dq (15) sin 2q 8 = ______ 3/2 q + _____ + c 2 (15)
(
)
8 = ______ 3/2 (q + sinq + cosq) + c, (15)
where
4 1 q = tan–1 ____ ___ tan x + __ 4 ÷15
[ (
)]
237. We have, dx 1 __ Ú _____________ 2 4 sin x ____ + sec x 2
)
2
sin x dx 2 sin x cosx = Ú ___________ dx Ú ___________ sin4x + cos4x sin4x + cos4x
Divide the numerator and the denominator by cos4x we get
(
( )
fi sec2 x dx = dt
1 = – ______ + c (t + 2)
dx = Ú _____________ (sinx + 2 sec x)2
÷15 2 fi sec x dx = ____ sec q dq 4 2
238. The given integral is
Let tan x = t
)]
÷15 2 ____ sec q 1 4 = __ Ú _________ sec4q 4 15 2 ___ 16
236. We have,
)
___
dt 1 Let tan2x = t = __ Ú _____ 2 t2 + 1 fi 2 tan x sec2 x dx = dt 1 = __ tan–1 (t) + c 2
(
___
sec2x dx 1 = __ Ú ________________ 2 tan x 4 tan2x + ____ + 1 2
1 2tan x sec2x dx = __ Ú __________ 2 tan4x + 1
)
___
tan x sec2x dx = Ú _________ tan4x + 1
(
sin x cos x dx Ú ___________ sin4x + cos4x
sec2x dx 1 = __ Ú ___________ 2 tan x 4 ________ 2 + sec2x
___
235. We have,
1.67
2
sec x __________ dx Ú 2 tan x tan4x + 1 Let tan2x = t fi 2 tan x sec2x dx = dt
dt = Ú __2 + 1 t = tan–1 (t) + c
= tan–1 (tan2x) + c
1.68 Integral Calculus, 3D Geometry & Vector Booster 239. The given integral is
dx Ú ______________ (2sinx + 3cosx)2
Divide the numerator and the denominator by cos2x, we get sec2x dx Ú ___________ (2 tan x + 3)2 Let 2 tan x + 3 = t fi 2 sec2 x dx = dt 1 dt = __ Ú __2 2 t 1 = – __ + c 2t 1 = – __ (2 tan x + 3) + c 2
2
2
dx sec x dx sec x dx 2 = Ú _________ = Ú _________ Ú ________ 2 2 + cos x 1 + 2 sec x 2tan2x + 3
Let tan x = t fi sec2 x dx = dt
dt = Ú ______ 2 2t + 3
÷ 2 = __ ÷ 3 tan __
( ÷ ) ( ÷ ) ( ÷ )
__
__
2 __ tan x + c 3
–1
sinx sin x dx = Ú _____________ dx Ú _____ sin 3x 3 sin x – 4 sin3x dx = Ú _________ 3 – 4sin2x
sec2x dx = Ú _____________ 3 sec2x – 4 tan2x
sec2x dx = Ú ________ 3 – tan2x
dt = – Ú _____ t2 – 3
Let tan x = t sec2 x dx = dt
| | __
sinx = Ú _____________ dx 3 sin x – 4 sin3x
dx = Ú _________ 3 – 4 sin2x
sec2x dx = Ú _____________ 3 sec2x – 4 tan2x
sec2x dx = Ú ________________ 3 + 3 tan2x – 4 tan2x
sec2x dx = Ú ________ 3 – tan2x Let tan x = t fi sec2 x dx = dt
dt = – Ú _____ 2 t –3
t–÷ 3 – 1 = ____ __ log ________ + c 2÷3 t+÷ 3
t+÷ 3 1 = ____ __ log ________ 2÷3 t – ÷3
| | | | | | __
__
241. We have,
3
cosec x sinx dx = Ú _____ dx Ú _______ cosec x sin3x
tan x + ÷ 3 1 = ____ __ log ___________ + c 2÷3 tan x – ÷3 243. The given integral is
__
2 2 = __ tan–1 __ t + c 3 3
|
__
dt __ = Ú _________ 3 2 2 t + __ 2
240. The given integral is
|
__
tan x – ÷3 1__ = – ____ log ___________ + c 2÷3 tan x + ÷ 3 242. The given integral is
t – ÷3 1__ = – ____ log ________ + c 2÷3 t+÷ 3
sec3x cosx dx = Ú _____ dx Ú _____ secx cos3x
cos x = Ú _____________ dx 4 cos3x – 3 cos x dx = Ú _________ 4 cos2x – 3 Divide the numerator and the denominator by cos2x we get, sec2x dx = Ú _________ 4 – 3 sec2x
sec2x dx = Ú _____________ 4 – 3 – 3 tan2x
sec2x dx = Ú __________ 1 – 3 tan2x
Let tan x = t fi sec2 x dx = dt dt = Ú ______ 1 – 3t 2
Indefinite Integrals
| | | |
dt 1 = __ Ú __________ __ 3 t2 – ( 1/÷3 )2
1 t – ___ __ 3 ÷ 1 = ____ __ log ______ + c 1 2÷3 t + ___ __ 3 ÷ __
244. We have,
( ) ( )
3 – dt 1 1 = __ Ú ________ fi __ sec2 __ dx = dt 2 t2 – 2t – 2 2 2
dt 1 = – __ Ú _____________ 2 2 (t – 1)2 – ( ÷__ 3 )
(t – 1) – ÷ 3 1__ = – ____ log ___________ __ + c 2÷3 (t – 1) + ÷ 3
__ x tan __ – 1 – ÷3 2 1__ = – ____ log ________________ __ + c x 2÷3 tan __ – 1 + ÷ 3 2
)
( )
( ) ( )
2dt = Ú _________ 2 t + 4t + 1 dt = 2 Ú _____________ __ (t + 2)2 – ( ÷ 3 )2
|
__
|
(t + 2) – ÷ 3 2 = ____ __ log __________ __ + c 2÷3 (t + 2) + ÷ 3
|
( (
( ) ) ( ) )
__
245. We have,
dx Ú _____________ sin x + cos x + 1
(
) (
dx = Ú _____________________________ x x __ 2 tan 1 – tan2 __ 2 2 __________ + __________ x x + 1 1 + tan2 __ 1 + tan2 __ 2 2
( ) ( ) ( ) ( ) x sec ( __ ) dx 2 _____________________ = Ú x x ( 2 tan ( __ ) – tan ( __ ) + 2 ) 2 2 dt x 1 = __ Ú _________ Let ( __ ) = t 2 2t – t + 2 2
)
2
2
( (
|
( ) ) ( ) )
dx Ú _________ 1 + 2 sin x dx = Ú _________________ 2tan(x/2) 1 + 2 ___________ 1 + tan2 (x/2)
(
)
sec2 (x/2)dx = Ú ____________________ tan2 (x/2) + 4tan(x/2) + 1
Let tan(x/2) = t
x 1 fi __ sec2 __ dx = dt 2 2
2dt = Ú _________ t2 + 4t + 1 dt = 2 Ú _____________ __ (t + 2)2 – ( ÷ 3 )2
| |
__
( )
|
(t + 2) – ÷ 3 ___________ __ + c (t + 2) + ÷ 3
1 = ___ __ log 3 ÷
(tan(x/2) + 2) – ÷ 3 1 = ___ __ log _________________ __ + c 3 (tan(x/2) + 2) + ÷ 3 ÷
__
|
245. The given integral is dx dx = __________________ Ú ________ 3cosx + 4 Ú 1 – tan2 (x/2) 3 ___________ + 4 1 + tan2 (x/2)
(
)
(1 + tan2 (x/2)) = Ú __________________________ dx 4 + 4tan2 (x/2) + 3 – 3tan2 (x/2)
sec2 (x/2) = Ú ___________ dx tan2 (x/2) + 7
2
|
|
(1 + tan2 (x/2))dx = Ú ____________________ 1 + tan2 (x/2) + 4tan(x/2)
|
__
x tan __ + 2 – ÷3 2 1 = ___ __ log ________________ __ + c x 3 ÷ tan __ + 2 + ÷3 2
|
244. The given integral is
x Let tan __ = t 2 x fi sec2 __ dx = 2dt 2
( ) ( )
2 tan 2 1 + 2 __________ x 2 __ 1 + tan 2 x 2 __ sec 2 = Ú ___________________ x x dx 2 __ 1 + tan + 4 tan __ 2 2
(
dx dx = ________________ Ú _________ x 1 + 2 sin x Ú __
( )
3 tan x – 1 ÷ 1 __ = ____ __ log __________ + c 2÷3 ÷3 tan x + 1
1.69
Let t = tan(x/2)
fi
( )
x 2dt = sec2 __ dx 2
1.70 Integral Calculus, 3D Geometry & Vector Booster
(
)
2dt = Ú _____ 2 t +7
1 1 1 1 = __ Ú _____ – ______ dt + __ Ú sec2 (x/2)dx 2 t2 – 1 (t + 1)2 2
t 2 = ___ __ tan–1 ___ __ + c 7 ÷7 ÷
)
t–1 1 __ 1 1 = __ log ____ + ______ + tan(x/2) + c 2 2 t+1 (t + 1)
tan(x/2) 2 __ = ___ __ tan–1 ______ + c 7 ÷7 ÷
cos x – 1 1 1 1 = __ __ log ________ + _________ 2 2 cos x + 1 (cosx + 1)
( )
(
246. The given integral is dx Ú _____________ sinx + cosx + 1
dx = Ú ___________________________ 2tan(x/2) 1___________ – tan2 (x2) ___________ + + 1 1 + tan2 (x/2) 1 + tan2 (x/2)
( ) ( ) ( )
x Let tan __ = t 2 x 1 fi __ sec2 __ dx = dt 2 2 x fi sec2 __ dx = 2dt 2
2dt = Ú _____ 2t + 2
(
= Ú dt/t + 1
x = log tan __ + 1 + c 2
|
( )
(1 + tan2 (x/2))dx = Ú ____________________ tan2 (x/2) – 4tan(x/2) + 1
sec2 (x/2)dx = Ú ____________________ tan2 (x/2) – 4tan(x/2) + 1
Let tan(x/2) = t x fi sec2 __ dx = 2dt 2
dt = 2 Ú _________ 2 t – 4t + 1 dt = 2 Ú _____________ __ (t – 2)2 – ( ÷ 3 )2
|
|
__
)
|
249. The given integral is
1 + sinx dx Ú ____________ sinx(1 + cosx)
dx dx = Ú ____________ + _________ sinx(1 + cosx) Ú (1 + cosx)
sin x dx = Ú _________________ + (1 – cos2x)(1 + cosx)
dt = Ú ___________ + 2 (t – 1)(1 + t)
( )
x 1 __ Ú sec2 __ dx 2 2
( )
x 1 __ Ú sec2 __ dx 2 2
Let t = cos x
fi dt = – sin x dx
__
(tan(x/2) – 2) – ÷ 3 = log ________________ __ + c (tan(x/2) – 2) + ÷ 3
247. The given integral is
(
( )
(t – 2) – ÷ 3 1 = 2 × __ log __________ __ + c 2 (t – 2) + ÷ 3
= log |t + 1| + c
)
)
dx = Ú _________________ 2tan(x/2) 1 – 2 ___________ 1 + tan2 (x/2)
(1 + tan2 (x/2))dx = Ú _________________________________ 1 + tan2 (x/2) + 2tan(x/2) + 1 – tan2 (x/2) sec (x/2)dx = Ú ___________ 2tan(x/2) + 2
| | ) | |
+ tan(x/2) + c 248. The given integral is dx Ú ________ 1 – 2 sin x
2
( (
dt = Ú ___________ + (t – 1)(t + 1)2
( )
x 1 __ Ú sec2 __ dx 2 2
dx Ú ________________ 3 sin x + 4 cos x + 5
dx = Ú _________________________________ 2tan(x/2) 1___________ – tan2 (x/2) ___________ 3 + 4 + 5 1 + tan2 (x/2) 1 + tan2 (x/2)
(
) (
)
(1 + tan2 (x/2))dx = Ú ____________________ tan2 (x/2) + 6tan(x/2) + 9 2dt = Ú _________ t 2 + 6t + 9 2dt = Ú ______ (t + 3)2
Let tan(x/2) = t x fi sec2 __ dx = 2dt 2
( )
Indefinite Integrals
2 = – ______ + c (t + 3)
dx 1 = ___ __ Ú _______________ 1__ 1 2 ___ ÷ sinx + ___ __ cosx 2 ÷2 ÷
2 = – ___________ + c (tan(x/2) + 3)
dx 1 = ___ __ Ú _________ p 2 sin x + __ ÷ 4 p 1__ ___ = Ú cosec x + __ dx 4 2 ÷
(
250. The given integral is
dx Ú ___________ cos x + cos a dx = Ú _________________ , where k = cos a 2 1 – tan (x/2) ___________ + k 1 + tan2 (x/2) [1 + tan2 (x/2)]dx = Ú __________________________ 1 – tan2 (x/2) + k[1 + tan2 (x/2)]
(
)
sec2 (x/2)dx = Ú __________________________ 1 – tan2 (x/2) + k[1 + tan2 (x/2)]
( ) ( )
2dt = Ú ________________ 2 2 (1 – t ) + k(1 + t )
x Let tan __ = t 2 x fi sec2 __ dx 2 = 2 dt
2dt = Ú _______________ (1 + k) + (k – 1)t2
2dt 1 = ______ __________ (1 – k) Ú (1 + k) ______ – t2 (1 – k)
2dt 1 ______ = ______ ___________ – t2 (1 – k) Ú (1 + k) 2 ______ (1 – k)
( ÷
)
| | | ÷ ÷
______
(1 + k) ______ +t (1 – k) 1 1 ______ + c = ______ × ________ ______ log __________ (1 – k) (1 + k) (1 + k) 2 ______ ______ –t (1 – k) (1 – k)
1 = ____ log sina
|
÷
cot(a/2) + tan(x/2) ________________ + c cot(a/2) – tan(x/2)
251. The given integral is
)
( ) x p 1 = ___ log tan ( __ + __ ) + c 2 8 | | ÷2 __
253. We have,
1.71
dx __ Ú ____________ ÷3 sinx + cosx dx 1 __ = __ Ú ______________ 2 ___ ÷3 1 __ sinx + cosx 2 2 dx 1 _________ __ = Ú p 2 sin x + __ 6 p 1 = __ Ú cosec x + __ dx 2 6 p x ___ 1 __ __ = log tan + + c 2 2 12
(
|
)
( ) ( ) |
257. We have, 2 sin x + cos x = l (3 sin x + 2 cos x) + m(3 cos x – 2 sin x) = (3l – 2m) sin x + (2l + 3m) cos x Comparing the co-efficients of sin x and cos x, we get 3l – 2 m = 2 and 2l + 3 m = 1 On solving, we get, l = 4/5 and m = – 3/ 5 4 3 sin x + 2 cos x \ __ Ú _____________ dx 5 3 sin x + 2 cos x 3 3 cos x – 2 sin x + __ Ú _____________ dx 5 3 sin x + 2 cos x
(
)
3 3 cos x – 2 sin x 4 = __ Ú dx – __ Ú _____________ dx 5 5 3 sin x + 2 cos x
3 4 = __ x – __ log |3 sin x + 2 cos x| + c 5 5
258. We have,
dx Ú ______________ tanx + 4cotx + 4
4sin x cos x = Ú _______________________ dx sin2x + 4cos2x + 4sin x cos x
2 sin x 1 = __ Ú __________ dx 2 sin x + cos x
1 (sin x + cos x) + (sin x – cos x) = __ Ú _________________________ dx 2 sin x + cos x
252. We have,
dx Ú __________ sinx + cosx
sin x dx Ú __________ sin x + cos x
1.72 Integral Calculus, 3D Geometry & Vector Booster
1 (1 + cos x – sin x = __ Ú _______________ dx 2 sin x + cos x)
1 1 = __ x + __ log |sin x + cos x| + c 2 2
259. The given integral is + 3 cos x _____________ dx Ú 2 sin x 3 sin x + 4cosx
= (3l – 4m)sin x + (4l + 3m) cos x Comparing the co-efficients of sin x and cos x, we get (3l – 4m) = 2, (4l + 3m) = 3
On solving, we get 18 1 l = ___ , m = ___ 25 25
The given integral reduces to 18 1 ___ (3 sin x + 4 cos x) + ___ (3 cos x – 4 sin x) 25 25 Ú ____________________________________ dx (3 sin x + 4 cos x)
(
cos x 1 dx = Ú __________ dx Ú _______ 1 + tan x sin x + cos x 2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
1 = __ (x + log |sin x + cos x|) + c 2 263. The given integral is
= l(3 sin x + 4 cos x) + m(3 cos x – 4 sin x)
we have 2 sin x + 3 cos x
1 = __ (x + log |sin x + cos x|) + c 2 262. The given integral is
)
18 1 (3cosx – 4sinx) = Ú ___ + ___ _____________ dx 25 25 (3sinx + 4cosx) 18 1 = ___ x + ___ log |3 sin x + 4 cos x| + C 25 25
1 dx Ú _______ 1 – tan x
cos x = Ú __________ dx cos x – sin x
2 cos x 1 = __ Ú __________ dx 2 cos x – sin x
1 (cos x – sin x) + (cos x + sin x) = __ Ú _________________________ dx 2 cos x – sin x
(cosx + sinx) 1 = __ Ú 1 + ___________ dx 2 cosx – sinx
1 = __ (x – log |cos x – sin x|) + c 2
(
)
264. The given integral is
260. The given integral is
1 dx Ú _______ 1 + cot x
sin x dx Ú __________ sin x – cos x
cos x = Ú __________ dx sin x + cos x
2 sin x 1 = __ Ú __________ dx 2 sin x – cos x
2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
1 (sin x – cos x) + (sin x + cos x) = __ Ú __________________________ dx 2 sin x – cos x
1 (cos x + sin x) + (cos x – sin x) = __ Ú __________________________ dx 2 sin x + cos x
cos x + sin x 1 = __ Ú 1 + __________ dx 2 sin x – cos x
(cos x – sin x) 1 = __ Ú 1 + ___________ dx 2 sin x + cos x
1 = __ (x + log |sin x + cos x|) + c 2
cos x dx Ú __________ sin x + cos x
265. We have, 3 sin x + 2 cos x + 4 = l(3 cos x + 4 sinx+ 5)
2 cos x 1 = __ Ú __________ dx 2 sin x + cos x
+ m(– 3 sin x + 4 cos x) + n
1 (cos x + sin x) + (cos x – sin x) = __ Ú __________________________ dx 2 sin x + cos x
(cosx – sinx) 1 = __ Ú 1 + ___________ dx 2 sinx + cosx
+ (5 l + n) Comparing the co-efficients of sin x and cos x we get 4 l – 3 m = 3, 3 l + 4 m = 2 and 5 l + n = 4
(
)
1 1 = __ x + __ log |sin x – cos x| + c 2 2 261. The given integral is
(
)
(
)
= (4 l – 3 m) sin x + (3 l + 4 m) cos x
Solving, we get
18 1 1 l = ___ , m = – ___ , n = __ 25 25 5
Thus,
+ 2 cos x + 4 ________________ dx Ú 3 sin x 3 cos x + 4 sin x + 5
(
(
))
(
)
x – tan–1 (4/3) 2 + ___ tan–1 ____________ + c. 2 25
269. We have,
|
271. We have,
dx 2 + __ Ú ________________ 5 4 sin x + 3 cos x + 5 18 1 = ___ x – ___ log |4 sin x + 3 cos x + 5| 25 25
|
1.73
49(sinx – cosx) – 5 1 = – ___ log _________________ + c 40 4(sinx – cosx) + 5
18 4 cos x – 3 sin x 1 = Ú ___ + ___ ________________ dx 25 25 4 sin x + 3 cos x + 5
Indefinite Integrals
Ú (cos x – sin x)(2 + 3 sin x 2x)dx
dx Ú ____________ cos x + cosec x dx = Ú __________ 1 cos x + ____ sin x sin x dx = Ú ___________ cos x sin x + 1 2 sin x dx = Ú _____________ 2 cos x sin x + 2
(sin x + cos x) + (sin x – cos x) = Ú __________________________ 2cos x sin x + 2
(sinx + cosx) = Ú ______________ dx 2 cos x sin x + 2
= Ú (cos x – sin x)(2 + 3(–1 + 1 + sin 2x))dx
= Ú (cos x – sin x)(–1 + 3(sin x + cos x)2)dx
(sin x – cos x) + Ú _____________ dx 2 cos x sin x + 2 (sin x + cos x) (sin x – cos x) = Ú ___________ dx + Ú ___________ dx 2 + sin 2 x 2 + sin 2 x
= Ú (3t 2 – 1) dt Let t = sin x + cos x
fi dt = (cos x – sin x)dx
3
= (t – t) + c = (sin x + cos x)3 – (sin x + cos x) + c 270. We have,
sin x + cos x dx Ú ___________ 9 + 16 sin 2 x
sin x + cos x = Ú ___________________ dx 9 + 16(1 – 1 + sin 2x)
sin x + cos x = Ú ___________________ dx 25 – 16(sin x – cos x)2
Let sin x – cos x = t
fi (cos x + sin x)dx = dt
dt = Ú ________ 25 – 16t2 dt 1 = – ___ Ú ________ 2 16 2 5 __ t – 4
( )
| | |
5 t – __ 1 1 4 = – ___ × _____ log _____ + c 16 5 5 2 × __ t + __ 4 4 4t – 5 1 = – ___ log ______ + c 40 4t + 5
|
(sin x + cos x) = Ú _______________ dx 3 – (sin x – cos x)2
(sin x – cos x) + Ú ________________ dx 1 + (sin x + cos x)2
|
__
|
(sin x – cos x) – ÷ 3 1__ __ = – ____ log _________________ 2÷3 (sin x – cos x) + ÷ 3 –1
– tan (sin x + cos x) + c 271. The given integral is dx Ú ____________ cos x + cosec x
sin x cos x = Ú ___________ dx sin x + cos x
2 sin 2x 1 = __ Ú __________ dx 4 sin x + cos x
2 2 1 (sin x + cos x) – (sin x – cos x) = __ Ú ___________________________ dx 4 sin x + cos x
(sin x – cos x) 1 = __ Ú (sin x + cos x) – ___________ dx 4 sin x + cos x
1 1 = __ (cos x – sin x) – __ (sin x – cos x)3 + c 4 3
[
[
]
]
272. The given integral is
Ú (sin x + cos x)(2 + 3 sin 2 x)dx = Ú (sin x + cos x)[2 + 3{1 – (1 – sin 2x)}]dx
1.74 Integral Calculus, 3D Geometry & Vector Booster
= Ú (sin x + cos x)(5 – 3(sin x – cos x)2)dx
Let sin x – cos x = t fi (cos x + sin x)dx = dt
= Ú (5 – 3t2)dt 3
= (5t – t ) + c = (5(sin x – cos x) – (sin x – cos x)3) + c
273. The given integral is
Ú (sin x – cos x)(3 – 4 sin 2 x)dx = Ú (sin x – cos x)(7 – 4(1 + sin 2 x)) = Ú (sin x – cos x)(7 – 4(sin x + cos x)2)dx = Ú (cos x – sin x)(4(sin x + cos x)2 – 7)dx
Let sin x + cos x = t fi (cos x + sin x)dx = dt
= Ú (4t2 – 7)dt
4t3 = ___ – 7t + c 3
(
)
(
)
4(sin x + cos x)3 = _____________ – 7(sin x + cos x) + c 3 274. The given integral is
cos x – sin x dx Ú ( ___________ 3 + 2 sin 2 x )
cos x + sin x = Ú _________________ dx 1 – 4(sin x – cos x)2
Let sin x – cos x = t fi (cos x + sin x)dx = dt
dt = Ú ______ 1 – 4t2
dt 1 = – __ Ú _________ 4 t2 – (1/2)2
2t – 1 1 = – __ log ______ + c 2 2t + 1
1 = __ log 2
| |
|
|
2(sin x – cos x) – 1 ________________ + c 2(sin x – cos x) + 1
276. The given integral is
+ sin x 2 cos x dx Ú (____________ 9 + 16 sin 2x )
3 1 __ (cos x + sin x) + __ (cos x – sin x) 2____________________________ 2 = Ú dx 9 + 6 sin 2x 3 (cos x + sin x) 1 (cos x – sin x) = __ Ú ___________ dx + __ Ú ___________ dx 2 2 (9 + 6 sin 2x) 9 + 6 sin 2x (cos x + sin x) 3 = __ Ú __________________ dx 2 15 – 6(sin x – cos x)2 (cos x – sin x) 1 + __ Ú __________________ dx 2 (3 + 6(sin x + cos x)2)
(cos x – sin x) = Ú ______________ dx 1 + 2(1 + sin 2x)
Let sin x – cos x = t and sin x + cos x = z fi (cos x + sin x) dx = dt and (cos x – sin x) dx = dz
(cos x – sin x) = Ú _________________ dx 1 + 2(sin x + cos x)2
3 dt dz 1 = __ Ú _______ + __ ______ 2 15 – 6t2 2 Ú 6z2 + 3
dt dz 1 1 = __ Ú ______ + __ ______ 2 5 – 2t2 6 Ú 2z2 + 1
dt dz 1 1 __ + __ = – __ Ú __________ Ú ______ 4 t2 – ( ÷5 / 2 )2 6 2z2 + 1
__ t – ( ÷5 /2 ) 1 1 __ + ____ = – __ log __________ __ tan–1 ( z÷2 )+ C 4 t + ( ÷5 /2 ) 6÷2
where
sin x – cos x = t and sin x + cos x = z
Let sin x + cos x = t fi (cos x + sin x)dx = dt
dt = Ú ______ 1 + 2t2 dt 1 = __ Ú _________ 2 ___ 1__ 2 2 + t 2 ÷ 1__ –1 ( __ ) ___ = tan t ÷2 + c 2 ÷ 1__ –1 __ = ___ tan ( ÷2 ( sin x + cos x) ) + c 2 ÷ 275. The given integral is
( )
+ sin x __________ dx Ú ( cos x 5 – 4 sin 2x )
(
)
cos x + sin x = Ú ___________________ dx 5 – 4(1 – (1 – sin 2x))
|
__
277. We have,
2
x + 1 4 dx Ú ______ x + 1
(
)
1 1 + __ 2 x = Ú _______ dx 1 2 x + __ 2 x
|
Indefinite Integrals
(
)
1 1 + __ 2 x = Ú ______________ dx 1 2 1 x – __ x + 2 ◊ x ◊ __ x
(
)
(
)
1 Let x – __ x = t
1 fi 1 + __ 2 dx = dt x
(
dt __ Ú _________ 2 ( )2
=
t 1 = ___ __ tan–1 ___ __ + c 2 ÷2 ÷
1 1 1 = ___ __ tan–1 ___ __ x – __ x + c 2 ÷2 ÷
(
))
(
)
Ú tan–1 ( x – __ 1x ) – __ 23 tan–1 (x3) + c
2
– 3x + 1 dx Ú x__________ 4 2 x + x + 1
(x2 – x + 1) – 2x
x + __ 2 x
=
dx – Ú _________ 2x dx Ú ________ 2 4 2
=
dx dt Let x2 = t __ , Ú ________ Ú ______________ 2 2 3 2
( )
1 1 – __ 2 x = Ú ______________ dx 2 1 1 x + __ x – 2 ◊ x ◊ __ x
(
)
(
)
1 Let x + __ x = t 1 fi 1 – __2 dx = dt x
( )
t – ÷ 2 1 __ = ____ __ log ______ + c 2÷2 t + ÷2
__ 1 __ x + – 2 ÷ x 1 = ____ __ log | ____________ __ + 1 2÷2 x + __ x + ÷ 2
| | __
| | ( (
279. We have,
) )
(x2 – x + 1) x + x + 1
x +x+1
x +x +1
x +x +1
( ) ( )
t + t + 1
fi 2x dx = dt
( ) ( )
1 1 x + __ t + __ 2__ –1 _____ 2 2 2 –1 _____ ___ ___ = tan __ + __ tan __ + c 3 ___ ÷3 ___ ÷ ÷3 ÷3 2 2
2x __ +1 2 = ___ __ tan–1 ______ + 3 ÷3 ÷
2x +__ 1 2x2 __ +1 2 2 = ___ __ tan–1 ______ + ___ __ tan–1 ______ + c 3 ÷3 ÷3 ÷3 ÷
( ) ( )
(
)
2t +__ 1 2 ___ __ tan–1 _____ + c 3 ÷3 ÷
(
)
281. We have,
4
(x2 + 1)2 – 2x2 = Ú _____________ dx 6 x + 1 (x2 + 1)2 – 2x2 = Ú _________________ dx (x2 + 1)(x4 – x2 + 1) (x + 1)
2
(x – x + 1)
x +1
dx – 2 Ú _____ x dx Ú ___________ 4 2 6
( ) ÷ 2
dx – x ___________ __________ Ú 1 1 _____ 2 4 + x2 1 + x + x
( ) ) ÷ ( ( ) ) ÷(
1 – x2 1 – __ 2 dx x _____________ = Ú _____________________ 1 1 2 2 __ __ x x + x x x + 2 + 1 x
1 1 – __ 2 dx x ___________ = – Ú __________________ 1 1 2 __ __ x + x x + 2 + 1 x
x + 1
2
x + x + 1
÷ 1 x + __ + ___ 2 2
x + 1 dx Ú ______ 6
=
(
dx Ú _________ 2x dx Ú __________ 4 2 4 2
1 1 + __ 2 2 2 x __________ dx – __ Ú _____ 3x dx 2 6 3 x +1 1 x – __ x + 1
=
= Ú _________ dt __ t2 – ( ÷ 2 )2
=
( ) ) ( Ú
( )
)
dx Ú _______________ 4 2
=
x +1
=
1 2 1 – __ x = Ú ________ dx 1 2
(
2 – 1 dx Ú x______ 4 x + 1
1 1 + __ 2 2 x 2 = Ú ___________ dx – __ Ú _____ 3x dx 6 3 x +1 1 x2 + __ 2 – 1 x
(x – x + 1)
280. We have,
278. We have,
=
t + ÷2
( )
2
2 dx – __ Ú _____ 3x dx Ú ___________ 4 2 6 3
)
(x2 + 1)
1.75
(
(
)
)
1.76 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 dt = – Ú _______ _____ Let x + __ x = t 2 t ÷ t – 1 1 fi 1 – __ 2 dx = dt x = cosec–1 (t) + c 1 = cosec–1 x + __ x + c x = sin–1 _____ 2 + c x +1 282. We have,
( )
(
(
)
÷x + x + x
( ( Ú (
)
=
)
)
)
÷(
)
) ÷(
)
)
( ) ( )
(
2 tdt Ú _______ 2
) ÷( ) 1 Let ( x + __ x + 1 ) = t
( )
=
_____ tdt Ú ________ 2 2
=
v dv Ú ________ 2
2
= 2 tan–1 t + c
1/2 1 = 2 tan–1 x + __ x + 1 + c
2
÷x + 1
x +1
dx x2 – 1 = Ú _______ × ___________ __________ 1 1 __ 2 x x + x x x2 + __2 x 2 dx x – 1 ___________ _________ = Ú × __________ 1 1 x2 x + __ x x2 x2 + __2 x
÷ ( ) ( ) ÷ ( )
(
)
fi 2t dt = 2v dv
dv Ú _______ 2
(v + 2) v 1 = ___ __ tan–1 ___ __ + c 2 ÷2 ÷
( )
( ) ( ÷ ( ) )
(t + 1)t
dx – 1 × _______ _____ Ú x_____ 2 4
Let t2 – 2 = v2
(v + 2)v
1 1 1 __ = ___ __ tan–1 ___ __ x2 + 2 + c 2 ÷2 ÷ x
= Ú ______ 2 dt (t2 + 1)
t ÷t – 2
=
283. We have,
( )
2 ÷ t 2 –__ 1 = ___ __ tan–1 ______ + c 2 ÷2 ÷
)
)
284. We have,
1 fi 1 – __2 dx = dt x
(
=
(
1 Let x + __ x = t 1 fi 1 – __ 2 dx = dt x
___________
1 1 – __ 2 x dx = Ú __________ × ____________ ___________ 1 1 __ __ x + x + 2 x + x + 1
t÷t – 2
)
_____
dx x2 – 1 ____________ 2 × ____________ ___________ 1 x(x + 2x + 1) x + __ x + 1
( (
dt ______ Ú _______ 2
dx x2 – 1 = Ú ____________ × ____________ ___________ 1 1 x2 x + __ x + 2 x + __ x + 1
=
)
÷(
) ÷(
(
dx x2 – 1 = Ú _______ × _____________ ____________ 1 (x + 1)2 2 x x + __ x + 1 dx x2 – 1 = Ú ________ × ____________ ___________ 2 1 x(x + 1) __ x + x + 1
÷ (
)
dx x x +– 11 ) × ___________ __________ Ú (_____ 3 2
( )
1 1 – __ 2 dx x = Ú _______ × ____________ ___________ 1 1 2 __ x + x x + __ x – 2
2
x – 1 ___________ Ú ______________ 3 4 2
dx x ÷2x – 2x + 1 2
x – 1 dx _____________ = Ú ______________ 2 1 3 __ x 2 – __ 2 + 4 x x
) ÷ ( Ú ) ÷( ) ( Ú ) ÷( ( (
x2 – 1 dx ___________ = ______________ 2 1 __ x5 2 – __ 2 + 4 x x 1 1 __ 3 – __ 5 x x ___________ = ____________ dx 2 1 __ __ 2 – 2 + 4 x x
)
2 1 Let 2 – __2 + __ 4 = t2 x x 4 4 fi __ 3 – __5 dx = 2t dt x x
1 2t dt = __ Ú ____ t 4 1 = __ × t + c 2 1 2 1 1/2 = __ × 2 – __ 2 – __ 4 + c 2 x x
(
)
)
Indefinite Integrals
285. We have,
(
4
x –1 _________ Ú ____________ 2 4 2
=
)
dx x ÷x + x + 1
( ( ÷
4
x –1 _____________ Ú ________________ 1
)
dx x x x + __ 2 + 1 x 2
2
2
( ( )) ( ( ( ) ) )
)
( )
÷
÷
= Ú ___ t dt t
=
=t+c
1/2 1 = x + __ 2 + 1 + c x
Ú dt
(
)
2
– 1 dx x – 3x Ú __________ 4 2 x + x + 1
3 x2 – 1 = Ú _________ dx – __ Ú _________ 2x dx 4 2 x4 + x2 + 1 x + x2 + 1
( )
1 1 – __ 2 x 3 2x = Ú __________ dx – __ Ú __________ dx 1 2 x4 + x2 + 1 2 __ x + 2 + 1 x 1 1 – __ 2 x 3 = Ú ___________ dx – __ Ú _________ 2x dx 2 x4 + x2 + 1 1 2 __ 1 – x – 1
(
( ) )
(
)
1 Put x + __ x = t and x2 = z
1 fi 1 + __ x dx = dt and 2x dx = dz 2
(
)
3 dt – __ Ú ________ Ú _____ 2 2 2 dz
t –1
z +z+1
(
( ) ( )
÷ (
))
÷(
))
1 x2 x + x2 1 + __ 2 x
)
) | (
| |
dx _________ Ú _________________
(
(
)
(
1 x + __ x – 1 __ 2(x2) + 1 1 __ __________ __ = log – ÷ 3 tan–1 ________ + c 2 1 ÷3 x + __ x + 1
dx ________________ _______ 1 2 x x + x 1 + __ 2 x 1 = 1 + __ 2 = t2 x =
|
2z + 1 3 t–1 1 2 __ = __ log ____ – __ × ___ __ tan– 1 ______ + c 2 t+1 2 ÷ 3 ÷3
x ( x + ÷ 1 + x )
( Ú (
= log |t + 1| – t + c 1 1/2 1 1/2 = log 1 + 1 + __ 2 – 1 + __ 2 + c x x
1 – 1 ) dt Ú ( ____ t+1
| |
)
2
dx _____ Ú _____________ 2 2 =
=
286. We have,
=
( ) ( ) ( )
1 Let x2 + __ 2 + 1 = t2 x 2 fi 2x – __ 3 dx = 2t dt x 1 fi x – __ 3 dx = t dt x
=
287. We have,
1 x – __ 3 x ___________ = Ú ____________ dx 1 2 + 1 x2 + __ x
– t dt Ú ____ 1+t
x4 – 1 _____ x3 ____________ = Ú ___________ dx 1 2 __ x + 2 + 1 x
1.77
(
288. The given integral is dx 4 = Ú ___________ sin x + cos4x
=
4
sec x 4 dx Ú ________ tan x + 1
Ú
(1 + tan2x) sec2x ______________ dx tan4x + 1
)
1 Put 1 + __ 2 = t2 x 2 fi – __3 dx = 2t dt x dx ___ fi 3 = – t dt x
Put tan x = t fi sec2 x dx = dt =
and then solve it. 289. The given integral is
)
2
+1 dt Ú t______ 4 t +1
dx 1 2 4 = __ Ú _______ dx Ú _____ x + 1 2 (x4 + 1)
1.78 Integral Calculus, 3D Geometry & Vector Booster
2 2 1 (x + 1) – (x – 1) = __ Ú _______________ dx 2 (x4 + 1)
1 = __ Ú 2
(x2 + 1) 1 _______ dx – __ Ú 4 2 (x + 1)
x2 – 1 _______ dx (x4 + 1)
and then you do it. 290. The given integral is 2
2
x dx 2x 1 4 = __ Ú _____ 4 dx Ú _____ 2 x +1 x +1
2 2 1 (x + 1) + (x – 1) = __ Ú ________________ dx 2 x4 + 1
2
1 (x + 1) 1 = __ Ú _______ dx + __ Ú 2 2 x4 + 1
( ) 4
=
2
x ÷x + x + 1 4
=
x –1 ______________ dx Ú ________________ 1
=
x –1 ____________ dx Ú ______________ 1
=
Ú
÷ ( ÷(
2
4
)
x3 x2 + __ 2 + 1 x 1 x – __ 3 x ____________ ____________ dx 1 2 x + __ 2 + 1 x
÷(
) ( ) ( )
1 Put x2 + __ 2 + 1 = t2 x
1 x – __3 dx = t dt x
= Ú dt
=t+c
(x2 + 1)2 – 2x2 __________________ dx (x2 + 1) (x4 – x2 + 1)
Ú
=
3x 2 dx – __ Ú _______ 3 2 dx Ú ___________ 4 2 3 (x – x + 1) (x ) + 1
(x2 + 1) 2 ___________ dx – __ tan– 1 (x3) + c 3 (x4 – x2 + 1)
=
(x2 + 1)
÷
__________
2
dx dx 2 = Ú ___________ Ú __________ 1 3/4 5 __ x (1 + x4)3/4
(
dx dx 2 = Ú ___________ Ú __________ 5 4/5 1 4/5 6 __ x (1 + x )
(
) ( )
x 1 + 4 x
(
= – Ú dt
= – t + c
1 1/5 = – 1 + __ 5 + c x
(
296. The given integral is
–1 ___5 dx = t3 dt x = – Ú dt
x ÷1 + x
)
1 Put 1 + __ 5 = t5 x 5 fi – __6 dx = 5t4 dt x 1 fi – __6 dx = t4 dt x
1 Let 1 + __ 4 = t4 x
= – t + c
)
x 1 + 5 x
and then you do it. 293. The given integral is
)
x2 x2 x2 + __ 2 + 1 x
1 = x2 + __ 2 + 1 + c x 295. The given integral is
(x + 1) – 2x
)
4
=
(
dx Ú _____________ (x6 + 1)
Ú
)
x –1 __________ dx Ú _____________ 2 4 2
(x – 1) _______ dx x4 + 1
x +1 6 dx Ú ______ x +1 2
294. The given integral is
2
and then you do it. 291. The given integral is 2 (x_______________ + 1) – (x2 – 1) 2dx _____ = dx Ú x4 + 1 Ú x4 + 1 2 (x2 + 1) (x – 1) _______ = Ú _______ dx – dx Ú 4 4 x +1 x +1 and then you do it. 292. The given integral is
(
1 1/4 = – 1 + __ 4 + c x
(
)
2
x _____ –1 dx Ú ________ 4
=
(x2 – 1)
_______ dx Ú __________ 1 2 2
÷
x x + __ 2 x
)
Indefinite Integrals
Ú
=
( )
1 1 – __ 2 x _______ ____________ dx 1 2 x + __ x – 2
÷(
)
(
)
1 Let x + __ x = t 1 fi 1 – __2 dx = t4 dt x
(
Ú
dt ______ _____ 2 ÷ t – 2
)
=
= log t + ÷t 2 – 2 + c
_____
|
| (
)
|
÷
______
|
1 1 = log x + __ x + x2 + __ 2 + c x
Ú
=
÷
2 __ 2 x 1 = __ Ú __________ __________ dx 2 1 2 __ x + 2 + 3 x
) ( )
1 1 1 + __ 2 – 1 – __ 2 x x 1 _________ = __ Ú _________________ dx 2 1 2 __ x + 2 + 3 x 1 = __ Ú 2
(
)
( )
1 1 1 – __ 2 1 + __ 2 x x 1 __________ __________ dx __________ dx – __ Ú ___________ 2 1 1 2 __ x2 + __ 2 + 3 x + 2 + 3 x x
÷
(
÷
)
( )
1 1 1 + __ 2 1 – __ 2 x x 1 1 ___________ ___________ = __ Ú ____________ dx – __ Ú _____________ dx 2 2 2 1 2 __ x + 1x + 1 x – x + 5
÷( ) 1 1 = __ log ( x – __ x ) + x 2 ÷
|
÷(
__________ 2
|
1 + __ 2 + 3 x
298. The given integral is
)
( )
)
1 fi 1 – __2 dx = dt x
(
dt = – Ú _______ _____ 2 t÷t + 3
t dt = – Ú ________ _____ 2 2 t÷ t + 3
)
Let t2 + 3 = v2 fi 2t dt = 2v dv
v dv = – Ú ________ 2 (v – 3)v
dv = – Ú _______ 2 (v – 3)
v– ÷ 3 1__ __ = – ____ log _______ + C 2÷3 v+÷ 3
÷t 2 + 3 – ÷ 3 1__ _____ = – ____ log ___________ __ + c 2 2÷3 ÷t + 3 + ÷ 3
__ 1 x2 + __ 2 + 1 – ÷3 x 1__ ____ ________________ = – log _________ __ + c 2÷3 1 2 __ x + 2 + 1 + ÷ 3 x
| |
(
)
__
_________
|
|
dx –1 ) × ___________ ___________ Ú ( x_____ x 4 2 2
÷x + 3x + 1
=
=
( ) ÷ Ú ( ) (
dx 1 2 × __________ __________ Ú 1 – __ 1 x 2
x + __ 2 + 3 x dx 1 1 – __ 2 × ____________ _______ 1 2 x x + __ + 1 x
| (
÷
)
÷
)
__________
|
1 1 = log x + __ x + x2 + __ 2 + 3 + c x
300. The given integral is
2
÷x + x + 1
|
_____
÷ ÷
dx +1 × ___________ __________ Ú x______ 2 4 2 1–x
|
299. The given integral is
1 1 – __ tan–1 x + __ x + c 2
(
1 Let x – __ x = t
__
x x + __ 2 + 3 x
÷
)
dx __________ Ú ____________ 1 2 2
(
) ÷(
(
dx ____________ __________ 4 x ÷x + 3x2 + 1
÷
)
1 1 + __ 2 x dx = – Ú _______ × ____________ __________ 1 1 2 __ x – x x – __ x + 3
297. The given integral is
(
1.79
xx(x2x + 1) (ln x + 1)
dx Ú _________________ (x4x + 1)
Let xx = t
fi xx(1 + ln x) dx = dt
1.80 Integral Calculus, 3D Geometry & Vector Booster
=
=
2
t +1 4 dt Ú ______ t +1
(Ú )
1 1 + __ 2 t _______ dt 1 2 t + __ 2 t
Ú
=
x dx Ú __________ (1 + x2)1006
=
x dx Ú ______________ 1 1006 2012 __
( )
1 1 + __ 2 t __________ dt 1 2 __ t – + 2 t
=
1 1 1 = ___ __ tan– 1 ___ __ t – __ + c t 2 ÷2 ÷
1 1 = ___ __ tan– 1 ___ __ (x x – x– x) + c ÷2 2 ÷
( )
=
( ( ) ) ( )
2
=
1 1 __ 3 – __ 5 x x ____________ ___________ dx 2 1 __ __ 2 – 2 + 4 x x
1 = __ Ú 2 1 = __ Ú 2 1 = __ t 2
dt 1 = – __ Ú ____ 2 t1006
t– 1005 1 = – __ × ______ +C 2 – 1005
1 = __________ + C 2010 × t1005
÷
__________
(
2009
(2x + 1) dx
=
(2x + 1) dx
Ú _______________ 4 __ 1 3/2 3 __
(
)
)
x 1 + x + 2 x
2x4 – 2x2 + 1 = ____________ +c 2x2 x dx Ú __________ (1 + x2)1006
)
Ú _____________ (x2 + 4x + 1)3/2
4 __ 4 dx = 2t dt x
302. The given integral is
(
303. We have,
1 2 1 __ = __ 2 – __ 2 + 4 + c 2 x x
÷
1 = _________________ + C 1 1005 __ 2010 × 1 + 2 x
=
Ú
1 2 __ 2 + __ 3 x x dx _____________ 4 __ 1 3/2 __ 1 + x + 2 x
(
)
(
)
(
1 __ 3 dx = – t dt x
4 1 Let 1 + __ x + __ 2 = t2 x 4 2 fi – __2 – __ 3 dx = 2t dt x x
+c
( )
1 Let 1 + __2 = t x 2 __ fi – 3 dx = dt x dx ___ dt ___ fi 3 = – 2 x
dt
)
tdt ___ t
____________
1 __ 4 = t2 x
1 1 1 fi __ 3 – __ 5 dx = __ tdt 2 x x
) ÷( ) ( Ú ) ÷( ( ) ( ) ( ) __ x5 2 – __ 2 + 4 x x
4 fi __ 2 – x
(x – 1)
____________ dx Ú ______________ 2 1
2 Let 2 – __ 2 + x
(
x2 – 1 ____________ dx Ú _______________ x3 ÷2x 4 – 2x2 + 1 =
Ú
)
301. The given integral is
(
2009
1 + 2 x dx ____________ 1 1006 3 x 1 + __ 2 x x
2009
2 fi __ 2 + x t dt = – Ú ___ t3 dt = – Ú __ 2 t 1 = __ + c t
)
Indefinite Integrals
304. We have,
Ú
1 = __________ __________ + c 4 __ 2 1 + __ x + 2 x
÷
(
____
____
)
cot x – ÷tan x ÷ ____________ dx 1 + 3 sin 2x
=
1 – tan x 1 ____ dx Ú _____ ( _____________ 1 + 6 sin x cos x ) ÷tan x
( (
)
=
1 – tan x 1 ____ _____________ sec2x dx Ú _____ ÷tan x sec2x + 6 tan x
=
1 – tan x 1 ____ ________________ sec2x dx Ú _____ ÷tan x tan2x + 6 tan x + 1
=
1–t 2t dt Ú __ 1t _________ t4 6t2 + 1
( Ú (
)
2
2
)
( )
1 __ 2 – 1 t = 2 Ú _________ dt 1 2 t + __ 2 + 6 t 1 __ 2 – 1 t _______ = 2 Ú + 4 dt 12 t + __ t dz 1 = 2 Ú _____ 2 Let z = t + __ t z +4 1 2 fi dx = 1 – __ 2 dx t z 1 = 2 × __ tan– 1 __ + c 2 2
( ( ) ) ( )
( )
(
305. We have,
=
2t dt 4 Ú _____ t +1
=
Ú _________________ t4 + 1
=
+ Ú _________ 4 Ú ________ t4 + 1 t +1
2
(t2 – 1) + (t2 + 1) dt (t2 – 1) dt
(t2 + 1) dt
=
(Ú )
( )
=
(Ú )
) ( Ú
1 1 1 – __ 2 dt 1 + __ 2 dt t t _________ + Ú _________ 1 1 2 __ t + 2 t2 + __ 2 t t
1 1 – __ 2 dt t ___________ + 1 2 __ t + – 2 t
1 1 + __ 2 dt t __________ 1 2 __ t – + 2 t
( )
( )
| |
( ( ) )
( ) ( )
__ 1 1 t + __ – ÷2 t – __ t t 1 1 – 1 ___ ______ __ = ____ __ log ___________ + c __ + __ tan 1 2÷2 2 2 ÷ ÷ __ t + + ÷2 t
| | ( (
) )
__ 1 x2 + __ 2 – ÷2 x 1 = ____ __ log _____________ __ 1 2 2÷2 __ x + 2 + ÷ 2 x
( ) ) (
1 x2 – __ 2 x 1 __ tan–1 ________ __ + ___ + c 2 ÷2 ÷
306. We have,
= tan–1
( )
( ) 1 1 ( __ ( t + __ ) ) + c t 2
t ◊ 2t dt 4 Ú ______ t +1
)
1–t = 2 __________ dt t4 + 6t2 + 1
=
1.81
)
____ 1 ____ = tan–1 __ ÷ tan x + ÷ cot x + c 2
____
dx Ú ÷tan x
Put tan x = t2
fi
fi
____
____
– ÷cot x ) dx Ú (÷tan x
( ) tan x – 1 = Ú ( _______ dx tan x ) =
____
1 – _____ ____ dx Ú ÷tan x ÷tan x ____ ÷
Put
tan x = t2
sec2x dx = 2t dt
fi
sec2x dx = 2t dt
2t dt dx = _____ 2 sec x
fi
2t dt dx = _____ 2 sec x
2t dt = ________ 1 + tan2x
2t dt = ________ 1 + tan2x
2t dt = _____ 1 + t4
2t dt = _____ 1 + t4
| |
1.82 Integral Calculus, 3D Geometry & Vector Booster
=
= =
=
=
( ) Ú ( ) Ú ( ) 2
t –1 4 2t dt Ú ______ t +1
Ú
Ú
( )
1 1 – __ 2 t __________ dt 2 1 __ t + – 2 t
=
1 1 – __ 2 t _____ dt 1 2 t + __ 2 t
311.
|( )|
Ú
(
____ ÷cot x –
)
1 ____ _____ dx = cot x ÷
__
____
2t dx = – ______ dt 1 + t4
|
= 2x + 2
2
(
t2 + 1 = 2x + 2 Ú ______ 4 dt Let tan x = t2 t + 1 fi sec2 dx = 2t dt
2
– cosec x dx = 2t dt
____
+1 ________ ____ + 2 dx Ú tan x ÷tan x
)
Let cot x = t
=
____
2t 4 dt Ú ( t 2 – __ 1t ) × ______ 1+t
( ) 2
t –1 = 2 Ú ______ dt t4 + 1
( )
1 1 – __ 2 t = 2 Ú _____ dt 1 t2 + __ 2 t
( ( ) )
1 1 – __ 2 t = 2 Ú ___________ dt 1 2 __ t + – 2 t
| | ( )
__ 1 t + __ – ÷2 t 1 = 2 × ___ __ log ___________ __ + c 1 ÷2 t + __ + ÷ 2 t
( )
2t fi dx = _____ 4 dt 1+t
Ú
1 1 + __ 2 t _____ dt 1 t2 + __ 2 t
Ú
1 1 + __ 2 t __________ dt 1 2 __ t – + 2 t
cot x – 1 ____ dx Ú ________ ÷cot x
____
+ ÷cot x + 2) dx Ú (÷tan x
= 2x + 2
(
|
( )
(÷tan x + ÷cot x ) – ÷ 2 1 __ + c = ____ __ log ___________________ ____ ____ 2÷2 (÷tan x + ÷ cot x ) + ÷ 2
307. Do yourself
__
____ ÷
|
____
) +1 = 2x + Ú (________ tan x dx tan x )
( ) ( ) ____
____
____
4 + ÷ cot x ) 2 dx Ú (4÷tan x
=
( ( ) )
|
__
309. Do yourself 310. Do yourself
__ 1 t + __ – ÷2 t 1 = ____ __ log ___________ __ + c 1 2÷2 t + __ + ÷ 2 t
308.
(÷cot x + ÷tan x ) – ÷ 2 __ + c = ÷2 log ___________________ ____ ____ (÷cot x + ÷ tan x ) + ÷ 2
2t dt t2 – 1 ____ ______ 4 t t +1 t2 – 1 ______ dt t4 + 1
( )
__ 1 t + __ – ÷2 t = ÷2 log ___________ __ + c 1 t + __ + ÷ 2 t __
( ( ) )
( ( ) )
1 1 1 = 2x + 2 × ___ __ tan–1 ___ __ t – __ + c t 2 ÷2 ÷
(
)
__ 1 = 2x + ÷2 tan–1 ___ __ (tan x – cot x) + c 2 ÷
312. We have,
2
dx sec x ____ 4 = Ú ___________ ____ dx ____ Ú ______________ (÷sin x + ÷ cot x ) (÷tan x + 1)4
Put tan x = t2
fi dt Ú ______ (t + 1)4
=
4 = – _______ +c (t + 1)3
4 = – ___________ + c ____ (÷tan x + 1)3
sec2 x dx = 2t dt
Indefinite Integrals
Integration by Parts 313. We have,
Ú x ex dx = x Ú ex dx – Ú (1 ◊ ex) dx
1 = x log (x2 + 1) – 2 Ú 1 – _____ 2 dx 1+x
= x log (x2 + 1) – 2(x – tan–1 x) + c
Ú x sin x dx = x Ú sin x dx – Ú (1 – cos x)dx
= x Ú sin x dx + Ú cos x dx
= – x cos x + sin x + c
x = tan–1 (x) Ú dx – Ú _____ dx 1 + x2 1 = x tan–1 (x) – __ log |1 + x2| + c 2
320. We have,
Ú x2 sin x dx = x2 Ú sin x dx – Ú (2x – cos x) dx
= – x2cos x + 2 Ú cos x dx
= – x cos x + 2x Ú cos x dx – 2 Ú (1 sin x) dx
= – x2cos x + 2x sin x + 2 cos x + c
2
–x dx Ú cos–1 1______ 1 + x2
= Ú (2 tan–1 x) dx
= 2 Ú tan–1 x dx
1 = 2 tan–1 x Ú dx Ú ______ 2 ◊ x dx 1+x
2x 1 = 2 x tan–1 x – __ ______ dx + c 2 1 + x2
1 = 2 x tan–1 x – __ log |x2 + 1| + c 2
( (
2
316. We have,
Ú log x dx = Ú (log x ◊ 1) dx 1 = log x Ú dx – Ú __ x ◊ x dx
= log x Ú dx – Ú 1 ◊ dx
= x log x – x + c
)
__
Ú e x dx ÷
Let x = t2 fi dx = 2 t dt
Ú (log x)2 dx = Ú {(log x)2 ◊ 1} dx
(
)
= 2 Ú t et dt
= 2 t Ú et dt – Ú (1 ◊ et) dt
1 = (log x)2 Ú dx – Ú 2 log x ◊ __ x ◊ x dx
= 2 (t et – et) + c
= x (log x) – 2 Ú log x dx + c
÷x = 2 (÷x e – e ÷ x ) + c
= x (log x)2 – 2 (x log x – x) + c
2
Ú log (x2 + 1) dx = Ú (log (x2 + 1) ◊ 1) dx 1 = log (x2 + 1) Ú dx – Ú _____ 2 ◊ 2x ◊ x dx x +1
(
( )
x2 = x log (x2 + 1) – 2 Ú _____ 2 dx x +1
)
__
__
__
322. We have,
318. We have,
) ) ) )
(
317. We have,
( Ú (
321. We have,
( )
( )
315. We have,
Ú tan–1 x dx = Ú {tan–1 x ◊ 1}dx
314. We have,
)
319. We have,
= xex – ex + c
(
1.83
x + sin x dx Ú ( ________ 1 + cos x )
x sin x = Ú ________ dx + Ú ________ dx 1 + cos x 1 + cos x x x 2 sin __ cos __ 2 2 x ________ ______________ = Ú x dx + Ú dx x 2 cos2 __ 2 cos2 __ 2 2
( ) ( ) ( ) ( ) x x 1 = __ Ú x sec ( __ ) dx + Ú tan ( __ ) dx 2 2 2 2
1.84 Integral Calculus, 3D Geometry & Vector Booster
[
( ( ) ) ]
( )
x x 1 = __ x Ú sec2 __ dx – Ú 2 tan __ dx 2 2 2
( )
x + Ú tan __ dx 2 x x x 1 = __ 2x tan __ – Ú tan __ dx + Ú tan __ dx + c 2 2 2 2
[
( ) ]
( )
( )
( )
x = x tan __ + c 2 323. We have,
( ÷ )
1–x dx Ú tan _____ 1+x
fi dx = – sin q dq
( ÷
_________
)
1 – cos q = – Ú tan _________ sin q dq 1 + cos q –1
( ÷
__________
(
) ] 4 1 = __ p [ q Ú sin 2q dq + __ Ú cos 2q dq ] – x + c 2 4 – q cos 2q = __ p [ ( ________ ) + __ 14 sin 2q ] – x + c 2 4 __ 1 1 __ = __ p [ – 2 (q (1 – 2 sin q)) + 2 sin q cos q ]– x + c 4 __ 1 1 __ = __ p [ – 2 (sin x (1 – 2x )) + 2 x ÷1 – x ]– x + c 2
–1
_____
2
2
325. We have,
Let x = cos q
______
–1
[
cos 2q 4 = __ p q Ú sin 2q dq – Ú 1 ◊ – ______ dq – x + c 2
)
2
x dx Ú ______________ (x sin x + cos x)2
x cos x = Ú ____________ ◊ x sec x dx x sin x + cos x x cos x = x sec x Ú ____________ dx x sin x + cos x
2 sin2 (q/2) = – Ú tan–1 _________ sin q dq 2 cos2 (q/2)
q = – Ú tan–1 tan __ sin q dq 2
(sec x + x sec x tan x) – Ú – _________________ dx x sin x + cos x
1 = – __ Ú q sin q dq 2
x sec x = ___________ + sec x dx + c x sin x + cos x Ú
1 = – __ q Ú sin q q – Ú (1 – cos q) dq 2
x sec x = ___________ + log |sec x + tan x| + c x sin x + cos x
1 = – __ [– q cos q + sin q] + c 2
326. We have,
( ( ) )
[
(
]
1 = __ [q cos q – sin q] + c 2
1 = __ [ x cos–1 x – ÷1 – x2 ] + c 2
_____
324. We have, __
(
(
)
)
Let x = sin2q
fi dx = sin 2q dq
–1
x tan x dx Ú _________ (1 + x2)3/2
Let x = tan q
fi dx = 2 sec2q dq
tan q a tan–1 (tan q) = Ú _______________ ◊ sec2q dq sec3q
= Ú q cos q dq
= q Ú cos q dq – Ú (1 . sin q) dq
= q sin q + cos q + c
x 1 _____ = _______ ◊ tan–1 x + ______ ______ + c. 2 – x2 ÷ 1 + x ÷ 1
__
–1 sin–1 ÷x – cos ÷x dx Ú ________________ –1 __ –1 __ sin ÷x – cos ÷x __ p –1 __ __ sin–1 ÷x – – sin ÷x 2 = Ú _____________________ dx p __ 2 p 2 –1 __ __ __ = p Ú 2 sin ÷x – dx 2 __ 4 = __ p Ú sin–1 ÷x dx – Ú dx
4 = __ p Ú sin–1 (sin q) sin 2q dq – x + c 4 = __ p Ú q sin 2q dq – x + c
)
327. We have,
_____
x dx Ú sin–1 ( ÷_____ a + x )
( ÷
Put x = a tan2q fi dx = 2a tan q sec2q dq
__________
=
Ú sin
)
a tan2q __________ 2a tan q sec2q dq a + a tan2q
–1
1.85
Indefinite Integrals
= 2a Ú sin–1 (sin q) tan q sec2q dq
Let
x = t2 dx = 2t dt
= 2a Ú q tan q sec2q dq
= 2 Ú t sin t dt
= 2 t Ú sin t dt + Ú cos t dt + c
= 2 [– t cos t + sin t] + c
= 2 [– ÷x cos ÷x + sin ÷x ] + c
Let tan q = t
fi sec2q dq = dt
= 2a Ú t tan–1 t dt
t2 1 __ = 2a tan–1 t Ú t dt – Ú _____ 2 dt 1+t 2
[ [
2
( ) ] Ú ( )]
t 1 1 = 2a __ tan–1 t – __ 1 – _____ 2 dt 2 2 1+t
= a [t2 tan–1 t – t + tan–1 t] + c
= a [(t2 + 1) tan–1 t – t] + c
= a [(tan2q + 1)q – tan q] + c
[ (
__
÷ ÷ ]
x2 1 = log (1 + x) Ú x dx – __ Ú ______ dx 2 (x + 1)
2 x2 1 (x – 1) + 1 = __ log (1 + x) – __ Ú __________ dx 2 2 (x + 1)
x2 1 1 = __ log (1 + x) – __ Ú x – 1 + _____ dx 2 2 x+1
x2 1 x2 = __ log (1 + x) – __ __ – x + log |x + 1| + c 2 2 2
x = log (1 + x) Ú dx – Ú _____ dx x+1
(
)
= x log (1 + x) – (x – log|x + 1|) + c
x – sin x dx Ú 1________ – cos x x dx sin x = Ú ________ – _______ dx 1 – cos x Ú 1 – cos x 2 sin (x/2) cos (x/2)
x dx – Ú ________________ Ú _________ 2 sin2 (x/2) 2 sin2 (x/2) dx
=
x 1 = __ Ú x cosec2 (x/2) dx – Ú cot __ dx 2 2
1 = __ x cosec2 (x/2)dx + 2 Ú cot (x/2) dx 2 Ú
( )
[
– Ú cot (x/2) dx + c 1 = __ (– 2x cot (x/2)) + Ú cot (x/2) dx 2
– Ú cot (x/2) dx + c
__
)
Ú (sin–1 x)2 dx –1 2x sin x ______ = (sin–1 x)2 Ú dx – Ú ________ dx ÷ 1 – x2
–1 x sin x ______ = x (sin–1 x)2 – 2 Ú _______ dx ÷ 1 – x2
= x (sin–1 x)2 – 2 Ú q sin q dq, Let sin–1 x = q
= x (sin–1 x)2 – 2 Ú q sin q dq
= x (sin–1 x)2 – 2 [q sin q dq + q cos q dq] + c
= x (sin–1 x)2 – 2 [– q cos q dq + cos q]
= x (sin–1 x)2 – 2 [– ÷1 – x2 sin–1 x + x] + c
= x (sin–1 x)2 + 2 Î÷1 – x2 sin–1 x – x˚ + c
______
333. The given integral is x – sin x dx Ú ________ 1 – cos x x sin x = Ú _______ dx – Ú ________ dx 1 – cos x 1 – cos x
= – cot (x/2) + c
Ú sin ÷x dx
(
______
]
330. The given integral is
)
x _____ ______ dx = dq ÷ 1 – x2
(
329. The given integral is
Ú x log (1 + x) dx
332. The given integral is
Ú log (1 + x) dx 1 = x log(1 + x) – Ú 1 – _____ x+1
__
__
__
331. The given integral is
x x x = a __ a + 1 tan–1 __ a – __ a + c 328. The given integral is
)
__
x = Ú ________ x dx – 2 sin2 __ 2
( )
Ú
( ) ( ) ( )
x x 2 sin __ cos __ 2 2 ______________ dx x 2 sin2 __ 2
1.86 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) x x 1 = __ [ x Ú cosec ( __ ) dx – 1 ◊ ( – 2 cot ( __ ) ) dx ] 2 2 2 x – Ú cot ( __ ) dx 2 x x 1 = __ [ x ( – 2 cot ( __ ) ) + Ú ( 2 cot ( __ ) ) dx ] 2 2 2 x – Ú cot ( __ ) dx + c 2 x = – x cot ( __ ) + c 2
x x 1 = __ Ú x cosec2 __ dx – Ú cot __ dx 2 2 2 2
334. The given integral is
Ú x sin3x dx cos3x = x Ú sin3x dx – Ú _____ – cos x dx 3
( ) )
( (
)
( (
= Ú (2 tan–1 x) dx
= 2 Ú tan–1 x dx
x = 2 tan–1x Ú dx – Ú _____ dx 1 + x2
]
[
]
1 = 2 x tan–1x – __ log | 1 + x2 | + c 2 338. The given integral is
) )
[
x3 sin–1 (x2)
______ dx Ú _________ 4
÷1 – x
–1 2 (x ) × x2 1 2x sin______ = __ Ú ______________ dx 2 ÷ 1 – x4
Let sin–1 (x2) = t 2x ______ fi ______ dx = dt ÷ 1 – x4
cos3x cos3x = x _____ – cos x – Ú _____ – cos x dx 3 3
cos3x sin3x 1 = x _____ – cos x – __ sin x – _____ 3 3 3
1 = __ Ú t sin t dt 2
= Ú [t Ú sin t dt + Ú cos t dt] + c
+ sin x + c 335. The given integral is
(
)
sec 2x – 1 dx Ú x _________ sec 2x + 1 1 – cos 2 x = Ú x _________ dx 1 + cos 2 x
(
)
= Ú x tan x dx
=
= x Ú sec2x dx – Ú tan x dx – x + c
= x tan x – log |sec x| – x + c
336. The given integral is
( )
2x 2 dx Ú sin–1 1_____ +x
= Ú (2 tan–1 x) dx
= 2 Ú tan–1x dx
= 2 tan–1x Ú dx – Ú _____ x dx 1 + x2
[ [
]
]
1 = 2 x tan–1x – __ log | 1 + x2 | + c 2 337. The given integral is 2x 2 dx Ú tan–1 1_____ –x
( )
1 = __ [– t cos t + sin t] + c 2 __________ 1 = __ – sin–1 (– x2) ÷1 – x4 + x2 + c 2
[
]
339. The given integral is
2
Ú x(sec2x – 1) dx = Ú x sec2x dx – Ú dx
sec x (2 + sec x)
dx Ú ______________ (1 + 2 sec x)2
2 cos x + 1 = Ú __________ (cos x + 2)2
cos x (cos x + 2) + sin x = Ú ___________________ (cos x + 2)2
cos x sin x = Ú __________ dx + Ú __________ (cos x + 2) (cos x + 2)2
sin x 1 = _________ cos dx – Ú __________ (cos x + 2) Ú (cos x + 2)2
sin x + Ú __________ (cos x + 2)2 sin x = ________ + c cos x + 2 340. The given integral is
1–x dx. Ú e_____ x +x (1 + ex) – (x + ex) = Ú _______________ dx (ex + x)
+c
Indefinite Integrals
(
)
(1 + ex) = Ú _______ – 1 dx (ex + x)
= log |(ex + x)| – x + c
341. We have,
Here, f (x) = sin x fi f ¢(x) = cos x Thus,
Ú e (sin x + cos x) dx = e sin x + c x
x
342. We have,
x
x e dx Ú (x_______ + 1)2
Ú
ex (x + 1 – 1) ____________ dx (x + 1)2
=
x+1 1 = Ú e x _______ – _____ dx 2 x +1 (x + 1)
(
343. We have,
)
(
(
e x x = _________ + c (1 + x2)1/2
(
)
2
x +1 dx Ú ex _______ (x + 1)2
( ( (
)
x2 – 1 + 2 = Ú e x _________ dx (x + 1)2
(x + 1)(x – 1) + 2 = Ú e x ________________ dx (x + 1)2
(x – 1) _______ 2 = Ú e x ______ + dx (x + 1) (x + 1)2
(x – 1) = e x ______ + c (x + 1)
(
) )
)
)
( ( ) )
}
ex = __ x + c
347. The given integral is x dx Ú ex (x_______ + 1)2
(
)
(
)
sin 2 x 2 = Ú e x _________ + _________ dx 1 + cos 2 x 1 + cos 2 x
(
{
1 1 1 2 dx = Ú ex __ x + – __2 dx Ú ex __ 1x – __ x x
2 + sin 2 x dx Ú e x _________ 1 + cos 2 x
346. The given integral is
1 1 = – Ú e x _____ – _______ dx x + 1 (x + 1)2 ex = – _____ + c x+1
x 1 = Ú e x _________ + _________ dx (1 + x2)1/2 (1 + x2)3/2
Ú ex [ f (x) + f ¢(x)] dx = ex f (x) + c
)
345. We have,
Ú ex (sin x + cos x) dx
We know that,
(
1.87
)
(
)
)
(x + 1) – 1 = Ú ex __________ dx (x + 1)2
1 1 = Ú ex ______ – _______ dx (x + 1) (x + 1)2
ex = Ú ______ + c (x + 1)
(
)
2 sin x cos x 2 = Ú e x ______ 2 + _________ dx 2 cos x 2 cos2x
= Ú e x (sec2 x + tan x) dx
= Ú e x (tan x + sec2 x) dx
sin x 1 = Ú ex _______ – _______ dx 1 – cos x 1 – cos x
2 sin (x/2) cos (x/2) 1 = Ú ex _________ – ________________ dx 2 2 sin (x/2) 2 sin2 (x/2)
x
= e tan x + c
344. We have,
348. The given integral is
(
)
+ x + x3 dx Ú ex 1_________ (1 + x2)3/2
(
)
1 + x (1 + x2) = Ú e x ____________ dx (1 + x2)3/2
x 1 = Ú e x _________ + _________ dx (1 + x2)3/2 (1 + x2)1/2
(
)
1 – sin x dx Ú ex { ________ 1 – cos x }
(
)
( (
)
)
1 = Ú ex __ cosec2 (x/2) – cot (x/2) dx 2 1 = Ú ex (cot (x/2) + – __ cosec2 (x/2) dx 2 x x __ = – e cot + c 2
( )
349. The given integral is
(
)
1.88 Integral Calculus, 3D Geometry & Vector Booster
2 + sin 2 x dx Ú ex { _________ 1 + cos 2 x }
(
)
ex = _______ + c (x + 1)2
sin 2 x 2 = Ú e x _________ + __________ dx 1 + cos 2 x 1 + cos 2 x 2 sin x cos x 2 = Ú e x ______ 2 + _________ dx 2 cos x 2 cos2 x
354. The given integral is 1 1 – ______ dx Ú ____ log x (log x)2
= Ú e x (sec2x + tan x) dx
= Ú e (tan x + sec x) dx
= e x tan x + c
(
)
x
(
(
)
)
2 – 1) + 2 x2 + 1 x _______ x (x __________ e d x = e dx Ú (x + 1)2 Ú (x + 1) 2
(
)
(
)
(x – 1) = e ______ + c (x + 1) x
351. The given integral is
log x __________ dx Ú (1 + log x)2 t
t e = Ú ______ dt (t + 1)2
Let log x = t dx fi dt = ____ log x e
[(t + 1) – 1]et = Ú ____________ dt (t + 1)2
1 1 = Ú et ____ + – ______ dt t+1 (t + 1)2
[
(
e = ____ + c t+1 x = ________ + c log x + 1
Ú e x ( log x + __ 1x ) dx = e x log x + c
(
)
x–1 dx Ú e (x_______ + 1)3
(
)
(x + 1) – 2 = Ú ex __________ dx (x + 1)3
1 2 = Ú ex _______ + – _______ dx (x + 1)2 (x + 1)3
[
(
)]
(
)
1–x dx Ú ex ______ 1 + x2 2
( (
)
(1 – x)2 = Ú ex ________ dx (1 + x2)2
(1 + x2 – 2x) = Ú ex ___________ dx (1 + x2)2
– 2x 1 = Ú ex _______ + _______ dx (1 + x2) (1 + x2)2
ex = _______ + c (1 + x2)
[
(
)
)]
356. The given integral is
–x dx Ú ex ( 11 _____ + x) 2
( ( (
)
(1 – x)2 = Ú ex _______2 dx (1 + x)
(1 + x2) – 2x = Ú ex ___________ dx (1 + x)2
x2 – 1 – 2 (x – 1) = Ú ex _______________ dx (x + 1)2
2 (x – 1) x2 – 1 = Ú e x _______ – ________ dx 2 (x + 1) (x + 1)2
2 (x – 1) x2 – 1 = Ú e x _______ – ________ dx 2 (x + 1) (x + 1)2
2(x – 1) x–1 = Ú ex _____ – _______ dx x+1 (x + 1)2
353. The given integral is x
dx fi dt = ____ log x e
t
352. The given integral is
)]
Let t = log x
355. The given integral is
(x – 1) _______ 2 = Ú ex ______ + dx (x + 1) (x + 1)2
( )
}
1 1 = Ú e t __ – __ dt, t t2 et = __ + c t x = ____ + c log x
2
350. The given integral is
{
( ( ( ( ( (
)
)
) ) )
) 2 (x + 1 – 2) x – 1 = Ú e _____ ) – ___________ dx x+1 (x + 1) ) 4 = Ú e ( _____ dx ( xx +– 31 ) – (x_______ + 1) ) x
2
x
2
Indefinite Integrals
(
)
x–3 = Ú ex _____ + c x+1 357. The given integral is
(
x +1 _______ dx Ú ex (1 + x)2
( (
x – 1 + 2 = Ú ex _________ dx (x + 1)2
x2 – 1 2 = Ú e _______ + _______ dx (x + 1)2 (x + 1)2
x–1 2 = Ú e _____ + _______ dx x+1 (1 + x)2
x–1 = ex _____ + c x+1
x
x
Ú ecos
( ( )
(
)
( ( (
) (
)
)
= esin x (x + sec x) + c
(
)
1 362. Ú e x log x + __ 2 dx x _____
)
Let cos–1 x = t
dx ______ fi ______ = – dt ÷ 1 – x2
)
{ (
(
)
)}
1 1 __ 1 __ = Ú ex log x – __ x + x + x2 dx 1 = e x log x – __ x + c 1 363. Ú log (log x) + ______ 2 dx (log x)
+ 1) + ÷1 – x2 ) x (x ______ _______________ dx (x + 1)2 ÷ 1 – x2
( (
))
1 = et sin–1 t + ______ _____ + c ÷ 1 – t2
358. The given integral is
(
t 1 1 = Ú et sin–1 t + ______ _____ – ______ _____ + ________ dt 2 2 (1 – t2)3/2 ÷1 – t ÷ 1 – t
)
( )
–1
fi cos x dx = dt
t = Ú et sin–1 t – ________ dt (1 – t2)3/2
)
2
)
x – tan x = Ú esin x _______ cos x dx cos2x Let sin x = t
)
2
(
(
{
)
(
}
)
1 = Ú et log t + __ 2 dt, t
Let
t = log x
fi
dx dt = ____ logx . e
1 + cos t + sin t = – Ú et _____________ dt (1 + cos t)2
sin t 1 = – Ú et _________ + _________ dt (1 + cos t) (1 + cos t)2
1 = Ú et log t – __ + t
et = – _______ + c 1 + cos t
1 = et log t – __ + c t
e cos x = – _____ + c 1+x
1 = x log (log x) – ____ + c log x
)
–1
359. We have,
Ú e (2sec x – 1) tan x dx 2
x
= Ú e x (2 sec2x tan x – tan x) dx = Ú e x [(sec2x – tan x) + (2 sec2x tan x – sec2x)] dx = Ú e x (sec2x – tan x) + c 360. We have,
Ú ex (log (sec x + tan x) + sec x) dx
= ex log (sec x + tan x) + c 361. We have,
(
3
)
x cos x – sin x Ú e sinx ____________ dx cos2x
1.89
= Ú esin x (x cos x – sec x tan x) dx
(
[ (
)
(
(
) ( __ 1t + __ t1 ) ] dt 2
)
)
x4 + 2 364. Ú ex _________ dx (1 + x2)5/2
(
)
(1 + x2)2 + 1 – 2x2 = Ú e x ________________ dx (1 + x2)5/2 x 1 = Ú ex _______ _____ – ________ 2 (1 – x2)3/2 ÷1 + x
( {
{
}
})
x 1 – 2x2 + _________ + _________ dx 2 3/2 (1 + x2)5/2 (1 + x )
(
)
x 1 = ex _______ ______ + ________ + c 2 3/2 2 ÷ 1 + x (1 – x ) esin x (x cos3x – sin x) 365. Ú _________________ dx cos2x
= Ú esin x (x cos x – sec x tan x) dx
1.90 Integral Calculus, 3D Geometry & Vector Booster
Let sin x = t fi cos x dx = dt dt dt dx = ____ cos x = ______ _____ – t2 ÷ 1
fi
(
)
t = Ú e sin t – ________ dt (1 – t2)3/2 t
( (
–1
)
)
1 _____ = et sin–1 t – ______ + c – t2 ÷ 1
( (
(
)
3
x – x – 2 2 dx Ú ex _________ (x + 1)2
)
( {
}
{
})
2
1–x – 2x + _______ + _______ 2 2 dx 2 2 (x + 1) (x + 1)
( (
)
x 1 = Ú ex _______ 2 + _______ + c (x + 1) (x2 + 1)
)
x+1 = ex _______ + c (x2 + 1)
368. We have, e2x (sec2x + 2 tan x) dx = Ú e2x (2 tan x + sec2x) dx
= Ú e2x [2 cos x + (– sin x)] dx
= e2x 2 cos x + c
= e 2 tan x + c
(
( ) sin 2 x 1 = Ú e ( _________ + _________ dx 1 + cos 2 x 1 + cos 2 x ) 1 + sin 2 x = Ú e2x __________ dx 1 + cos 2 x 2x
(
)
2 sin x cos x 1 = Ú e2x __________ + _________ dx 2 1 + 2cos x 2 cos2x
1 = Ú e2x __ sec2x + tan x dx 2
1 = __ Ú e2x (sec2x + 2 tan x) dx 2
1 = __ Ú e2x (2 tan x + sec2x) dx 2
e2x 2 tan x = ________ + c 2
(
)
)
(
________
)
x – __ ÷1 – sin x e 2 _________ dx
Ú
1 + cos x
(
)
x – __ cos(x/2) – sin (x/2) = Ú e 2 ________________ dx 2 cos2 (x/2)
x – __ 1 x x x 1 = Ú e 2 __ sec __ – __ sec __ tan __ dx 2 2 2 2 2
( ( ) ( ) ( ) ) x x x 1 1 = Ú e ( – __ sec ( __ ) + __ sec ( __ ) tan ( __ ) ) dx 2 2 2 2 2 x – __ 2
( )
x – __ x = Ú e 2 sec __ + c 2
2 sin 4 x – 4 dx Ú e2x ___________ 1 – cos 4 x
(
1 + sin 2 x dx Ú e2x ( 1_________ + cos 2 x )
2x
369. We have,
Ú e2x (– sin x + 2 cos x) dx
Ú e3x (3 sin x + cos x) dx = e3x sin x + c
372. We have,
367. We have,
= e2x cot 2x + c
)
x 1 = Ú ex _______ + _______ 2 2 (x + 1) (x + 1)
x (x2 + 1) + (x2 + 1) + (1 – 2x2) – 2x = Ú e ________________________________ dx (x2 + 1)2
= Ú e2x (2 cot 2 x + (– 2 cosec22x)) dx
(x3 + x) + (x2 + 1) + (1 – 2x2) – 2x = Ú ex _______________________________ dx (x2 + 1)2 x
371. We have,
= esin x (x – sec x) + c
)
4 sin 2 x cos 2 x _______ 4 = Ú e2x ____________ – 2 dx 2 sin22x 2sin 2x
370. We have,
t 1 1 = Ú et sin–1 t + ______ _____ – ______ _____ – ________ dt 2 2 (1 – t2)3/2 ÷ 1 – t ÷1 – t
366. We have,
(
373. We have,
)
2 sin 4 x 2 = Ú e x _________ – _________ dx 1 – cos 4 x 1 – cos 4 x
Ú e2 x [2 × log (sec x + tan x) + sec x] dx = e2 x (log (sec x + tan x)) + c
1.91
Indefinite Integrals
= Ú e– 2t sin t dt,
374. We have,
Ú e sin 3 x dx x
ex = ______ 2 2 (1 ◊ sin 3 x – 3 cos 3 x) + c 1 +3 ex = ___ (sin 3 x – 3 cos 3 x) + c 10
375. We have,
Ú e4x cos 3 x dx e4x = ______ 2 2 (4 cos 3 x + 3 sin 3 x) + c 4 + 3
4x
e = ___ (4 cos 3 x + 3 sin 3 x) + c 25
Ú e2x sin 3 x dx
e2x = ___ (2 sin 3 x – 3 cos 3 x) + c 13
e– 2 log x = ______ [2 sin (log x) + cos (log x)] + c 5 1 = – ___ 2 [2 sin (log x) + cos (log x)] + c 5x
x x = __ ÷ 4 – x2 + 2 sin–1 __ + c 2 2
382. We have,
1 dx = 3 Ú ( __ dx Ú ÷1 – 9x ÷ 3 ) – x2 2
( ÷
( ) x ( ___ ) ) + c 1/3
( ÷
x 1 1 = 3 __ __ – x2 – ___ sin–1 2 9 18
e = ______ (– cos x + sin x) + c (1 + 1)
383. We have, _____
– x
e = ___ (– cos x + sin x) + c 2
)
1 __ ______ 9 x __ x 1 2 __ __ = 3 – x – sin–1 ___ + c 2 9 2 1/3 ______
– x
______
_____
| + c Ú ÷x 2 + 1 dx = __ 2x ÷ x2 + 1 + __ 12 log | x + ÷x 2 + 1
384. We have, _______
Ú e– x cos (3x + 4) dx
e2x = _______ [2 cos (3x + 4) + sin (3x + 4)] + c 3(4 + 1)
379. We have,
2 + 1 dx Ú ÷3x
÷ x 1 1/3 =÷ 3 ( __ x + __ + ___ log x + x + __ 2÷ 3 2 | ÷ 13 | )+ c ______
__ 1 = ÷3 Ú x2 + __ dx 3
Ú e
x
=
=
=
=
380. We have, 1 sin (log x) dx Ú __ x3
______ 2
385. We have,
1 __ Ú ex (2cos2x) dx 2 1 __ Ú ex (1 + cos 2 x) dx 2 1 __ Ú (ex + ex cos 2 x) dx 2 ex 1 __ ex + __ (cos 2 x + 2 sin 2 x) + c 2 5
[
2
2
cos x dx
______
__
( )
________
2
378. We have,
______
x x 4 – x2 dx = __ ÷ 4 – x2 + __ sin–1 ( __ ) + c Ú ÷4 2 2 2
Ú e– x cos x dx
_____
_______
377. We have,
e– 2t = ____ (– 2 sin t – cos t) + c 5 e– 2t = – ____ (– 2 sin t + cos t) + c 5
______
e2x = ______ (2 sin 3 x – 3 cos 3 x) + c (4 + 9)
= Ú e– 2t sin t dt
381. We have,
376. We have,
Let t = log x dx fi dt = ____ logx e
]
______
dx Ú ÷x 2 – 9 ______
______
|
|
9 x = __ ÷x 2 – 9 – __ log x + ÷ x 2 – 9 + c 2 2
386. We have,
_______
2 – 1 dx Ú ÷4x
________
÷ ( )
1 2 = 2 Ú x2 – __ dx x
x 1/4 1 ___ 1 = 2 __ x2 – __ – log x + x2 – __ + c 2 4 2 4
( ÷
______
|
÷
______
|)
1.92 Integral Calculus, 3D Geometry & Vector Booster
( ÷
_____
1 = x x – __ – 4
2
|
÷
_____
1 __ log x + x2 – 4
|)
1 __ + c 4
387. We have,
______________
= Ú ÷– {(x – a)2 – a2} dx
= Ú ÷a 2 – (x – a)2 dx
(x – a) ________2 __ a2 x – a = ______ ÷2ax – x + sin–1 _____ a + c 2 2
___________
__________
+ 3 dx Ú ÷x 2 + 2x _______________ (x2 + 2x + 1) + 2 dx
= Ú ÷
=
=
________
Ú ÷
|
c
__________
|
_______________
= Ú ÷– (x 2 + 4x – 3) dx = Ú ÷
= Ú ÷7 – (x + 2)2 dx
_______
___________
=
dx Ú ÷x – 4x2
(x + 2) __________ ______ ÷3 – 4x – x2 + 2
( )
x+2 7 __ sin–1 _____ __ + c 2 ÷7
________________
Ú
=
2
2
fi 2dx = dt
÷( ) t 1 1 = __ ___ – t + ___ sin (4t) + c 4 ÷16 32 1 ( 2x – __ ) 4 ___ 1 1 = _______ – ( 2x – __ ) ÷16 4 4 1 1 + ___ sin [ 4 ( 2x – __ ) ] + c 32 4
( 2 )
__________ x+1 2 _____ ÷ x + 2x + 3
|
1 1 2 = __ Ú __ – t2 dt 2 4 _______
__________
2
|
388. We have,
–1
_____________ 2
__________ ÷3 – 4x – x2 dx
Ú
= Ú ÷– (x 2 + 4x – 3) dx
= Ú ÷– {(x + 2)2 – 7} dx
–1
____________
_____________
392. We have,
__________ 7 – (x + 2)2 dx
__________
= Ú ÷
(x + 2) _________2 __ x +__ 2 7 = ______ ÷3 – 4x – x + sin–1 _____ + c 2 2 ÷7
( )
389. We have, ________
– x2 dx Ú ÷2ax __________
2
________
1 + __ log (x + 1) + ÷ x 2 + 2x + 3 + c 2
2
______________ __
= Ú ÷(x + 1)2 + ( ) 2 dx ÷2
1 2 1 2 __ = Ú __ – 4 x – dx 4 8
________
__________ ÷x 2 + 2x + 3 dx
÷( ) ( ) 1 1 = Ú ( __ ÷ 4 ) – 4 ( 2x – __ 4 ) dx 1 1 1 = __ Ú ( __ – t dt, Let ( 2x – __ ) = t 2 ÷ 4) 4 _______________
387. We have,
|
391. We have,
______________ – {(x2 + 2)2 – 7} dx
___________
= Ú ÷(x + a)2 – a2 dx
x2 dx Ú ÷3 – 4x –
+ x2 dx Ú ÷2ax
(x + a) ________ = ______ ÷ x 2 + 2ax 2 ________ a2 + __ log (x + a) + ÷ x 2 + 2ax + c 2
388. We have,
)
390. We have,
______________ __ (x + 1)2 + ( ) 2 dx ÷2 __________ log (x + 1) + ÷ x 2 + 2x + 3 +
|
(
= Ú ÷– (x 2 – 2ax) dx
+ 4 dx Ú (2x + 1) ÷x 2 + 3x __________ = Ú ((2x + 3) – 2) ÷ x 2 + 3x + 4 dx __________ = Ú ((2x + 3)) ÷ x 2 + 3x + 4 dx __________ – 2 Ú ÷x 2 + 3x + 4 dx
Let x2 + 3x + 4 = t2 fi (2x + 3) dx = 2t dt
÷(
___________________
) ( )
9 3 2 = Ú (2t) t ◊ dt – 2 Ú x + __ + 4 – __ dx 2 4
Indefinite Integrals
÷( 2 ) ( 2 )
= 2 Ú t dt – 2 Ú 2
__________________ __ ÷7 2 3 2 __ ___ x + + dx
( ) [ ( )
_________
__________
_________ 2
__________
)
|]
3 7 + __ log x + __ + ÷x 2 + 3x + 4 + c 4 2
[ ( )
395. We have,
__________ 3 __ ÷ x2 + 3x + 4
1 – 2 __ x + 2 2
| (
_________
__________
)
|]
392. We have, __________
__________
2
__________
= Ú (2x + 3) ÷ x 2 + 3x + 4 dx – 2 Ú x2 + 3x + 4 dx 2
Let x + 3x + 4 = t
396. We have,
÷( 2 ) ( 2 )
_______________ __ ÷7 2 3 2 ___ x + __ + dx
( ) 7 – [ __ log ( x + __ 4 | 32 ) + ÷x
_________
|]
__________ 2 + 3x + 4 +
c
_________
1 = __ Ú (2x – 2 ______ 1 = __ Ú ((2x + 1) – 11) ÷ x 2 + x dx 2 ______ 1 = __ Ú (2x + 1) ÷ x 2 + x dx 2
÷( ) ( )
_____________
11 1 2 1 2 – ___ Ú x + __ – __ dx 2 2 2 and then you do. 394. We have,
÷( ) ( )
÷5 2 1 1 2 __ + __ Ú ___ – x – dx 2 2 2 and then you do it.
Partial Fractions 397. We have,
dx Ú (3x – 2) ÷x 2 + x + 1
( ) 3 4 = __ Ú ( 2x – __ ) ÷ x 2 3
_________
2 = 3 Ú x – __ ÷ x2 + x + 1 dx 3 _________ 2 + x + 1 dx
(2x + 1)dx
Ú ____________ (x + 2)(x + 3)
(2x + 1) A B fi ____________ = ______ + ______ (x + 2)(x + 3) (x + 2) (x + 3)
_________
1 = – __ Ú (1 – 2x) ÷1 + x – x2 dx 2
_______________ __
______ 10) ÷x 2 + x dx
1 = – __ Ú (– 2x) ÷1 + x – x2 dx 2 _________ 1 = – __ Ú ((– 2x) – 1) ÷1 + x – x2 dx 2
______
1 = __ Ú (2x) ÷1 + x – x2 dx 2
dx Ú (x – 5) ÷x 2 + x
dx Ú x ÷1 + x – x2 _________
393. We have,
_________
__________ 2 (x2 + 3x + 4)3/2 3 = _______________ – x + __ ÷x 2 + 3x + 4 3 2
2
and then you do it.
2
fi (2x + 3) dx = 2t dt
= 2 Ú t2 dt – 2 Ú
_________ 2
____________
= Ú (2x + 3 – 2) ÷x 2 + 3x + 4 dx
( ) 3 = 2 Ú ( (2x – 1) + __ ) ÷ x – x – 2 dx 2 3 = 2 Ú ( (2x – 1) + __ ) ÷ x – x – 2 dx 2 3 + 3 Ú (x ÷ – 1) – ( __ 2 ) dx 1 = 2 Ú 2x + __ ÷x 2 – x – 2 dx 2
_________ 2
+ 4 dx Ú (2x + 1) ÷x 2 + 3x
dx Ú (4x + 1) ÷x 2 – x – 2
_________
3 7 + __ log x + __ + ÷ x 2 + 3x + 4 + c 2 4
__________________ __ 2 ÷ 2
and then you do.
2 = __ (x2 + 3x + 4)3/2 3
( ( ) ) 3 7 4 = __ Ú ( ( 2x + __ ) – __ ) ÷ x + x + 1 dx 2 3 3 3 7 1 + __ Ú ( x + __ ) + ( ___ ) dx 2 ÷ 2 2 3 7 4 = __ Ú 2x + __ – __ ÷ x2 + x + 1 dx 2 3 3
3 t3 1 = 2 __ – 2 __ x + __ ÷x 2 + 3x + 4 3 2 2
| (
1.93
fi
(2x + 1) A(x + 2) + B(x + 3) ____________ = _________________ (x + 2)(x + 3) (x + 2) (x + 3)
fi
2x + 1 = A(x + 3) + B (x + 2)
When
x = – 2, A = – 4 + 1 = – 3
1.94 Integral Calculus, 3D Geometry & Vector Booster and
x = – 3, B = – 6 + 1 = – 5
Thus,
Ú ____________ (x + 2)(x + 3)
dx dx = – 3 Ú _____ – 5 Ú _____ x+2 x+3
= – 3 log |x + 2| – 5 log |x + 3| + c
dx dx 1 1 + __ Ú ____________ – __ ____________ 6 (x + 2) (x + 3) 6 Ú (x + 1) (x + 2)
(2x + 1)dx
(
)
1 1 = Ú _____ – _____ dx x – 2 x – 1
= log |x – 2| – log |x – 1| + c
x–2 = log _____ + c x–1
| |
dx Ú _________________ (x + 1)(x + 2)(x + 3) ((x + 3) – (x + 1)) 1 = __ Ú __________________ 2 (x + 1)(x + 2)(x + 3) dx 1 = __ Ú ____________ – 2 (x + 1)(x + 2)
dx 1 __ Ú ____________ 2 (x + 2)(x + 3)
)
(
)
| | | | | ( ) ( ) |
x+1 x+2 1 1 = __ log _____ – __ log _____ + c 2 x+2 2 x+3
) (
)|
400. We have, dx Ú _______________________ (x + 1)(x + 2)(x + 3)(x + 4)
((x + 4) – (x + 1)) dx 1 = __ Ú _______________________ 3 (x + 1)(x + 2)(x + 3)(x + 4)
dx 1 = __ Ú __________________ 3 (x + 1)(x + 2)(x + 3)
(x + 1) – 2 = Ú ____________ dx (x + 1) (x – 2)
dx 2 = Ú ______ – ____________ dx (x – 2) Ú (x + 1) (x – 2)
dx 1 1 = Ú ______ – 6 Ú _____ – _____ dx (x – 2) x–2 x+1
= log |x – 2| – 6 log |x – 2| + 6 log |x + 1| + c
= 6 log |x + 1| – 5 log |x – 2| + c
1 ((x + 4) – (x + 2)) dx = – __ Ú __________________ 6 (x + 2)(x + 3)(x + 4)
)
2x – 1
dx Ú __________________ (x – 1) (x + 2) (x – 3)
(2x – 1) Now __________________ (x – 1) (x + 2) (x – 3)
(2x – 1) = A (x + 2) (x – 3)
+ B (x – 1) (x – 3) + C (x – 1) (x + 2)
When
x = 1,
x = – 2, then B = –1/3
and
x = 3,
then A = – 1/6 then C = 1/2
dx dx dx 1 1 1 – __ Ú ______ + __ ______ + __ ______ 6 (x – 1) 3 Ú (x + 2) 2 Ú (x + 3) 1 1 1 = – __ log |x – 1| + __ log |x + 2| + __ log |x + 3| + c 3 2 6 403. We have,
)
x k 1 _________ n – 1 + 2(n – 1) Ú _________ 2 n –1 – _______ dx 2 2 (x + k) (x + k) (x + k)n dx 1 = __ Ú ____________ – 6 (x + 2) (x + 3)
(
Thus, the given integral reduces to
dx 1 – __ Ú __________________ 3 (x + 1)(x + 2)(x + 3)
(
x–1
C A B ______ + ______ + ______ (x – 1) (x + 2) (x – 3)
x+3 x + 1 _____ 1 = __ log _____ 2 x+2 x+2
402. We have,
1 1 1 1 1 1 = __ Ú _____ – _____ dx – __ Ú _____ – _____ dx 2 x+1 x+2 2 x+2 x+3
| (
dx Ú ____________ (x + 1) (x – 2)
399. We have,
(
)|
401. We have,
dx Ú ____________ (x – 1)(x – 2)
) (
x + 2 _____ x+2 1 + __ log _____ + c x+3 x+1 6
398. We have,
| (
x + 2 _____ x+4 1 = __ log _____ x+3 x+3 6
1 dx __ Ú ____________ 6 (x + 3)(x + 4)
x3
dx Ú ____________ (x – 1) (x – 2) x3 = Ú __________ 2 dx x – 3x + 2
( Ú (
)
Indefinite Integrals
| | | |
7x – 6 = Ú x – 3 + __________ dx x2 – 3x + 2
t+2 = log _____ + c t+3
7x – 6 = x – 3 + ____________ dx (x – 1)(x – 2)
) )
sinq + 2 = log ________ + c sinq + 3
(
2
406. We have,
7x – 6 x = __ – 3x Ú ____________ dx 2 (x – 1) (x – 2)
1.95
(1 – cos x)
dx Ú _____________ cos x (1 + cos x)
7x – 6 A B Now, ____________ = ______ + ______ (x – 1) (x – 2) (x – 1) (x – 2)
(1 + cos x – 2 cos x) = Ú ________________ dx cos x (1 + cos x)
(7x – 6) = A (x – 2) + B (x – 1)
When
x = 1, then A = –1
dx dx = Ú ____ cos x – 2 Ú _________ (1 + cos x)
and
x = 2, then B = 8
1 – cos x = Ú sec x dx – 2 Ú ________ dx sin2x
= Ú sec x dx – 2 Ú (1 – cosec2 x – cosec x cost x) dx
= log |sec x + tan x| – 2(cosec x – cost x) + c
(
Thus, the given integral reduces to
(
) ( ( )
)
8 x2 –1 + ______ dx __ – 3x + Ú ______ 2 (x – 1) (x – 2) x2 = __ – 3x – log |(x – 1)| + 8 log |x – 2| + c 2
2x
2 dx Ú ______________ (x + 1) (x2 + 2)
Let x2 = t fi 2x dx = dt Thus, the given integral reduces to
dt 2 Ú ___________ (t + 1) (t + 2)
(
407. We have,
404. We have,
)
dx
Ú ___________ sin x – sin 2x
dx = Ú _______________ sin x – 2sin x cos x
dx = Ú ______________ sin x (1 – 2cos x)
sin x dx = Ú ______________ 2 sin x (1 – 2cos x)
sin x dx = Ú ___________________ (1 – cos2x) (1 – 2cos x)
1 1 – ____ dt = 2 Ú ____ t+1 t+2
)
(t + 1) = 2 log ______ + c (t + 2)
dt = Ú _____________ , where cos x = t (1 – t2) (1 – 2t)
dt = – Ú __________________ (t – 1) (t + 1) (2t – 1)
(x2 + 1) = 2 log _______ + c (x2 + 2)
1 Now, __________________ (t – 1) (t + 1) (2t – 1)
405. We have,
| |
| |
c A B ______ + ______ + _______ (t – 1) (t + 1) (2t – 1)
cosq dq Ú __________________ (2 + sinq ) (3 + sinq)
– 1 = A (t + 1) (2t – 1) + B (t – 1) (2t – 1) + c (t 2 – 1)
Let sinq = t
fi cosq dq = dt
Put
t =1,
t = –1, then B = 1/10
t =1/2, then C = 4/3
Thus, the given integral reduces to
dt Ú ___________ (t + 2) (t + 3)
(
)
1 1 = Ú _____ – _____ dt t+2 t+3
then A = –1/2
The given integral reduces to
dt 1 – __ Ú ______ + 2 (t – 1)
dt 1 ___ Ú ____ + 10 t + 1
dt 4 __ Ú _______ 3 (2t – 1)
1.96 Integral Calculus, 3D Geometry & Vector Booster 1 1 = – __ log |t – 1| + ___ log |t + 1| 2 10 4 + __ log |2t – 1| + c 3 1 1 = – __ log |cos x – 1| + ___ log |cos x + 1| 2 10
When Put and
x = – 2, A = – 3 x = 3, C = 7/5 x = 0, 9A – 6B + 2C = 1
fi
6B = 14/5 – 27 – 1
fi
B = – 21/5
4 + __ log |2cos x – 1| + c 3
2x + 1 Thus, _____________ dx (x + 2)(x – 3)2
408. We have,
dx = – 3 Ú _____ – x+2
dx
Ú _____________ (x + 1) (x + 1)2
dx dx 7 21 ___ Ú _____ + __ _______ 5 x – 3 5 Ú (x – 3)2
7 21 = – 3 log |x + 2| – ___ log |x – 3| – _______ + c 5 5 (x – 3) C 1 A B Now, ____________ = ______ + ______ + _______ (x + 1) (x – 1)2 (x + 1) (x – 1) (x – 1)2 415. We have, (2x + 3) dx 1 Ú (x_____________ _____________ fi 2 + 1) (x2 + 4) (x + 1) (x – 1) (2x + 3) Bx + C A Now, _____________ = ______ + _______ A (x – 1)2 + B (x2 – 1) + C (x – 1) 2 2 (x + 1) = _____________________________ (x + 1)(x + 4) (x + 4) (x + 1) (x – 1)2 fi 2x + 3 = A(x2 + 4) + Bx (x + 1) + C (x + 1) fi 1 = A (x – 1)2 + B (x2 – 1) + C (x + 1) When x = –1, A = 1/5 When x = –1, A = 1/4 and x = 0, 4 A + C = 3 x = 1, C = 1/2 fi C = 3 – 4 A = 3 – 4/5 = 11/5 and x = 0, A – B + C = 1 Also x = 1, 5A + 2B + 2 C = 5 fi B = 3/4 – 1 = –1/4 fi 2B = 5 – 5A – 2C = 5 – 1 – 22/5 = – 2/5 dx Thus, 2 Ú _____________ fi B = – 1/5 (x + 1) (x + 1) dx dx dx 1 1 1 = __ Ú ______ – __ ______ + __ _______ 4 (x + 1) 4 Ú (x – 1) 2 Ú (x – 1)2 1 1 1 = __ log |x + 1| – __ log |x – 1| – _______ + c 4 4 2 (x – 1) 409. We have,
2x + 1 dx Ú _____________ (x + 2)(x – 3)2
2x + 1 _____________ (x + 2)(x – 3)2
C A B = ______ + ______ + _______ (x + 2) (x – 3) (x – 3)2
2x + 1 _____________ (x + 2) (x – 3)2 fi
(2x + 3) dx Thus, Ú _____________ (x + 1)(x2 + 4) dx xdx dx 1 1 11 = __ Ú _____ – __ _____ + ___ Ú _____ 5 x + 1 5 Ú x2 + 4 5 x2 + 4 1 1 1 = __ log |x + 1| – ___ |x2 + 1| + __ tan–1 (x) + c 10 5 5 416. We have,
(3x – 2) dx
Ú _____________ (x – 1)(x2 + 9)
(3x – 2) Bx + C A Now, ____________ = ______ + _______ (x – 1)(x2 + 9) (x – 1) (x2 + 9) fi
(3x – 2) = A (x2 + 9) + Bx (x – 1) + C (x – 1)
A (x – 3)2 + B (x2 – 9) + C (x + 2) = _____________________________ (x + 2) (x – 3)2
When
x = 1, A = 1/10
and
x = 0, 9 A – C = – 2
2x + 1 = A (x – 3)2 + B (x2 – 9) + C (x + 2)
fi
C = 2 + 9 A = 2 +9/10 = 29/10
Indefinite Integrals
Also
x = –1, 10A + 2B – 2 C = – 5
421. We have,
fi
2 B = 2 C – 10A – 5 = 29/5 –1 – 5
fi
2B = –1/5
x dx Ú _____________ (x2 – 1)(x2 + 1)
fi
B = –1/10
Put
x2 = t
Thus,
(3x – 2) dx Ú _____________ (x – 1)(x2 + 9)
2
t A Now, ___________ = _______________ (t – 1)(t + 1) (t – 1) + B (t + 1) A(t + 1) + B (t – 1) t fi ___________ = _________________ (t – 1)(t + 1) (t – 1)(t + 1)
29 dx xdx dx 1 1 = ___ Ú _____ – ___ _____ + ___ _____ 10 x – 1 10 Ú x2 + 9 10 Ú x2 + 9 fi 1 1 Put = ___ log |x – 1| – ___ log |x2 + 9| 10 20 Put
( )
29 x + ___ tan–1 __ + c 30 3 417. We have, 2x – 1 dx Ú _____________ (x + 1)(x2 + 2)
Bx + C 2x – 1 A = ______ + _______ Now, _____________ (x + 1) (x2 + 2) (x + 1) (x2 + 2)
(2x – 1) = A (x2 + 2) + Bx (x + 1) + C (x + 1)
When
x = –1, then A = –1
x = 0, then C = 1 and x = 1, then B = 1 Thus, the given integral reduces to
t = A (t + 1) + B (t – 1) t = 1, A = 1/2 t = –1, A = –1/2 2
Thus,
x dx Ú _____________ 2 (x – 1)(x2 + 1)
dx 1 1 = __ Ú _____ – __ _____ 2 x2 – 1 2 Ú x2 + 1
x–1 1 1 = __ log _____ – __ tan–1 (x) + c 4 x+1 2
| |
426. We have,
(x2 + 3)(x2 + 1)
dx Ú ______________ (x2 – 1)(x2 + 2)
x2 = t
Put
(t + 3)(t + 1) __________ t2 + 4t + c Thus, ___________ = 2 (t – 1)(t + 2) t +t–2
dx dx x – Ú ______ + _______ dx + Ú _______ (x + 1) Ú (x2 + 2) (x2 + 2)
( )
x 1 1 __ tan–1 ___ __ + c = – log |x + 1| + __ log |x2 + 2| + ___ 2 2 ÷2 ÷ x _____________ dx 418. Ú (x + 1)(x2 + 4)
1.97
(
) (
)
– 3t + 5 – 3t + 5 = 1 + ________ = 1 + ___________ 2 (t + 1)(t – 2) t +t–2 – 3t + 5 A B Now, ___________ = ____ + ____ (t + 1)(t – 2) t – 1 t – 2 fi
(– 3t + 5) = A(t – 2) + B(t – 1)
Bx + C x A Now, Ú ____________ = ______ + _______ (x + 1)(x2 + 4) (x + 1) (x2 + 4)
When
t = 1, then A = –2
and
t = 2, then B = –1
x = A(x2 + 4) + Bx (x + 1) + C(x + 1)
Therefore, the given integral reduces to
Let and Thus, the
dx dx – Ú _____ Ú dx – 2 Ú x_____ 2 –1 x2 – 2
x – ÷2 x–1 1 __ = x – log _____ – ____ __ log ______ + c x+1 2÷2 x + ÷2
x = –1, then A = –1/5 x = 0, then C = 4/5 x = 1, then B = 1/5 given integral reduces to
427. We have,
dx dx x 1 1 4 – __ Ú ______ + __ _____ + __ _____ 5 (x + 1) 5 Ú x2 + 4 5 Ú x2 + 4
( )
x 1 1 2 = – __ log |(x + 1)| + __ log |x2 + 4| + __ tan–1 __ + c 2 5 5 5 419. Do yourself 420. Do yourself
| |
| |
(x2 + 1)(x2 + 2)
dx Ú ______________ (x2 + 3)(x2 + 4)
x2 = t (t + 1)(t + 2) Now, ___________ = (t + 3)(t + 4) Put
t2 + 3t + 2 __________ t2 + 7t + 12
__
1.98 Integral Calculus, 3D Geometry & Vector Booster
(
) (
)
(
)
4t + 10 4t + 10 = 1 – __________ = 1 – ___________ 2 (t + 3)(t + 4) t + 7t + 12
Thus,
6 2 2 – ______ dt Ú 1 + ______ (t + 2) (t2 + 4)
4t + 10 A B Again, ___________ = ______ + ______ (t + 3)(t + 4) (t + 3) (t + 4)
__ t t = t + ÷ 2 tan–1 ___ __ – 3 tan–1 __ + c 2 2 ÷
(4t + 10) = A (t + 4) + B (t + 3)
When
t = – 4, then B = 6
and
t = – 3, then A = – 2
Hints
and
2
( )
( )
1 – __ (sec x ◊ tan x + log |sec x + tan x|) + c 2
2I = sec3 x tan x 1 – __ (sec x tan x + log |sec x + tan x|) + c 2 1 I = __ (sec3 x tan x) 2 1 – __ (sec x tan x + log |sec x + tan x|) + c 4
( ) )
solutions
fi
Ú tan x ◊ sec x dx 4
( )
__ x2 x2 = (x2 + ÷2 tan–1 ___ __ – 3 tan–1 __ + c 2 2 ÷ 428. Do yourself.
(
Let tan x = t fi sec2 x dx = dt = Ú t4dt
fi
t5 = __ + c 5 5 (tan x) = ______ + c 5
2. Let I = Ú sec x dx 3
4. Ú tan2 x sec4 x dx = Ú tan4 x (1 + tan2 x) sec2 x dx
Let tan x = t
fi sec2xdx = dt
= Ú t2 (1 + t2) dt
= Ú (t4 + t2) dt
= sec x ◊ tan x – Ú (sec x ◊ tan x) dx
t 5 t3 + c = __ + __ 3 5
= sec x ◊ tan x – Ú (sec x (sec2 x – 1)) dx
tan5 x _____ tan3 x + + c = _____ 3 5
= sec x ◊ tan x – Ú sec3x dx + Ú sec x dx
= sec x ◊ tan x – I + log |sec x + tan x| + c
= Ú tan3 x (1 + tan2 x)2 sec2 x dx
fi 2I = sec x ◊ tan x + log |sec x + tan x| + c 1 fi __ [sec x ◊ tan x + log |sec x + tan x|] + c 2
= Ú t3 (1 + t2) 2 dt,
= Ú sec2 x ◊ sec x dx = sec x Ú sec x dx – Ú (sec x ◊ tan x) tan x dx 2
2
3. Let I = Ú sec5 x dx
( (
)
)
5. Ú tan3 x sec6 x dx
Let tan x = t fi sec2x dx = dt
= Ú t3 (t4 + 2t2 + 1) dt
= Ú sec3 x ◊ sec2 x dx
= Ú (t7 + 2t5 + t3) dt
= sec3 x Ú sec2 x dx – Ú (3 sec3 x ◊ tan x) ◊ tan x dx
t8 t3 __ t4 = __ + __ + + c 8 3 4
= sec x ◊ tan x – 3 Ú sec x (sec x – 1) dx
= sec3 x ◊ tan x – 3 Ú sec5 x dx + 3 Ú sec3 x dx
tan8 x _____ tan3 x _____ tan4 x = _____ + + + c 8 3 4
= sec3 x ◊ tan x – I
3
3
2
( (
6. Ú tan3 x sec5 x dx
)
)
Indefinite Integrals
1.99
= Ú sec4 x ◊ tan2 x sec x ◊ tan x dx
– Ú (– cot x) (– cosec x ◊ cot x) dx
= Ú sec4 x (sec2 x – 1) sec x ◊ tan x dx
= – cosec x ◊ cot x – Ú (cosec2 x – 1) cosec x dx
= – cosec x ◊ cot x – I – Ú cosec3 x dx + Ú cose x dx
= – cosec x ◊ cot x – I + log |cosec x – cot x| + c
Let sec x = t
fi secx ◊ tan xdx = dt
= Ú t4 (t2 – 1) dt,
= Ú (t6 – t4) dt
fi
2I = – cosec x cot x + log |cosec x – cot x| + c
t7 t5 = __ – __ + c 7 5
fi
1 1 I = – __ cosec x cot x + __ log |cosec x – cot x| + c 2 2
sec7 x _____ sec5 x = _____ – + c 7 5
( (
7. Ú sec x dx
8. Ú sec9 x dx
9. We have,
)
12. Ú cot2 x cosec4 x dx
)
4
Ú cosec2 x ◊ cot2 x dx
Let cot x = t 2
fi cosec x dx = – dt
= – Ú t dx 3
t = – __ + c 3 3 (cot x) = – ______ + c 3 10. Let I = Ú cot3 x cosec3 x dx Let cosec x = t fi – cosec x ◊ cot x dx = dt fi cosec x ◊ cot x dx = – dt
2
= Ú cot3 x ◊ cosec3 x dx
= Ú cot2 x (1 + cot2 x) cosec2 x dx = – Ú t2 (1 + t2) dt, = – Ú (t4 + t2) dt
(
)
t5 t3 = – __ + __ + c 3 5 x x 5 __ = – cot + cot3 __ + c 3 5
(
)
13. Ú tan– 5 x ◊ sec6 x dx
sec6 x = Ú _____ dx tan5 x
(1 + tan2 x)2 sec2 x = Ú _______________ dx tan5 x
(1 + t2)2 = Ú _______ dt t5
t4 + 2t2 + 1 = Ú __________ dt t5
( (
)
= Ú cot2 x ◊ cot x ◊ cosec3 x dx
= Ú (cosec x – 1) ◊ cosec x (cosec x ◊ cot x) dx
= – Ú (t2 – 1) t2 dt
= Ú (t4 – t2) dt
t5 t3 = __ – __ + c 5 3
14. Ú cot3 x ◊ cosec– 8 x dx
cosec5 x _______ cosec3 x = _______ – + c 3 5
2
( (
)
11. Let I = Ú cosec3 x dx
= Ú cosec2 x ◊ cosec x dx = cosec x Ú cosec2 x dx
2
)
Let cot x = t fi cosec2x dx = dt
( ( (
)
)
1 2 __ 1 = Ú __ + __ 3 + 5 dt t t t 1 1 = log |t| – __ 2 – ___ 4 + c t 4t 1 1 = log |tan x| – _____ 2 – ______ 4 + c tan x 4 tan x
)
cot3 x = Ú _______ dx cosec8 x
cos3 x = Ú _____ dx sin5 x
(1 – sin2 x) cos x = Ú _____________ dx sin5 x
)
1.100 Integral Calculus, 3D Geometry & Vector Booster
( )
1 – t2 = Ú _____ dt, Let t = cos x t5 dt = – sin xdx
( ( (
)
1 1 = Ú __ 5 – __ 3 dt t t 1 1 = ___ 2 – ___ 4 + c 2t 4t 1 1 = ______ 2 – ______ + c 2 sin x 4 sin4 x
| |
t 1 = __ log ____ + c 4 t+1
x4 1 = __ log _____ 4 + c 4 x +1
| | | |
x3 1 dx 20. Ú ________ = __ log _____ 3 + c x +1 x (x4 + 1) 3
)
dx dx 21. Ú __________ = Ú ___________ 1 6/7 8 x2 (x7 + 1)6/7 x 1 + __ 7 x
( )
)
15. Ú cosec5 x dx
16. Ú cosec7 x dx
17. We have,
dx dx = Ú _________ Ú ________ 7 8 – 7
x (x + 1)
x (1 + x )
Put
1 + x– 7 = t x– 8
fi
dx 1 fi ___8 = – __ dt 7 x
– 7 dx = dt
1 dt = – __ Ú __ 7 t 1 = – __ log |t| + c 7 1 __ = – log |1 + x– 7| + c 7
t6 dt = – Ú ____ t6 = – t + c
(
dx dx = Ú ___________ Ú x__________ 2 3/4 5 (1 + x) x (1 + x– 4)3/4
Let 1 + x– 4 = t4
fi – 4x– 5 dx = 4t3 dt dx fi ___5 = – t3 dt x
dt = – Ú t3 __3 t = – Ú dt
)
(
)
1 1/3 dx dx 1 + __ 22. Ú ___________ = Ú ___ 3 3 3 1/3 4 x x (1 + x ) x
(
t2 dt = – Ú ____ t
= – Ú t dt t2 = – __ + c 2 1 1 2/3 = – __ 1 + __ 3 + c 2 x
(
)
23. We have,
= – t + c = – (1 + x– 4)1/4 + c
dt 1 = __ Ú ______ , Let x4 = t 4 t (t – 1) 4x3dx = dt
(
)
1 Put 1 + __ 3 = t3 x 3 fi – __4 dx = 3t2 dt x dx fi ___4 = – t2 dt x
dx x4 _________ 19. Ú ________ = dx Ú x (x5 + 1) x5 (x5 + 1)
)
1 1/7 = – 1 + __ 7 + c x
18. We have,
(
1 Let 1 + __ 7 = t7 x 7 fi – __6 dx = 7t6 dt x dx ___ fi 6 = – t6 dt x
)
1 1 ____ 1 = __ Ú __ – dt 4 t t+1
2
x dx Ú _______ (x + 3)2
Let x + 3 = t fi dx = dt (t – 3)2 = Ú ______ dt t2 t2 – 6t + 9 = Ú _________ dt t2 6 9 = Ú 1 – __ + __ 2 dt t t
(
)
1.101
Indefinite Integrals
(
)
9 = t – 6 log |t| – __ + c t
9 = (x + 3) – 6 log |x + 3| – ______ + c (x + 3)
(
)
24. We have,
3
x dx Ú ________ (2x + 3)2
Let 2x + 3 = t
fi
2 dx = dt
fi
dx = 1/2 dt
fi
( )
3–t x = ____ 2
(
2
3
)
27 1 27 ___ = ___ Ú ___ – + 9 – t dt t 16 t2
t2 27 1 = ___ – ___ – 27 log |t| + 9t – __ + c t 2 16
( Ú (
)
)
4 1 (t + 2) = ____ Ú ______ dt 243 t3
t4 + 8t2 + 24t2 + 32t + 15 1 = ____ Ú ______________________ dt 243 t3
x2 25. Ú _______ dx (x + 2)3
Let (x + 2) = t
=
=
=
fi dx = dt (t______ – 2)2 Ú t3 dt t2 – 4t + 4 dt Ú _________ t3 Ú __ 1t – __ t42 + __ t43 dt 4 2 log |t| + __ – __ + c t t2 4 2 log |(x – 2)| + ______ – _______ + c (x + 2) (x + 2)2
(
26. Ú x2 (ax + b)2 dx
)
)
(
)
(
)
15 1 24 32 ___ = ____ Ú t + 8 + ___ + ___ 2 + 3 dt t 243 t t 2 32 15 1 t = ____ __ + 8t + 24 log |t| – ___ – ___2 + c t 243 2 2t
(
)
where t = (3x – 2) 29. We have,
dx dx = Ú __________ Ú __________ +2 3 5 3x ______ x3 (3x + 2)3
(
)
x x
Let ax + b = t 1 fi dx = __ a dt b fi x = t – __ a
Let 3x – 2 = t dt fi dx = __ 3 t____ +2 and x = 3
where t = 2x + 3
(
dt 1 = __ Ú ________ where (x3 + 1) = t 3 t2 (t3 – 1)
=
)
x4 28. Ú ________ dx (3x – 2)3
1 (27 – 27t + 9t – t ) = ___ Ú _________________ dt 16 t2
(
b2 1 = __ 2 (ax + b) – 2 log |(ax + b)| – _______ + c (ax + b) a
2x = 3 – t
=
)
and then use partial fractions.
3 1 (3 – t) = ___ Ú ______ 16 t2 dt
(
dx x2 27. Ú _________ 3 2 = Ú _________ dx x (1 + x ) x3 (1 + x3)2
3–t 3 ____ 2 1 = __ Ú ______ dt 2 t2
2 1 (t – b) = __ 2 Ú ______ dt a t2 2 2 1 (t – 2bt + b = __ 2 Ú ____________ dt a t2 1 = __ 2 Ú (1 – 2b/t + b2/t2) dt a b2 1 = __ 2 t – 2b log |t| – __ + c t a
( (
)
2 Let 3x + __ x = t 2 fi 3 + __ x = t 2 fi – __2 dx = dt x dx dt fi ___2 = – __ 2 x
)
2 Also, __ x = t – 3 t–3 1 ____ fi __ x = 2
1.102 Integral Calculus, 3D Geometry & Vector Booster
3 1 (t – 3) = ___ Ú ______ dt 16 t3
3 2 1 (t – 3t + 3t – 27) = – ___ Ú _________________ dt 16 t3
3 3 ___ 27 1 = – ___ Ú 1 – __ + __ 2 – 3 dt t 16 t t
3 27 1 = – ___ t – 3 log |t| – __ + ___ 2 + c t 16 2t
( (
(
dt 1 = – __ Ú ________ 2 ____ 2 3 3 t t –1
( )
3 1 (t – 1) = – ___ Ú ______ dt 16 t3
)
)
)
3x + 2 where, ______ x = t
(
(
)
x x
( ) b fi ( a + __ x ) = t
b + ax Let ______ x = t
b fi – __2 dx = dt x dx dt fi – ___2 = __ b x –a 1 t____ Also, __ x = b
3 1 (t – a) = – __4 Ú ______ dt b t3
( (
)
3a 3a2 __ a3 1 = – __4 Ú 1 – ___ + ___ 2 – 3 dt t b t t 3a2 1 = – __4 t – 3a log |t| – ___ + t b b where a + __ x = t
)
a3 ___ 2 + c 2t
)
dx dx 31. Ú _________ = Ú _________ x+2 3 5 _____ x2 (x + 2)3 x x
( )
)
)
dx dx 32. Ú __________ 2 = Ú _________ 3 a + bx 2 5 ______ x (a + bx) x x
(
)
(
)
a + bx Let ______ x = t
fi
a – __2 dx = dt x
dx fi ___2 = – __ a x a Also, x = ____ t–b
dt 1 ________ = – __ a Ú ____ a 3 2 t t–b 3 1 (t – b) = – __4 Ú ______ dt a t2
( Ú (
)
1 t3 – 3t2b + 3t b2 – b2 = – __4 Ú _________________ dt a t2
)
3b2 __ b3 1 = – __4 t – 3b + ___ – 2 dt t a t
(
)
b2 1 t2 = – __4 __ – 3bt + 3b2 log |t| + __ + c t a 2
(
( )
x+2 Put _____ x = t
fi
)
1 __ 3 dt t
)
a + ax where ______ x = t
3 __ 2 – t
( )
3 2 2 3 1 (t – 3t a + 3ta – a ) = – __4 Ú ___________________ dt b t3
( (
3 1 = – ___ Ú 1 – __ + t 16
2 where t = x + __ x
dx dx = Ú __________ Ú __________ b + ax 2 5 ______ x3 (b + ax)2
(
)
3 1 1 = – ___ t – 3 log |t| – __ + ___ 2 + c t 16 2t
30. We have,
(
1 t3 – 3t2 + 3t – 1 = – ___ Ú ______________ dt 16 t3
2 – __2 dx = dt x
dx dt fi ___2 = – __ 2 x 2 Also, 1 + __ x = t 2 fi x = ____ t –1
dx 33. Ú _________ 2 x (1 + x2)3
(
)
dx dx ______ a – bx 2 ___ 34. Ú _________ = Ú x x2 (a – bx)2 x2
(
)
a – bx Let ______ = t x
a fi __2 dx = – dt x
Indefinite Integrals
dx dt fi ___2 = – __ a x a Also, __ x = t + b a fi x = ____ t+b
dt 1 ________ = – __ a Ú ____ a 2 2 t t+b
( )
)
)
)
a – bx where t = ______ x dx dx 35. Ú __________ 2 = Ú __________ 4 2x + 1 3 x (2x + 1) x7 ______ x 2x + 1 Let ______ = t x
(
)
(
)
dx fi ___2 = – dt x 1 Also, 2 + __ x = t 1 fi x = ____ t–2
( )
(
)
80 ___ 2 + t
( )
dt = – Ú __3 × (t – 1)5 t
(t – 1)5 = – Ú ______ dt t3
(t5 – 5t4 + 10t3 – 10t2 + 5t – t5) = – Ú ___________________________ dt t3 10 5 __ 1 = – Ú t2 – 5t + 10 – ___ + __ 2 – 3 dt t t t
(
)
80 ___ – 40 + 10t – t2 dt t
(
( )
)
dx dx _____________ = Ú _______________ Ú _______________ 4 3 5 (x – 1)3/4 (x + 2)5/4 ÷(x – 1) (x + 2)
dx = Ú ______________ 3/4 x – 1 _____ (x + 2)2 x+2
( )
x–1 Let _____ = t x+2
dx fi _______ = dt (x + 2)2
16 ___ 80 t3 = – ___ + + 80 log |t| – 40t + 5t2 – __ + c 2 t 3 t 2x + 1 where t = ______ x
1 dt Thus, __ Ú ___ 3 t3/4
36. We have,
)
5 t3 5t2 1 = – Ú __ – ___ + 10t – 10 log |t| – __ + ___ 2 + c t 3 2 2t x–1 where, t = _____ x–2
(t – 2)5 dt = – Ú ________ t3 (2 – t)5 dt = Ú ________ t3 32 – 80t + 80t2 – 40t3 + 10t4 – t5 = Ú ____________________________ dt t3
(
dx = Ú _____________ x–1 2 _____ (x – 2)7 x–2
37. We have,
dt = – Ú ______ 5 t3 1 ____ t–2
32 = Ú ___ 3 – t
b2 1 = – __3 t + 2b log |t| – __ + c t a
(
(x – 2) ◊ 1 (x – 1) ◊ 1 fi _______________ dx = dt (x – 2)2 – dx fi _______ = dt (x – 2)2
2t – 1 Also, x = _____ t–1 1 fi (x – 2) = ____ t–1 dx Thus, Ú _____________ (x – 1)3 (x – 2)4
2b b2 1 = – __3 Ú 1 + ___ + __ 2 dt t a t
(
( )
x–1 Let _____ = t x–2
2 1 (t + b) dt = – __3 Ú ________ a t2 2 2 1 (t + 2bt + b ) = – __3 Ú ____________ dt a t2
(
1.103
dx Ú _____________ (x – 1)3 (x – 2)4
1 = __ Ú t– 3/4 dt 3 1 __ = (4t1/4) + c 3
( )
4 x – 1 1/4 = __ _____ + c 3 x+2
1.104 Integral Calculus, 3D Geometry & Vector Booster dx 38. Ú _____________ (x – 1)3 (x – 2)2 dx = Ú _____________ x+2 2 (x – 1)5 _____ x–1
dx dx ______________ 42. Ú ________________ = Ú _______________ 4/5 5 4 6 (x + 1) (x + 3)6/5 + 1) (x + 3) ÷ (x
( ) x+2 Put ( _____ = t x – 1)
(x – 1) ◊ 1 – (x + 2) ◊ 1 fi __________________ dx = dt (x – 1)2
fi
dx fi _______ = (x – 1)2
x+2 Also, _____ = t x–1
fi
fi
fi
3 – _______ dx = dt (x – 1)2
(
dt ___ – 3
)
x + 2 = tx – t t+2 x = _____ t–1 3 t+2 x – 1 = _____ – 1 = ______ t–1 (t – 1)
dt 1 = – __ Ú ________ 3 ____ 3 3 2 t t–1
( )
3 1 (t – 1) dt = – __4 Ú ________ 3 t2 3 2 1 (t – 3t + 3t – 1) dt = – __4 Ú ________________ 3 t2
( (
)
3 1 1 = – __4 Ú – 3 + __ – __ dt t t2 3 1 t2 1 = – __4 __ – 3t + 3 log |t| + __ + c t 2 3 x + 2 where t = _____ x–1
)
dx dx 39. Ú _____________ 5 = Ú _____________ 4 x–3 4 (x – 3) (x – 2) (x – 2)9 _____ x–2
( )
dx = Ú ______________ x + 1 4/5 2 _____ (x + 3) x+3
( )
x–3 Put _____ = t and then you do it x–2
( )
(
(
)
)
x+1 Put _____ = t5 and then you do it. x+3 43. We have,
2
dx x dx 3 = Ú __________ Ú _________ 3 x (2 + 3x ) x (2 + 3x3)
3x2 dx 1 = __ Ú __________ 3 x3 (2 + 3x3)
dt – __2 t 1 = __ Ú ________ 3 __ 3 1 2 + __ t t
dt 1 = – __ Ú _____ 3 2t + 3
1 = – __ Ú log |2t + 3| + c 6
1 2 = – __ log __ 3 + 3 + c 6 x
(
| |
44. We have,
)
4
5x dx dx 1 = __ Ú __________ Ú _________ 5 5 5 x (3 + 5x ) x (3 + 5x5)
1 Let x5 = __ t
dt fi 5x4 dx = – __2 t
dt 1 = – __ Ú _______ 5 (3t + 5) 1 = – ___ log |3t + 5| + c 15 3 1 = – ___ log __ 5 + 5 + c 15 x
| |
dx x dx 45. Ú _________ 2 = Ú __________ 2 x (2 + 3x ) x (2 + 3x2)
x – 3 3/2 dx 40. Ú _______________ = Ú dx _____ (x – 2)5 3/2 7/2 x–2 (x – 3) (x – 2)
Let
x2 = t
Put (x – 3/x – 2) = t2 and then you do it.
fi
1 x dx = __ dt 2
dx dx _____________ 41. Ú _______________ = Ú ______________ 5 7 x – 2 5/2 6 – 2) (x – 5) (x – 5) _____ ÷ (x x–6 x–2 Put _____ = t2 and then you do it. x–6
(
( )
)
dt 1 = __ Ú ________ 2 t (2 + 3t)
3 1 1 ______ = __ Ú __ – dt 4 t 2 + 3t
(
)
Indefinite Integrals
1 = __ (log |t| – log |2 + 3t|) + c 4 t 1 = __ log ______ + c 4 2 + 3t
( |
dx 4x3 dx 1 49. Ú _________ = __ Ú __________ 4 4 x (3x + 1) 4 x (3x4 + 1)
|)
x2 1 = __ log ______ + c 4 2 + 3x2
3x2 dx dx 1 46. Ú _________ 3 = __ Ú __________ x (3 + 4x ) 3 x3 (3 + 4x3) Let x3 = t
fi
2
3x dx = dt
dt 1 = __ Ú ________ 3 t (3 + 4t)
(
)
1 1 __ 4 = __ Ú __ – + 4t dt 9 t 3 1 = __ (log |t| –log |3 + 4t|) + c 9 t 1 = __ log ______ + c 9 3 + 4t
( |
( |
|)
|)
x3 1 = __ log _______ + c 9 3 + 4x3
dt 1 = __ Ú ________ 4 t (3t + 1)
3 1 1 _____ = __ Ú __ – dt 4 t 3t + 1
t 1 = __ log _____ + c 4 3t + 1
x4 1 = __ log ______ 4 + c 4 3x + 1
Let x3 = t
fi
Let
(x2 – 1) = t
fi
2x dx = dt
Also, x2 = t + 1
3x2 dx = dt
2 1 (t – 1) dt = __ Ú ______ 2 t3
1 t2 – 2t + 1 dt = __ Ú __________ 2 t3
1 1 __ 2 1 – + __ dt = __ Ú __ 2 t t2 t3
dt 1 = – __ Ú _______ 3 t (5t – 2)
5 1 1 __ = __ Ú __ – – 2 dt 6 t 5t
( |
|)
x3 1 = __ log ______ 3 + c 6 5x – 2
dx 4 3 dx 1 48. Ú _________ 4 = __ Ú _________ x (1 – 4x ) 4 x4 (1 – 4x4)
)
)
1 2 – = __ log |t| + __ t 2
(
)
1 ___ 2 + c 2t
)
(x2)4 4x dx
9
x 1 = __ Ú _________ Ú ___________ 2 5 (2x + 3) dx 4 (2x2 + 3)5
Let 2x2 + 3 = t
Thus,
fi 4x dx = dt
dt 1 = – __ Ú _______ 4 t (4t – 1)
1 1 _____ 4 = __ Ú __ – dt 4 t 4t – 1
t 1 = __ log _____ + c 4 4 –1
| |
( (
dt 1 = __ Ú _______ 4 t (1 – 4t)
)
(
1 2 1 2 = __ log |x2 – 1| + _______ – _________ + c 2 (x – 1) 2 (x2 – 1)2 51. We have,
(
| |
Ú x5 (x2 – 1)4 dx
dt 1 = __ Ú _______ 3 t (2 – 5t)
)
| |
50. We have,
(
(x2)2 ◊ 2x x5 1 _______ __ dx = dx Thus, Ú ________ Ú 2 (x2 – 1)4 (x2 – 1)4
3x2 dx dx 47. Ú _________ 3 = Ú _________ 3 x (2 – 5x ) x (2 – 5x3)
( ) 1 = __ ( log _____ 6 | 5t t– 2 | ) + c
x3 1 = __ log ______ 3 + c 4 4x – 1
|)
( |
| |
1.105
( )
t –3 4 ____ dt – 3)4 dt 2 1 1 (t________ ___ __ Ú ________ = Ú 4 64 t5 t5
4 3 2 1 (t – 12t + 54t – 108t + 81 = ___ Ú ________________________ dt 64 t5
(
1 1 = ___ Ú __ – 64 t
12 ___ 2 + t
54 ___ 3 – t
108 ____ 4 + t
)
81 ___ 5 dt t
1.106 Integral Calculus, 3D Geometry & Vector Booster
(
)
36 81 1 12 27 ___ = ___ log |t| + ___ – ___ 2 + 3 – ___ 4 + c t 64 t t 4t
where 2x2 + 3 = t x3 52. Ú ________ dx 2 (x + 1)4
Let (x2 + 1) = t fi 2x dx = dt
2 1 x (2x) dx = __ Ú ________ 2 (x2 + 1)4 1 (t – 1) = __ Ú ______ 2 t4 dt
(
)
1 1 1 = __ – ___2 + ___ 3 + c 2 2t 3t
1 __ 4 dt t
1 1 1 = __ – _________ + _________ + c 2 2 (x2 + 1)2 3 (x2 + 1)3
)
x5 53. Ú ________ dx 2 (x – 3)4
Let
(x2 – 3) = t
fi
2xdx = dt
2 2 1 (x ) (2x) dx = __ Ú __________ 2 (x2 – 3)4 2 1 (t + 3) dt = __ Ú ________ 2 t4
(
)
)
3 3 __ 2 – __ 3 + c t t 2 3 (x – 3) 3 3 1 = __ _______ – ________ 2 2 – ________ 2 3 + c 2 3 (x – 3) (x – 3) )
(
)
x7 54. Ú _________ dx 2 (3x – 2)4
2 3 1 (x ) (6x) dx = __ Ú __________ 6 (3x2 – 2)4
( )
1 1 6 ___ 12 8 = ____ Ú __ + __ 2 + 3 + __ 4 dt 162 t t t t
6 6 __ 1 2 = ____ log |t| – __ – __ – + c t t2 t3 162
(
(
) )
2 4 4 (x ) (6x) dx = __ Ú __________ 3 (3x2 – 2)5
( )
t+2 4 ____ dt 3 4 = __ Ú ________ 3 t5
Let 3x2 – 2 = t fi 6xdx = dt
+ 2)4 dt 4 (t________ = ____ Ú 243 t5
t4 + 8t3 + 24t2 + 32t + 16 4 = ____ Ú _______________________ dt 243 t5
8 24 ___ 32 16 4 = ____ t + __ 2 + ___ 3 + 4 + ___ 5 dt 243 Ú t t t t
( (
(
2
)
)
)
8 12 ___ 32 4 t 4 __ = ____ – __ – ___ 2 – 2 – __ 4 + c t 243 2 t 2t t
where t = 3x2 – 2 56. Do yorself. 57. We have,
9 6 1 = __ Ú t2 + __ 3 + __ 3 dt 2 t t 1 t3 = __ __ – 2 3
2 1 (t + 6t + 9) dt = __ Ú ____________ 2 t4
3 + 6t2 + 12t + 8) dt 1 (t__________________ = ____ Ú 162 t 4
)
(
8x6 55. Ú ________ dx (3x2 – 2)5
1 1 = __ Ú __ – 2 t3
(
3 1 (t + 2) dt = ____ Ú ________ 162 t4
where t = 3x2 – 2
(
t+2 3 _____ dt 3 1 = __ Ú _________ Let 3x2 – 2 = t 4 6 t fi 6xdx = dt
2 sin x + 3 dx Ú ___________ (3 sin x + 2)2
2 sec x tan x + 3 sec2 x = Ú __________________ dx (3 tan x + 2 sec x)2
Let 2 sec x + 3 tan x = t
fi (2 sec x tan x + 3 sec2 x) dx = dt
dt = Ú __2 t 1 = – __ + c t 1 = – ______________ + c (2 sec x + 3 tan x)
58. Do yourself 59. Do yourself
1.107
Indefinite Integrals
60. We have,
(
)
4 cos x + 3 dx Ú ___________ (3 cos x + 4)2
(
)
(4 cosec x cot x + 3 cosec2 x = Ú ______________________ dx (3 cot x + 4 cosec x)2
dx dt fi ___________ = _____ (2 + 3 cos x)2 5 sin x
3 – 2t Also, cos x = ______ 3t – 2
__ _____
Let 3 cot x + 4 cosec x = t fi – (3 cosec2 x + 4 cosec x cot x) dx = dt dt = – Ú __2 t 1 __ = + c t 1 = ________________ + c (3 cot x + 4 cosec x)
1 dt = __ Ú ____ 5 sin x
– 2) 1 (3t _____ = ____ __ _______ dt Ú 2 5 ÷5 ÷ t – 1
1 __ = __ ÷ 5 3÷t 2 – 1 – 2 log t + ÷t 2 – 1 + c 5
(
_____
)
fi 4ex + 6e– x = ex (9l + 9m) + e– x (– 6l + 6m)
fi
__ _____ 2 ÷________ 7 ÷ t – 1
cos x =
(3 – 4t)
(
dx 1 dt = – __ Ú ____ Ú ___________ 7 cos x (3 + 4 sin x)2 3 – 4t 1__ ______ = – ____ _____ dt Ú 7÷7 ÷t 2 – 1 _____ _____ 1__ = – ____ (3 log (t + ÷ t 2 – 1 ) – 4 ÷t 2 – 1 )+c 7÷7
4 + 3 sin x where, _________ x=t 3 + 4 sin 64. Do yorself. 65. Do yorself.
5 13 9ex + 6e– x = – ___ Ú dx + ___ Ú __________ dx 18 18 (9ex – 6e– x)
5 13 = – ___ x + ___ log |(9ex – 6e– x)| + c 18 18
70. Do yorself. 71. Do yorself. 72. We have,
(
)
)
4ex + 6e– x Thus, Ú _________ x dx 9e – 4e– x
Thus,
dx 66. Ú __________ = t (2 + 3 cos x)2
|)
68. Do yorself. 69. Let 4ex + 6e– x = l (9ex – 6– x) + m (9ex + 6e– x)
3t – 4 Also, sin x = ______ 3 – 4t
|
4 + 3 sin x Let _________ = t fi 9 (l + m ) = 4 and – 6 ( l – m ) = 6 3 + 4 sin x dx dt 5 13 fi ___________ = – ____ cos x fi l = – ___ , m = ___ 18 18 (3 + 4 sin x)2
_____
(
2 cos x + 3 where, t = _________ 2 + 3 cos x 67. Do yorself.
dx Ú ___________ (3 + 4 sin x)2
2 ÷5 ÷ t – 1 sin x = ________ (3t – 2)
fi
61. Do yorself. 62. Do yorself. 63. We have,
dx Let __________ = t (2 + 3 cos x)2
5 sin x fi ___________ dx = dt (2 + 3 cos x)2
dx ______________ Ú _______________ 5 3
÷(x + 2) (x + 1)
dx ______________ = Ú ________________ x+1 3 8 _____ (x + 2) x+2 dx = Ú ______________ x + 1 3/2 4 _____ (x + 2) x+2
÷
(
)
( ) x +1 Put ( _____ = t x + 2)
2
dx fi ______ = 2t dt (x + 2)2
1.108 Integral Calculus, 3D Geometry & Vector Booster
Also,
2t2 – 1 x = ______ 1 – t2
fi
2t2 – 1 1 x + 2 = ______ + 2 = _____ 2 1 – t2 1 –t
Thus,
dx Ú _______________ + 1 3/2 4 x _____
( )
(
)
8t8 – 12t6 + 6t4 – t2 = – 2 Ú _________________ dt (3t2 – 2) and then you do it
(x + 1) 74. Ú ______________ dx (x + 2) (x + 3)3/2
(x + 2) x+2
(1 – t2)4 = Ú _______ × 2 t dt t3
(x + 2) – 1 = Ú ______________ dx (x + 2) (x + 3)3/2
1 – 4t2 + 6t4 – 4t6 + t8) = 2 Ú ____________________ dt t2
dx dx = Ú ________ – Ú _____________ (x + 3)3/2 (x + 2) (x + 3)3/2
1 = 2 Ú __ 2 – 4 + 6t2 – 4t4 + t6 dt t
t7 1 4 = 2 – __ – 4t + 2t3 – ___ 5 + __ + c t 7 5t
(
)
(
(
÷
_____
)
)
x+1 where t = _____ x+2
(
)
dx x + 2 1/2 ___ 73. Ú ______ ◊ x 2x + 3
dx dx = Ú ________ 3/2 – Ú ____________ x + 2 _____ (x + 3) (x + 3)2 x+3
( ) x+2 Put ( _____ = t x + 3)
(
)
(x + 3) ◊1 – (x + 2) ◊ 1 fi __________________ ◊ dx = dt (x + 3)2
dx fi _______ = dt (x + 3)2
x+2 Put ______ = t2 2x + 3
dt 2 _____ = – ______ – Ú __ t x + 3 ÷
(2x + 3 – 2x – 4) fi _______________ dx = 2t dt (2x + 3)2
x+2 2 _____ = – ______ – log _____ + c x+3 x + 3 ÷
dx fi ________ = – 2t dt (2x + 3)2
dx 2–x 2 75. Ú _______ 2 3 _____ dx = 2 Ú ____________________ 5/3 2 + x (2 – x) (2 – x) (2 + x/2 – x)1/3
– dx fi ________ = – 2t dt (1 – 2t2)2
dx fi _________ = 2t dt (2x2 – 1)2
÷
_____
3t2 – 2 Also, x = ______2 1 – 2t
( )
(2 – x) ◊1 + (2 + x) ◊1 fi __________________ dx = dt (2 – x)2
dx dt fi _______ 2 = __ 4 (2 – x)
1 – 2t2 = Ú t ◊ ______ 2t (2t2 – 1)2 dt 2 3t – 2
t2 (1 – 2t2)3 = 2 Ú _________ dt 3t2 – 2)
t2 (1 – 6t2 + 12t4 – 8t6) = 2 Ú ___________________ dt (3t2 – 2)
4
6
4
(t – 6t + 12t – 8t ) = 2 Ú __________________ dt (3t2 – 2)
( )
( )
2
dx = 2 Ú _______________ 2 + x 1/3 (2 – x)2 _____ 2–x 2+x Put _____ = t 2–x
dx = 2t (2t2 – 1)2 dt
fi
| |
1 dt = __ Ú ___ 2 t1/3 1 = __ t2/3 + c 3
(
)
1 2 + x 2/3 = __ _____ + c 3 2–x
dx 76. Ú _______________ _____ _____ 3 x + 1 + ÷ x + 1 ÷
Let x + 1 = t6
fi
dx = 6t5 dt
1.109
Indefinite Integrals
6t5 dt = Ú _____ t3 + t2 3
dx 79. Ú ________ x1/2 + x1/3
Let x = t6 fi dx = 6t5 dt
t = 6 Ú ____ dt t+1
(t3 + 1) – 1 = 6 Ú __________ dt t+1
t5 = 6 Ú ______ dt 3 t + t2
(t3 + 1) dt = 6 Ú ______ dt – 6 Ú ____ t+1 t+1
t3 = 6 Ú _____ dt t + 1
(t3 + 1) – 1 = 6 Ú _________ dt t+1
1 = 6 Ú (t2 – t + 1) – ____ dt t+1
t3 t2 = 6 __ – __ + t – log |t + 1| + c 3 2
(t + 1) (t2 – t + 1) dt = 6 Ú ________________ dt – 6 Ú ____ t + 1 t+1 1 = 6 Ú ((t2 – t + 1) – ____ ) dt t+1
(
t3 = 6 __ – 3
)
t2 __ + t – log |t + 1| + c 2
_____
(
)
(
)
(
)
where t = 6÷ x + 1
where t = (x)1/6
dx 77. Ú _________________ _____ x + 1 – (x + 1)1/4 ÷
x ÷ __ 80. Ú _______ dx 4 3 ÷ x + 1
Put x = t4
fi dx = 4t3 dt
__
Let (x + 1) = t4
fi dx = 4t3 dt
2 × t3 = 4 Ú t______ dt 3 t + 1 t5 = 4 Ú _____ dt t3 + 1
4t3 dt = Ú _____ t2 – t
t2 dt = 4 Ú ____ t–1
(
t2 = 4 Ú t2 – _____ 3 dt t +1
t3 1 = 4 Ú __ – __ log |t3 + 1| + c 3 3
)
1 = 4 Ú t + 1 + ____ dt t–1
(
)
2
t = 4 __ + t + log |t – 1| + c 2
(x + 1)2 = 4 _______ + (x + 1) + log |x| + c 2
(
( Ú (
3
)
__
÷x 81. Ú ________ dx __ 3 __ x + ÷ x ÷
6
Let x = t fi dx = 6t5 dt 2
(
)
where t = x1/4
)
1 + x1/2 – x1/3 78. Ú ____________ dx 1 + x1/3
(
) )
t3 × t5 = 6 Ú ________ t3 + t2
t6 dt = 6 Ú ____ t+1
Let x = t6 fi dx = 6t5 dt
1+t –t 5 = 6 Ú _________ t dt 1 + t2
t8 – t7 + t5 = 6 __________ dt 1 + t2
t+1 = 6 Ú t – t – t + t – t – t – 1 + _____ dt t2 + 1
(t6 – 1) + 1 = 6 Ú _________ dt t+1
t7 t6 __ t5 t4 __ t3 t2 = 6 __ – __ – + __ – – __ – t 7 6 5 4 3 2 1 + __ log |t2 + 1| + tan– 1 (t) + c 2 __ where t = 6÷ x
1 = 6 Ú 1 + t + t2 + t3 + t4 + t5 + ____ dt t +1
t2 t3 __ t4 t5 __ t6 = 6 t + __ + __ + + __ + + log |t + 1| + c 2 3 4 5 6
( [
6
5
4
3
)
2
]
(
(
(
where t = x1/6
)
)
)
1.110 Integral Calculus, 3D Geometry & Vector Booster dx_____ 82. Ú ____________ (x – 1) ÷x + 3
x – 3 = t2
Let
fi dx = 2t dt Also, x – 1 = t2 – 3 – 1 = t2 – 4
= Ú _______ 2t dt (t2 – 4) t
dt = 2 Ú _______ (t2 – 22)
t–2 1 = __ log ____ + c 2 t+2
x + 3 – 2 ÷ 1 = __ log __________ _____ + c 2 ÷x + 3 + 2
|
_____
Let x + 2 = t2
fi dx = 2t dt
= 2 tan– 1 (t) + c
= 2 tan– 1 (÷x + 2 )+c
_____
t 2__ – 1 ___ = ___ tan __ + c 3 ÷3 ÷
x2 2 = ___ __ tan– 1 ___ __ + c 3 ÷3 ÷
Let x2 = t fi 2xdx = dt
( (
( (
)
)
)
)
x = 2 __ – x2 + log |x2 + 1| + c 2
t dt = 2 Ú ________ (t2 + 2) t = 2 Ú dt (t2 + 2)
Let (x – 2) = t2 fi 2xdx = dt
x2 = t
fi
2xdx = dt
( )
( )
t dt = Ú ______ t 2 t +5 _____ 2
( )
2x + 1 = t2 dx = t dt
t2 – 1 Also, x = ____ 2 t2 + 5 t2 – 1 fi x + 3 = ____ + 3 = _____ 2 2 dt = 2 Ú ______ 2 (t + 5) t 2__ – 1 ___ = ___ tan __ + c ÷5 ÷5
( )
( ÷
______
t2 = 2 __ – t + log |t + 1| + c 2 4
t2 dt = 2 Ú ____ t+1 2 (t – 1) + 1 = 2 Ú _________ dt t+1 1 = 2 Ú (t – 1) + ____ dt t +1
Let
dx______ 87. Ú _____________ (x + 3) ÷2x + 1
dx _____ 85. Ú _______ x ÷x – 2
__
÷x 84. Ú _____ dx x +1
dt = 2 Ú ______ (t2 + 3)
Let
t dt = 2 Ú _______ (t2 + 1) t dt = 2 Ú ______ (t2 + 1)
x–2 = ÷2 tan _____ + c 2 dx 86. Ú _________ __ (x + 3) ÷x
– 1
t dt = 2 Ú _______ 2 (t + 3) t
dx_____ 83. Ú ____________ (x + 3) ÷x + 2
______
__
| |
|
( ) ( ÷( ) )
__ 1 = ÷2 tan– 1 ___ __ + c 2 ÷
)
2x + 1 2 = ___ __ tan– 1 ______ + c 5 ÷5
dx 88. Ú ________ _____ x2 ÷x – 1
Let
x – 1 = t2
fi
dx = 2t dt
Also, x = t2 + 1
2t dt = Ú ________ (t2 + 1) 2t
dt = 2 Ú _______ 2 (t + 1)2
dt Now, Ú ______ t2 + 1
Indefinite Integrals
1.111
1 1 = _____ 2 Ú dt – Ú ________ × (– 2t) ◊ t dt t +1 (t2 + 1)2
x dx x dx 91. Ú __________________ _____ = Ú _________________ _____ (x2 + 2x + 2) ÷x + 1 ((x + 1)2 + 1 ÷x + 1
(t2 + 1) – 1 t = _____ 2 + 2 Ú _________ dt t +1 (t2 + 1)2
dt dt t = _____ 2 + 2 Ú _______ – 2 Ú _______ t +1 (t2 + 1) (t2 + 1)2
(t2 – 1)t dt = 2 Ú _________ (t4 + 1)t
t2 – 1 = 2 Ú ______ 4 dt t +1
1 – (1/t2) = 2 Ú _________ 2 dt t + (1/t)2
1–1 _____ t2 __________ dt = 2 Ú 1 2 t + __ – 2 t
__ 1 t + __ – ÷2 t 1 = ___ __ log ___________ __ + c 1 ÷2 t + __ + ÷ 2 t
dt t fi 2 Ú ________ = ______ + 2 tan– 1 t + c (t2 + 1)2 t2 + 1 _____
_____ x – 1 ÷ = ______ + 2 tan– 1 (÷x – 1 )+c x
dx 89. Ú _____________ _____ 2 (x – 4) ÷x + 1
Let (x + 1) = t2
fi dx = 2tdt
Let (x + 1) = t2 fi dx = 2tdt
( )
(
)
( ( ) )
| | ( (
t dt = 2 Ú ____________ ((x + 1)2 – 4) t
dt = 2 Ú __________ 2 (t + 2t – 3)
where t = ÷x + 1
_____
dt = 2 Ú ___________ (t + 3) (t – 1)
1 1 1 = __ Ú ____ – ____ dt 2 t–1 t+3
t– 1 1 = __ log ____ + c 2 t+3
x + 1 – 1 ÷ 1 _____ = __ log __________ + c 2 ÷x + 1 + 3
(
)
| |
dx 90. Ú _________ __ 2 (x + 1) ÷x
|
Let
x = t2
fi
dx = 2tdt
t dt = 2 Ú _______ 4 (t + 1) t
dt = 2 Ú ______ 3 (t + 1)
2 dt = Ú ______ (t4 + 1)
(t2 + 1) – (t2 – 1) = Ú _______________ dt (t4 + 1)
( Ú (
dx 92. Ú _________ __ 2 (x – 1) ÷x
|
_____
) )
Let x2 = t fi 2xdx = dt
t dt = 2 Ú _______ (t4 – 1)t dt = 2 Ú ______ 4 (t – 1) dt = 2 Ú ____________ (t2 – 1) (t2 + 1) 1 1 = _____ 2 – _____ dt t – 1 t2 + 1
(
| | | |
)
t–1 1 = __ log ____ – tan– 1 (t) + c 2 t+1
__ x – 1 ÷ 1 = __ log ______ – tan– 1 (÷x ) + c __ 2 ÷x + 1
__
x dx 93. Ú ________________ _____ 2 (x – 2x + 2) ÷x – 1
) )
2 (t2 + 1) (t – 1) ______ = ______ – dt 4 4 (t + 1) (t + 1) and then you do it.
x dx = Ú __________________ _____ 2 ( (x – 1) + 1 ) ÷ x – 1 Let (x – 1) = t2
fi dx = 2t dt
2
(t + 1) 2t dt = Ú __________ (t4 + 1) t
1.112 Integral Calculus, 3D Geometry & Vector Booster
(t2 + 1) dt = 2 Ú ________ t4 + 1)
1 1 + __ 2 dt t = 2 Ú _________ 1 2 __ t + 2 t
1 where t = _____ x+1
( ) ( ) (Ú )
dx __________ 95. Ú _________________ (x + 2) ÷x 2 + 2x + 2
1 1 + __ 2 dt t = 2 __________ 1 2 __ t – + 2 t
( )
( )
dz 1 = 2 Ú _____ Let z = t – __ 2 t z +2 __ z __ + c =÷ 2 tan– 1 ___ 2 ÷
( )
( ( ) ) (
__ 1 1 = ÷2 tan– 1 ___ __ t – __ + c t 2 ÷
__ 1 _____ 1 = ÷2 tan– 1 ___ __ (÷ x – 1 – ______ _____ + c 2 ÷x – 1 ÷
)
dx_____ 94. Ú _____________ (x + 1) ÷x 2 + 1 1 Let (x + 1) = __ t
1 fi dx = – __2 dt t 1–t 1 Also, x = __ – 1 = ____ t t
1 – __2 dt t ___________ = Ú ______________ 1 1 2 __ 1 – __ + 1 t t
÷(
dt ____________ = – Ú _____________ 2 – 2t + 1) ÷ (2t dt 1__ ___________ = – ___ Ú __________ 2 ÷ 1 2 __ t – t + 2
)
dt 1__ ________________ = – ___ Ú _______________ 2 ÷ 1 2 1 2 t – __ + __ 2 2
÷ ( ( ) ( ) ) 1 1 1 = – ___ log ( t – __ ) + t – t + __ + c ÷ 2 2| | 2 1 1 1 = – ___ log ( t – __ ) + t – t + __ + c ÷ 2 2| 2 | _________
__ ÷ __
÷
1 Put (x + 1) = __ t
fi
2
dt ______ = – Ú _______ 2 ÷ t + 1
= – log t + ÷t 2 + 1 + c
1 1 2 + 1 + c = – log __ x + __ x
÷
_____
|
|
÷
|
______
|
dx _____ 96. Ú ________ x ÷x 2 + 4 1 Put x = __ t
fi
1 – __2 dt t _____ = Ú ________ 1 1 __ __ 2 + 4 t t
dx ______ = – Ú _______ 2 + 1 ÷ 4t
dt 1 _________ = – __ Ú __________ 2 t 2 + (1/2)2
1 1 = – __ log t + t2 + __ + c 2 4
1 1 1 1 = – __ log __ x + __ 2 + __ + c 2 4 x
÷
|
|
÷
______
÷
|
______
|
1 Put (x – 1) = __ t
fi
1 dx = – __2 dt t
÷
dx_____ 97. Ú ____________ (x – 1) ÷x 2 + 4
_________ 2
1 dx = – __2 dt t
1 – __2 dt t = Ú________ _____ 1 1 __ __ + 1 t t2
)
dt = – Ú ____________ ___________ – 1)2 + t2 ÷ (t
÷(
1 __ 2 dt t = Ú – ______________ ____________ 2 1 1 __ __ + 1 + 4 t t
÷(
)
1 dx = – __2 dt t
Indefinite Integrals
dx ___________ = – Ú _____________ + t)2 + 4t2 ÷ (1
dt __________ = – Ú ___________ 2 + 2t + 1 ÷ 5t
÷
dt 1__ ______________ = – ___ Ú _____________ ÷5 1 2 2 2 t + __ + __ 5 5
÷( ) ( ) 1 = – ___ log ( t + __ | 15 ) + ÷ t __ 25 t + __ 15 | + c ÷5 ________
2
__
1 where t = ______ (x – 1) dx _____ 98. Ú ______________ (2x – 1) ÷x 2 + 1
(
1 (2x – 1) = __ t dt dx = – ___2 2t
dt _____________ = – Ú ______________ ÷(1 – 2t)2 – 36t2
dt ___________ = – Ú ____________ 32t2 ÷ 1 – 4t –
fi
(
)
1 1 + 1 Also, x = __ __ 2 t
÷ (
)
dt 1__ _______________ = – ____ Ú ______________ 4÷2 t ___ 1 2 – t – __ – 8 32
÷ {
}
dt 1__ _____________________ = – ____ Ú ____________________ 4÷2 3 2 1 2 ___ – t – – ___ 16 16
÷ { (
) ( ) }
dt 1__ ________________ = – ____ _______________ Ú 2 4÷2 3 1 2 ___ ___ – t – 16 16 1 t – ___ 16 1__ – 1 _______ ____ = – sin + c 3 4÷2 ___ 16
÷( ) ( ) ( )
( )
dt ___ 2 2t ____________ = – Ú _______________ 2 1 __ 1 __ 1 __ + 1 + 1 t 4 t
dt ___________ = – 2 Ú _____________ + t)2 + 4t2 ÷ (1
dt __________ = – 2 Ú ___________ 2 + 2t + 1 ÷ 5t
dx ______ 100. Ú _________ 2 x ÷x 2 – 1
dt 2__ ___________ = – ___ Ú __________ 2t __ 1 ÷5 2 __ t + + 5 5
Put
1 x = __ t
fi
1 dx – __2 dt t
÷ (
)
÷
dt 2__ ______________ = – ___ Ú _____________ 2 ÷5 1 2 2 __ t + + __ 5 5
÷( ) ( ) 2 = – ___ log ( t + __ | 15 ) + ÷ t + __ 25 t + __ 1t | + c ÷5 __________
2
__
1 where t = _______ (2x – 1) dx _____ 99. Ú ______________ (3x + 2) ÷x 2 – 4
)
1 1 Also, x = __ __ – 2 3 t
Put
fi
dt ___ 2 3t ____________ = – Ú _______________ 2 1 1 __ 1 __ __ – 2 – 4 t 9 t
1 (3x + 2) = __ t dt ___ dx = – 2 3t
Put
dx 1__ ___________ = – ___ Ú __________ 2 1 ÷5 2 __ t + t + __ 5 5
1.113
(
)
–1 1__ – 1 16t = – ____ sin ______ + c 3 4÷2
1 – t2 1 Also, x2 – 1 = __ 2 – 1 = _____ t t2 1 – __2 dt dx t _____ _____ Thus, Ú ________ = Ú ________ 2 2 – t2 x ÷x – 1 1 1_____ __ 2 2 t t
÷
– t dt _____ = Ú ______ ÷ 1 – t2
= ÷1 – t2 + c
_____
1.114 Integral Calculus, 3D Geometry & Vector Booster _______
÷ ( )
t dt______ = – Ú _____________ 2 (1 – t ) ÷1 + 2t2
1 2 = 1 – __ x + c
÷ 1 – x2 = ______ + c x
v dv 1 = __ Ú ___________ Let (1 + 2t2) = v2 2 4 _____ v –1 – 1 v 2 fi 4tdt = 2v dv
_____
dx_____ 101. Ú ____________ 2 (x + 1 ÷x 2 + 2 1 Put x = __ t dt fi dx = __ 2 t
dt – __2 t ______ = Ú ______________ 1 1 __ 2 + 1 __ 2 + 2 t t t dt______ = – Ú ______________ 2 (t + 1) ÷1 + 2t2
( ) ÷
dv______ 1 = – __ Ú _____________ , where t2 = v 2 (v + 1) ÷1 + 2v fi 2tdt = dx
Let
dv = z dz 2
z –1 Also, v = _____ 2
(1 + 2v) = z2
z dz 1 = – __ Ú ___________ 2 z2 – 1 _____ + 1 z 2
(
= cot– 1 (z) + c
= cot– 1 (÷1 + 2v) +c
)
dv 1 = __ Ú _______ 2 (v2 – 3)
v –÷ 3 1__ __ = ____ log ______ + c 4 ÷3 v+÷ 3
÷1 + 2t2 – ÷ 3 1 ______ = ____ __ log ____________ __ + c 2 4 ÷3 ÷1 + 2t + ÷3
| |
|
__
______
__
| | ÷ ÷
______
|
__ 1+2 _____ – ÷3 2 x 1__ ____ ___________ = log ______ + c __ 4 ÷3 1+2 _____ + 3 ÷ x2 dx ______ 103. Ú _____________ 2 (1 + x ) ÷1 – x2 1 Put x = __ t
dt fi dx = – __2 t
dt __ 2 t ______ = – Ú _____________ 1 1 1 + __ 2 1 – __ 2 t t
t dt _____ = – Ú _____________ 2 (1 + t ) ÷t 2 – 1
)
dz = – Ú _______ (z2 + 1)
(
( ) ÷
v dv = – Ú ________ , 2 (v + 2) v
______
_______
Let (t 2 – 1) = v fi 2tdt = dx
= cot– 1 (÷1 + 2t2) +c
2 = cot– 1 1 + __ 2 + c x
dv = – Ú _______ (v2 + 2)
v 1__ – 1 ___ = – ___ cot __ + c 2 ÷2 ÷
÷t 2 – 1 1 __ = ___ __ cot– 1 ______ + c 2 ÷2 ÷
1 1 1 = ___ __ cot– 1 ___ __ __ – 1 + c 2 ÷2 x2 ÷
( ÷ ) ______
dx______ 102. Ú _____________ 2 (x – 1) ÷x 2 + 2
fi
dt __ 2 t ______ = – Ú _____________ 1 1 __2 – 1 __ 2 + 2 t t
( ) ÷
( ) _____
1 Let x = __ t
( )
dt dx = – __2 t
( ÷ ) _____
x dx_____ 104. Ú _____________ 4 (x – 1) ÷x 4 + 3
Let x2 = t fi
2xdx = dt
1.115
Indefinite Integrals
dt ______ 1 = __ Ú _____________ 2 (t2 – 1) ÷t 2 + 3
dy ___ 2 y 1 1 ______ = – __ Ú ______________ Let t = __ y 2 __ 1 1 2 – 1 __ 2 + 3 dy y y fi dt = – ___ 2 y
( ) ÷
(
)
t fi _______ _____ dx = dt 2 ÷ x + 1
1 1 2 Also, x2 + 1 = __ t – __ + 1 t 4
fi ÷ x2
y dy 1 ______ = – __ Ú _______________ 2 (1 – y2) 1 + 3y2
v dv 1 = – __ Ú ________ Let (3y2 + 1) = v2 2 4 (4 – v ) v fi 3ydy = 2vdu
2
2
(
_____
)
( ( ) )
1 1 __ t + __ t 2 ________ x = dt t
dv 1 = __ Ú _______ 4 (v2 – 4)
fi
d
v–2 1 = ___ log _____ + c v+2 16
fi
1 t2 + 1 dx = __ ______ 2 dt 2 t
=
|
|
_______ 3y2 + 1 – 2 1 ___ ______ + log ___________ 16 3y2 + 1 +2
÷ ÷
| | ÷ ÷
______
( )
t2 + 1 1 1 __ Ú t10 _____ dt = __ Ú t8 (t2 + 1) dt 2 2 t2
dx __________ 105. Ú __________________ 2 (x – 1) ÷x 2 + 4x + 5
1 Let x = __ t
fi
dt dx = – __2 t
dt – __2 t __________ = Ú _________________ 1 1 __ 4 __ __ 2 – 1 2 + + 5 t t t
( ) ÷
t dt __________ = Ú __________________ 2 (t – 1) ÷5t 2 + 4t + 1
_____
1 = __ Ú (t10 + t8) dt 2
1 t11 = __ ___ + 2 11
______
(x + ÷ x 2 + 1 )=t
Let
1 _____ × 2x fi 1 + ________ dx = dt 2 ÷x 2 + 1
x+÷ x 2 + 1 fi __________ _____ dx = dt ÷ x 2 + 1
(
(
)
t9 __ + c, 9
______
5 107. Ú (x – ÷x 2 + 4) dx
)
_____
)
_____
Put (x – ÷x 2 + 4 )=t
1 _____ × 2x fi 1 – ________ dx = dt 2 ÷x 2 + 4
x fi 1 – _______ _____ dx = dt 2 ÷ x + 4
10
(
where (x + ÷x 2 + 1 )=t
106. Ú (x + ÷x + 1 ) dx 2
_____
( )
Thus, the given integral reduces to
c
3 __ 2 + 1 – 2 x 1 ______ = ___ log ___________ + c 16 3 __ 2 + 1 + 2 x
÷ x 2 + 1 Thus, dx = _______ dt t
| |
_______
÷
___________
______
( ) 1 1 + 1 = __ ( t – __ ) + 4 t 2÷ 1 1 = __ ( t + __ ) t 2÷ 1 1 = __ ( t + __ ) t 2
(
)
(
(
)
_____
)
2 + 4 ÷ x _____ fi x – _______ dx = – dt 2 ÷ x + 4 t fi _______ _____ dx = – dt 2 ÷ x + 4
(
)
( ( ) )
t fi _________ dx = – dt t __ 2 __ – + t 2 fi
(
1 t dx = __ __ + t 2
)
2 __ dt t
1.116 Integral Calculus, 3D Geometry & Vector Booster fi
( )
1 2 dx = __ + __ dt 2 t2
{ ( ) } Ú ( ) ( )
3 = __ ( t– 8/3 – t– 2/3 ) + c 4 _____
where t = (x + ÷x 2 – 4 )
1 2 = Ú t5 __ + __ dt 2 t2
110. Do yourself.
5
t __ + 2t3 dt 2
=
t6 t4 = ___ + __ + c 12 2
dx_____ dx ______ 111. Ú ____________ = Ú _____________ 2 2 9 3 x (x – ÷ x + 9 x 1 – 1 + __ 2 x
( ÷ )
_____
where t = (x – ÷x 2 + 4 ) 108. Ú (x +
______ )n dx ÷ 1 + x2
______
Put (x + ÷ x 2 + 1) =t
( ) 2
1 t +1 dx = __ _____ dt 2 t2
fi
( )
t2 + 1 1 = __ Ú tn ______ 2 dt 2 t
1 = __ Ú tn– 2 (t2 + 1) dt 2
1 = __ Ú (tn + tn– 2) dt 2
tn– 1 1 tn + 1 = __ _____ + _____ + c 2 n+1 n–1
(
t 1 = __ Ú ____ dt 9 t–1
1 (t – 1) + 1 = __ Ú _________ dt 9 t–1
( (
)
)
1 1 = __ Ú 1 + ____ dt 9 t–2 1 = __ (t + log |t – 1|) + c 9
÷
9 where t = 1 + __ 2 x
______
Put (x + ÷ x 2 – 4) =t
1 ______ × 2x fi 1 + ________ dx = dt 2 ÷x 2 – 4
(
)
( (
)
(
fi
1 t __ 2 dx = __ __ – dt t 2 t
fi
1 2 dx = __ – __ dt 2 t2
( )
Let 2 + 3x = t x
fi
2 x = ____ t –3
fi
– 2 dt dx – ______ (t – 3)2
( )
)
t – 5/3 = Ú ____ – 2t– 11/3 dt 2
_______ – 2 dt (t – 3)2 _____________ = Ú 2 1/2 3/2 3/2 ____ t x t–3
– 2 dt _______ (t – 3)2 _________________ = Ú 3/2 2 1/2 3/2 ____ 2 ____ t t –3 t –3
= – Ú t– 3/2 dt
2 = __ _ + c t ÷
)
1 2 __ – __ 2 t2 _______ = Ú 5/3 dt t
)
x fi 1 + _______ ______ dx = dt 2 ÷ x – 4 t fi ______ _____ dx = dt 2 ÷ x – 4
(
dx 112. Ú ____________ x1/2 (2 + 3x)3/2
dx _____ 109. Ú _____________ (x + ÷ x 2 – 4 )5/3
)
______
where t = (x + ÷ x + 1 )
t 1 = – __ Ú ____ dt 9 1 – t
2
)
_____
(
9 Put 1 + __ 2 = t2 x dx t ___3 = – __ dt 9 x
( )
( ) ( )
( )
1.117
Indefinite Integrals
2 = _______ ______ + c 2______ + 3x x
÷
115. Ú dx/x1/3 (2x + 1)5/3
dx 113. Ú ____________ x2/3 (2 + 3x)4/3
Put (2x + 1) = t x
Let (2 + 3x) = t x
1 fi x = ____ t–2
fi x (t – 3) = 2
dt fi dx = – ______ (t – 2)2
2 fi x = ______ (t – 3) – 2 fi dx = ______ dt (t – 3)2
2 ______ 2 (t – 3) __________________ = – Ú dt 2 2/3 4/3 ____ 2 4/3 ____ t t–3 t–3
( )
2 = – Ú __ × t– 4/3 dt 4
1 = – __ Ú t– 4/3 dt 2
3 = __ (t – 4/3) + c 2
( )
( (
3 = __ (t– 2/3) + c 2
)
Put (3x – 1) = t x 1 fi x = – ______ (t – 3) 1 fi dx = ______ 2 dt (t – 3)
1 ______ (t – 3)2 = Ú __________________ dt _________________ 3/4 5/4 1 5/4 1 ____ t ____ t –3 t –3
( ) ( )
= Ú t – 5/4 dt
= – 4t– 1/4 + c
4 = – ___ + c 4 _ t ÷
dx Ú _____________ (x – 1)3 (x + 2)4 dx = Ú _______________ 3 (x – 1) ______ (x + 2)7 (x + 2)
(
)
( )
x–1 Put _____ = t x+2
3 fi _______ dx = dt (x + 2)2
2t + 1 Also, x = _____ 1–t
fi
3 x + 2 = ____ 1–t
dt 1 = __ 6 Ú ________ 1 5 3 t3 ____ t –1
( )
5 1 (t – 1) dt = __ 6 Ú ________ t3 3 5 – 5t4 + 10t3 – 10t2 + 5t – 1) dt 1 t____________________________ __ = Ú t3 36
4 _______ = – ________ + c 1 4 3 – __ x
÷(
( ( ) )
3 1 = __ _________ + c 2 1 2/3 2 + __ x
dt = Ú ___ t5/4
( )
117. Do yourself. 118. We have,
= – Ú t– 5/3 dt
( )
116. Do yourself
dx 114. Ú ____________ 3/4 x (3x – 1)5/4
3 1 = __ _________ 1/3 + c 2 2 3 + __ x
)
dt ______ 2 (t – 2) ___________________ = – Ú 1 1/3 5/3 ____ 1 5/3 ____ (t) t –2 t –2
)
(
10 1 = __ 6 Ú t2 – 5t + 10 – ___ + t 3
5 __ 2 – t
)
1 __ 3 dt t
1.118 Integral Calculus, 3D Geometry & Vector Booster
(
)
5 5 1 t3 __ 1 = __ 6 __ – t2 + 10t – 10 log |t| – __ + ___ 2 + c t 2t 3 3 2 x–1 where t = _____ x+2
( )
dx dx 119. Ú _____________ 2 = Ú ______________ 3 x–2 2 5 _____ (x – 1) (x – 2) (x – 1) x–1 x–2 Put _____ = t x–1
( )
(
1 dt = __ Ú ___ 3 t3/4 4 1 = __ ___ _ + c 3 4÷t
( )
( ( ) ) ÷
4 _______ 1 = __ _________ + c 3 4 _____ x–1 x–2
)
(x – 1) ◊ 1 – (x – 2) ◊ 1 ___________________ dx = dt (x – 1)2 dx _______ = dt (x – 1)2 t–2 Also, x = _____ t–1 t– 2 1 (x – 1) = ____ – 1 = – ______ t–1 (t – 1) dt = Ú ________ 1 3 2 ____ t t–1
( )
dx dx 122. Ú _________ = Ú ______________ x 2 _____ x2 (x + 5)4 (x + 5)6 x+5
(
)
( )
x Put _____ = t x+5
5 dx fi _______ = dt (x + 5)2
dx fi _______ = __ 5 (x + 5)2
dt 1 = __ 5 Ú ________ 1 4 2 5 t ____ 1–t
( )
(t – 1)3 dt = Ú _________ t2
4 1 (t – 1) dt = __ 5 Ú _________ t2 5
(t3 – 3t2 + 3t – 1) = Ú _______________ dt t2
4 3 2 1 (t – 4t + 6t – 4t + 1) dt = __ 5 Ú ______________________ t2 5
3 1 = Ú t – 3 + __ – __ dt t t2
t2 1 = __ – 3t + 3 log |t| + __ + c t 2
1 4 = __ 5 Ú t2 – 4t + 6 – __ + t 5
1 t3 1 = __ 5 __ – 2t2 + 6t – 4 log |t| – __ + c t 3 5
( (
(
)
)
)
x–2 where t = _____ x–1
(
dx dx ______________ 121. Ú _______________ = Ú ________________ 3/4 4 3 5 (x – 1) (x + 2)5/4 ÷(x – 1) (x + 2)
( )
)
1 __ 2 dt t
)
)
x where t = _____ x+5
120. Do yourself.
dx = Ú _______________ x – 1 3/4 _____ (x + 2)2 x+2
( (
dx 123. Ú _______________ 3/2 (x – 1) (x + 1)5/2
dx = Ú _______________ x – 2 3/2 _____ (x + 1)2 x+1
( )
( )
x – 1 Put _____ = t x+2
x–1 Let _____ = t x+1
(x + 2) ◊ 1 – (x – 1) ◊ 1 fi __________________ dx = dt (x + 2)2
(x + 1) ◊ 1 – (x – 1) ◊ 1 fi ___________________ dx = dt (x + 1)2
3 fi _______ dx = dt (x + 2)2
2 fi _______ dx = dt (x + 1)2
dx dt fi _______ = __ 3 (x + 2)2
dx dt fi _______ 2 = __ 2 (x + 1)
(
)
Indefinite Integrals
1 dt = __ Ú ___ 2 t3/2
1 = __ Ú t– 3/2 dt 2 1 = – ___ + c t ÷ 1 _______ = – ________ + c x –1 _____ x+1
÷( ) x+1 = – _____ ÷ x – 1 + c
Let x3 = t
fi
Again let
3x2 dx = dt
dt 1 ______ = __ Ú ________ 3 t ÷3t + 4
Also,
v2 – 4 t = ______ 3
(
)
dv 2 = __ Ú _______ 3 (v2 – 4)
v–2 2 1 = __ × __ log _____ + c 3 2 v+2
fi
– 30x5 dx = 2t dt
fi
t dt x5 dx = – ___ 15
t dt 1 = – ___ Ú _______ 15 2_____ – t2 t 5 1 = __ Ú dt (t2 – 2) 3
( )
| |
|
_______
|
3x 3 + 4 – 2 ÷ _______ 1 = __ log ____________ + c 3 3 ÷3x + 4 + 2
|
__
|
– ÷2 ÷ 2 – 5x6 1 = __ log _____________ _______ __ + c 3 ÷2 – 5x6 + ÷2
3t + 4 – 2 ÷ 1 ______ = __ log ___________ + c 3 ÷3t + 4 + 2
Let 3x – 2 = t2 fi 3 dx = 2t dt
t dt 2 = __ Ú ________ 2 3 _____ t +2 t 3
( )
dt = 2 Ú ______ 2 t +2
__ t =÷ 2 tan– 1 ___ __ + c 2 ÷
__ 3x – 2 =÷ 2 tan– 1 ______ + c 2
( )
( ÷ ) _____
3
dx x _______ _______ 125. Ú _________ = Ú __________ dx 4 4 x ÷5x + 3 x ÷5x 4 + 3
Let 5x4 + 3 = t2
fi
20x3 dx = 2t dt
fi
t dt x dx = ___ 10
Put (2 – 5x6) = t2
|
| | __
dx ______ 127. Ú ________ x ÷3x – 2
( )
_______
t – ÷3 1 __ = ____ __ log ______ + c 4 ÷3 t + ÷3
__
v dv 2 = __ Ú _________ 2 9 v_____ –4 ◊ v 3
|
dt 1 = __ Ú _______ 2 (t2 – 3)
t – ÷2 1 __ = __ log ______ + c 3 t + ÷2
3 dt = 2v dv
______
3t + 4 = v2
fi
| |
)
dx x5 _______ _______ 126. Ú _________ = Ú __________ dx x ÷2 – 5x6 x6 ÷2 – 5x6
(
_______
2 dx x_______ dx _______ 124. We have, Ú _________ = Ú __________ 3 3 x ÷3x + 4 x ÷3x 3 + 4
t dt 1 = ___ Ú ________ 10 2 __ 3 t – t 5
where t = ÷ 5x 4 + 3
_____
1.119
dx x8 _______ _______ 128. Ú _________ = Ú __________ dx x ÷3x 9 – 2 x9 ÷3x 9 – 2
Let 3x9 – 2 = t2
fi 27x9 dx = 2t dt
2t dt fi x8 dx = ____ 27
3
1.120 Integral Calculus, 3D Geometry & Vector Booster
t dt 2 = ___ Ú _______ 27 _____ t2 + 2 t 3
dx dx 131. Ú __________ = Ú ___________ c 3 (c + dx2)3/2 x d + __ 2 3/2 x
dt 2 = __ Ú ______ 9 (t2 + 2)
(
( )
__
( ) ( ______ ÷3x 2– 2 ) + c
2 ÷ t = ___ tan– 1 ___ __ + c 9 ÷2
2 ÷ = ___ tan– 1 9
__
______
Let 2x10 – 3 = t2
fi
10x9 dx = 2t dt
fi
t dt x9 dx = ___ 5
t dt 1 = __ Ú _______ 2 5 _____ t +3 t 10 dt = 2 Ú ______ (t2 + 3)
( )
t 2 = ___ __ tan– 1 ___ __ + c 3 ÷3 ÷
2 = ___ __ tan– 1 3 ÷
10
t dt = – Ú _________ 2 (2t + 3)3/2
1 Let x = __ t 1 fi dx = – __2 dt t
Again let (2t2 + 3) = v2
fi
1 = __ c × 1 = __ c ×
132. Do yourself
1 __ + c t 1 _________ ________ + c c d + __ 2 x
÷ (
)
1 v dv = – __ Ú ____ 2 v3 1 dv = – __ Ú ___2 2 v 1 = ___ + c 2v 1 _________ = _______ + c 2 ÷2t 2 + 3 x = _________ _______ + c 2 ÷2 + 3x2
2t dt = v dv
( ) ( )
2 Let __ 4 – 5 = t2 x 8 fi – __5 dx = 2t dt x dx t dt fi ___5 = – ___ 4 x
dt 1 __ = – __ c Ú t2
( )
dx 130. Ú __________ (2 + 3x2)3/2
x dx x dx dx 134. Ú _________ = Ú __________ = Ú ___________ 3/2 2 3/2 2 6 _____ 5 __ (2 – 5x4)3/2 x 4 x 4 – 5 x –5 x
_______
t dt 1 ___ = – __ c Ú t3
133. Do yourself
( ) –3 ( ________ ) + c ÷ 2x 3
)
2c fi – ___3 dx = 2t dt x dx t dt fi ___3 = – ___ c x
9
(
c Let d + __ 2 = t2 x
dx x9 _______ _______ 129. Ú __________ = Ú ___________ dx x ÷2x 10 – 3 x10 ÷2x 10 – 3
)
1 t dt = – __ Ú ___ 4 t3
1 dt = – __ Ú __2 4 t
1 = __ + c 4t
1 2 = __ __ – 5 + c 4 x4
÷(
_______
)
dx dx 135. Ú x2 _________ = Ú ___________ 3/2 1 7 __ (1 – 4x6)3/2 x 6 – 4 x
( )
( )
1 Let __ 6 – 4 = t2 x 6 fi – __7 dx = 2t dt x dx t dt fi ___7 = – ___ 3 x
1 t dt = – __ Ú ___ 3 t3
1 dt = – __ Ú __2 3 t 1 = __ + c 3t
÷
1 Put x – 2 = __ t 1 fi dx = – __2 dt t Thus, the given integral reduces to 1 – __2 t t ______ Ú ________ ______ dt = – Ú ________ dt 2 1 __ 1 __ 1 + 3t ÷ 2 2 + 3 t t
Again, let 1 + 3t2 = v2
fi
3t dt = v dv
v dv Now, the integral is – Ú ____ v
= – v + c
= ÷1 + 3t2 + c
_______
÷
3 = 1 + _______ + c (x – 2)2
1 Let (x + 1) = __ t fi
1 __ 2 dt t_____ = – Ú _________ 1 __ 1 __ 3 2 + 3 t t
t2 dt ______ = – Ú ________ ÷ 1 + 3t2
÷
1 dx = – __2 dt t
÷
_____
|
÷
_____
|)
|
1 1 + ____ __ log t + t2 + __ + c 3 3÷3
dx 1 __________ = __ Ú __________________ 2 (x – 2)3÷x 2 – 4x + 5 dx 1 __________ = __ Ú __________________ 2 (x – 2)3 (x – 2)2 + 1
÷
1 Let (x – 2) = __ t
fi
dt __ 2 t______ = – Ú __________ 1 1 __ 3 __ 2 + 1 t t
t2 _____ = – Ú _______ dt ÷ t 2 + 1
(t2 + 1) – 1 ______ = – Ú __________ dt ÷ t 2 + 1
dx dx __________ ___________ 137. Ú __________________ = Ú ___________________ 3 2 3 (x + 1) ÷ x + 2x + 4 (x + 1) ÷(x + 1)2 + 3
|
__ t2 1 __ 1 1 = – ÷3 __ t2 + __ + log t + t2 + __ 2 3 6 3
dx _____________ 138. Ú _____________________ 3 (x – 2) ÷ 4x 2 – 16x + 20
__________
( ÷
÷ ( )
1 where t = ______ (x + 1)
÷
÷( ( ) )
__ dt 1 2 1__ ___________ = – ÷3 Ú t2 + ___ __ dt + ____ _________ Ú 3 3÷3 ÷ 1__ 2 2 ___ t + 3 ÷
= – Ú dv
)
________
______
dx __________ = Ú __________________ 2 (x – 2) ÷(x – 2)2 + 3
1 1 = – __ Ú ÷(1 + 3t2) – ________ ______ dt 3 ÷1 + 3t2
dx __________ Ú __________________ 2 2 (x – 2) ÷ x – 4x + 7
____________
1 1 = __ __ – 4 + c 3 x6
136. We have,
)
+ 3t2) –1 1 (1 ______ = – __ Ú __________ dt 3 ÷ 1 + 3t2
______
(
(
1.121
Indefinite Integrals
1 dx = – __2 dt t
( )÷
( (
)
_____
)
1 = – Ú ÷t 2 + 1 – ______ _____ dt 2 ÷ t + 1 _____ _____ t 2 = – __ ÷ t + 1 + log t + ÷ t 2 + 1 2
[
|
|
|
|]
– log t + ÷ t 2 + 1 + c
1 where t = ______ (x – 2)
dx _____________ 139. Ú _________________________ 2 (x – 6x + 9) ÷4x 2 – 24x + 20
_____
dx 1 __________ = __ Ú _________________ 2 (x – 3)2 ÷x 2 – 6x + 5
1.122 Integral Calculus, 3D Geometry & Vector Booster
dx 1 ___________ = __ Ú ___________________ 2 (x – 3)2 (x – 4 ÷ – 3)2 1 Let (x – 3) = __ t dt fi dx = __ 2 t
dt __ 2 t______ = –Ú ___________ 3 1 1 __ __ 2 – 4 t t
1 = – ___ Ú dv 12 v = – ___ + c 12 _______
2 + 1 ÷ 6t = – _______ +c 12
÷
____________
( ) ÷
t2 ______ = – Ú _______ dt ÷ 1 – 4t2
1 1 = __ Ú (1 – 4t2) – _______ ______ dt 4 ÷ 1 – 4t2
1 1 = __ 1 – 4t2 – _______ ______ dt 4 ÷1 – 4t2
6 ________ + 1 (2x + 1)2 _____________ = +c 12
dx dx __________ __________ 141. Ú ___________________ = Ú __________________ 3 2 3 (x + 1) ÷ x + 2x – 4 (x + 1) ÷(x + 1)2 – 5
( Ú ( ÷
) )
______
Let
fi
1 (x + 1) = __ t 1 dx = – __2 dt t
dt __ 2 t______ = – Ú ___________ 3 1 1 __ __2 – 5 t t
t2 dt ______ = – Ú ________ ÷ 1 – 5t2
1 where t = ______ (x – 3)
5t2) – 1 1 1 –______ dt = __ Ú ___________ 5 ÷1 – 5t2
dx ___________ 140. Ú ________________________ 2 (4x + 4x + 1)÷4x 2 + 4x + 7
1 1 – _______ ______ dt = __ Ú ÷1 – 5t2 5 ÷1 – 5t2
÷( )
________
dt 1 1 2 1 ________ = __ Ú __ – t2 dt – __ Ú _________ 2 2 4 1 2 2 __ – t 2
÷( )
[ ÷
_____
]
1 t __ 1 1 1 = __ __ – t2 + __ sin– 1 (2t) + __ sin– 1 (2t) + c 2 2 4 8 4
dt ___ 2 2t_____ = – Ú ___________ 2 1 __ 1 __ 2 + 6 t t
t dt 1 _______ = – __ Ú ________ 2 ÷6t 2 + 1
1 Let (2x + 1) = __ t dt ___ fi dx = – 2 2t
______
÷( )
)
__ dt 1 2 1 __________ dt – ____ __ =÷ 5 Ú ___ __ – t2 Ú _________ 2 ÷5 5÷5 1 ___ __ – t2 ÷5 __
[ ÷
__
÷( )
]
__ t 1 1 = ÷5 __ __ – t2 + ___ sin– 1 (t ÷5 ) 2 5 10 __ 1__ – ____ (sin– 1 t ÷5 ) + c 5÷5
(2x + 3) __________ 142. Ú __________________ dx (3x + 4)÷x 2 + 2x + 4
Again, let 6t2 + 1 = v2
fi
12t dt = 2v dv
fi
1 t dt = __ v dv 6
)
1 where t = ______ (x + 1)
1 v dv = – ___ Ú ____ 12 v
(
( ) ÷
(
_________
dx ___________ = Ú ____________________ 2 (2x + 1) ÷(2x + 1)2 + 6
( ) ÷
Let 2x + 3 = A (3x + 4 ) + B Comparing the co-efficients of x and the constant term, we get fi
A = 2/3, B = 1/3
Indefinite Integrals
dx dx 2 __________ 1 __________ = __ Ú ____________ + __ Ú __________________ 2 3 (x 3 + 1) + 3 (3x + 4) (x + 3 ÷ ÷ + 1)2
|
2 = __ log (x + 1) + ÷ x 2 + 2x + 4 3 3 ____________ dt ___ + – ____ , Ú ___________ 2 28 ÷28t ÷ – 2t + 1
where 3x + 4 = 1/t fi
dt 3dx = __ 2 t
|
2 = __ log (x + 1) + 3
__________ ÷ x 2 + 2x + 4
| (
|
) ÷
___________
|
3 t 1 1 ___ ___ + – ____ log t – ___ + t2 – ___ + + c 28 14 28 28 ÷
1 where t = ______ 3x + 4
|
dt __ 2 t_____ 1 – Ú _________ , Let (x + 1) = __ t 1 1 __ __ + 8 t t2 dt dx = – __2 t
( )÷
|
__________ ÷ x 2 + 2x + 9 –
|
__________ dt 1__ ____________ __________ ÷ x 2 + 2x + 9 ____ 2÷2 2 1__ 2 ____ t + 2÷2
= 2 log (x + 1) +
= 2 log (x + 1) +
(4x + 7) __________ 144. Ú _________________ dx (x + 2)÷x 2 + 4x + 7 4 (x + 2) – 1 ___________ = Ú __________________ dx (x + 2)÷(x + 2)2 + 3 dx dx ___________ ___________ = 4 Ú ____________ – Ú _________________ 2 ÷(x + 2) + 3 (x + 2)÷(x + 2)2 + 3 __________
|
|
= 4 log (x + 2) + ÷ x 2 + 3x + 7 dt __ 2 t ______ – Ú __________ , Let (x + 2) = 1 1 __ __ 2 + 3 t t dt fi dx = __ 2 t
( )÷
|
__________
|
__________
1 __ t
|
dt ______ = 4 log (x + 2) + ÷ x 2 + 3x + 7 – Ú _______ 2 + 1 ÷ 3t
dx dx __________ __________ = 2 Ú ____________ + Ú _________________ 2 ÷(x + 1) + 8 (x + 1)÷(x + 1)2 + 8
|
|
1 where t = ______ (x + 1)
|
2 (x + 1) + 1 ___________ = Ú _________________ dx (x + 1)÷(x + 1)2 + 8
= 2 log (x + 1) +
÷
|
dt 1__ ___________ = 4 log (x + 2) + ÷ x 2 + 3x + 7 – ___ Ú _________ 3 2 ÷ 1 2 t + ___ __ 3 ÷ __________ = 4 log (x + 2) + ÷ x 2 + 3x + 7
(2x + 3) __________ 143. Ú _________________ dx (x + 1)÷x 2 + 2x + 9
__________ +9 ÷ x 2 + 2x
|
______
1__ 1 – ____ log t + t2 + __ + c 8 2÷2
dx dx 2 __________ 1 __________ = __ Ú ___________ + __ Ú __________________ 3 ÷x 2 + 2x 3 + 4 (3x + 4)÷x 2 + 2x + 4
|
|
= 2 log (x + 1) + ÷ x 2 + 2x + 9
2 1 __ (3x + 4) + __ 3 3 ____________________ ____________ 2 (3x + 4)÷x + 2x + 4 dx
__________
__________
|
Thus, the given integral is reduces to
1.123
|
dt ______ Ú _______ 2 + 1 ÷ 8t
|
Ú
÷ ( )
|
÷
______
|
÷ ( )
|
1__ 1 – ___ log t + t2 + __ + c 3 3 ÷
1 where t = ______ (x + 2) x2 + __________ 4x + 2 145. Ú _________________ dx (x + 1)÷x 2 + 2x + 3 Put x2 + 4x + 2
= l (x + 1) (2x + 2) + m (x + 1) + n
= l (2x2 + 4x + 2) + m (x + 1) + n
= 2 l x2 + (4l + m) x + (2l + m + n)
Comparing the coefficients of the like terms of both the sides, we get L = 1/2, M = 2, N = – 1. Thus, the given integral reduces to (x + 1) (2x + 2) 1 __________ __ Ú _________________ dx 2 (x + 1)÷x 2 + 2x + 2
1.124 Integral Calculus, 3D Geometry & Vector Booster (x + 1) __________ + 2 Ú _________________ dx (x + 1)÷x 2 + 2x + 2
__
__
3 3/2 = Ú 3÷ x (1 + ÷ x + 3x + x ) dx
= Ú (x1/3 + 3x5/6 + 3x4/3 + x11/6) dx dx _________________ __________ – Ú (x + 1)÷x 2 + 2x + 2 9 3 18 6 = __ x4/3 + ___ x11/6 + __ x7/3 + ___ x17/6 + c 7 4 11 17 + 2) dx dx 1 (2x ___ __________ _________ = __ Ú ___________ + 2 Ú ___________ 2 2 ÷x 2 + 2x 149. Ú 3÷ x 2 (3 + x– 2/3)– 2 dx + 2 + 2 ÷ x + 2x
(
dx __________ – Ú _________________ Here, a = – 2 fi a negative integer (x + 1)÷x 2 + 2x + 2 Let x = t3
__________
=÷ x 2 + 2x + 2
|
)
+ 2 log (x + 1) +
fi dx = 3t2 dt.
__________ ÷ x 2 + 2x + 2
|
– sec– 1 (x + 1) + c
x2 + __________ 5x + 6 146. Ú _________________ dx (x + 2)÷x 2 + 5x + 4
= Ú t2 (3 + t– 2)– 2 ◊ 3t2 dt
t4 = 3 Ú ________ dt (3 + t– 2)2 dt = 3 Ú _________ 2 1 2 3t + __ 3
(
1 1 __ (x + 2) (2x + 5) + __ (x + 2) 2 2 ________________________ __________ = Ú dx 2 (x + 2)÷(x + 2) + 1
+ 5) dx dx 1 (2x 1 __________ __________ = __ Ú ___________ + __ Ú ___________ 2 ÷x 2 + 2x + 2 5 + 2)2 +1 ÷ (x
dx Now, I = Ú _______ x2 + 1/3
__________
_________
|
|
1 = ÷x 2 + 2x + 5 + __ log (x + 2) + ÷ x 2 + 4x + 5 + c 2
dt 1 = __ Ú ________ 2 3 2 __ 1 t + 3
(
)
– 2x ◊ x 1 = _________ 2 ◊ x – Ú _________ dx (x + 1/3) (x2 + 1/3)2 (x2 + 1.3) – 1/3 1 = _________ 2 ◊ x + 2 Ú _______________ dx (x + 1/3) (x2 + 1/3)2
x2 +__________ 10x + 6 147. Ú _________________ dx 2 (x + 2)÷x + 4x + 9
(
)
dx dx x 2 = ________ 2 + 2 Ú _________ – __ Ú _________ + c (x + 1/3) (x2 + 1/3) 3 (x2 + 1/3)2
1 __ (x + 2) (2x + 4) + 6 (x + 2) – 10 2 ____________ = Ú _____________________________ dx (x + 2)÷x 2 + 4x + 10)
dx x 2 = _________ 2 + 6 tan– 1 (3x) – __ Ú _________ + c 3 (x2 + 1/3)2 (x + 1/3)
(2x + 4) dx 1 ___________ ___________ = __ Ú ____________ dx + 6 Ú ____________ 2 2 ÷x 2 + 4x + 10 10 ÷ x + 4x +
x 2 = _________ 2 + 6 tan– 1 (3x) – __ I + c 3 (x + 1/3)
10 dx 5 x __________ – Ú _________________ fi __ I = _________ 2 + 6 tan– 1 (3x) + c 3 (x + 2)÷(x + 2)2 + 6 (x + 1/3)
___________
|
___________
|
= ÷ x 2 + 4x + 10 + 6 log (x + 2) + ÷ x 2 + 4x + 10
|
______
__
|
÷t 2 + 6 – ÷6 1__ ______ – ____ log ____________ __ + c 2 2 ÷6 + ÷ 6 ÷t + 6
where t = (x + 2) 148. We have,
)
__
__
Ú 3÷x (1 + ÷ x )3 dx
3x 18 fi I = __________ 2 + ___ tan– 1 (3x) + c 5 5 (x + 1/3)
(
_______ __
)
x ÷ 1 + __4÷ 150. Ú ________ dx 3 4 ÷ x 3 1 – __ + 1 __ b + 1 4 4 Here, _____ = _______ = __ = 1 g 1 1 __ __ 4 4
Indefinite Integrals 1 __
Let (1 + x 4 ) = t2
1 – __3 fi __ x 4 = 2t dt 4
Thus, Ú x
(1 + x ) dx
= 8 Ú t dt
8t3 = ___ + c 3
1 3/2 __ 8 = __ 1 + x 4 + c 3
– 3/4
1/4 1/2
)
fi
x4 (1 – t2) = 1
fi
1 x4 = ______ 2 (1 – t )
dx ______ Thus, Ú _________ 7 x÷ 1 + x4
( )
1 1 + x4 = __ _____ – 2 x4
x = t6
fi
dx = 6t5 dt
6t5 = Ú _________ dt 2 t (2 + 3t2)
(
)
(2 + 3t2) – 2 = 2 Ú __________ dt (2 + 3t2)
(
)
2 = 2 Ú 1 – ________ dt (2 + 3t2)
( ÷ ) ÷ ( ÷ ) ( ÷ ) __
__
3 3 __ tan– 1 __ t + c 2 2 __
__
2÷2 3 = 2 (x1/6) – ____ __ tan– 1 __ x 1/6 + c 2 3 ÷
1 1 + x4 3/2 __ _____ + c 3 x4
Here, a = Р10 ΠI
Let x = t4
fi
dx = 4t3 dt
_________ __
÷
7 3 4 154. Ú 3÷x × ( 1 + ÷ x ) dx
__
dx 152. Ú ____________ = Ú x– 1/2 (1 + x1/4)– 10 dx __ 4 __ 10 x ( ÷ x + 1) ÷
4t3 dt = Ú _________ 2 t (1 + t)10
Let
4 = 2t – __ × 3
( ) ( ÷ ( ) ) ______
dt 4 __ = 2 Ú dt – __ Ú _________ 3 2 2 2 __ t + 3
t 1 __ _______ 2 (t2 – 1)2 _________ = Ú dt 1 3 _____ 2 t 1–t
t3 1 = __ t – __ + c 2 3
)
Here, a = Р2 ΠI
Let 1 + x4 = x4 t2
)
153. Ú x– 1/2 (2 + 3x1/3)– 2 dx
1 = __ Ú (1 – t2) dt 2
)
9 8 = 4 _________ 1/4 9 – _________ 1/4 8 + c (x + 1) (x + 1)
– 3 1 = ___ – __ = – 2 = Integer. 2 2
)
9 8 = 4 _______ – _______ + c (t + 1)9 (t + 1)8
b+1 – 7 + 1 __ 1 Here, _____ + p = _______ – g 4 2
(
(t + 1) – 1 = 4 Ú __________ dt (t + 1)10
( ( (
dx ______ 151. Ú _________ 72 x ÷ 1 + x4
t dt = 4 Ú _______ (t + 1)10
1 1 = 4 Ú _______ – _______ dt (t + 1)9 (t + 1)10
2
(
1.125
1 __ + 1 b_____ +1 3 Here, = _____ = 1 ΠI g 4 __ 3
Let (1 + x4/3) = t7 4 fi __ x1/3 dx = 7t6 dt 3 21 fi x1/3 dx = ___ t6 dt 4 21 = ___ Ú (t6 ◊ t) dt 4
1.126 Integral Calculus, 3D Geometry & Vector Booster
21 t8 = ___ × __ + c 4 8
4/3 8/7 21 (1 + x ) = ___ × __________ +c 4 8
1 fi x4 = ______ 2 (t – 1)
2t fi 4x3 dx = – _______ dt (t2 – 1)2
155. Ú x– 6 (1 + 2x3)2/3 dx
t dt fi x3 dx = – ________ 2 2 (t – 1)2
b +1 Р6 + 1 __ 2 Here, _____ + a = ______ + = Р1 ΠI g 3 3
dx ______ = Ú __________ 11 2 x ÷ 1 + x4
1 = – __ Ú (t2 – 1)2 dt 2
1 = – __ Ú (t4 – 2t2 + 1) dt 2
Let 1 + 2x3 = x3t3
fi x3 (t2 – 2) = 1
1 fi x3 = ______ 3 (t – 2)
3t2 dt fi 3x2 dx = – _______ (t3 – 2)2
t2 dt + x4 fi x2 dx = – _______ 2 where t = 1_____ 3 (t – 2) x4
(
÷
______
(
2 3× __
3
_______ __
)
(tx) 3 = Ú ______ dx x6
÷1 + 4÷ x __ 158. Ú ________ dx 4 3 ÷ x
t2x2 dx = Ú ______ x6
t2 dt t2 = Ú ________ × – _______ 3 1 (t – 2)2 ______ 3 2 t – 2)
(
)
= – Ú t4 dt
t5 = – __ + c 5
( )
(
)
1 1 + 2x3 5/3 = – __ ______ + c 5 x3
156. Do yourself. dx _____ 157. Ú __________ 11 2 x ÷ 1 + x4
)
2t3 1 t5 ___ = – __ __ – + t + c 2 5 3
Let (1 + x1/4) = t3 1 fi __ x– 3/4 dx = 3t2 dt 4 dx fi ___ = 12t2 dt x3/4
= 12 Ú t × t2 dt
= 12 Ú t3 dt
t4 = 12 __ + c 4
= 3t4 + c
= 3 (1 + x1/4)4/3 + c
( )
dx _________ 159. Ú ______________ 2 (1 + ÷ x + x + 1 )
= Ú x– 11 (1 + x4)– 1/2 dx
Here a = 1 > 0, therefore we put the Euler first substitution, i.e.
= Ú x– 11 (1 + x4)– 1/2 dx
_________
÷ x2 + x + 1 = t – x
b+ 1 Р11 + 1 __ 2 Here, _____ + a = _______ + = Р2 ΠI g 4 4
fi (x2 + x + 1) = (t – x)2
Put (1 + x4) = x4t2
fi (x2 + x + 1) = t2 – 2tx + x2
fi x4 (t2 – 1) = 1
fi (1 + 2t) x = t2 – 1
Indefinite Integrals
( )
t2 – 1 fi x = _____ 2t + 1
dt = – 2 Ú _____ t2 – 1
2 (t2 + t + 1 fi dx = ___________ dt (2t + 1)2
t – 1 1 = – 2 × __ log _____ + c 2 t+1
t – 1 = log _____ + c t+1
_________
Now, 1 = ÷ x 2 + x + 1 = 1 + t – x
( )
t2 – 1 = 1 + t – _____ 2t + 1 dx _________ Thus, Ú _______________ 2 ( 1 + ÷ x + x + 1 )
(
= – (x – 2) (x – 5) = (x – 2) (5 – x), so we can apply Euler third
2 (t + t + 1) = Ú _________________ dt 2 (t + 3t + 2) (2t + 1)
substitution, i.e. ___________
÷ 7x – 10 – x2 = (x – 2) t
) dt dt 2 – __ ( Ú ____ – 2 _____ 3 t + 2 Ú 2t + 1 )
8 dt dt dt 4 = 2 – 2 Ú ____ + __ ____ + __ _____ t + 1 3 Ú t + 2 3 Ú 2t + 1
_____________
fi ÷– (x – 2) (x – 5) = (x – 2) t fi (5 – x) = (x – 2) t2
)
8 2 = 2 – 2 log |t + 1| + __ log |t + 2| + __ log |2t + 1| 3 3 2 – __ (log |t + 2| – log |2t + 1|) + c 3 _________
where
t = x + ÷x 2 + x + 1
dx _________ 160. Ú _____________ 2 x+÷ x – + 1 Here, c = 1 > 0, therefor we put the Euler second substitution, i.e. _________ ÷ x2 – x + 1 = t x
(x2 – x + 1) = t2x2 – 2t x + 1
fi
2t – 1 x = ______ t2 –1
fi
– 2 (t2 – t + 1) dx = ____________ dt t2 – 1)2
dx _________ Thus, Ú _____________ 2 x+÷ x – x + 1
5 + 2t2 fi x = _______ 1 + t2 6t dt fi dx = – _______ 2 (t + 1)2
(
– 2 (t2 – t + 1) ____________ 2 2 (t – 1) = Ú _____________ dt t2 – t + 1 _________ 2 (t – 1)
)
5 + 2t2 3t Now, (x – 2) t = _______ – 2 t = _____ 2 2 t +1 t +1 dx ___________ Thus, Ú x ____________ – 10 – x2 ÷ 7x
– 6t x = Ú _______ × _______ dt (x – 2) t (1 + t2)2
5 + 2t2 _______ 2 (1 + t ) – 6t = Ú ________ × _______ dt 2 2 3t (1 + t ) _______ (1 + t2)
5 + 2t2 = – 2 Ú _______ dt (1 + t2)2
2 (1 + t2) + 3 = – 2 Ú ___________ dt (1 + t2)2
dt dt = – 4 Ú _______ – 6 Ú _______ (1 + t2) (1 + t2)2
–1
fi
)
Here, 7x – 10 – x2 = – (x2 – 7x + 10)
2
(
_________
x dx ___________ 161. Ú ____________ – 10 – x2 ÷ 7x
( )
| |
1 + ÷x 2 – x + 1 where, t = _____________ x
2(t2 + t + 1) ___________ 2 (2t + 1) = Ú ____________ dt t2 – 1 ______ 1 + t – 2t + 1
(
| |
1.127
1.128 Integral Calculus, 3D Geometry & Vector Booster
(
)
fi
(x – 1) (x – 2) = (x – 1)2 t2
fi
(x – 2) = (x – 1) t2
fi
x (1 – t2) = 2 – t2
– 10 – x2 ÷ 7x where, t = ____________ (x – 2)
fi
dx _________ 162. Ú _____________ 2 x–÷ x – x + 2
t2 – 2 x = ______ 2 t –1
fi
(t2 – 1) (2t) – (t2 – 2) 2t dx = ____________________ dt (t2 – 1)2
fi
2t (t2 – 1 – t2 + 2) dx = ________________ dt (t2 – 1)2
fi
2t dt dx = _______ 2 (t – 1)2
t 1 ______ 1 = – 4 tan– 1 (t) – 6 __ + __ tan– 1 t + c 2 t2 + 1 2
3t = – 7 tan (t) – ______ 2 + c t +1 – 1
___________
_________
Put ÷x 2 – x + 2 = t + x fi (x2 – x + 2) = (t + x)2 fi (x2 – x + 2) = t2 + 2tx + x2 fi (2t + 1) x = 2 – t2
( )
(2x + 1) (– 2t) – (2 – t2) 2 fi dx = ______________________ dt (2t + 1)2
2t dt _______ 2 (t – 1)2 _________________ = Ú dt t2 – 2 1 _____ 2 × – _______ t –1 (t2 – 1)
– 4t2 – 2t + 2t2 – 4 fi dx = _________________ dt (2t + 1)2
2t dt = – Ú _______ 2 (t – 2)
= – log|(t2 –2)| + c
– 2t2 – 2t – 4 fi dx = ____________ dt (2t + 1)2
+ 2 ÷ x 2 – 3x where t = ___________ (x – 1)
( )
2 – t2 fi x = _____ 2t + 1
__________
dx ______ 165. Ú __________ x+÷ x 2 – 1
1 2t2 + 2t + 4 = Ú __ × ___________ dt t (2t + 1)2
]
)
(x – ÷ x 2 – 1 = Ú ___________ dx (x2 – x2 + 1)
= Ú (x – ÷x 2 – 1 ) dx
x2 x ÷x 2 – 1 __ 1 = __ – ________ – log x + ÷x 2 – 1 + c 2 2 2
1 t +t+2 = 2 Ú __ × _________ dt t (2t + 1)2
9 dt 7 _______ dt dt = 2 2 Ú __ – __ Ú – __ ________ t 2 (2t + 1) 2 Ú (2t + 1)2
9 7 = 2 2 log |t| – __ log |2t + 1| – ________ + c 4 4 (2t + 1)
[
(
______
______
_________
where t = ( ÷x 2 – x + 2 – x ) 163. Do yourself. dx __________ 164. Ú _____________ 2 x ÷x – 3x + 2
fi
2
(
______
|
______
Here, x2 – 3x + 2 = (x – 1) (x – 2)
______
(x – ÷x 2 – 1 ) ______ = Ú ______________________ ______ dx (x + ÷ x 2 – 1 ) (x – ÷ x 2 – 1 )
2
__________ Let ÷ x2 – 3x +2
( )
x dx ______ 166. Ú __________ = Ú x (x – ÷x 2 – 1 ) dx 2 x+÷ x – 1 ______
= Ú (x2 – x ÷x 2 – 1 ) dx
x3 = __ – Ú x÷x 2 – 1 dx 3
= (x – 1) t 2 2
(x – 3x + 2) = (x – 1) t
________
______
|
)
Indefinite Integrals
8 cos6 x ◊ sin x ___ = __________ + cos4 x ◊ sin x 9 63
Let x2 – 1 = t2
xdx = tdt
x = __ – Ú t2 dt 3
170. We have,
x3 = __ – 3
Ú cos8 x dx
2 3/2 x3 (x – 1) = __ – _________ +c 3 3
3
t3 __ + c 3
(
Ú sin x dx 5
cos x sin4 x __ 4 cos x sin2 x __ 2 = – _________ – – _________ + I1 3 3 5 5
(
9 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x 10 80 63 cos3 x ◊ sin x __ 5 + ___ __________ + I2 80 6 6
)
(
4
cos x sin x ___ 4 = – _________ – (cos x sin2 x + 2 cos x) + c 5 15
Ú sin x dx
6
(
)
63 cos3 x ◊ sin x ___ 5 + ___ __________ + (x + sin x ◊ cos x) + c 80 12 6
171. We have, Ú tan5 x dx
5 cos x sin5 x __ = – _________ + I4 6 6
( (
)
tan4 x = _____ – I3 4
5 cos x ◊ sin x __ 3 cos x sin x __ = _________ + – __________ + Ú sin2 x dx 4 4 6 6
tan4 x _____ tan2 x = _____ – + I1 4 2
5 cos x sin5 x ___ = – _________ – cos x ◊ sin3 x 24 6
tan4 x _____ tan2 x = _____ – + Ú tan x dx 4 2
(
)
5 cos x ◊ sin3 x __ 3 cos x sin5 x __ = – _________ + – __________ + I2 4 4 6 6 5
(
)
3
3 + __ (x – sin x cos x) + c 4
169. We have,
8 cos6 x ◊ sin x __ = __________ + I5 9 9
(
)
8 48 cos6 x ◊ sin x ___ = __________ + cos4 x ◊ sin x + ___ I3 9 63 63 6
8 cos x ◊ sin x ___ = __________ + cos4 x ◊ sin x 9 63
tan4 x _____ tan2 x = _____ – + log |sec x| + c 4 2 172. We have,
8 cos4 x ◊ sin x __ 6 cos6 x ◊ sin x __ = __________ + __________ + I3 7 7 9 9
)
Ú tan6 x dx
Ú cos7 x dx
)
9 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x 10 80
168. We have,
)
9 63 cos7 x ◊ sin x ___ = __________ + cos5 x ◊ sin x + ___ I4 10 80 80
cos x sin4 x __ 4 = – _________ + I3 5 5
)
9 cos5 x ◊ sin x __ cos7 x ◊ sin x ___ 7 = __________ + __________ + I4 10 10 8 8
(
48 cos2 x ◊ sin x __ 4 + ___ __________ + sin x + c 63 5 5
9 cos7 x ◊ sin x ___ = __________ + I6 10 10
167. We have,
1.129
( (
)
tan5 x = _____ – I4 5
)
tan5 x _____ tan3 x = _____ – + I2 3 5 tan5 x _____ tan3 x = _____ – + Ú tan2 x dx 3 5
(
tan5 x _____ tan3 x = _____ – + tan x – x + c 3 5 2 48 cos x ◊ sin x __ 4 + ___ __________ + I1 173. We have, 63 5 5 Ú cot7 x dx.
(
)
)
tan5 x _____ tan3 x = _____ – + Ú (sec2 x – 1) dx 3 5
1.130 Integral Calculus, 3D Geometry & Vector Booster
cot6 x = – _____ – I5 + c 6
cot6 x cot4 x = – _____ – – _____ – I3 + c 4 6
(
6
)
5 sec5 x ◊ tan x __ = __________ + I5 6 6
5 sec3 x ◊ tan x __ 3 sec5 x ◊ tan x __ = __________ + __________ + I3 4 4 6 6
5 15 sec5 x ◊ tan x ___ = __________ + sec3 x ◊ tan x + ___ I3 24 24 6
5 sec5 x ◊ tan x ___ = __________ + sec3 x ◊ tan x 24 6
4
cot x _____ cot x = – _____ + + I3 + c 4 6
cot6 x _____ cot4 x cot2 x = – _____ + + – _____ – I1 + c 4 2 6
cot6 x _____ cot4 x _____ cot2 x = – _____ + – – I1 + c 4 2 6
cot6 x _____ cot4 x _____ cot2 x = – _____ + – – log |sin x| + c 4 2 6
(
)
(
(
cot5 x – _____ – I4 + C 5 cot5 x _____ cot3 x = – _____ + + Ú cot2 x dx + c 3 5
179. We have, sin 3x Ú _____ dx sin x 2 sin (3 – 1) x = ___________ + I3–2 (3 – 1)
= sin 2x + I1
cot5 x _____ cot3 x = – _____ + + Ú (cosec2 x – 1) dx + c 3 5
= sin 2x Ú dx
cot5 x _____ cot3 x = – _____ + – cot x – x + c 3 5
= sin 2x + x + c
180. We have, Ú_____ sin 5x dx sin x 2 sin (5 – 1)x = ___________ + I5–2 (5 – 1)
175. We have, Ú sec2 x dx
sec x ◊ tan x __ 1 = _________ + I1 2 2
sec x ◊ tan x __ 1 = _________ + log |sec x + tan x| + C 2 2
176. We have, Ú sec4 x dx
2
sec x tan x __ 2 = _________ + I2 3 3
sec2 x tan x __ 2 = _________ + tan x + c 3 3
3 sec x ◊ tan x __ = __________ + I3 4 4
2 sin (3 – 1)x 1 = __ sin 4x + ___________ + I1 2 (3 – 1)
2 = __ sin 4x + sin 2x + x + c 3
3
sec x ◊ tan x = __________ 4
(
3
1 = __ sin 4x + I3 2
sin 6x Ú _____ dx sin x 2 sin (6 – 1)x = ___________ + I6–2 (6 – 1)
Ú sec5 x dx
181. We have,
177. We have,
)
3 x 1 + __ sec x ◊ tan __ + __ log |sec x + tan x| + c 4 2 2
Ú sec7 x dx
2 = __ sin 5x + I4 5 2 2 = __ sin 5x + __ sin 3x + I2 3 5
2 2 = __ sin 5x + __ sin 3x + Ú _____ sin 2x dx 3 5 sin x
2 2 = __ sin 5x + __ sin 3x + 2 Ú cos x dx 3 5
2 2 = __ sin 5x + __ sin 3x + 2 sin x + c 3 5
178. We have,
)
15 sec x ◊ tan x __ 1 + ___ _________ + log |sec x + tan x| + c 24 2 2
174. We have,
)
Indefinite Integrals
182. We have, Ú _____ sin 8x sin x 2 sin (8 – 1) x = ___________ + I6 (8 – 1)
2 2 = __ sin 7x + __ sin 5x + I4 7 5 2 2 2 = __ sin 7x + __ sin 5x + __ sin 3x + I2 7 3 5 2 2 2 = __ sin 7x + __ sin 5x + __ sin 3x + 2 sin x + c 3 5 5
183. We have Ú _______ dx (x2 + 2)2
(
x 1 2.2 – 3 + __ _______ I2–1 = _________ 2.2 (x2 + 2) 2 2.2 – 2
x 1 Ú _______ dx = __ (x2 + 2) + __ 4 4 (x2 + 2)
)
x 1 + __ I2 = __________ 12 (x2 + 3)2 4
x x 1 11 + __ __________ + __ __ I1 = __________ 12 (x2 + 3)2 4 2.3 (x2 + 3) 3 2
)
x x + _________ = __________ 12 (x2 + 3)2 24 (x2 + 3)
( )
x 1 + _____ __ tan– 1 ___ __ + c 24 ÷2 ÷2
185. We have, x + 1 Ú ____________ dx (x2 + 3x + 2)2
3 1 x + __ – __ 2 2 1 ____ ___________ + log + C 1 3 1 __ __ 2 ◊ x + + __ 2 2 2
3 x + __ 2 1 1 = – ____________ – __ ___________ 2 2 2 3 1 __ 2 (x + 3x + 2) x + – __ 2 4
( ) ]
x+1 + log _____ + c x+2
x3 1 = (log x) ◊ __ – __ I 3 3 2,0
x3 = (log x) ◊ __ – 3
x3 x3 = (log x) ◊ __ – __ + C 3 9
1 __ Ú x2 dx 3
187. We have,
(
[ ( ( ) ) ( (( )) )] [ ( ( ) )
Ú x2 log x dx
Ú _______ dx (x2 + 3)2 x 1 2.3 – 3 + __ _______ I3–1 = ________________ 2 (3 – 1) 3 (x2 + 3)2 3 22.3 – 2
3 x + __ 2 1 2 1 __________ __ __ = – (x + 3x + 2) – 2 2 3 2 __ 1 __ x + – 2 4
[ ( ) ( ) ]
186. We have,
( )
(
dx 1 1 = – ____________ – __ Ú _______________ 2 2 (x + 3x + 2) 2 3 2 1 22 __ x + – __ 2 2
x x 1 __ tan– 1 ___ __ + c = __ (x2 + 2) + ____ 4 4 ÷2 ÷2 184. We have,
)
(2x + 2) 1 = __ Ú ____________ dx 2 2 (x + 3x + 2)2 1 (2x + 1) – 1 = __ Ú ____________ dx 2 (x2 + 3x + 2)2 (2x + 3) 1 1 dx = __ Ú ___________ dx – __ Ú ___________ 2 (x2 + 3x +2)2 2 (x2 + 3x + 2)2
1.131
Ú x2 (log x)2 dx
x3 2 = (log x)2 __ – __ I 3 3 2,1
x3 = (log x)2 __ – 3
x3 2 x3 x3 = (log x)2 __ – __ log x) __ – __ + c 3 3 3 9
(
)
2 __ Ú x2 log x dx 3
(
)
188. We have, Ú x3 (log x)2 dx
x4 2 = (log x)2 __ – __ I 4 4 3,1
x4 1 x4 1 = (log x)2 __ – __ (log x) __ – __ I 4 2 4 4 3,0
x4 = (log x)2 __ – 4
x4 = (log)2 __ – 4
( ( (
)
x4 1 __ (log x) __ – 2 4
x4 1 __ (log x) __ – 2 4
)
1 __ Ú x3 dx 4
)
x4 ___ + c 16
1.132 Integral Calculus, 3D Geometry & Vector Booster
[ [ [
189. We have, Ú x2 (1 – x)3 dx 3
x3 (1 – x)3 __ 1 = _________ + I2,2 2 6
(
3
3
2
)
Hints
1. We have,
cos 2x – cos x dx Ú ___________ 1 – cos x
and
solutions
( Ú (
)
1 – 3sin2x cos2x = Ú ______________ dx sin2x cos2x
sin2x + cos2x – 3sin2x cos2x = _______________________ dx sin2x cos2x
= Ú (sec2x + cosec2x – 3)dx
= (tan x – cot x – x) + c
2cos2x – cos x – 1 = Ú _______________ d x 1 – cos x
(cos x – 1)(2cos x + 1) = Ú ___________________ dx 1 – cos x
= – Ú (2cos x + 1) dx
cos 2x Ú _________ dx cos2x sin2x
= – (2sin x + x) + c
cos2x – sin2x = Ú ___________ dx cos2x sin2x
= Ú (cosec2x – sec2x)dx
= – (cot x + tan x) + c
cos 5x + cos 4x Ú _____________ d x 1 – 2 cos 3x
sin3x (cos 5x + cos 4x) = Ú ___________________ dx sin 3x – sin 6x
9x 3x 3x x 2sin ___ cos ___ 2cos ___ cos __ 2 2 2 2 = Ú ____________________________ dx 9x 3x – 2cos ___ sin ___ 2 2
( ) ( ) ( ) ( ) ( ) ( ) 3x x = – Ú 2cos( ___ )cos ( __ ) dx 2 2
= – Ú (cos 2x + cos x) dx
sin 2x = – _____ + sin x + c 2
4. We have,
2. We have,
(
)
3. We have,
(
)]
Note: Q-190 to 197, do yourself.
(
)
sin6x + cos6x Ú ___________ 2 sin x cos2xdx
)]
3 2 x3 (1 – x)3 __ x3 1 x (1 – x) 2 x3 ___ = _________ + _________ + __ __ + + c 2 12 6 5 5 4
x (1 – x) 1 x (1 – x) 2 = _________ + __ _________ + __ I2,1 2 6 5 5
)]
3 2 x3 (1 – x)3 __ 1 x (1 – x) 2 x3 __ 1 = _________ + _________ + __ __ + Ú x2 dx 2 4 6 5 5 4
3
x (1 – x) 3 = _________ + _________ I (2 + 3 + 1) (2 + 3 + 1) 2,2
3
( (
3 2 x3(1 – x)3 __ 1 x (1 – x) 2 x3 __ 1 = ________ + _________ + __ __ + I2,0 2 4 6 5 5 4
5. We have,
sin 2x Ú __________ dx sin 5x sin 3x
sin (5x – 3x) = Ú ___________ dx sin 5x sin 3x
sin 5x cos 3x – cos 5x sin 3x = Ú ______________________ dx sin 5x sin 3x
= Ú (cot 3x – cot 5x)dx
1 1 = __ log |sin 3x| – __ log |sin 5x| + c 3 5
6. We have,
Ú___________ sin x dx sinx + cos x
)
Indefinite Integrals
2sin x 1 = __ Ú ___________ dx 2 sin x + cos x
b = – __ a log |t| + k
1 (sin x + cos x) + (sin x – cos x) = __ Ú _________________________ dx 2 sinx + cos x
b – x = – __ a log|a e + c| + k
sin x – cos x 1 = __ Ú 1 + __________ dx 2 sin x + cos x
1 = __ (x – log|sin x + cos x|) + c 2
(
)
10. We have, dx Ú ______ 1 + ex
7. We have,
dx Ú _____________ sec x + cosec x
exdx = Ú _________ x x e (e + 1)
dt = Ú ______ , t(t + 1)
sin x cos x = Ú ___________ dx sin x + cos x
1 2 sin x cos x = __ Ú ___________ dx 2 sin x + cos x
1 (1 + sin2x) – 1 = __ Ú _____________ dx 2 sin x + cos x
2 dx 1 (sin x + cos x) 1 = __ Ú_____________ dx – __ Ú __________ 2 sin x + cos x 2 sin x + cos x
fi exdx = dt
(
)
1 1 = Ú __ – ____ dt t t+1 t = log ____ + c t+1
| |
| |
ex = log _____ x + c e +1
11. We have, x+9 Ú _______ dx x3 + 9x
dx 1 1 = __ Ú (sin x + cos x)dx – __ Ú __________ 2 2 sin x + cos x
x + 9 + x2 – x2 = Ú _____________ dx x(x2 + 9)
dx 1 1__ _________ = __ Ú (sin x + cos x)dx – ____ Ú p 2 2÷2 sin x + __ 4 p x __ 1 1__ __ ____ __ = (sin x – cos x) – log tan + + c 2 2 8 2÷2
x + (9 + x2) – x2 = Ú _______________ dx x(x2 + 9)
dx dx x _______ = Ú _____ + Ú ___ dx x – Ú (x2 + 9) x2 + 9
x 1 1 = __ tan– 1 __ + log|x| – __ log|x2 + 9| + c 3 3 2
|
( (
) )|
8. We have,
Ú tan 3x tan 2x tan x dx = Ú (tan 3x – tan 2x – tan x) dx
1 1 = __ log |sec 3x| – __ log |sec 2x| 3 2
Ú sin x – cos x/ex + sin xdx
– log|sec x| + c
(ex + sin x) – (ex + cos x) = Ú _____________________ dx ex + sin x)
1 – (ex + cos x) = Ú _____________ x dx (e + sin x)
= (x – log|ex + sin x|) + c
9. We have,
b Ú _______ dx a + c ex
be– x = Ú ________ dx c e– x + c
dt b __ = – __ a Ú t ,
( )
12. We have
Let ex = t
1.133
(
)
13. We have, cos x – sin x + 1 – x Ú _________________ d x |ex + sin x + x Let a e– x + c = t x
fi – a e dx = dt
(cos x – 1) – (sin x + x) = Ú ____________________ dx ex + sin x + x
1.134 Integral Calculus, 3D Geometry & Vector Booster
[
cos(x – a)cos(x – b) 1 = _________ __________________ dx cos(b – a) Ú sin(x – a)cos(x – b)
(ex + cos x – 1) – (ex + sin x + x) = Ú ___________________________ dx ex + sin x + x = log|ex + sin x + x| – x + c
]
sin(x – a)sin(x – b) + _________________ dx sin(x – a)cos(x – b)
1 = _________ (cot(x – a) + tan(x – b))dx cos(b – a) Ú
14. We have, sin(x + a) Ú _________ dx sin(x + b)
1 = _________ [log |sin(x – a)| – log(sec{x – b})] + c cos(b – a)
sin(t – b + a) = Ú ____________ dt sin t
1 = _________ [log |sin(x – a)cos(x – b)|] + c cos(b – a)
sin(t + (b – a)) = Ú _____________ dt sin t
17. We have, ____
sin t.cos(a – b) + cos t.sin(a – b) = Ú ____________________________ dt sin t
÷tan x Ú________ dx sin x cos x
= cos(a – b) Ú dt + sin(a – b) Ú cot dt
sec2x ÷tan x = Ú __________ dx tan x
2 sec x ____ = Ú _____ dx tan x ÷
= (x + b)cos(a – b) + sin(a – b)log |sin (x + b)| + c
2t dt = Ú ____ , t
dx 15. We have Ú ________________ sin(x – a)sin(x – b)
= t cos(a – b) + sin(a – b)log |sin t | + c
sin(b – a) 1 = ________ ________________ dx sin(b – a) Ú sin(x – a)sin(x – b)
= 2t + c
= 2÷tan x + c
]
cos(x – a)sin(x – b) – _________________ dx sin(x – a)sin(x – b)
1 = ________ [cot(x – b) – cot(x – a)]dx sin(b – a) Ú 1 = ________ [log|sin(x – b)| – log|sin(x – a)|] + c sin(b – a)
( |
|)
sin(x – b) 1 = ________ log _________ + c sin(b – a) sin(x – a) 16. We have, dx Ú ________________ sin(x – a)cos(x – b) cos(b – a) 1 = _________ Ú _________________ dx cos(b – a) sin(x – a)cos(x – b) cos ( (x – a) – (x – b) ) 1 = _________ ___________________ dx cos(b – a) Ú sin(x – a)cos(x – b)
fi sec2xdx = 2tdt ____
dx _____ Ú _______ x÷x 4 – 1
[
Let tan x = t2
18. We have,
sin ( (x – a) – (x – b) ) 1 = ________ ___________________ dx sin(b – a) Ú sin(x – a)sin(x – b) sin(x – a)cos(x – b) 1 = ________ __________________ dx sin(b – a) Ú sin(x – a)sin(x – b)
____
3 x_____ dx = Ú ________ 4 4 x÷ x – 1
dt 1 = __ Ú t _______ 2 (t2 + 1)t
1 = __ Ú______ dt 2 (t2 + 1)
1 = __ tan– 1(t) + c 2
1 ) + c = __ tan– 1 ( ÷x 4 – 1 2
_____
19. Let dx Ú _______ 2 (x + 1)
(
)
2x◊x 1 = _______ 2 Ú dx – Ú – ________ dx (x + 1) (x2 + 1)2
(x2 + 1) – 1 1 = _______ 2 Ú dx – 2Ú __________ dx (x + 1) (x2 + 1)2
(
)
Indefinite Integrals
(
)
2t + 1 1 = ___ __ tan– 1 ______ __ + c 3 ÷3 ÷
dx dx x = _______ 2 + 2 Ú _______ – 2 Ú _______ (x + 1) (x2 + 1) (x2 + 1)2
1.135
(
)
2x2 + 1 1 = ___ __ tan– 1 _______ __ + c 3 ÷3 ÷
dx x fi 2 Ú ________ 2 = _______ 2 + 2 tan– 1x – tan– 1x + c 2 (x + 1) (x + 1)
dx x 1 fi Ú ________ = _______ + __ tan– 1x + c (x2 + 1)2 (x2 + 1) 2
22. Let I = Ú sec3x dx
20. We have,
= Ú (sec2x ◊ sec x)dx
= sec x Ú sec2x dx – Ú (sec x ◊ tan x ◊ tan x) dx
= sec x Ú sec2x dx – Ú (sec x ◊ tan2 x)dx
= sec ◊ x tan x – Ú (sec x (sec2x – 1))dx
= sec ◊ x tan x – Ú sec3x dx + Ú secx dx
fi
2I = sec x ◊ tan x + Ú secx dx
fi
2I = sec x ◊ tan x – log|sec x + tan x| + c
fi
1 I = __ (sec x ◊ tan x + log|sec x + tan x|) + c 2
÷ Ú ÷
_________
sin(x – a) Ú _________ dx sin(x + a) ____________________
sin(x – a) _________ sin(x – a) = _________ × dx sin(x + a) sin(x – a) sin(x – a) ____________ = Ú _____________ dx 2 2x – sin a ÷ sin sin x cos a – cos x sina ___________ = Ú ____________________ dx 2 2 x – sin a ÷ sin sin x ____________ = cos a Ú _____________ dx 2 2 x – sin a ÷ sin
cos x ____________ – sin a Ú _____________ dx 2 2 x – sin a ÷ sin
23. Let I = Ú cosec3x dx
sin x ____________ = cos a Ú _____________ dx 2 2 a – cos x ÷ cos
cos x ____________ – sin a Ú _____________ dx 2 2 x – sin a ÷ sin
(
= Ú cosec3x ◊ cosec x dx
= cosec x Ú cosec2x dx –
Ú (– cosec x ◊ cot x ◊ – cot x)dx
)
cos x = – cos a sin – 1 _____ cosa
= – cosec x ◊ cot x – Ú cosec x ◊ cot2 x dx
= – cosec x ◊ cot x – Ú cosec x(cosec2x – 1)dx
21. We have,
= – cosec x ◊ cot x – Ú cosec3x dx + Ú cosec x dx
x Ú __________ dx x4 + x2 + 1
fi 2I = – cosec x ◊ cot x + Ú cosec x dx
|
– sin a log sin a +
____________ 2 2x – sin a + ÷ sin
1 dx = __ Ú ___ + t + 1, 2 t2
Let x2 = t
fi 2xdx = dt
( ) __
|
3 2 ÷ dx 1 = __ Ú _______ 2 + ___ 2 2 1 t + __ 2 1 t + __ 1 2 2 – 1 __ = __ × ___ __ tan _____ + c 2 ÷ 3 3 ÷ ___ 2
( )
( )
c
fi 2I = – cosec x ◊ cot x + log|cosec x – cot x| + c 1 fi I = __ (– cosec x ◊ cot x + log|cosec x – cot x| + c 2 24. We have, x2 – 1 ___________ Ú _____________ dx x÷x 4 + 3x2 + 1
x2 – 1 ________ = Ú ______________ dx 1 2 4 x x + __ 2 + 3 x
÷(
)
1.136 Integral Calculus, 3D Geometry & Vector Booster
( )÷(
________
)
1 1 2 = Ú 1 – __ 2 x + __ x – 2 + 3 x
dt ______ = Ú _______ , 2 ÷ t + 1
1 Let x + __ x = t
( )
1 fi 1 – __2 dx = dt x
|
_____ ÷t 2 + 1 +
|
= log t +
1 1 2 __ = log x + __ 3 + c x + x + x2 +
| (
25. We have,
c
÷(
)|
dx Ú _____________ x4 + 18x2 + 81 = Ú _______ dx (x2 + 9)2
Now, Ú ________ dx (x2 + 9)2
(
( )
27. We have, Ú sin 4x etan x dx 2
____________
)
( ) ( ) x x = __ ( 2tan ( __ ) ) + c 2 2 x = x tan ( __ ) + c 2
x x x x = __ Ú sec2 __ dx – Ú tan __ dx + Ú tan __ dx 2 2 2 2
)
= Ú 2sin2x . cos2x.etan x dx
1 – tan2x tan2 x 2 tan x = Ú 2 ________ ◊ ________ ◊ e dx 2 1 + tan x 1 + tan2x
1 – tan x = 2Ú __________ ◊ 2tan x ◊ sec2x ◊ etan x dx (1 + tan2x)3
2
2
2
Let tan2x = t
fi
et(1 – t) = 2 Ú ________ dt (1 + t)3
2tan x sec2x dx = dt
– 2x ◊ x 1 = _______ 2 Ú dx – Ú _______ dx (x + 9) (x2 + 9)2
(x2 + 9) – 9 x = _______ 2 + 2 Ú __________ dx (x + 9) (x2 + 9)2
(t + 1) – 2 = – 2 Ú et _________ dt (t + 1)3
x = _______ 2 + 2 Ú _______ dx – 18 Ú ________ dx 2 (x + 9) (x + 9) (x2 + 9)2
1 2 = – 2 Ú et _______ – ______ dt (t + 1)2 (t + 1)3
t = – 2 Ú _______ e + c (t + 1)2
(
)
dx fi 19Ú ________ (x2 + 9)2
( ) x ( __ ) + c 3
( )
(
)
(
)
x x x 2 1 = _______ 2 + __ tan– 1 __ – __ tan– 1 __ + c 3 3 3 (x + 9) 3
28. We have,
x 1 = _______ 2 + __ tan– 1 (x + 9) 3
2x + 2 ____________ Ú sin– 1 _____________ dx 2 + 8x + 13 ÷ 4x
fi
x x 1 dx = _________ + ___ tan– 1 ( __ ) + c Ú ________ 2 2 3 19(x2 + 9) 57
(x + 9)
26. We have,
(
)
x + sin x Ú ________ dx 1 + cos x
(
)
x sin x = Ú ________ + ________ dx 1 + cos x 1 + cos x
2sin (x/2)cos(x/2) x = Ú _________ + _______________ dx 2 2cos (x/2) 2cos2(x/2)
(
(
)
( )
( ) )
x x x = Ú __ sec2 __ + tan __ dx 2 2 2
(
)
(
)
2x + 2 ____________ = Ú sin1 ______________ dx + 8)2 + 32 ÷ (2x
Let (2x + 2) = 3tanq
fi 2dx = 3sec2q dq
(
)
3tanq 3 = __ Ú sin– 1 ______ sec2q dq 2 3secq
3 = __ Ú q sec2q dq 2
3 = __ (q Ú sec2q dq – Ú tanq dq ) 2
3 = __ (q tanq – log|secq |) + c 2
Indefinite Integrals
(
( ) ( ÷ ( ) ) ) 3 2x + 2 __ ( ______ – log (4x + 8x + 13) + c 3 ) 4 ___________
3 2x + 2 2x + 2 2x + 2 2 = __ ______ tan – 1 ______ – log 1 + ______ + c 2 3 3 3
= (x + 1) tan– 1
2
29. We have,
2 = Ú _________ cos x dx + Ú __________ sin x dx (2 + cos x) (2 + cos x)2
2 1 = Ú _________ Ú cos x dx – Ú __________ sin x dx (2 + cos x) (2 + cos x)2
sin2x + Ú __________ dx (2 + cos x)2
x2 Ú _____________ dx (x sinx + cos x)2
sin x = _________ + c (2 + cos x)
(x cosx)(x sec x) = Ú _____________ dx (x sin x + cos x)2
(x cosx) = x sec x Ú ______________ dx (x sin x + cos x)2
32. We have,
x sec x = – ____________ + sec2x dx x sin x + cos x Ú
( ) 1 + sin2q = Ú cos(2q ) log ( ________ dq cos2q ) 1 + sin 2q = log (_________ cos 2q ) Ú cos (2q ) dq
x sec x = – ______ + cos x + tanx + c x sin x
sin (2q ) ______ 2 – Ú _______ ◊ dq cos 2q 2
cosq + sinq Ú cos(2q ) × log ___________ dq cosq – sinq
(sec x + x sec x ◊ tan x) – Ú – _________________ dx x sin x + cos x
(
30. We have,
x sec2x + tan x = Ú x _____________ dx (x tan x + 1)2
x sec2x + tan x = x2 _____________ dx (x tan x + 1)2
( Ú (
) )
2
(
33. We have,
(
(
)
)
x x cos x = – _________ + 2 Ú _____________ dx (tan x + 1) (x sin x + cos x) x2 = – _________ + 2log|x sin x + cos x| + c (tan x + 1)
31. We have, sec x (2 + sec x)
dx Ú ______________ 2
1 2cos x + = Ú __________ dx (2 + cos x)2
2cos x + cos2x + sin2x = Ú ___________________ dx (2 + cos x)2
cos x(2 + cos x) + sin2x = Ú ____________________ dx (2 + cos x)2
(1 + 2 sec x)
)
x4 + 2 Ú ex _________ dx (1 + x2)5/2
1 – Ú 2x. – __________ dx (x tan x + 1) 2
)
(2q ) __ 1 + sin 2q 1 = log _________ × sin ____ – log |sec 2q | + c 2 2 cos 2q
x2(x sec2x + tan x) dx Ú ________________ (x tan x + 1)2
1.137
4 (x + 1) + 1 = Ú ex __________ dx (1 + x2)5/2
(x2 + 1)2 + (1 – 2x2) = Ú ex __________________ dx (1 + x2)5/2
(x2 + 1)2 (1 – 2x2) _________ = Ú ex _________ + dx (1 – x2)5/2 (1 + x2)5/2
(1 – 2x2) 1 _________ = ex _________ + dx (1 + x2)1/2 (1 + x2)5/2
x 1 = Ú ex _________ + _________ dx (1 + x2)1/2 (1 + x2)5/2
( Ú ( (
)
)
)
(
)
(1 – 2x2) x __________ + Ú ex _________ + dx (1 + x2)3/2 (1 + x2)5/2
(
)
x 1 = ex _______ ______ + _________ + c 2 (1 + x2)3/2 ÷ 1 + x
1.138 Integral Calculus, 3D Geometry & Vector Booster 34. We have,
36. We have,
(
)
Ú cot– 1(1 + x + x2) dx
x cos3x – sin x Ú esin x ____________ dx cos2x = Ú e
(x cosx – sec x ◊ tan x) dx
= Ú e
sin x ◊ cos x x cosx – _________ dx cos3x
sin x
sin x
(
(
)
)
t = Ú et sin– 1t – ________ dt, (1 – t2)3/2
Let t = sin x
fi dt = cosx dx
[
(
)]
t 1 1 = Ú et sin– 1t – ______ _____ + _______ _____ – ________ dt 2 2 (1 – t2)3/2 ÷ 1 – t ÷1 – t
(
)
1 _____ = e sin t – _______ + c ÷ 1 – t2 t
– 1
dx Ú ____________ cosx + cosec x
= Ú tan– 1(1 + x)dx + Ú tan– 1x dx
)
Let 1 + x = t fi dx = dt
= Ú tan– 1t dx +
Ú tan– 1x dx
1 = t tan– 1t – __ log |t2 + 1| + 2 1 x tan– 1x – __ log |x2 + 1| + c 2 1 = (x + 1)tan– 1(x + 1) – __ log|(x + 1)2 + 1| 2 1 x tan– 1x – __ log |x2 + 1| + c 2
37. We have,
= Ú ____________ sin x dx sin x ◊ cos x + 1
= Ú ______________ 2sin x dx 2sin x ◊ cos x + 2
= Ú _________ 2sin x dx sin 2x + 2
(sin x + cos x) + (sin x – cos x) = Ú __________________________ dx sin 2x + 2
Ú tan– 1 (1 + x + x2) dx
(
p = Ú __ – tan– 1(1 + x + x2) dx 2
(sin x + cos x) (sin x – cos x) = Ú ____________ dx + Ú _____________ dx sin 2x + 2 2 + sin 2x + 2x
Let sin x – cos x = t and sin x + cos x = v fi (cos x + sin x)dx = dt and (cos x – sin x)dx = dy
= Ú _________ dt – Ú _________ dv 1 + t2 + 2 v2 – 1 + 2
dv dt = Ú _____ – Ú ______ 2 2 v +1 t –3
| | __
(1 + x) – x = Ú tan– 1 __________ dx 1 + x(1 + x)
35. We have,
)
1 = Ú tan– 1 ________ dx 1 + x + x2
= esin x(x – sec x) + c
( (
t – ÷3 1__ __ = – ____ log ______ – tan– 1(v) + c 2÷3 t +÷ 3 __
(sin x – cos x) – ÷ 3 1__ __ = – ____ log| ________________ 2÷3 (sin x – cos x) + ÷ 3 – tan– 1(sin x + cos x) + c
)
p 1 = __ x – (x + 1)tan– 1(x + 1) – __ log|(x + 1)2 + 1| 2 2 38. We have,
1 – x tan– 1x – __ log |x2 + 1| + c 2
sin x Ú _____ dx sin 4x sin x = Ú ____________ dx 2 sin 2x cos 2x 1 dx = __ Ú _______________ 4 cos x (2cos2x – 1) 1 cos x dx = __ Ú __________________ 4 (1 – sin2x)(1 – 2sin2x) 1 dt = __ Ú _____________ , 4 (1 – t2)(1 – 2t2)
Let t = sin x
fi
1 dt = __ Ú ______________ 4 (t2 – 1)(2t2 – 1)
dt = cos x dx
Indefinite Integrals
(
) ( )
1 1 2 = __ Ú _____ – _______ dt 4 t2 – 1 2t2 – 1
dv dv 3 3 = __ Ú __________ – __ _______ 2 (v2 – v + 1) 2 Ú (v3 + 1)
| | | |
1 1 dt ___ 1 2 = __ Ú _____ dt – __ Ú __ __ 2 2 4 t – 1 4 t ÷2
2
| |
( ) ( )
(
)
2v – 1 1 1 = ___ __ tan– 1 ______ __ – __ log |v3 + 1| 2 3 ÷3 ÷
__
v–1 3 + __ Ú _________ dv 2 v2 – v + 1
t ÷2 – 1 t – 1 ____ 1 1 __ = __ log _____ – __ log _______ + c 8 t + 1 4÷2 t ÷2 + 1
(
)
2v __ –1 1 1 = ___ __ tan– 1 ______ – __ log|v3 + 1| 2 3 ÷3 ÷
where t = sin x. 39. We have, Ú (x
3m
2m
+x
2m
m
+ x )(2x
+ 3x + 6)
dx
= Ú (x3m – 1 + x2m – 1 + xm – 1) × x × (2x2m + 3xm + 6)1/m dx
= Ú (x3m – 1 + x2m – 1 + xm – 1)(2x3m + 3x2m + 6xm)1/m dx 3m
2m
+ 3x
3 (2v – 1) – 1 + __ Ú __________ dv 2 v2 – v + 1
1/m
m
2
dv 3 3 v – (v – 1) __ = __ Ú _______________ – __ Ú ___________ dv 2 2 2 2 3 ÷ 1 (v3 + 1) __ ___ v – + 2 2
1 t – ___ __ 2 ÷ t–1 1 1 1__ = __ × __ log _____ – ____ log ______ + c 4 2 t + 1 4÷2 1__ ___ t + 2 ÷
| |
(
Let 2x
+ 6x = t
fi 6m(x3m – 1 + x2m – 1 + xm – 1)dx = dt
)
1__ – 1 1 1 = ___ tan 2v – ___ __ – __ log |v3 + 1| 3 3 2 ÷ ÷
3 1 = __ log|v2 – v + 1| – __ log |v3 + 1| + c 2 2
1 = ___ Ú t1/m dt 6m 1 = ________ t m + 1 + c 6(m + 1)
where v = t2 = (tan x)2/3
1 = ________ (2x3m + 3x2m + 6xm) m – 1 + c 6(m + 1)
ex = Ú _________ dx ex(ex – 1)2
40. We have,
dt = Ú _______ Let ex = t t(t – 1)2
dx 41. We have Ú _______ x 2 (e – 1)
_____
dx Ú 3÷tan x
fi exdx = dt
3
Let tan x = t
fi sec 2x dx = 3t2 dt
( )
2v – 1 3 1__ – 1 ______ + __ log|v2 – v + 1| – ___ tan __ + c 2 ÷3 ÷3
m
2
3 = Ú _____ 3t dt t6 + 1
(
)
1 1 1 = Ú __ – ____ + ______ dt t t – 1 (t – 1)2
t 1 = log ____ – ____ + C t–1 t–1
ex 1 = log ______ x – ______ x + C e –1 e –1
2
3t 3t fi dx = _____ 2 dt = _____ 2 dt sec x 1+t
| |
| |
42. We have
3 t 2.2t = __ Ú _____ dt 2 t6 + 1 v 3 = __ Ú _____ d v, 2 v3 + 1
Let t2 = v
fi
(v + 1) – 1 3 = __ Ú ________________ dv 2 (v + 1) (v2 – v + 1)
1.139
2t dt = dx
tan– 1x Ú ______ dx x4
[
( ) ]
dx 1 1 = tan– 1x Ú ___4 – Ú ______ 2 × – ___ 3 dx x 1+x 3x
x dx tan– 1x __ 1 = ______ + Ú ________ 3 4 3 (– 3x ) x (1 + x2)
1.140 Integral Calculus, 3D Geometry & Vector Booster
dt tan– 1x __ 1 = ______ + Ú _______ , 3 6 (– 3x ) t(t – 1)2
Let x2 + 1 = t fi
(
2x dx = dt
)
tan– 1x __ 1 1 _____ 1 1 = ______ + Ú __ – + _______ dt 3 t t – 1 6 (– 3x ) (t – 1)2 – 1
( | |
)
tan x __ t 1 1 = ______ + log ____ – ____ + c t–1 t–1 (– 3x3) 6 – 1
( | | )
3 = 4 log|t3 + 1| – __ log|t2 – t + 1| 2
( )
__ 2t – 1 1 + ___ __ tan– 1 ______ __ + c, Let t = 3÷x 3 ÷3 ÷
45. We have, dx Ú ___________ 2sin x + sec x
cos x = Ú _________ dx sin 2x + 1
2
tan x __ x + 1 __ 1 1 = ______ + log ______ 2 – 2 + c 3 6 (– 3x ) x x
43. We have,
2cos x 1 = __ Ú _________ dx 2 sin 2x + 1
dx Ú _______________ 4 ____ _____ (÷cos x + ÷ sin x )
1 (sin x + cos x) – (sin x – cos x) = __ Ú _________________________ dx 2 sin 2x + 1
sec2x = Ú ___________ 4 dx ____ (1 + ÷ tan x )
= Ú _______ 2t dt , (1 + t)4
1 (sin x + cos x) 1 (sin x – cos x) = __ Ú ____________ dx – __ Ú ____________ dx 2 2 sin 2x + 1 sin 2x + 1 Let tan x = t2 fi
sec2xdx = dt
(t + 1) – 1 = 2 Ú _________ dt (1 + t)4
(
( ) x p 1 1 = ____ log tan ( __ + __ ) – __ (sin x + cos x) + c 2 8 2 | | 2 ÷2
)
1 1 = 2 Ú _______ – _______ dt (1 + t)3 (t + 1)4
1 1 = – 2 Ú ________ – _______ + c 3(1 + t)2 4(t + 1)3
1 1 = – 2 ____________ 2 – ____________ ____ ____ 3 + c 3(1 + ÷ tan x ) 4(÷tan x + 1)
44.
– 2/3
dx 1 1 (sin x – cos x) = __ Ú ____________ – __ Ú ____________ dx 2 (sin x + cos x) 2 (sin x + cos x)2 p 1 1 (sin x – cos x) = ____ __ cosec x + __ dx – __ Ú ____________ dx Ú 4 2 (sin x + cos x)2 2÷2
( Ú(
1 (sin x + cos x) 1 (sin x – cos x) = __ Ú ____________ dx – __ Ú ____________ dx 2 2 (sin x + cos x) 2 (sin x + cos x)2
)
__
)
( ) – 2
+x 1+t 2 dx 3 Ú ______ t dt, Ú 1_______ 1+x 1 + t3 Let x = t3 fi dx = 3t2 dt
dt t2 = 3 Ú _____ 3 dt + 3 Ú _____ 1+t 1 + t3 t2 – (t2 – 1) = log|t3 + 1| + 3 Ú __________ dt t3 + 1 (t – 1) = 4 log|t3 + 1| – 3 Ú ________ dt 2 t – t + 1 3 (2t – 1) – 1 = 4 log|t3 + 1| – __ Ú __________ dt 2 t2 – t + 1 3 3 dt = 4 log|t3 + 1| – __ log|t2 – t + 1| + __ Ú ________ 2 2 t2 – t + 1
46. We have, 4 + 1 ______ Ú x dx 6 x +1
(x2 + 1)2 – 2x2 = Ú _____________ dx (x6 + 1) (x2 + 1)2 x2 = Ú _______ dx – 2 Ú _______ dx 6 6 (x + 1) (x + 1) (x2 + 1) x2 = Ú ___________ dx – 2 Ú _______ 6 dx 4 2 (x + x + 1) (x + 1)
(
)
(
)
1 1 + __ 2 x 3x2 2 = Ú ___________ dx – __ Ú _________ dx 3 ( (x3)2 + 1 ) 1 x2 + __ 2 – 1 x
(
)
1 1 + __ 2 x 3x2 2 = Ú ____________ dx – __ Ú _________ dx 2 3 3 ( (x )2 + 1 ) 1 x + __ x + 1
( (
)
)
Indefinite Integrals
(
)
1 2 – 1 3 __ = tan–1 x – __ x – 3 tan (x ) + c 47. We have,
Ú dx(e
2
x
– 1)
= Ú_______ dt , t(t – 1)2
Let t = ex
fi dt = ex dx
(t2 – 1) + 1 t = ______ 2 + 2Ú _________ dt (t – 1) (t2 – 1)2
dt dt t = ______ 2 + 2Ú ______ + 2Ú _______ 2 2 (t – 1) (t – 1) (t – 1)2
)
| |
| |
t t–1 1 = – ______ – __ log ____ + c t+1 (t2 – 1) 2
_____
where t = ÷ x + 1 50. We have,
1 1 1 = Ú __ – ______ + ______ dt, t (t – 1) (t – 1)2
x4 – 1 _________ Ú ____________ dx x2÷ x 4 + x2 + 1
1 1 = log ____ – ____ + C t –1 t– 1
| |
x4 – 1 ________ = Ú ______________ dx 1 3 2 x x + __ 2 +1 x 1 x – __ 3 dx x ________ = Ú ____________ 1 x2 + __ 2 +1 x
÷( ) ( ) ÷( )
x
e 1 = log _____ x – ______ + C e –1 ex – 1 48. We have, dx_____ Ú ____________ (x – 1)÷x + 2
= Ú _______ 2t dt , (t2 – 3)t
Let x + 2 = t2
fi dx = 2t dt
1 Let x2 + __ 2 + 1 = t2 x 1 fi 2x – 2 __3 dx = 2tdt x
t dt = Ú ___ , t
(
dt = 2Ú __________ __ 2 ( t – (÷3 )2 )
= Ú t dt
t–÷ 3 2 __ = ____ __ log ______ + c 2÷3 t + ÷3
=t+c
|
1 = x2 + __ 2 + 1 + c x
| | __
|
__
x2 – 1 ____________ Ú ______________ dx x3÷ 2x 4 – 2x2 + 1
dx 2 _____ Ú ________ x ÷x + 1 2t dt = Ú ________ , 2 (t – 1)2t dt = 2Ú _______ (t2 – 1)2
dt Now, Ú _______ (t2 – 1)
Let x + 1 = t2
fi dx = dtdt
÷
51. We have,
49. We have,
)
__________
(x + 2) – ÷ 3 1 __ + c = ___ __ log ___________ ÷3 (x + 2) + ÷ 3
)
dt dt t fi 2Ú _______ = – ______ – Ú ______ (t2 – 1)2 (t2 – 1) (t2 – 1)
exdx = Ú _________ x x 2 e (e – 1)
(
(
1.141
(
)
2t ◊ t 1 = ______ 2 Ú dt – Ú – _______ dt (t – 1) (t2 – 1)2
x2 – 1 __________ = Ú ____________ dx 2 1 __ x5 2 – __ 2 + 4 x x
(
÷
)
1 1 5 dx __ 3 – __ x x __________ = Ú ___________ 2 1 __ __ 2 – 2 + 4 x x
1 = __ Ú 2
÷
t dt ___ , t
2 Let 2 – __2 + x
1 __ 4 = t2 x
2 fi ___ – 25
)
(
4 ___ dx = 2tdt 25
1.142 Integral Calculus, 3D Geometry & Vector Booster
1 = __ t + c 2
1 2 1 __ = __ 2 – __ 2 + 4 + c 2 x x
÷
__________
52. We have,
Ú
fi
Also, t2 = 1 – sin 2q
fi
=
dt ________ = Ú __________ ÷ 5 + 1 + t2
dt _________ = Ú ___________ __ ( )2 – t2 ÷ ÷6
Ú
÷
t = sin– 1 ___ __ + c ÷6
sinq –__cosq = sin– 1 __________ + c ÷6
______
cos2x _____ 2 d x sin x
÷
____________ 2
2
=
x – sin x ___________ dx Ú cos sin2x
=
Ú
________
÷cot 2x – 1 dx
( )
(
se2x Ú _____________ dx (sec x + tan x)9/2
__________ 2 ÷cosec x – 2 dx
Ú
=
2 cosec x–2 __________ = Ú ___________ dx 2 x – 2 ÷ cosec
cosec2x 2 __________ __________ = Ú ___________ dx – Ú ___________ dx 2 2 x – 2 ÷cosec x – 2 ÷ cosec
cosec2x 2sin x ________ _________ = Ú _________ dx – Ú __________ dx 2 x – 1 ÷1 – 2sin2x ÷ cot
2
cosec x 2sin x ________ _________ = Ú _________ dx – Ú __________ dx 2 x – 1 ÷2cos 2x – 1 ÷ cot
sec x.sec x = Ú _____________ dx (sec x + tan x)9/2
fi cosec2xdx = dt and – sin xdx = dx.
...(i)
Let
sec x + tan x = t
fi
sec x (sec x + tan x) dx = dt
fi
dt sec xdx = __ t
1 1 Also sec x – tan x = ___________ = __ (sec x + tan x) t
1 1 Thus, sec x = __ t + __ dt t 2
( )
Let cot x = t and cos x = v
)
54. We have
sin 2q = 1 – t2
_____
cos2x ÷ ______ dx sin x
(cosq + sinq)dq = dt
The given intergral (i) reduces to
dv dt = – Ú _______ _____ + 2 Ú ________ _______ 2 ÷2 v 2 – 1 ÷ t – 1
( )
__ dv dt _____ _________ = – Ú ______ + ÷ 2 Ú ___________ 2 1__ 2 2 ___ ÷ t – 1 v – 2 ÷
÷ ( )
1 1 dt __ t + __ __ t t 2__________ Ú 9/2 t
2 1 (t + 1) = __ Ú ______ dt 2 t13/2
1 1 1 = __ Ú ___ + ____ dt 2 t9/2 t13/2
1 2 2 = __ ____ + ______ + c 2 7t7/2 11t11/2
53. We have,
cosq + sinq __________ Ú ___________ dq ÷ 5 + sin(2q )
1 1 = ____ 7/2 + ______ 11/2 + c 7t 11t
1 1 = ______________ + ________________ + c 7(sec x – tan x)7/2 11(sec x + tan x)11/2
_____
|
________
|
= – log cot x + ÷cot 2x – 1
|
÷
______
|
__ 1 = – log|t + ÷ t 2 – 1 |+÷ 2 log v + v2 – __ + c 2
__
|
÷
_________
|
1 + ÷2 log cos x + cos x – __ + c 2
Let
2
sinq – cosq = 1
( Ú (
(
)
)
)
Indefinite Integrals
55. We have,
tan2q ___________ Ú ____________ dq 6 6 q + cos q ÷ sin
sin2x = tan x log(1 + sin2x) – 2Ú ________ dx 1 + sin2x
sin2q __________ = Ú ________________ dq 3 2 cos2q 1 – __ sin 2q 4
÷
(
2sin2q __________ = Ú ________________ dq cos2q ÷3cos 2 2q – 1
dt ______ = Ú ________ , t÷3t 2 – 1
Let t = cos2q
fi dt = 2sin2q dq
)
1 = tan x log(1 + sin2x) – 2Ú 1 – ________ dx 1 + sin2x
Ú sec2x log(1 + sin2x)dx
2sin x ◊ cos x = log(1 + sin2x) Ú sec2x dx – Ú tan x ◊ __________ dx (1 + sin2x)
2sin2q __________ = Ú ________________ dq 2 cos2q ÷4 – 3sin 2q
log(1 + sin2x) 2. Ú ____________ dx cos2x
=
tan2q ______________ = Ú _______________ dq 2 q ÷ 1 – 3sin2q cos
2 = tan x log(1 + sin2x) – 2(x) + 2Ú _________ sec x dx 2 2tan x + 1
dt = tan x log(1 + sin2x) – 2(x) + 2Ú ______ 2t2 + 1
Let tan x = t
t dt = – Ú _________ ______ 2 t ÷ 3t 2 – 1
__
3.
Ú x 3 ( 1 + x3 )
2 – 1 __
2 – __
1 = – __ tan– 1v + c 9
1 = – __ tan– 1 ( ÷3t 2 – 1 ) + c 9
1 = – __ tan– 1 ( ÷3cos 2 2q – 1 ) + c 9
dx dx = Ú __________ x2/3(1 + x2/3) Let x1/3 = t dx fi ____ = dt 3x2/3
______
__________
x2 + 6 1. Ú _______________ dx (x sin x + 3cos x)6
(x + 6)x = Ú _________________ dx 3 (x sin x + 3x2cos x)6 3
(x + 6)x cos x.(x sec x) = Ú ___________________ dx (x3 sin x + 3x2cos x)6 2
x(x + 6) cos x = (x3sec x)Ú _________________ dx + Ú sec2x dx 3 (x sin x + 3x2cos x)6 – x sec x = _____________ + tan x + c x sin x + 3 cos x
dx fi ___ = 3dt x2/3
dt = 3Ú _____ 2 1+t
= 3tan– 1(t) + c
= 3tan– 1(x1/3) + c
4
2
__
= tan x log(1 + sin2x) – 2(x) + 2÷2 t an– 1(t ÷2 ) + c where t = tan x
dv 1 = – __ Ú _____ , Let 3t2 – 1 = v2 9 v2 + 1 6 tdt = 2x dx
2
sec2x dx = dt
1.143
[
(
) ] [
(
)]
5p p 4. Ú 1 + tan ___ – x 1 + tan – ___ + x dx 16 16
= Ú 2dx
= 2x + c
(
p (1 + tan A)(1 + tan B) = 2, if A + B = __ 4 5. Útan (x – a ) ◊ tan (x + a ) ◊ tan 2x dx
)
= Ú(tan (2x) – tan (x – a ) – tan(x + a ))dx
1 = – __ log |cos (2x)| + log |cos (x – a )| 2
+ log |cos (x – a )| + c
1.144 Integral Calculus, 3D Geometry & Vector Booster
( ( ÷ ) ) _____
dt t Now, I1 = Ú ___________ t2 + 2t – 3 dt
1–x 6. Ú cos 2cot _____ dx 1+ x – 1
Let x = cosq
fi dx = – sinq dq
=
Ú
[ ( ÷
________
=
Ú
[ [ [ (
= – Ú cos (q ) dq
= – sinq + c ______ – ÷1 – x2 +
(
c
x (x + 1)dx
÷ { (
__________________________
(
)
}
)
1 1 (x2 – 1) x2 x2 + __ 2 + 2 x + __ x – 1 x = Ú _________________________________ dx x2(x2 + 2x + 1) ________________________
( )
)
)
}
1 1 1 1 – __ 2 x2 + __ 2 + 2 x + __ x – 1 x x = Ú _________________________________ 1 x + __ x + 2 1 Let x + __ x + 2 = t _________ ÷__________ t 2 + 2t – 3
( )
1 fi 1 – __ 2 dx = dt x
= Ú
t + _________ 2t – 3 dt = Ú ________________ 2 (t + 2)÷t + 2t – 3
dt (t + 2)
2
t(t + 2) – 3 _________ = Ú ________________ dt (t + 2)÷t 2 + 2t – 3
dt t _________ = Ú __________ _________ dt – Ú _______________ 2 (t + 2)÷t 2 + 2t – 3 ÷ t + 2t – 3
= I1 + I2 (say)
)
|
)
__
|
dx __________ Also, I2 = Ú _________________ (x + 2)÷x 2 + 2x – 3
2
(
(
1 where t = x + __ x
___________________ (x – 1)÷x 4 + 3x3 – x2 + 2x + 1 __________________________
( ) ÷{ (
)
– 1
Ú cos (p – q ) dq
7. Ú
(
(t + 1) – ÷2 1 1__ __ + c = __ log|t2 + 2t – 3| – ____ log __________ 2 2÷2 (t + 1) + ÷2
=
) )
(2t + 2) dt 1 = __ _________ dt – Ú ____________ 2 t2 + 2t – 3 (t + 1)2 – (2)2
q cos 2cot tan __ dq 2
=
)]
– 1
)
(2t + 2) dt 1 = __ _________ dt – Ú _________ 2 2 t2 + 2t – 3 t + 2t – 3
1 – cosq cos 2cot– 1 ________ dq 1 + cosq
{ ( ) } ] p q = Ú cos 2cot { cot ( __ – __ ) } ] dq 2 2 p q = Ú cos 2 __ – __ ) ] dq 2 2
( Ú ( Ú (
1 (2t + 2) – 2 = __ Ú __________ dt 2 t2 + 2t – 3
dt – __2 t = Ú _________________________ _____________________ 2 1 1 1 __ __ – 2 + 2 __ – 2 – 3 t t t
( ) ÷(
)
(
)
dt _______________________ = – Ú _________________________ 1 2 2 __ (1 – 2t) + 2t – 2 – 3t2 t
÷
(
)
dt = – Ú ___________________________ _________________________ 2 ÷(1 – 4t + 4t ) + 2t – 4t2 – 3t2 dt = – Ú _____________ ___________ ÷(1 – 2t + 3t2) dt = – Ú ________________ _______________ 2 + 1} ÷ – {3t + 2t dt = – Ú _________________ ________________ 2 1 – 3 t2 + __ t – __ 3 3
÷ {
}
dt = – Ú ___________________ _________________ 1 2 __ 4 – 3 t + __ – 3 9
÷ { ( ) }
dt 1__ _______________ = – ___ Ú _____________ 2 3 ÷ 2 1 2 __ __ – t + 3 3
÷( ) ( ) 3t + 1 1 = – ___ sin ( ______ + c 2 ) ÷3
– 1 __ 1 where t = x + __ x
(
)
(
p tan __ – x 4 __________________ 8. Ú _______________________ dx 2 3 cos x ÷tan x + tan2x + tan2x
)
Indefinite Integrals
(1 – tan x) sec2x dx _________________ = Ú __________________________ (1 + tan x)÷tan 3x + tan2x + tan x
(1 – t)dt _________ = Ú ________________ Let tan x = t (1 + t)÷t 3 + t2 + t fi sec2xdx = dt
(1 – t2 )dt _________ = Ú _________________ (1 + t)2 ÷ t 3 + t2 + t (1 – t2)dx ___________ = Ú _______________________ 1 2 (t + 2t + 1) t2 t + __ + 1 t
÷ (
)
( )
1 1 – __ 2 dt t __________ = Ú _____________________ 1 1 __ t + + 2 t + __ + 1 t t
(
(
)÷(
)
)
1 Let t + __ + 1 = v2 fi sec2xdx = dt t
– t2 )dt q 1 (1 __ ________ = – _______ + Ú 3 3 tan3q t3
q 1 1 = – _______ + __ Ú __ – 3 3 t3 3 tan q
q 1 ___ 1 = – _______ – __ 2 + log|t| + c 3 3 3 tan q 2t
q 1 ______ 1 = – _______ – __ 2 + log |sin q | + c 3 3 3 tan q 2sin q
( ) ( ) (
(1 – x sin x)ecos x = Ú ________________ dx xecos x( 1 – (xecos x)3 )
dt = Ú _______ , t(1 – t3)
=
dv = – 2Ú _______ 2 (v + 1)
dt Ú _______ t(1 – t3)
=
dt 3 Ú _______ t(t – 1)
3t2 1 = __ Ú ________ 3 3 dt 3 t (t – 1)
dv 1 = __ Ú _______ , 3 v(v – 1)
v –1 1 = __ log _____ + c v 3
t3 – 1 1 = __ log ______ 3 + c 3 t
(xex)3 – 1 1 = __ log ________ + c 3 (xex)3
= – 2 tan– 1(v) + c
( ÷
)
1 + c = – 2tan– 1 t + __ + 1 t 1 where t = x + __ x
9.
– 1
tan x 4 dx Ú ______ x Let x = tan q fi dx = sec2xdq
tan– 1(tan q )
Let t = xex
where v = t3
| |
| |
| |
=
sec2 q dq Ú __________ tan4q
=
q sec q 4 dq Ú _______ tan q
sec2q 1 dq = q Ú _____ 4 dq + __ Ú _____ 3 tan3q tan q
(1 + x cos x)esin x = Ú _______________ dx xesin x(1 – (xesin x )2)
q dq 1 = – ______ + __ Ú _____ 3 tan3q 3 tan3q
dt = Ú _______ , t(1 – t2)
q 1 cos3q = – _______ + __ Ú _____ dq 3 3 sin3q 3 tan q
dt = Ú _______ t(1 – t2)
– sin2q)cosq q 1 (1 __ _____________ = – _______ + dq 3Ú 3 tan3q sin3q
2t dt 1 = – __ Ú ________ 2 t2(t2 – 1)
2
)
(1 – x sin x) 10. Ú ____________ dx x(1 – x3e3 cos x)
v dv = – 2Ú ________ 2 (v + 1)v
________
1 __ dt t
where tan q = x
1.145
(1 + x cos x) 11. Ú _____________ dx x(1 – x2e2sin x)
where t = xesin x
1.146 Integral Calculus, 3D Geometry & Vector Booster
dv 1 = – __ Ú _______ , where v = t2 2 v(v – 1)
v 1 = __ log _____ + c 2 v –1
t2 1 = __ log _____ 2 + c 2 t –1
(xesin x)2 1 = __ log __________ sin x 2 + c 2 (xe ) – 1
| |
| |
|
14.
( ) (
( )
)
|
( )
1 – Ú x sec2 __ x dx + c
( ) 1 = x tan ( __ x ) + c
( )
( )
1 1 1 = x3 tan __ x + Ú x sec2 __ x dx – Ú x sec2 __ x dx + c 3
cos x = Ú _________ dx sin 2x + 1
2cos x 1 = __ Ú _________ dx 2 sin 2x + 1
1 (sin x + cos x) + (cos x – sin x) = __ Ú _________________________ dx 2 (sin x + cos x)2
dx cos x – sin x 1 1 = __ Ú ____________ + __ ____________ 2 (sin x + cos x) 2 Ú (sin x + cos x)2
p 1 1 cos x – sin x = ____ __ dx + __ Ú ____________ dx Ú cosec x + __ 4 2 (sin x + cos x)2 2÷2
__________
÷3cos 2x – 1 15. Ú ___________ dx cos x 3cos 2x –1 _________ = Ú ______________ dx cos x ÷cos 2x – 1 3(2cos2x – 1) __________ = Ú _______________ dx cos x ÷3cos 2x – 1 dx cos x _______________ _________ = 6Ú ________________ dx – 4Ú _______________ dx 2 cos x 3cos2x – 1 ÷ – 2sin x) – 1 ÷ 3(1
( ) 1 1 = ____ log tan ( __ + c (sin x + cos x) | 2x + __ p 8 ) | – ___________ 2÷2
__ cos x dx cosx _________ = 3÷2 Ú ____________ __________ dx – 4Ú ___________________ 2 – 3sin x (1 – sin2x)÷1 – 3sin2x ÷ (1
__
dx _________________ 13. Ú ______________________ cos x ÷sin(2x + a) + sin a
Ú ( 3x2tan ( __ 1x ) – x sec2 ( __ 1x ) ) dx
1 1 1 = tan __ x Ú (3x2)dx – Ú sec2 __ x – __2 (x3) dx x
dx 12. Ú___________ 2sin x + sec x
tan x cos a + sin a ______ = _______________ + c 2cos a ÷
__ cos x dx cos x ____________ _________ = ÷6 Ú ______________ dx – 4Ú ___________________ 2 1 2 (1 – sin2x)÷1 – 3sin2x ___ __ – sin x 3 ÷
÷( )
__ __ cos x dx _________ = ÷ 6 s in– 1(÷3 s in x) – 4Ú ___________________ (1 – sin2x)÷1 – 3sin2x
dx _____________________________ = Ú __________________________________ cos x÷sin(2x)cos a + cos(2x)sina + sin a
__ __ dt______ = ÷ 6 sin– 1(÷3 sin x) – 4Ú ______________ 2 (1 – t )÷1 – 3t2
dx ____________________________ = Ú __________________________________ cos x÷sin (2x) cos a + (1 + cos (2x) sin a
dx _________________________ = Ú _______________________________ cos x ÷2sin x cos x cos a + 2cos2x sina
where t = sin x __ __ y dy ______ = ÷ 6 s in– 1(÷3 sin x) + 4Ú _____________ 2 (y – 1)÷y 2 – 3
1 dx = ___ __ Ú _____________________ ________________ 2 2 cos x÷tan x cos a ÷ + sin a
sec2x dx 1 _______________ = ___ __ Ú ________________ 2 ÷tan x cos a ÷ + sin a
Let tan x cos a + sin a = t
where t = 1/y
__ __ v dv = ÷ 6 sin– 1(÷3 sinx) + 4 Ú ________ 2 , v2 = y2 – 3 (v + 2)v 2
__ __ dv = ÷ 6 s in– 1(÷3 s inx) + 4 Ú _______ 2 (v + 2)
( )
2t dt 1 ____ = _______ __ Ú t 2 cos a ÷
__ __ __ v = ÷ 6 s in– 1(÷3 sin x) + 2÷2 tan– 1 ___ __ + c 2 ÷
1 = _______ __ Ú dt 2 cos a ÷
y2 – 3 = ÷ 6 s in– 1(÷3 sin x) + 2÷2 tan– 1 _____ + c 2
( ÷ ) ______
__
__
__
Indefinite Integrals __
__
( ÷
_________
__
)
= – Ú ________ 2t dt (t2 + 1)t3
cos2x – 3 = ÷6 s in (÷3 s inx) + 2÷2 t an ________ + c 2 – 1
– 1
( ) [ ( )
( ) ]
dt = – 2Ú ________ 2 (t + 1)t3
2x 2x 1 16. Ú _____ 8 cos– 1 _____ 2 + tan– 1 _____ 2 dx 1–x 1+x 1–x
(
(
)
(
)
1 = – 2Ú – __ – tan– 1t + c t
p dx = __ Ú _______ 2 (1 – x8)
( ÷
_________
( ÷
[
]
] ]
)
(x2 + 1) – (x2 + 1) p 1 1 = __ Ú ______ 2 + ______ 2 + Ú _______________ dx 8 1–x 1+x x4 + 1
)
(x2 + 1) (x2 + 1) _______ 1 1 ______ 2 + ______ 2 + Ú _______ – 4 dx 1–x 1+x x +1 x4 + 1
4 cos = Ú _______________ x dx sin6x(1 + cot5x)3/5
4 2 x dx = Ú _____________ cos x cosec 5 3/5 (1 + cot x)
and then you do it.
4 = Ú _________ t dt , (1 + t5)3/5
x2 – x3 17. Ú __________________ dx (x + 1)(x3 + x2 + x)3/2
5t4dt 1 = – __ Ú _________ 5 (1 + t5)3/5
x2(1 – x) = Ú ___________________ dx 3/2 1 (x + 1)x3 x + __ x + 1
)
(1 – x) = Ú __________________ dx 3/2 1 x(x + 1) x + __ x + 1
(
)
(1 – x2) = Ú ___________________ dx 3/2 1 x(x + 1)2 x + __ x + 1
(
)
2
(1 – x ) = Ú ________________________ dx 3/2 1 2 x(x + 2x + 1) x + __ x + 1
(
)
(1 – x2) = Ú ________________________ dx 3/2 1 1 2 __ x x + x + 2 x + __ x + 1
(
) (
( )
)
) (
)
Let (1 + t5) = y5
fi
4 1 5y dy = – __ Ú _____ 3 5 y
= – Ú y dy
y2 = – __ + c 2
1 = – __ (1 + t5)2/5 + c 2
1 = – __ (1 + cot5x)2/5 + c 2
(
)
1 Let x + __ x + 2 = t2 1 fi 1 – __ 2 dx = 2t dt x
( )
Let cot x = t
5t4dt = 5y4dy
( )
sin3x dx 19. Ú _________________________________ 4 2 (cos x + 3cos x + 1) tan– 1(sec x + cos x)
1 1 – __ 2 dx x = Ú ______________________ 3/2 1 1 __ x + x + 2 x + __ x + 1
(
)
4 x cos 18. Ú ___________________ 3 5 sin x(sin x + cos5x)3/5
p dx dx = __ Ú _______ 4 + Ú ______ 4 4 (1 – x ) x +1
(
)
1 1 = 2 ________ _______ + tan– 1 x + __ x + 1 + c 1 x + __ x 1
p dx = __ Ú _____________ 2 (1 – x4)(1 + x4)
p = __ 8
)
1 1 = – 2Ú __ 2 – _____ 2 dt t t +1
p dx = Ú _______ 8 __ – 2 tan– 1x + 2 tan– 1x dx (1 – x ) 2
[ ( [ Ú (
1.147
(1 – cos2x) sin x dx = Ú _________________________________ (cos4x + 3cos2x + 1) tan– 1(sec x + cos x) (1 – t2)dt = – Ú ______________________ , where cos x = t 1 4 2 (t + 3t + 1)tan= 1 t + __ t
( )
1.148 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) ( ) ) ( Ú
dy 1 = 2Ú _______ , y = t + __ 2 t 4y – 2
dy = Ú _______ 2 2y – 1
) ( ) dy 1 = Ú ____________ , where ( t + __ ) = y t (y + 1)tan y
dy 1 = __ Ú ___________ __ 2 2 y – (1/÷2 ) 2
y÷2 – 1 1 = ____ __ log _______ __ + c 2÷2 y÷2 + 1
– 1 = Ú ___ dv v , where tan (y) = v
= Ú ___ dv v
1 __ t + __ ÷ 2 – 1 t 1 = ____ __ log ____________ + c 1 __ 2÷2 __ t + ÷ 2 + 1 t
1 1 – __ 2 dt t = Ú _____________________ 1 1 2 __ t + 2 + 3 tan= 1 t + __ t t
1 1 – __ 2 dt t = ______________________ 2 1 1 = 1 __ __ t + + 1 tan t + t t
( ( )
2
1 = log tan– 1 t + __ + c t
= log |tan– 1(cos x + sec x)| + c
|
( ) |
dx 4 + 3sin 2x
)
tan x – 1 = Ú ______________________ dx ____ 2tan x tan x 4 + 3 ________ ÷ 1 + tan2x
( Ú ( Ú (
))
)
(tan x – 1)sec2x = Ú _______________________ dx ____ tan x ( 4 + 4tan2 x + 6tan x ) ÷ 2
x2 = – __________ + 2 Ú ___________ x cos x dx (x tan x + 1) x sin x + cos x
x2 = – __________ + 2log|x sinx + cos x| + c (x tan x + 1)
÷
___________
)
(t – 1)2t dt = ____________ dt, t(4t4 + 6t + 4)
)
where tan x = t
(t2 – 1)dt = 2 _____________ dt t(4t4 + 6t2 + 4)
(
|
(x sec2x + tan x) = x2 Ú _____________ dx + 2Ú ________ x dx x tanx + 1 (x tan x + 1)2 x2 x = – __________ + 2Ú ________ dx (x tan x + 1) x tanx + 1
( (
|
__
tan x) x2(x sec2x + 21. Ú _______________ dx (x tan x + 1)2
____ ____ ÷____________ cot x – ÷ tan x
(
|
( ) ( )
(tan x + cot x)÷2 – 1 1 __ = ____ __ log __________________ + c 2÷2 (tan x + cot x)÷2 + 1
= log |v| + c
Ú
|
– 1
20.
| | __
( ) ( ( ) ) ) ( Ú
)
1 1 – __ 2 dt t = 2Ú ______________ dt 1 4 t2 + __ 2 + 6 t
( ( )) ) ( ( )
1 1 – __ 2 dt t = 2 ___________________ dt 1 2 4 t + __ – 2 + 6 t
{ ( ) }
1 1 – __ 2 dt t = 2Ú _____________ dt 1 2 4 t + __ – 2 t
( ( ) )
sin x – sin3x 22. Ú ___________ dx 1 – sin3x
÷
_____________
sin x(1 – sin2x) = Ú _____________ d x 1 – sin3x
cos x÷sin x = Ú _________ ________ dx 3 ÷ 1 – sin x
cos x÷sin x = Ú ____________ ___________ dx 3/2 2 x) ÷ 1 – (sin
____
____
Let (sin3/2x) = t 3 fi __ sin1/2x cos x dx = dt 2 2 fi sin1/2x cos x dx = __ dt 3
dt 2 _____ = __ Ú ______ 3 ÷ 1 – t2
Indefinite Integrals
x+2 dx Ú ex ( _____ x + 4) 2
2 = __ sin– 1(t) + c 3
25.
2 = __ sin– 1(sin3/2x) + c 3
(x + 2)2 = Ú ex _______ dx (x + 4)2
x2 + 4x + 4 = Ú ex __________ dx (x + 4)2
x2 + 4x 4 = Ú ex _______2 + _______ dx (x + 4) (x + 4)2
x(x + 4) 4 = Ú ex _______ + _______ dx (x + 4)2 (x + 4)2
x
{
}
{1 + (1 – x)}2 ______ = Ú ex _____________ dx (1 – x)÷1 – x2
(
{ Ú { ÷
÷
_____
} }
)
( ( (
2
e (2 – x) ______ 23. Ú ____________ dx (1 – x)÷1 – x2
)
) )
1 + x 1 ______ = Ú ex _____________ + _____ dx 2 (1 – x)÷1 – x 1 – x
1 + x _____________ 1 ______ = ex _____ + dx 1 – x (1 – x)÷1 – x2
x 4 = Ú ex ______ + _______ dx (x + 4) (x + 4)2
1+x = ex _____ + c 1–x
x = ex ______ + c (x + 4)
______
( ÷ ) _____
24.
26. Ú
÷sin(x + a )cos (x – b )
Put (x – b ) = t dx = dt
=
dt _______________________ Ú _________________________ 3 ÷cos t(sin t ◊ cosq + cos t sinq )
=
dt ______________________ ____________________ 4 ÷cos t(tan t ◊ cosq + sinq )
=
Let (tant ◊ cosq + sinq ) = v
fi sec2t cosq dt = 2v
2v fi sec2t dt = _____ cosq
= 2secq ÷cosq tan t + sinq + c
= 2sec(a + b )
t fi 4x7 dx = – ______ dt (t2 – 1)
÷
=
(x4t) = Ú ____ × x7dx x20
t x7 = Ú ___ dx x16
c
8
+x dx Ú 1_____ x13
_____________
______________________________ ÷cos(a + b )tan(x – b ) + sin(a + b ) +
1 fi x8 = ______ 2 (t – 1) 2t 7 fi 8x dx = – ______ dt (t2 – 1)
______
Let (1 + x8) = x8t2
2
dx
x13
= Ú{ x– 13(1 + x8)1/2 }dx
÷
2v = ____ + c cosq
)
sec t dt ________________ Ú __________________ (tan t ◊ cosq + sinq )
v dv 2 = _____ ____ cosq Ú v
(
2
)
b+1 – 13 + 1 __ 1 Here, _____ + a = _______ + = – 1 g 8 2
dt ________________ = Ú __________________ 3 ÷sin(t + b + a )cos t dt = Ú ______________ _____________ , (given) 3 + q )cos t ÷ sin(t
Ú
(
______ _______ ÷1 + x8
dx __________________ Ú ____________________ 3
1.149
t t 1 = – __ Ú _______ × – _______ 2 dt 2 4 _______ 1 (t – 1) 2 2 (t – 1)
1 = – __ Ú t2 dt 4
t3 = – ___ + c 12
1.150 Integral Calculus, 3D Geometry & Vector Booster
( )
ex(x + 1) = Ú ___________ dx xex(1 + xex)2
dx 27. Ú ___________ 3 sin x + cos3x
dt = Ú _______ , t(1 + t)2
dx = Ú ________________________ (sin x + cos x)(1 – sin x cos x)
=
t 1 = log ____ + ______ + c t+1 (t + 1)
1 1 + x8 3/2 = – ___ _____ + c 12 x8
(
)
sin x + cos x 1 2 = __ Ú ____________ – _____________ dx 3 (sin x + cos x) (1 – sin x cos x)
( ) ( )
)|
x p 2 2 = ____ __ log tan __ + __ – __ tan– 1 (sin x – cos x) + c 2 8 3 3÷2 ______
( x + ÷ 1 + x2 ) 28. Ú _____________ dx ______ ÷ 1 + x2 3
dx _________ = Ú __________ 2 + 3cot x ÷
Let 2 + 3cot x = t2
fi – 3cosec2x dx = 2t dt
2t dt fi dx = – _______ 4 3(t + 1)
______
Let (x + ÷ 1 + x2 ) = t
x fi 1 + _______ ______ dx = dt ÷ 1 + x2
______ x+÷ 1 + x2 ___________ fi ______ dx = 2 ÷1 + x
t fi _______ ______ dx = dt ÷ 1 + x2
dx dx fi _______ ______ = ___ t 2 ÷ 1 + x
(
)
( (
(
dt = Út3 × __ t
= Út2dt
t3 = __ + c 3
)
)
)
dt
dt 2 = – __ Ú ______ 3 (t4 + 1)
( Ú ( )
(x + ÷1 + x2 )3 = ____________ +c 3
) Ú ( )
2 2 1 (t + 1) – (t – 1) = – __ Ú _______________ dt 3 (t4 + 1)
2 2 1 (t + 1) 1 (t – 1) = – __ _______ 4 dt + __ ______ dt 3 (t + 1) 3 (t4 + 1)
( ) ( )
( ) ( ) ) ( Ú
( ) ( ) ) ( Ú
1 1 1 + __ 2 1 – __ 2 t t 1 1 = – __ Ú _______ dt + __ Ú _______ dt 3 3 1 1 2 __ t + 2 t2 + __ 2 t t
(
) ( )
1 1 1 + __ 2 1 – __ 2 t t 1 1 = – __ __________ dt + __ _______2 – 2 dt 2 3 3 1 1 __ __ t – + 2 t + t t
( )
_____
(x + 1) 29. Ú _________ dx x(1 + xex)2
÷
sin x 30. Ú ____________ dx 2sin x + 3cosx
t dt 2 = – __ Ú _______ 3 (t4 + 1)t
| |
____________
sin x + cos x 2 ___________ 2 = ____ __ – __ Ú ________________ dx Ú dx p 3 3÷2 (sin x + __ (1 + (sin x – cos x)2) 4 p sin x + cos x 2__ 2 ____ = dx – __ Ú ________________ dx Úcosec x + __ 4 3 (1 + (sin x – cos x)2) 3÷2
(
dt dt dt – Ú ______ – ______ Ú __ t (t + 1) Ú (1 + t)2
where t = xex
dx sin x + cos x 2 1 = __ Ú ____________ – __ _____________ dx 3 (sin x + cos x) 3 Ú (1 – sin x cos x)
|
where xex = t
( ) 1 ( t + __ ) – ÷2 t 1 1 __ ____ ___________ ( t – ) + log + c t 1 6÷2 ( t + __ ) + ÷2 t
| | __
1__ – 1 = – ____ tan 3÷2
_________
where t = ÷ 2 + 3cot x cos 4x + 1 31. Ú __________ dx cot x – tan x
__
__
Indefinite Integrals
2cos2x 2x dx = Ú ___________ cos2x – sin2x ___________ sin x cos x
2cos2x 2x dx = Ú ___________ 2cos 2x _________ 2sin x cos x
= Ú cos 2x sin 2x dx
1 = __ Ú 2cos 2x sin 2x dx 2
1 = __ Ú sin (4x) dx 2
1 = – __ cos (4x) + c 8
cos x + cos x 32. Ú ____________ dx sin2x + sin4x 3
5
cosx (cos2x + cos4x) = Ú _________________ dx (sin2x + sin4x)
(1 – t2) + (1 + t4)2 = Ú _______________ dt, where sin x = t t4 + t2
2 – 3t2 + t4 = Ú __________ 4 dt t + t2
=
– 4t2 – 4 + 6 = 1 + ____________ dt t4 + t2
– 4t2 – 4 + 6 = Ú 1 + ___________ dt t2(t2 + 1)
4(t2 – 1) 6 = Ú 1 – ________ + ________ 2 2 dt 2 2 t (t + 1) t (t + 1)
4 1 1 = Ú 1 – __ 2 + 6 __ 2 – ______ 2 dt t t t +1
2 1 = 1 + __ 2 – 6 ______ 2 dt t t +1
2 = 1 – __ – 6tan– 1(t) + c t
= sin x – 2cosec x – 6tan– 1(sin x) + c
( Ú (
2
)
(
dx 33. Ú ___________ sin6x + cos6x
(1 + tan2x)2sec2x = Ú _______________ dx tan6x + 1
(1 + t2)2 = Ú _______ dt, t6 + 1
(1 + t2)2 = Ú ________________ dt (t2 + 1)(t4 – t2 + 1)
(1 + t2) = Ú ___________ dt (t4 – t2 + 1)
1 1 + __ 2 t = Ú ___________ dt 1 2 __ t + 2 – 1 t
1 1 + __ 2 t = Ú ____________ dt 2 1 t – __ + 1 t
(
(
sec6x dx = Ú ________ tan6x + 1
(
( ) (
where t = tan x
)
)
( ( ) ) 1 = tan ( t – __ ) + c t – 1
= tan– 1(tan x – cot x) + c
34. Let f (x) = ax2 + bx + c
– 4t + 2 dt Ú 1 + ________ t4 + t2
( Ú (
1.151
Given f (0) = f (1) = – 3 = 3 f (2)
)
On solving, we get
)
x2 – x – 3 Thus, I = Ú _______________ dx (x – 1)(x2 + x + 1)
)
( )) ( ) ) )
a = 1, b = – 1, c = – 3
x2 – x – 3 Now, _______________ (x – 1)(x2 + x + 1)
Bx + C A = ______ + __________ (x – 1) (x2 + x + 1)
Solving, we get
A = – 1, B = 2, C = 2
The given integral reduces to
2x + 2 – Ú ______ dx + _________ dx (x – 1) x2 + x + 1
dx dx 2x + 1 = – Ú ______ + _________ dx + Ú _________ 2 (x – 1) Ú x2 + x + 1 x +x+1
1.152 Integral Calculus, 3D Geometry & Vector Booster
|
dx = – log|(x – 1)| + log (x2 + x + 1) + Ú ________ 2 x + x +1
sec4x dx = Ú ____________ sec4x – 3tan2x
(x2 + x + 1) dx __ = log __________ + _______________ (x – 1) 2 3 2 ÷ 1 __ ___ x + + 2 2
(1 + tan2x)sec2x = Ú __________________ dx (1 + tan2x)2 – 3tan2x
(1 + t2)dt = Ú ____________ (1 + t2)2 – 3t2
t2 + 1 = Ú _________ dt t4 – t2 + 1
1 1 + __ 2 t = Ú __________ dt 1 2 __ t + 2 – 1 t
1 1 + __ 2 t = Ú ___________ dt 1 2 t + __ + 1 t
1 = tan– 1 t – __ + c t
= tan– 1(tan x – cot x) + c
| |
| |
|
( ) ( )
(
)
(x2 + x + 1) 2x + 1 2 = log ___________ + ___ __ tan– 1 ______ __ + c (x – 1) ÷3 ÷3
Integer Type Questions
(
( ( ) )
1. The given integral is
+ cos 2x ____________ dx Ú cos x 2cos x – 1
2 x + cos x – 1 dx ________________ = Ú 2cos 2cos x – 1
(2cosx – 1)(2cosx + 1) = Ú ____________________ dx 2cos x – 1
= Ú (cosx + 1)dx
= sin x + x + c
Clearly, A = 1, B = 1
)
(sin x + cos x) _____________ dx (1 – sin x cos x)
)
(sin x + cos x) dx 2 2 = __ Ú ___________ + __ Ú _______________ dx 3 (sin x + cos x) 3 (2 – 2sin x cos x) __
(sin x + cos x) 2 ÷ dx 2 = ___ Ú __________ + __ Ú ________________ 2 dx p 3 sin x + __ 3 (1 + (sinx – cosx) 4 __ 2 ÷ x 3p 2 = ___ log tan __ + ___ + __ tan– 1(sin x + cos x) + c 3 2 8 3
(
|
(
)
)|
Clearly, L = 3, M = 8, N = 2 and P = 2 Hence, the value of L + M – N – P = 7. 3. The given integral is dx Ú ___________ sin6x + cos6x
dx = Ú ______________ 1 – 3sin2x cos2x
4. The given integral is
x+1 Ú _____ dx x3 + x
dx = Ú _________________________ (sin x + cos x)(1 – sin x cos x)
(
( )
Hence, the value of L + M + 4 = 4.
dx Ú ___________ sin3x + cos3x
(
( ( ) )
Clearly, L = 1, M = – 1
Hence, the value of A + B + 3 = 5. 2. The given integral is
1 2 = __ Ú ____________ + 3 (sin x + cos x)
)
x+1 = Ú ________ dx x(x2 + 1) dx dx = Ú _______ + ________ (x2 + 1) x(x2 + 1)
| |
x2 1 = tan– 1x + __ log ______ 2 + c 2 x +1 Clearly, L = 2, M = 2 and N = 2
(
)
L+M + N Hence, the value of __________ = 2 2
5. The given integral is
(
) 2sin x 1 = __ Ú ( ___________ dx 2 sin x + cos x )
sin x Ú ___________ dx sin x + cos x
(
)
1 (sin x + cos x) + (sin x – cos x) = __ Ú ________________________ dx 2 sin x + cos x
Indefinite Integrals
(
)
1.153
(cos x – sin x) 1 = __ Ú 1 – ____________ dx 2 sin x + cos x
= tan– 1(x3 + x2 + 1) + c
1 = __ Ú (x – log|sin x + cos x|) + c 2
Hence, the value of L + M + N + P + Q = 8.
Thus, L = 1, M = 3, N = 2, P = 1 and Q = 1 cos4x 8. Ú ________________ 6 dx sin x(1 + cot5x)3/5
1 1 Clearly, A = __ and B = – __ 2 2
Hence, the value of A + B + 1 is 1
4 2 x dx = Ú _____________ cos x cosec 5 3/5 (1 + cot x)
(
)
x2 6. The given integral is Ú ________ dx (2x + 3)2
Let 2x + 3 = t
fi 2dx = dt
fi dx = 1/2dt
(
3–t fi x = _____ 2
)
x2 dx = Ú ________ (2x + 3)2
3–t 2 _____ 1 2 dt = __ Ú _______ 2 t2
( )
2x = 3 – t
2 1 (3 – t) dt = __ Ú _______ 8 t2
(
2 1 (t – 6t + 9) dt = __ Ú __________ 8 t2 6 9 1 = __ Ú 1 – __ + __2 dt t 8 t
(
)
(
)
9 1 = __ Ú (2x + 3) – log (2x + 3) – _______ + c 8 (2x + 3) (3 + 2x) __ 9 1 = ________ + log |3 + 2x | – ________ + c L M 8(2x + 3) Clearly, L = 8 and M = – 8 Hence, the value of L + M + 4 = 4. 7. We have,
(
2
)
3x + 2x Ú _________________________ dx 6 5 x + 2x + x4 + 2x3 + 2x2 + 2 3x2 + 2x = Ú _____________________________ dx 6 4 1 + ( x + x + 1 + 2(x5 + x3 + x2) ) 3x2 + 2x = Ú _______________ dx 1 + (x3 + x2 + 1)2
where cot x = t
5t4dt 1 = – __ Ú _________ 5 (1 + t5)3/5
Put (1 + t5) = y5
fi 5t4dt = 5y4dy
4 1 5y dy = – __ Ú _____ 5 y3
= – Ú y dy
y2 = – __ + c 2
1 = – __ (1 + t5)2/5 + c 2
1 = – __ (1 + cot5x)2/5 + c 2
1 tan5x + 1 2/5 = – __ ________ + c 2 tan5x
( )
(
)
Thus, L = 2, M = 2 and N = 5
)
9 1 + c = __ Ú t – logt – __ t 8
t4dt = – Ú _________ , (1 + t5)3/5
Hence, the value of L + M + N = 9 13 ___ + 1 b_____ + 1 ______ 2 9. Now, = = 3 g 5 __ 2 Put (1 + x5/2) = t3 5 __ x3/2 = 3t2 dt 2 6 x3/2 = __ t2dt 5 Thus, the given integral reduces to
6 __ Ú (t3 – 1)2 ◊ t3/2 ◊ t2 dt 5
6 = __ Ú (t6 – 2t3 + 1)t7/2 dt 5
6 = __ Ú (t19/2 – 2t13/2 + t7/2) dt 5
1.154 Integral Calculus, 3D Geometry & Vector Booster
(
)
6 t21/2 2t15/2 = __ ______ – ______ + 5 (21/2) (15/2)
t9/2 _____ + c (9/2)
5/2 21/2 2(1 + x5/2)15/2 (1 + x5/2)9/2 6 (1 + x ) = __ __________ – ___________ + __________ + c (21/2) (9/2) 5 (15/2)
(
)
(
)
4 4 = ___ (1 + x5/2)21/2 ___ 8 (1 + x5/2)15/2 + ___ (1 + x5/2)9/2 + c 35 15 25 Clearly, L = 4, M = – 1 and N = 4
Qustions asked in Past IIT-JEE Examinations
(
)
sin x = Ú __________ dx sin x – cos x
2sin x 1 = __ Ú __________ dx 2 sin x – cos x
1 (sin x – cos x) + (sin x + cos x) = __ Ú _________________________ dx 2 sin x – cos x
(sin x + cos x) 1 = __ 1 + ___________ dx 2 sin x – cos x
1 = __ (x + log |sin x – cos x|) + c 2
(
4. We have,
÷
( ) x x = Ú 1 + 2 sin ( __ ) cos ( __ ) dx 4 4 ÷ _________________
)
)
2. We have,
( ( ) ( ) ) x x = 4( – cos ( __ ) + sin ( __ ) ) + c 4 4 x x = Ú sin __ + cos __ dx 4 4
5. We have,
x2 _____ Ú ______ dx ÷1 – x (1 – t2)2 = Ú _______ × – 2 t dt, where 1 – x = t2 t
)
= – 2Ú (1 – t2)2dt = – 2Ú (1 – 2t2 + t4)dt
dt 1 = __ Ú _____ 2 1 + t2
1 1 = __ tan– 1(t) + c = __ tan– 1(x2) + c 2 2
3. We have, 2
(
Put a + bx = t fi
)
t5 2 = – 2 t – __ t3 + __ + c 3 5
(
)
(1 – x)5/2 2 = – 2 (1 – x)1/2 – __ (1 – x)3/2 + ________ + c 3 5
1 dx = __ dt b t____ –a also x = b Thus, the given integral reduces to
)
_________
)
( Ú (
x dx Ú ________ (a + bx)2
)
x __ Ú 1 + sin dx 2
x dx Ú _____ 1 + x4 2x 1 = __ Ú _______ dx 2 1 + (x2)2
(
a2 1 = __3 t – 2a log |t| – __ + c t b a2 1 = __3 (a + bx) – 2a log |(a + bx)| – _______ + c (a + bx) b
dx Ú _______ 1 – cot x
( )
)
(
1. We have,
(
1 t2 – 2ta + a2 = __3 Ú __________ dt b t2
(
10. Do yourself.
2 1 (t – a) = __3 Ú _______ dt b t2
2a a2 1 = __3 Ú 1 – ___ + __ 2 dt t b t
Hence, the value of L + M + N = 7
( )
t–a 2 ____ b 1 _______ × __ dt Ú t2 b
6. We have,
Ú (elog x + sin x) cos x dx
= Ú (elog x◊ sin x + sin x cos x) dx
= Ú (x◊ sin x + sin x cos x) dx
Indefinite Integrals
(
)
sin2x = Ú x◊ sin x + _____ dx 2
cos 2x = x Ú sin x dx + Ú cos x dx – _____ + c 4
cos 2x = – x cos x + sin x – _____ + c 4
) ( (
)
x+1 – 2 = Ú ex ________ dx (x + 1)2
1 –2 = Ú ex _______ 2 + _______ dx (x + 1) (1 + x)3
ex = _______ + c (x + 1)2
(
))
dx dx Ú __________ = Ú ___________ 2 4 3/4 1 3/4 x (x + 1) x5 1 + __ 4 x
( )
t3 = – Ú __3 dt t = – Ú dt
= – t + c
1 1/4 = – 1 + __ 4 + c x
(
)
)
______ __ 1 – ÷x ______ dx __ 1 + ÷x
1 + cos 2q = 2 Ú cos q – _________ dq 2
( ) sin 2q 1 = 2 Ú ( sin q – __ ( q + ______ + c 2 2 ))
= (2sinq – (q + sin 2q cosq )) + c
) ] + c = 2÷1 – x – [ ( cos– 1(÷x ) + ÷ x ÷ 1 – x
_____
__ _____
__
(
__
__
)
– 1 sin– 1÷ x – cos ÷ x Ú ________________ – 1 __ dx – 1 __ sin ÷x + cos ÷x
(
(
)
__ p – 1 __ __ sin– 1÷x – – sin ÷x 2 = Ú _____________________ dx p __ 2 __ p 2 __ = __ p Ú 2sin– 1÷ x – dx 2
(
)
)
(
4 = __ p Ú q Ú sin(2q ) dq +
cos 2q dq – Ú dx Ú ( ______ 2 )
)
where x = sin2q
Let x = cos q
fi ________
1 – cosq = Ú ________ × – sin (2q) dq 1 + cosq
q sin __ 2 = Ú _______ × 2sin (q ) cos (q ) dq q cos __ 2
( ) ( )
2
÷
= 2 Ú (cos q – cos 2q ) dq
4 = __ p Ú (q sin2q )dq – Ú dx
÷
9. We have – Ú
__ 4 = __ p Ú (sin– 1÷x ) dx – Ú dx
(
4 fi – __5 dx = 4t3 dt x 1 fi __5 dx = – t3 dt x
= 2 Ú (1 – cosq ) cos(q )dq
10. We have,
1 Let 1 + __ 4 = t4 x
8. We have,
2
x–1 Ú ex _______ dx (x + 1)3
2
7. We have,
(
( ) ( ) ( ) ( ) q = 4 Ú sin ( __ ) cos (q ) dq 2 q = 2 Ú ( 2sin ( __ ) ) cos(q ) dq 2
q sin __ 2 q q = 4 Ú _______ × sin __ cos __ cos (q )dq 2 2 q cos __ 2
1.155
dx = – sin 2q dq
(
)
q cos 2q ______ sin 2q 4 = __ p Ú – _______ + – x + c 2 4
(
__
__ _____
)
2 sin– 1(÷x ) (1 – 2x ) ÷x ÷1 – x 4 = __ p – ________________ + _________ – x + c 2 2
__ __ _____ 2 2 ) – x + c = __ p ( – sin– 1(÷x ) (1 – 2x ) + ÷ x ÷1 – x _____
cos2x ÷ 11. We have Ú ______ dx sin x
_________
2 x ÷ 1 – 2sin = Ú __________ dx sin x
1.156 Integral Calculus, 3D Geometry & Vector Booster
÷
( )
________
1 = Ú _______ dx 2 sin x – 2
1 1 + __ 2 t = 2Ú ______ dt 1 2 t + __ 2 t 1 1 + __ 2 t = 2Ú __________ dt 1 2 __ t – + 2 t
__________ 2 ÷cosec x – 2 dx
Ú
=
= Ú ÷ cot 2x – 1 dx
( ( ) )
________
Let cot x = secq
fi
– cosec2x dx = secq tanq dq
dz = 2Ú _____ 2 z +2
secq tanq = – _________ dq 1 + cot2x
z 1__ – 1 ___ = 2 ◊ ___ tan __ + c ÷2 ÷2
__ 1 1 = ÷2 tan– 1 ___ __ t – __ + c t 2 ÷
__ 1 = ÷2 tan– 1 ___ __ (tan x – cot x) + c 2 ÷
secq tanq dx = – _________ dq cosec2x
secq tanq = – _________ dq 1 + sec2x
secq = – ________ dq cos2q + 1
( )
( ( ) ) ( )
13. We have
(
)
4ex + 6e– x I = Ú _________ x dx 9e + 4e– x
tanq ◊ sinq = – Ú _________ dq cos2q + 1
sin2q = – Ú ______________ dq cosq (cos2q + 1)
2 = – Ú ______________ sin q ◊ cosq dq cos2q (cos2q + 1)
Comparing the co-efficients of ex and e– x, we get 3 4 m + n = __ and n – m = __ 9 2 Solving, we get,
t2 = – Ú ____________ dt, where sin q = t (1 – t2)(2 – t2)
n = 35/36 and m = – 19/36. The given integral reduces to
t2 = – Ú ____________ dt (t2 – 1)(t2 – 2) 1 = – Ú _____ 2 – _____ dt t2 – 2 t2 – 1
19 35 9ex + 4e– x – ___ Ú dx + ___ Ú _________ dx 36 36 9ex – 4e– x
( Ú ( )
4ex + 6e– x = m(9ex – 4e– x) + n(9ex + 4e– x)
)
( )
(
)
19 35 = – ___ x + ___ log | 9ex – 4e– x | + c 36 36
| |
2 1 = – _____ 2 dt + Ú _____ 2 dt t –2 t –1
19 9e2x – 4 35 = – ___ x + ___ log _______ + c 36 36 ex
t – ÷2 1 __ = – 2◊ ____ __ log ______ + 2÷2 t+÷ 2
19 35 35 = – ___ x + ___ log | 9e2x – 4 | – ___ log |ex| + c 36 36 36
19 = – ___ + 36
3 35 = – __ x + ___ log |9e2x – 4| + c 2 36
(
| |) | |
where t = sinq 12. We have,
____ ____ (÷tan x + ÷cot x ) dx
Ú
( ) t + 1 _____ 2t = Ú ( _____ dt t )t + 1 tan x + 1 = Ú ________ ____ dx tan x ÷ 2
4
( )
t2 + 1 = 2Ú ______ 4 dt t +1
__
t – 1 1 __ log ____ + c 2 t +1
(
(
)
35 35 ___ x + ___ log | 9e2x – 4 | + c 36 36
)
log(1 + x1/6) 1 14. Ú ________ 1/3 1/4 + ___________ __ dx 1/3 x +x x + x ÷ = I1 + I2 (say) dx Now, I1 = Ú _________ x1/3 + x1/4
Indefinite Integrals
(
)
t11 dt = 12 Ú _____ , where x = t12 4 t + t3
cos q + sin q I = cos 2q log ___________ dq cos q – sin q
t8 dt = 12Ú ____ t+1
(
)
8
(t – 1) + 1 = 12Ú __________ dt t+1
(
(
)
)
t8 t7 __ t6 t5 __ t4 t3 __ t2 = 12Ú __ + __ + + __ + + __ + + log|t + 1| + c 7 8 4 3 2 6 5
log (1 + x1/6) Let I2 = Ú ___________ dx x1/2 + x1/3
log(1 + t) 5 = 6Ú _________ t dt, where x = t 6 t3 + t2
t3log(1 + t) = 6 Ú __________ dt t+1
(t3 – 1 + 1)log(1 + t) = 6 Ú __________________ dt t+1
(
)
log (1 + t) = 6 Ú (t2 – t + 1) log (1 + t) – _________ dt (1 + t)
(
)
) ( )
cosq + sinq _____ sin 2q = log __________ 2 cosq – sinq
2 2 1 sin(2q) ◊ 2 (sin q + cos q) = – __ Ú _____________________ ◊ dq 2 cos2q – sin2q
sin (2q) sin 2q cosq + sinq = _____ × log __________ – Ú _______ ◊ dq 2 cos (2q) cosq – sinq
( ) ( ) sin 2q cosq + sinq = ( _____ × log ( __________ 2 ) cosq – sinq )
( (
) (
(
)
3
)
cos3x + cos5x dx Ú ____________ 2 sin x + sin4x
(t – 1)log(1 + t) log(1 + t) = 6Ú _______________ – _________ dt t+1 (1 + t)
(
cosq – sinq cosq + sinq 1 – __ Ú sin (2q) ◊ ___________ + ___________ dq 2 cosq + sinq cosq – sinq
1 __ log | sec (2q) | + c 2 16. We have,
3
[
)
(
cos q + sin q = log ___________ cos (2q) dq cos q – sin q Ú
1 = 12Ú t7 + t6 + t5 + t4 + t3 + t2 + t + ____ dt t+1
1.157
2
)
cos2x + cos4x cos x dx = ____________ 2 sin x + sin4x
]
( Ú ( Ú ( Ú (
)
(1 – sin2x) + (1 – sin2x)2 = Ú _____________________ cos x dx sin2x + sin4x
t t __ – __ + t 3 2 t__2 – t2) + (1 – t2)2 + t – Ú _________ dt = (1 ________________ dt, where t = sin x 1+t 2 t2 + t4 – 6 (log (1 + t))2 + c 4 ((t + t2) + (2 – 4t2)) __________________ 3 2 = dt t t 2 4 = 6 log (1 + t) __ – __ + t – log (1 + t) t + t 3 2 (2 – 4t2) 3 2 ________ 2t – 3t + 6 = 1 + dt 2 4 + Ú ___________ dt t + t t+1
(
)
t3 = 6 log (1 + t) __ – 3
(
)
(
(
)
(
)
1 + Ú 2t2 + 5t + 5 + ____ dt t+1
(
)
)
( Ú (
)
4 2 = Ú 1 – ______ 2 + ________ 2 2 dt t + 1 t (t + 1)
(
))
4 1 1 = 1 – _____ 2 + 2 __2 – ______ dt t +1 t (t2 + 1)
t3 t2 = 6 log (1 + t) __ – __ + t – log (1 + t) 3 2
(
)
)
2 (2t2 – 1) = Ú 1 – _________ 2 2 dt t (t + 1)
t3 t2 = 6 log (1 + t) __ – __ + t – log (1 + t) 3 2
)
)
( (
)
where t = x1/6.
2 = t – 4 tan–1t – __ – 2 tan–1t + c t 2 –1 = t – 6 tan t – __ + c t
15. We have,
= (sin x – 6 tan–1(sin x) – 2cosec x) + c
2t3 5t2 + ___ + ___ + 5t + log (1 + t) + c 3 2
)
1.158 Integral Calculus, 3D Geometry & Vector Booster 1 1 + sin x 1 1 = – __ Ú _______ dx – __ Ú ________________ dx 5 cos2 x 5 __________ 2 tan (x/2) + 4 1 + tan2 (x/2) 1 = – __ Ú ( sec2 x + sec x tan x) dx 5 sec2 (x/2) 1 = – __ Ú _________________________ dx 5 ( 4 tan2 (x/2) + 2 tan (x/2) + 4 )
17. We have,
(
+1 Ú __________ dx x (1 + x ex)2
(x + 1)e x = Ú ___________ dx x ex(1 + x ex)2
dt = Ú _______ 2 , Where t = x ex t (1 + t)
(
1 1 1 = Ú __ – ____ – ______ dt t t + 1 (t + 1)2
1 1 = log ____ + ______ + c t+1 (t + 1)
x ex 1 = log _______ x + ________ x + c x e + 1 (x e + 1)
| |
|
dt 1 1 = – __ (tan x + sec x) – __ Ú _________ 5 5 2t2 + t + 2 x where tan __ = t 2 and then you do it.
)
( )
20. We have,
|
(
dx ____________ Ú ______________ ÷(x – p)3(x – q) dx = Ú _______________ (x – p)3/2(x – q)1/2 dx = Ú _______________ (x – q) 1/2 2 _______ (x – p) (x – p)
(
)
(x – q) Let ______ = t (x – p)
(x – p) ◊1 – (x – q) ◊ 1 fi __________________ dx = dt (x – p)2
dx dt fi _______ = ______ (x – p)2 (q – p)
_ 1 = ______ × 2÷t + c (q – p)
÷
_____
x–p 2 = _______ × 2 _____ x – q + c (q – p)
19. We have,
dx Ú _________________ (sin x + 4) (sin x – 1)
(
= 2 Ú (cos2x – sin2x)dx
= 2 Ú cos (2x)dx
sin 2x = 2 _____ + c 2
( )
= sin 2x + c
÷
)
1 1 = __ Ú________ 1 – ________ dx 5 (sin x – 1) (sin x – 4) 1 1 1 1 = __ Ú ________ dx – __ Ú _________ dx 5 (sin x – 1) 5 (sin x + 4) 1 1 1 1 = – __ Ú ________ dx – __ Ú _________ dx 5 (1 – sin x) 5 (sin x + 4)
______ _ 1 – ÷x dx ______ × ___ x __ 1 + ÷x
Let x = cos22q fi dx = – 2 cos x(2q) sin (2q)
÷
__________
dt 1 = ______ ___ (q – p) Ú t1/2
)
cos x – sin x = 2 Ú __________ × (cos x + sin x)2 dx cos x + sin x
21. Ú
)
The given integral reduces to
(
(
)
cos x – sin x Ú __________ (2 + 2sin 2x)dx cos x + sin x
18. We have,
)
1 – cos (2q) – 2cos (2q) sin (2q) = Ú __________ × ________________ dq 1 + cos (2q) cos2(2q) 2sin (q) cos (q) sinq _____________ = – 2Ú ____ × dq cosq cos(2q) 2sin2(q) = – 2 Ú _______ dq cos(2q)
(cos (2q) – 1) = 2 Ú ____________ dq (cos (2q)
= 2 Ú (1 – sec (2q))dq
1 = 2 q – __ log |sec (2q) + tan (2q)| + c 2
(
(
|
)
______
÷
|)
__ 1 1 1 __ ___ __ = 2 cos–1(÷x ) – log _ + 1 + x + c ÷x 2
22. We have, dx ____________ Ú ____________________ (2x – 7)÷x2 – 7x + 12
Indefinite Integrals
(
)
dx _____________ = 2Ú _____________________ (2x – 7)÷4x 2 – 28x + 48
2x + 2 ____________ Ú sin–1 ______________ dx 2 + 8x + 13 ÷ 4x
2 ◊ dx ___________ = Ú ___________________ (2x – 7) ÷ (2x – 7)2 –1
dt _____ = Ú _______ , where (2x – 7) = t t ÷t 2 – 1
Let (2x + 2) = 3 tanq
= sec–1(t) + c
–1
= sec (2x – 7) + c
x2 + 3x + 2 23. Ú _____________ dx (x2 + 1)2(x + 1)
x2 + 3x + 2) = Ú _____________ dx (x2 + 1)2(x + 1) (x + 1) (x + 2) = Ú _____________ dx (x2 + 1)2(x + 1)
(x + 2) = Ú ________ dx (x2 + 1)2
(x + 2) = Ú ________ dx (x2 + 1)2
dx 2x 1 = __ Ú ________ dx + 2 Ú ________ 2 (x2 + 1)2 (x2 + 1)2
dx 1 = – ________ 2 + 2 Ú ________ (x2 + 1)2 2(x + 1)
x 1 = – ________ + tan–1x + _____ + c 2 (x2 + 1) x2 + 1
)
3 = __ Ú (q sec2q)dq 2
3 = __ Ú (q tanq – Ú tanq dq) 2
3 = __ (q tanq – log |secq |) + c 2
3 = __ q tanq – log ÷1 + tan2q + c 2
|
(
_________
|)
3 3 = __ (q tanq) – __ log (1 + tan2q) + c 2 4 2x + 2 = (x + 1) tan–1 ______ 3 3 – __ log (4x2 + 8x + 13) + c 4
(
)
25. We have Ú (x3m + x2m + xm) (2x 2m + 3x m + 6)1/mdx = Ú (x3m – 1 + x2m – 1 + xm – 1) (2x 2m + 3x m + 6)1/m x dx
)
= Ú (x3m – 1 + x2m – 1 + x m – 1) (2x 3m + 3x 2m + 6xm)1/mdx Put (2x 3m + 3x 2m + 6x m) = t m
dx Let I1 = 2Ú ________ (x2 + 1)2
)
(
(
(
2x + 2 _____________ = Ú sin–1 _______________ ÷(2x + 2)2 + (3)2
(2 x + 2) _____________ Thus, ______________ = sin q + 2)2 + 32 ÷ (2 x
2
2
sec q dq = 2 Ú _______ , where x = tanq sec4q
6m (x 3m –1 + x 2m –1 + x m –1)dx = m t m – 1dt
dq = 2 Ú _____ sec2q
t m – 1 (x3m –1 + x 2m –1 + x m –1)dx = ____ dt 6 1 Thus, the given integral reduces to __ Ú t mdt 6
= Ú (2cos2q) dq
= Ú (1 + cos 2q) dq
1 t m + 1 = __ _____ + c 6 m+1
3m 2m m m + 1 1 (2x + 3x + 6x ) = __ ____________________ + c m+1 6
= (q + sinq + cosq) + c
x = tan–1x + _____ 2 + c x +1
(
24. We have,
)
1.159
( )
(
26. We have, x2 – 1 ____________ Ú ______________ dx x3÷ 2x 4 – 2x2 + 1
)
1.160 Integral Calculus, 3D Geometry & Vector Booster
x2 – 1 _____________ = Ú ________________ dx 2 1 3 4 __ x x 2 – __ 2 + 4 x x
) ÷ ( Ú ) ÷( ) ( Ú ) ÷( ( (
x2 – 1 ____________ = ______________ dx 2 1 5 __ x 2 – __ 2 + 4 x x
1 = ________ t n – 1 + c n (n – 1)
n – 1 ____ 1 = _______ (1 + nx n) n + c n (n – 1)
1 __ 1 = ________ (1 + nx n)( 1 – n ) + c n (n – 1)
28. We have
1 1 __ 3 – __ 5 x x ____________ = ______________ dx 2 1 __ 2 – __ 2 + 4 x x
J – I
)
2 1 Let 2 – __ 2 + __ 4 = t2 x x 4 4 fi __ 3 – __ 5 dx = 2t dt x x
)
1 2t dt = __ Ú ____ t 4
e xdx e– xdx _____________ = Ú ___________ – Ú e 4x + e 2x + 1 e– 4x + e– 2x + 1
(ex – e3x) dx = Ú ___________ e4x + e2x + 1
(1 – e2x) e xdx = Ú ____________ e4x + e2x + 1 Let ex = t fi e xdx = dt
1 = __ × t + c 2
(
)
1 – t2 = Ú _________ dt 4 t + t2 + 1
1 1 – __ 2 t = – Ú _________ dt 1 2 __ t + 2 + 1 t
1 1 – __ 2 t = Ú __________ dt 1 2 __ t + – 1 t
1 t + __ – 1 t 1 __ __________ = – log + c 2 1 t + __ + 1 t
t2 – t + 1 1 = – __ log ________ + c 2 t2 + t + 1
Similarly, we can write, (f0 f0 f 0....0 f )(x)
t2 + t + 1 1 = __ log _________ + c 2 t2 – t + 1
x = __________ . (1 + nx n)1/n
e2x + ex + 1 1 = __ log __________ + c 2 e2x – ex + 1
(
1 2 = __ × 2 – __ 2 + 2 x
)
1 1/2 __ 4 + c x
27. We have, ( f0 f) (x) = f ( f (x)) f (x) = __________ (1 + f (x))1/n x = __________ (1 + 2xn)1/n
Also, ( f0 f0 f ) (x)
= f (( f0 f ) (x))
( f 0 f )(x) = _____________ (1 + ( f 0 f (x))1/n
x = __________ (1 + 3xn)1/n
Thus, I = Ú x
n – 2
g (x)dx
29. We have,
(
( )
( ( ) )
| | ( ) ( )
| |
|
x = Ú x n – 2 __________ dx (1 + nx n)1/n
sec2x Ú ______________ dx (sec x + tan x)9/2
1 n – 2 n n = __ n Ú t dt, Where t = (1 + nx )
)
|
|
|
secx ◊ sec x = Ú ______________ dx (sec x + tan x)9/2
...(i)
Indefinite Integrals
Put sec x + tan x = t fi
sec x (sec x + tan x) dx = dt
dt sec x dx = __ t 1 1 ____________ Also sec x – tan x = = __ t (sec x + tan x) fi
( )
1 1 Thus, sec x = __ t + __ dt t 2 The given intergral (i) reduces to
( )
1 t + __ dt t 1 ________ __ t 2 __________ Ú t9/2
1.161
2 1 (t + 1) = __ Ú _______ dt 2 t13/2
1 1 = __ Ú ___ + 2 t9/2
1 2 2 = – __ ____ + ______ + c 2 7t7/2 11t11/2
1 1 = – ____ 7/2 + ______ 11/2 + c 7t 11t
1 1 = _______________ + ________________ + c 7(sec x + tan x)7/2 11(secx + tan x)11/2
(
( ( (
)
1 ____ 13/2 dt t
)
)
)
Chapter
2
Definite Integrals
Concept Booster 1. What
is
definite Integral?
Let f (x) be a function of x defined in the closed interval [a, b] and j (x) be another function such that j ¢(x) = f (x) for every x in the domain of f (x). Then b
Ú f (x) dx = [j (x) + c] ba = j (b) – j (a) a
is called the definite integral of the function f (x) over the interval [a, b]. Here, a is called the lower limit and b is called the upper limit.
3. Evaluation
of
definite integrals
by
substitution
Rules 1. When the variable in a definite integral is changed, the substitution in terms of new variable should be affected at three places. (a) in the integrand (f (x) will be changed) (b) in the differentials (dx will be changed) (c) in the limits (old limits will be changed)
4. Geometrical interpretation
of
definite integral
Consider the function y = f (x), where f (x) ≥ 0 for all b
x in [a, b]. The integral Ú is numerically equal to the area a
Notes: b
1. Ú is read as ‘the integral of f (x) from a to b’
bounded by the curve y = f (x), the x-axis and the lines x = a and x = b.
a
Y
2. To evaluate the definite integral, there is no need to keep the constant of integration.
C D
b
3. Geometrically, Ú f (x) dx represents the area bounded
by the curve y = f (x), the x-axis and the lines x = a and x = b. 4. Area bounded means, we shall use,
a
b
Ú | f (x)| dx, when the area lies below x-axis. a
O
A x=a
B x=b
X
b
Area of the region ABCD = Ú f (x) dx a
b
2. Evaluation
of
Definite Integrals
ules R 1. Simply find the indefinite integrals of the given function. 2. There is no need to keep the constant of integration. 3. Use the limits of integration.
In general, Ú f (x) dx represents an algebraic sum of the a
area bounded by the graph of the function y = f (x), x-axis and the lines x = a and x = b. The area above the x-axis is taken as the sum with the positive sign, while the area which is below x-axis is taken as the sum with the negative sign. So, the value of the integral may be positive, negative or zero.
2.2 Integral Calculus, 3D Geometry & Vector Booster
6. Evaluation of the limit of Newton-Leibnitz Formula
Y
O a
the
sum
using
From the definition of the definite integral, we have
A1
A3
A2
A5
A4
b
X
b
Ú f (x) dx a
b
Thus, Ú f (x) dx = A1 – A2 + A3 – A4 + A5 a
5. Definite integral
as the
limit
of
h Æ 0
r = 0
(
[ (
)
b – a n b–a = lim _____ n S f a + _____ n r n Æ • r = 0
1 y (x) __r = lim __ n S f n , n Æ •
where
(i)
r = j (x)
( )
 is replaced by Ú symbol
1 (ii) __ n is replaced by dx r (iii) __ n is replaced by x j (x) (iv) a = lim ____ n n Æ •
( ) y (x) and b = lim ( _____ n )
n Æ •
Divide the interval [a, b] into n equal sub-intervals denoted by [a, a + h], [a + h, a + 2h], [a + 2h, a + 3h], ...,
[a + (n – 1)h, a + nh]
7. Properties
of
Definite Integrals
Property I
b
b
a
a
Ú f (x)dx = Ú f (t)dt
Clearly,
b = a + nh
d Proof: Let ___ [g(x) + c] = f (x) dx
fi
nh = b – a
fi
fi
b–a h = _____ n
Now, Ú f (x) dx = g(x) + c|ba
Ú f (x) dx = g(x) + c b
a
b
Thus, Ú f (x) dx
a
= g(b) – g(a)
b
Also, Ú f (t)dt = g(t) + c|ba
= Area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b.
= Area of the region PQRSP between the curve, x-axis and the ordinates x = a and x = b.
Property II
= lim h [f (a) + f (a + h) + f (a + 2h) + ...
d Proof: Let ___ [g(x) + c] = f (x) dx
h Æ 0
+ f (a + (n – 1)h] n – 1
= lim h S f (a + rh) h Æ 0
a
Hence, the result
(
)
where nh = b – a which is known as the first principle of integration.
= g(b) – g(a)
b
a
a
b
Ú f (x)dx = – Ú f (x) dx
fi
r = 0
b – 1 n – 1 = lim _____ S f (a + rh), n n Æ • r = 0
)]
sum
Let f be a continuous real function on [a, b]. Assume that all the values taken by the function are non-negative, the graph of the function is a curve above the x-axis.
n
= lim h S f (a + rh)
Ú f (x)dx = g(x) + c b
Now, Ú f (x) dx = g(x) + c|ba a
a
Also,
= g(b) – g(a)
– Ú f (x) dx = – [g(x) + c]|ab b
Definite Integrals
= – [g(a) – g(b)]
Property V
= g(b) – g(a)
Ú f (x) dx = Ú f (a + b – x) dx
Hence, the result. Property III
b
b
a
a
Proof: We have
b
c
b
a
a
c
b
Ú f (x)dx = Ú f (x)dx + Ú f (x),
Ú f (a + b – x)dx
where
Let a+b–x=t a
a
a 0, (x3 + 16)4 " x Π(0, 1) Thus, f (x) is concave upaward.
(
1
)
331.
0
______ ÷ 1 + x3 dx
Ú 0 1
(
)
x4 1 £ x + __ 4 0
______
1£x£2
fi
1 £ x4 £ 16
fi
2 £ (1 + x4) £ 17
1 1 1 fi ___ £ _______ £ __ 17 (1 + x4) 2 1
dx 7 1 fi ___ £ Ú _____ £ ___ 17 0 1 + x4 24 •
(
0
5 fi 1 £ Ú ÷ 1 + x3 dx £ __ 4 0 334. We have,
0
x fi (1 – 0)f (0) £ Ú ______ 3 dx 0 x + 16
1
______
1
(1 – 0) £ Ú (e ) dx £ e(1 – 0) 1
1
fi Ú 1 dx £ Ú ÷ 1 + x3 dx £ Ú (1 + x3)dx
0
1
fi
2
)
(
)
tan– 1ax – tan– 1x 335. Let I = Ú _____________ dx x 0
1 – (0)(f (0) + f (1) • £ ________________ 2 dI 1.x 1 ___ fi = Ú _________ dx x 17 da 0 (1 + a2x2)x fi 0 £ Ú ______ 3 dx £ ___ 2 0 x + 16 • dx Hence, the result. = Ú ________ _____ 0 (1 + a2x2 3 Let f (x) = ÷ 3 + . • 3x2 dx 1 fi f ¢(x) = ________ _____ = __2 Ú _________ 3 a 0 (1/a2) + x2 2÷3 + x
fi
(
)
3x2 f ¢(x) = ________ ______ > 0, " x Œ (1, 3) 2÷3 + x3
Thus, f (x) increases in (1, 3). 3
______ 3 + x3) dx
So, (3 – 1)f (1) £ Ú (÷ 3
fi
4£
1
_____ (÷ 3 + x3 )dx
Ú 1
£
332. We have for every x in [0, 1], _______
÷ 4 – 2x
2
_________ £ ÷4 – x2 – x3
_____
£ ÷4 – x 2
1 1 1 ______ fi _______ £ ___________ _________ £ _______ ______ 2 2 3 ÷4 – x – x ÷ 4 – 2x2 ÷ 4 – x
)
)
(
)
|
• 1 = __2 (a tan– 1(ax)) 0 a
1 p = __ 2 a__ = 2 a
( )
£ (3 – 1)f (3)
___ 2÷30
( (
p ___ 2a
p I = __ ln a + c 2
fi
When a = 1, I = 0 fi c = 0 p Hence, I = __ ln a 2 p /2
(
)
dx 1 + a sin x ____ 336. Let I = Ú ln _________ , (|a| < 1). 1 – a sin x sin x 0
Definite Integrals
( Ú ( Ú (
p /2
) ) )
dI dx 2 sin x ____ fi ___ = Ú _________ , (|a| < 1) da 0 1 – a2sin2x sin x p /2
dx 2 sin x ____ = _________ 2 2 sin x 0 1 – a sin x
2 = _________ dx 0 1 – a2sin2x
p /2
p /2
(
2
p /2
•
(
xb log x d[I (b)] fi ______ = Ú ______ dx db log x 0
)
1
•
0
(
)
Let t = tan x
(
))
______
• 2 = ________ _______ × tan– 1(t÷1 – a2) 0 – a2) ÷ (1
(
)
p 2 = ________ _______ × __ 2 ÷(1 – a ) 2 p _______ = ________ ÷(1 – a2)
I = p sin– 1a + c
When a = 0, I = 0, then c = 0
(
)
dx 1 + a sin x ____ 336. Given I = Ú ln _________ , (|a| < 1) 1 – a sin x sin x 0 p /2
(
)
dI 1 – a sin x 2 sin x 1 fi ___ = Ú _________ × __________ × ____ d x da 0 1 + a sin x (1 – a sin x)2 sin x p /2
(
(
)
)
)
cot x p /2 –2 ______ = _______ tan– 1 _______ _____ ÷1 – a2 0 ÷ 1 – a2
p p 2 = _______ ______ × __ = _______ _____ 2 2 ÷1 – a2 ÷ 1 – a
Integrating, we get
When b = 0, then c = 0 Thus, I (b) = log(b + 1) 338. We have, x
2sec 2t (f (x))2 = Ú f(t). _______ dt 4 + tan t 0
f (x) . 2 sec2x 2f (x)f ¢(x) = __________ 4 + tan x sec2x fi f ¢(x) = _______ 4 + tan x Integrating, we get f (x) = log|(4 + tan(x))| + c When x = 0, then c = – log(4) f (x) = log|(4 + tan(x))| – log(4) p 5 fi f __ = log|(4 + 1)| – log(4) = log __ 4 4 339. Given, fi
( )
p /2
f (x) = Ú log(1 + x sin2q )/sin2q dx 0 p /2
fi
(
)
sin2q 1 f ¢(x) = Ú ___________ × _____ 2 dq 2 0 (1 + x sin q ) sin q
(
)
dq = Ú ___________ 0 (1 + x sin2q )
cosec2q dq = Ú ______________ 2 0 (cot q + (1 + x))
cot q p /2 1 _____ = – ______ tan– 1 ______ _____ 1 + x ÷1 + x 0 ÷
p _____ = _______ 2÷1 + x
p /2
2 cosec2x = Ú _____________ 2 dx 0 cot x + (1 – a2)
(
I (b) = log(b + 1) + c
p /2
2 cosec2x = Ú __________ dx 0 cosec2x – a2 p /2
)
2 dx = Ú _________ 0 1 – a2sin2x p /2
(
( )
Hence, I = p sin– 1a p /2
xb + 1 = _____ b+ 1 Integrating both sides w.r.t. b, we get
2dt 1 = _______ Ú ______________ 2 1 2 ________ (1 – a2) 0 t + _______ – a2) ÷ (1
= Ú (xb )dx
)
(
( )
1
2dt = Ú ___________ , 0 1 + (1 – a2)t2
fi
1
2 sec2x = Ú ______________ dx 0 1 + (1 – a2)tan2x
I = p sin– 1(a) + c When a = 0, I (a) = 0, then c = 0 Thus, I = p sin– 1(a) xb – 1 337. Given I (b) = Ú _____ dx 0 log x
2 sec x = Ú _____________ 2 dx 0 sec x – a2tan2x
2.91
(
(
(
Integrating, we get _________ f (x) p ÷1 + x + c When x = 0, f (0) = 0 , then c = – p
)
))
2.92 Integral Calculus, 3D Geometry & Vector Booster _____
_____
f (x) p ÷1 + x – p = p (÷1 + x – 1)
(
1
)
x cos a – 1 340. Let I = Ú ________ dx logex 0
(
1
)
x cos a log x dI fi ___ = Ú _________ dx da 0 logx
=
=
=
=
1
= Ú (x cos a )dx
0
(
)
x cos a + 1 1 ________ 1 = ________ = cosa + 1 0 1 + cos a Integrating, we get
I (a ) = log|(1 + cos a )| + c p When a = __ , I (a ) = 0, then c = 0 2 Thus, I (a ) = log|1 + cos a | 341. Do yourself 2
____________
0
÷
÷
___________
___________
2
2
£ Ú (1 + x)dx × Ú (1 + x )dx
0
) ÷(
÷(
4
0
)
x x5 2 = x + __ × x + __ 2 0 5 0 2 2
÷(
________
)
______ 32 = ÷(2 + 2) × 2 + ___ 5 ___ 42 = 2 × ___ 5 Hence, the maximum value of the given integral is
÷
÷
___
42 2 × ___ . 5 343. We have,
Ú ÷
÷
÷(
) ÷(
________
________
)
___
3 5 15 = __ __ = ___ ___ 2 4 8 15 Hence, the maximum value is ___ . 8 344. We have, p a + __ 2 (sin4x a p __ 2
Ú
÷
+ cos4x)dx
= Ú (sin4x + cos4x)dx 0
)
3 = Ú (__ + 4 0
( (
1 __ cos|4x|)dx 4
)
3 sin|4x| p /2 = __ x + ______ 4 16 0 3 p = __ × __ 4 2 3p = ___ 8
)
2p
1 = ___ Ú 2sin2x dx 4p 0
1 = ___ Ú (1 – cos2x) dx 4p 0
sin2x 2p 1 = ___ x – _____ 4p 2 0
2p
(
)
1 = ___ (2p – 0) 4p 1 = __ 2 2
÷ ÷ ÷ __
(
1 – cos4x 1 – _________ d x 4
0
x2 1 x4 1 = x + __ x + __ 2 0 4 0 __
)
2 sin2(2x) 1 – ________ d x 4
1
3 £ Ú (1 + x)dx Ú (1 + x )dx 0
Ú
)
sin2(2x) 1 – _______ dx 2
dx 1 346. Mean value = ______ Ú _____ (2 – 0) 0 ex + 1
÷
1
( Ú ( Ú
2p
1
(1 – 2 sin2x cos2x)dx
1 345. Mean Value = _______ Ú sin2x dx (2p – 0) 0
_____________ (1 + x)(1 + x3)d x 0 __________ ___________
________
________
Ú ÷(1 + x)(1 + x4) dx
Ú
p __ 2
342. We have,
p __ 2 0 p __ 2 0 p __ 2 0 p __ 2 0
2
e– xdx 1 ______ = __ Ú – x 20 e +1
( (
)
2 1 = – __ log|(e– x + 1)| 2 0 1 = – __ log|2(e– 2 + 1)| 2
( | ( ) | )
e2 1 = __ log 2 _____ 2 2 e+1 1
__ 1 347. Mean value = ______ Ú (3÷x ) dx (1 – 0) 0 1
__ 1 = ______ Ú (3÷x ) dx (1 – 0) 0
)
Definite Integrals
( )
1 3 1 = ______ × __ x4/3 (1 – 0) 4 0 3 = __ 4 2p 1 348. Mean value = _______ Ú (sin3x)dx (2p – 0) 0
= G (6 + 1) = 6! = 720 353. We have, 1
1 = ___ Ú (3sin x – sin|3x|)dx 8p 0
=
=
=
349. Mean value =
( (
)
cos|3x| 2p 1 ___ – 3cos x + ______ 8p 3 0 1 1 1 ___ – 3 + __ + 3 – __ 8p 3 3 0 1 dx 1 ______ Ú _____ (1 – 0) 0 1 + x2
)
) = ( tan p = __ 4 1 ex 1 350. Mean value = ______ Ú _____ x dx (1 – 0) 0 e + 1 (x) 10
– 1
( )
= (log|(ex + 1)|)10
( )
e+1 = log _____ 2
351. We have,
(
1
)
= lim (j – 1 – j log(j ))
= lim (j – 1) – lim ( j log(j ) )
•
j Æ 0 j Æ 0
( )
log(j ) = lim (j – 1) – lim ______ j Æ 0 j Æ 0 1 __ j
( )
1 __ j _____ = lim (j – 1) – lim j Æ 0 j Æ 0 1 – ___ 2 j
0
= G (11) = G (10 + 1) = (10)! 354. We have, •
= – 1 352. We have,
0 • 0
= G (7) = (3)! =6 355. We have,
j Æ 0
= Ú (e– tt n –1) dt,
(Let t = – log x)
0 •
= Ú (e– x t n – 1)dx
= G (n)
0
1
______
356. Let I = Ú x4 ÷1 – x2 dx 0 p /2
= Ú sin4q cos2q dq 0
•
1
0 •
= Ú e– xx7 – 1 dx
= G (7)
0
( ( ) )
= Ú (– log|x|)n – 1dx
0
= Ú e– xx4 – 1 dx
•
Ú e– xx6 dx
= Ú e– xx3 dx
= lim (j – 1) + lim (j ) j Æ 0
•
= Ú e– xx11 – 1 dx
0
= lim (x log(x) – x)1j
j Æ 0
0
= – Ú e– tt10 dt
1 n–1 Ú log __ x dx
j Æ 0
0
1
lim Ú log(x)dx j Æ 0 j
( ( ) )
1 10 Ú log __ x dx
2p
2.93
( ) ( ) ( ) ( ) ( )
2+1 4+ 1 G _____ G _____ 2 2 = _________________ 4 +2 +2 2G _______ 2 5 3 G __ G __ 2 2 = _________ 2G (4) 3 1 __ __ 1 __ __ ◊ __ ◊÷p × ÷ p 2 2 2 = ______________ 2×6 p = ___ 32
(Let sinq = x)
2.94 Integral Calculus, 3D Geometry & Vector Booster p /2
357. We have,
I = Ú sin8x cos4x dx
0
( ) ( ) ( ) ( ) ( ) ( )
358. We have,
______
Ú x6÷1 – x2 dx 0
p /2
= Ú sin6q cos2q dq 0
359. We have,
( ) ( ) ( ) ( ) ( )
6+1 2+1 G ____ G _____ 2 2 = ________________ 6+2+2 2G ________ 2 3 7 G __ G __ 2 2 _________ = 2G(5) 5 3 __ 1 __ 1 __ ◊__ ◊ ÷ p × __ p 222 2 = _____________ 2 × 24 5p = ____ 256
p /2
Ú sin4x cos6x dx 0
( ) ( ) ( ) 5 7 G( __ )G( __ ) 2 2 = _________
6+1 4 +1 G ____ G ____ 2 2 = ________________ 4+6+2 2G ________ 2
2G(6)
5 3 __ 3 1 __ __ 1 __ __ ◊ __ ◊ ÷p ◊ __ ◊ ÷ p 2 2 2 2 2 = ________________ 2 × 120 3p = ____ 512 1
360. We have,
______ x10 ÷1 – x2 dx
Ú 0
(Let sinq = x)
(Let sin q = x)
(
) ( ) ( ) ( ) ( )
10 + 1 2 +1 G ______ G _____ 2 2 = _________________ 10 + 2 + 2 2G __________ 2 3 11 G ___ G __ 2 2 __________ = 2G(7)
3 1 __ __ 3 1 __ 7 5 __ __ ◊ __ ◊ ◊ __ ◊ ÷p × ◊ __ ◊ ÷p 2 2 2 2 2 2 = _____________________ 2 ◊ (6)! 7p _____ = 2048
1
0
8+1 4+1 G ____ G ____ 2 2 ________________ = 8+4+2 2G ________ 2 9 5 G __ G __ 2 2 = _________ 14 ___ 2G 2
= Ú sin10q cos2q dq
p /2
9 7 5 3 1 __ 1 __ __ ◊ __ ◊ __ ◊ __ ◊ __ ÷ p ◊ __ ÷ p 2 2 2 2 2 2 = __________________ 2 × (6!) 21p = ____ 512
361. We have, p /2
p /2
____ dx Ú _____ ____ × Ú ÷sin x dx sin x 0 ÷ 0 p /2
p /2
0
0
1 __
= Ú sin–1/2x cos0x dx x × Ú sin2 x cos0x dx x
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 = ( G( __ ) ) 2
3 1 1 1 G __ G __ G __ G __ 4 2 4 2 = _________ × _________ 3 5 __ __ 2G 2G 4 4 3 1 1 1 G __ G __ G __ G __ 4 2 4 2 = _________ × _________ 3 1 1 2◊ ◊G __ 2G __ 4 4 4 2
= p 362. We have, •
Ú e– a x dx 2 2
0
•
1 = ___ Ú e– t t – 1/2dt, 2a 0
1 1 – __ 1 = ___ Ú e– t t 2 dt 2a 0
( )
•
363. We have,
( )
1 G __ 2 _____ = 2a __ p ÷ = ___ 2a p /2
Ú sin7x cos5x dx 0
(
) ( ) ( )
5+ 1 7+1 G _____ G _____ 2 2 = ________________ 7 + 5 + 2 2G _________ 2
(Let (ax)2 = t)
Definite Integrals
364. We have,
G(2)G(3) = ________ 2G(7) 3.2.1.2.1 = ___________ 2.6.5.4.3.2.1 1 = ____ 120 p /2
Ú sin10x dx 0
365. We have,
9 7 5 3 1 p = ___ ◊ __ ◊ __ ◊ __ ◊ __ ◊ __ 10 8 6 4 2 2 63p = ____ 512 p /2 0
366. We have,
6 4 2 = __ ◊ __ ◊ __ 7 5 3 16 = ___ 35
p /4
Ú 8 cos4x sin4x dx 0
p /4
1 = __ Ú 16 cos4x sin4x dx 2 0
1 = __ Ú (sin2x)4dx 2 0
1 = __ Ú (sin4t)dx 2 0
p /4
p /2
367. We have, p /2
1 3 1 p = __ ◊ __ ◊ __ ◊ __ 4 4 2 2 3p = ___ 64
5 3 1 __ 3 1 __ __ ◊ __ ◊ __ ÷ p ◊ __ ◊ __ ◊ ÷p 2 2 2 2 2 = _________________ 2 × (5!) 3p ____ = 512 370. We have, p /2 0
1
= Ú t7 dt,
(Let sin x = t)
0
( )
t8 1 = __ 8 0 1 __ = 8 371. We have, p /2
Ú sin7x cos5x dx
( ) ( ) ( )
0
5+1 7+1 G _____ G _____ 2 2 = _______________ 7+5+2 2G _________ 2 G(4)G(3) = ________ 2G(7) (3!) × (2!) = __________ 2 × (6!) 6×2 1 = _______ = ____ 2 × 720 120 372. We have, p /2
Ú sin x cos x dx 0
( ) ( ) ( ) ( ) ( )
6+1 4+1 G _____ G _____ 2 2 _______________ = 6+4+2 2G ________ 2 5 7 G __ G __ 2 2 = _________ 2G(6)
Ú sin7x cosx dx
Ú cos7x dx
6
5
(5◊3◊1)(4◊2) = ____________ (11◊9◊7◊5◊3◊1) 8 = ____ 693 p /2
368. We have, Ú sin5x cos3x dx 0
(4◊2)(2) = ________ (8◊6◊4◊2)
1 = ___ 24 p /2
369. We have, Ú sin6x cos4x dx 0
2.95
Ú sin11x dx 0
(
)(
)(
)(
)(
)
–9 – 3 11 – 5 11 – 7 11 11 – 1 11 = ______ ______ ______ ______ ______ 11 11 – 2 11 – 4 11 – 6 11 – 8 10 8 6 4 2 = ___ ◊ __ ◊ __ ◊ __ ◊ __ 11 9 7 5 3 256 = ____ 693 p /2 373. We have, Ú cos8x dx 0
( )(
)( )( )
p 8 – 1 _____ 8 – 3 _____ 8 – 5 _____ 8 – 7 __ = _____ ◊ 8 8–2 8–4 8–6 2 7 5 3 1 p = __ ◊ __ ◊ __ ◊ __ ◊ __ 8 6 4 2 3 35p ____ = 256
2.96 Integral Calculus, 3D Geometry & Vector Booster
1. Let I = Ú (|sin x| – |cos x|)dx
1 8 ×9 = __ × _____ 3 2
= 12
0
2p
10p
2p + p /4
= Ú (|sin x| – |cos x|)dx + Ú (|sin x| – |cos x|)dx 0
4. Let I = Ú ([sec– 1x] + [tan– 1x])dx 1
2p
p
sec 1
p /4
= 2Ú (|sin x| – |cos x|)dx + Ú (|sin x| – |cos x|)dx
= Ú ([sec– 1x] + [tan– 1x])dx
1
0
0
p /2
10p
+ Ú ([sec– 1x] + [tan– 1x])dx
p /4
sec 1
= 4Ú (sin x – cos x)dx + Ú (sin x – cos x)dx 0
0
= 4(– cos x –
sin x)p /2 0 +
(– cos x –
__
sec 1
= Ú (0 + 0)dx + Ú (1 + 0)dx
sin x)p /4 0
1
= 4 . 0 + (1 + ÷ 2 )
5p /12
5 Let I = Ú [tan x]dx 0
Hence, the value of the given integral __
= Î(1 + ÷ 2 ) ˚ = – 1 sq.u.
[ [ ( ) ] ] p x = Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2 p x + Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2 p x + Ú [ x[ 1 + cos( ___ ) ] + 1 ]dx 2
p x 2. Let I = Ú x 1 + cos ___ + 1 dx 2 – 2 – 1
– 2
1
2
0
1
= Ú [tan– 1x]dx + Ú [tan– 1x]dx
1
sec 1
= (10p – sec 1) sq. u.
__
= (1 + ÷ 2 )
10p
3
5p /12
+ Ú [tan– 1x]dx + Ú [tan– 1x]dx 2
3
= 0 + (tan– 12 – tan– 11)◊1 + (tan– 13 – tan– 12) ◊ 2
0
– 1 1
0
– 1
0
1
– 2
– 1
0
= Ú 1.dx + Ú 0.dx + Ú 1.dx
= 1( – 1– ( – 2)) + 0 + 1(1 – 0)
=1+1
= 2 sq.u. 3
5p p = ___ – __ – (tan– 13 + tan– 12) 4 4 = p + p – tan– 1( – 1) 9p = ___ sq.u. 4 p /4
6. Let I = Ú [sin x + {cos x + tan x + (sec x)}]dx 0
p Since 0 < x < __ 4 __ \ 1 < sec x < ÷ 2
( [ ] [ ] )
1 2 3. Let I = Ú [x] + x + __ + x + __ dx 3 3 0
(
1/3
2/3
1
0
1/3
2/3
= Ú 0. dx + Ú 1. dx + Ú 2. dx 4/3
5/3
2
1
4/3
5/3
7/3
8/3
3
2
7/3
8/3
Ú 6. dx + Ú 7. dx + Ú 5. dx
1 = __ (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) 3
fi
[sec x] = 1
fi
[tan x + [sec x]] = [tan x + 1]
+ Ú 3. dx + Ú 4. dx + Ú 5. dx
)
5p + ___ – tan– 13 ◊3 12
= [tan x] + 1 = 0 + 1 = 1
fi
[cos x + [tan x + [sec x]]]
= [cos x + 1] = [cos x] + 1 = 0 + 1 = 1
fi
[sin x + [cos x + [tan x + [sec x]]]]
= [sin x + 1] = [sin x] + 1 = 0 + 1 = 1 p /4
Thus, I = Ú [sin x + [cos x + [tan x + [sec x]]]dx 0
Definite Integrals
(
p m Now, __ p + n p + __ + 4 2 = 50 + 6 + 4
p /4
= Ú 1.dx
0
p = __ 4
p /2
10. Let
0
I = Ú tan(x2 – 6)dx + Ú tan(x2 + 18x + 75)dx
( ) ( ) ( ) ( )
p–1 p–2 p–3 = _____ + _____ + _____ 2 2 2 3p – 6 = ______ sq.u 2 sin(sin– 1b)
8. Let
| |
=
– 1
|
– 2
– 3
– 6
– 7
– 2
– 3
– 6
– 7
7
– 3
3
– 7
7
= Ú tan(x2 – 6)dx + Ú tan(y2 – 6)dx
|
= Ú tan(x2 – 6)dx + Ú tan(x2 – 6)dx – 3
cos(cos x) __________ dx sin(sin– 1x) cos(cos– 1a)
Ú
– 7
= Ú tan(x2 – 6)dx + Ú tan(x + 9)2 – 6)dx
cos(cos– 1x) = __________ dx Ú sin(sin– 1x) cos(cos– 1a) sin(sin– 1b)
)
= 60
7. Let I = Ú [sin– 1(cos x) + cos– 1(sin x)]dx
Graph:
2.97
3
7
7
3
3
= Ú tan(x2 – 6)dx + Ú tan(x2 – 6)dx
b
= 0
a
11. We have,
b x a
= Ú 1dx
=
= (b – a)
2n p
In = Ú |sin x|[sin x]dx – 2n p 2n p
= Ú |sin x|[sin x]dx
– 2n p
9. We have, 100
2n p
m = Ú sec– 1(secp x)dx
= Ú |sin x|(– 1 – [sin x])dx
0
– 2n p
2
= 50 × Ú sec– 1(secp x)dx
0
1 = 50 × __ × 2 × p 2 = 50p
2n p
fi 2In = – Ú |sin x|dx – 2n p
= Ú [sec (secp x)]dx + Ú [sec (secp x)]dx – 1
= – 2Ú |sin x|dx 0 2n p
1/2
2/p
3/p
1/p
2/p
+ Ú [sec– 1(secp x)]dx + Ú [sec– 1(secp x)]dx 1/p
2/p
1/2
( ) ( 6 1 = ( __ p – __ ) 2
1/p
fi In = – Ú |sin x|dx 0
p
3/p
= – 2nÚ |sin x|dx
2/p
= – 2n × 2 = – 4n
= Ú 0. dx + Ú 1. dx + Ú 2. dx + Ú 3. dx 0
2n p
1/p
– 1
1/2
– 2n p
– 2n p
0
0
– 2n p 2n p
3/p
2n p
= – Ú |sin x|dx – In
Also, n = Ú [sec– 1(secp x)]dx 1/2
2n p
= – Ú |sin x|dx – Ú |sin x|[sin x]dx
) (
)
3 2 1 1 2 1 = 0 + 1. __ p – __ + 2. __ p – __ p + 3. __ p – __ p 2
0
100
Now, S (In) n = 1
100
= S (– 4n) n = 1
2.98 Integral Calculus, 3D Geometry & Vector Booster
(
n = 1
= – 4(1 + 2 + 3 + ........... + 100)
( [ ] )
0
|sin x| m = Ú _______ dx x 1 – 2 __ p + __ 2 Let x = – y fi dx = – dy
( [ ] ) ( [ ] ) ( [ ] ) ( [ ] )
0
|sin(– y)| Thus, m = Ú ________ – y × – dy 1 ___ __ 2 p + 2
1 1 3. __ £ x2 + __ £ 1 fi 2 2 1 fi 0 £ x £ ___ __ 2 ÷ 3 1 4. 1 £ x2 + __ £ __ fi 2 2 1__ fi ___ £ x £ 1 2 ÷
2
|sin y| = – Ú _________ dy y 1 0 – __ – __ p 2
( [ ] )
– 1
1
– 1
0
p /2
2
(
1/÷2
__ 1/÷2
p /2
)
1
+ 2 Ú (cos– 1( – 1))dx + Ú __ cos– 1(0)dx 0
1/÷2
( ( ) ) ( ( ( ) ( )
p p 1 = 2 0 + __ 1 – ___ __ + 2 ___ __ + 2 2 2 ÷ ÷ p 1 = p 1 – ___ __ + 2 __ + 2 2 ÷
p ____ __ 2÷2
)
Dividing the numerator and the denominator by sin2 x, we get
)
1
(
8 + 7 cos x 15. Let I = Ú ___________ dx 0 (7 + 8 cos x)2
= 2 Ú (sin– 1(0))dx + Ú __ (sin– 1(1))dx 0
( )
( ( ) )
1
0
(
0
)
p p = – __ 0 – – __ 2 3 2 p = ___ 6
2
2
__
(
p = Ú __ dx – p /3 2
0
1/÷2
0
– 1
p = Ú – p + __ dx 2 – p /3
1
– 1
– p /3
( [ ] [ ] ) 1 1 = 2Ú ( sin [ x + __ ] + cos [ x – __ ] )dx 2 2 1 1 = 2Ú ( sin [ x + __ ] )dx + 2Ú ( cos [ x – __ ]dx ) 2 2 – 1
]
0
1
2
[
)) ) ( ( ( 1 1 = Ú ( – p + tan ( cos x – __ ) + cot ( cos x – __ ) )dx 2 2
1 1 I = Ú sin– 1 x2 + __ + cos– 1 x2 – __ dx 2 2 – 1 – 1
1 Ï ÔÔ0 : 0 £ x £ 2 1 5. x2 + __ = Ì 2 Ô1 : 1 £ x £ 1 ÔÓ 2 0
|sin x| = – Ú _______ dx x 1 0 __ p + __ 2 = – n fi m + n = 0. Hence, the result. 13. Let
1 __ £ x2 £ 1 2
2 1 I = Ú cot– 1 _________ + cot– 1 cos x – __ dx 2 cos x – 1 2 – p /3
|sin y| = – Ú _______ y dy 1 0 __ p + __ 2 2
1 0 £ x2 £ __ 2
14. Let
2
1 1 1 2. __ £ x2 + __ £ 1 + __ 2 2 2
0
|sin(y)| = – Ú ___________ × dy y 1 __ 2 – 1 – p + __ 2
)
Notes: 1. 0 £ x £ 1 fi 0 £ x2 £ 1
100 × 101 = – 4 × _________ 2 = – 20200
12. Given
) (
1 1 = p 1 – ___ __ + p 1 + ____ __ 2 2÷2 ÷ = 2p
100
= – 4S (n)
))
p 1 __ 1 – ___ __ 2 2 ÷
(
)
8 cosec2 x + 7 cosec x cot x I = Ú ______________________ dx 0 (7 cosec x + 8 cot x)2 Let
7 cosec x + 8 cot x = t
fi
(8 cosec2 x + 7 cosec x cot x)dx = – dt 7
( )
Thus, I = – Ú __ 2 • t 7 1 1 = __ = __ t • 7
( )
Definite Integrals
Put
x=0=y
fi
f (0) = 0
Put
y = – x, we get, f (x) – f (– x) = 0
fi
f (x) = f (– x)
\
f (x) is an even function.
– 1
3
Now, a = Ú (x – 1) f (x – 1)dx 2
1 2
( ( ) ) 1 – cos a – cos a = ( tan ( ________ – tan (__________ – 1sin a ) ) sin a ) 1 – cos a t = tan– 1 _____ sin a – 1 – cos a
16. Given f (x + y) = f (x) – f (y)
= Ú t f (t)dt, 2
(Let x – 1 = t)
– 1
( (
( ( ( ) ) (
a p __ a = __ + __ – 2 2 2
)
x3______ sin– 1x 19. Let I = Ú ________ dx 0 ÷1 – x2
1
= Ú y2f (y)dy,
(Let x + 1 = y)
– 2
p /2
0
( ( ( ( (
) )
4 (y – 2)2 + 1 )sin(y – 2) = Ú ___________________ dy 0 2(y – 2)2 – 7
2
)
(t2 + 1)sin t = Ú __________ 2 dt, – 2 2t – 7)
(Let y – 2 = t)
)
(y2 + 1) sin y = Ú __________ dy – 2 2y2 – 7 = 0, 1
( it is an odd function.)
(
(
cos3t = 0 – Ú _____ – cos t dt 3 0
1 = – __ Ú (cos3t)dt + Ú cos t dt 3 0 0
sin3t p /2 1 = – __ sin t – ____ + (sin t)p /2 0 3 3 0
1 1 = – __ 1 – ___ + 1 3 13
2 = 1 – __ 9
7 = __ 9
))
(
)
p /2
p /2
(
p /2
p /2
(
)
)
| |
(
)
)
sin a = Ú _________ 2 dt – 1 – cos a t + sin2a
1
(
)
(
1
Let x – cos a = t fi dx = dt 1 – cos a
( (
)
sin a = Ú ________________ 2 dx 2 – 1 x – cos a) + sin a
)
1 – x2 _______ 1 20. Let I = Ú _____2 × ______ dx 0 1+x ÷1 + x4
sin a 18. Let I = Ú ______________ 2 dx – 1 x – 2x cos a + 1 1
(
cos3t cos3t = t _____ – cos t – Ú _____ – cos t dt 3 3 0 0 p /2
(y2 – 4y + 5)sin(y – 2) 17. Let I = Ú ___________________ dy 0 2y2 – 8y + 1
2
)
= 2a
Thus, the value of 2a – b + 4 = 0 + 4 = 4
p /2
p /2 cos3t = t Ú sin3t dt 0 – Ú _____ – cos t dt 3 0
2
= 2Ú x2f (x)dx
(
– 2
dx Let sin– 1 x = t fi _______ ______ = dt ÷ 1 – x2
0
= Ú y2f (x)dx
4
)
Thus, I = Ú (t sin3t)dt
2
(
2
( ( ) ) )
0 1
= Ú x2f (x)dx
))
a a = tan– 1 tan __ + tan– 1 cot __ 2 2
a = __ 2
– 3
(
)
2sin2(a /2) 2cos2(a /2) = tan– 1 __________ + tan– 1__________ sin a sin a
0 2
Also, b = Ú (x + 1)2f (x + 1)dx
2.99
dx 1 – x2 = Ú ________ × _______ ______ 1 1 0 x2 x + __ x x2 + __ 2 x 1 __ 1 1 – 2 x dx = Ú _______ × ____________ __________ 1 1 2 0 x + __ x x + __ x – 2
(
( )
(
)
÷
) ÷(
)
)
2.100 Integral Calculus, 3D Geometry & Vector Booster
(
( )
)
1 1 Put x + __ x = t fi 1 – __ 2 dx = dt x
p /2
= 2Ú (log secq )dq 0
2
dt = – Ú _______ _____ • t ÷t 2 – 2
p /2
0
2
t dt = – Ú ________ _____ • t2 ÷t 2 – 2
Let t2 – 2 = y2 fi t dt = y dy
= – 2Ú (log cosq )dq
( ( ) )
p 1 = – 2 __ log __ 2 2
= p log 2
p = 8 __ log 2 8 = 8 m
__
(
)
÷2 dy = Ú _____ • y2 + 2
y ÷2 1__ – 1 ___ = – ___ tan __ ÷2 • ÷2
p 1__ __ = – ___ ( – 2 4 ÷ p = ____ __ 4÷2
2x 2x cos– 1 _____ 2 + tan– 1 _____ 2 1 + x 1 –x _________________________ 22. Let I = Ú __ dx ex + 1 – 1/÷3
(
21. Given, 1
(
( ) ) |
__
p __ ) 2
Hence, the result.
)
p /4
= Ú log (1 + tan q )dq
0
p /4
(
(
))
p = Ú log 1 + tan __ – q dq 4 0
( ) 1 + tan q + 1 – tan q = Ú log ( __________________ ) dq 1 + tan q
0
p /4
(
)
)
( ) p dx = __ Ú ( ______ 2 e + 1) p dx = __ Ú __ _____ x 2 – 1/÷3 e +1
...(i)
__
1/÷3
__ – 1/÷3
–x
__ 1/÷3
= p /2Ú __ (ex/ex + 1)dx
p /4
p /4
0
0
__
1/÷3
p = __ log 2 – m 4 p fi 2 m = __ log 2 4 p fi m = __ log 2 8 • log(1 + x2) Also, n = Ú _________ dx 0 1 + x2
(
)
( )
__
1/÷3
p p 2 = __ Ú __ dx = __ × ___ __ 2 – 1/÷3 2 3 ÷
p fi I = ____ __ 2÷3
( Ú ( ( Ú ( (
5p /4
) ) )
sin x + cos x 23. Let I = Ú __________ dx p ( x – __ ) – 3p /4 e 4 + 1 5p /4
2
cos x + sin x = __________ dx p p __ – x– __ – 3p /4 e 2 4 )+ 1
= Ú (log sec2q )dq
cos x + sin x = __________ dx p – x– __ – 3p /4 e 4 )+ 1
= Ú [log(1 + tan q )]dq
0
...(ii)
p 1 + ex fi 2I = __ Ú __ _____ dx 2 – 1/÷3 ex + 1
0 p /2
(
Adding Eqs (i) and (ii), we get
= Ú log(2)dq – Ú log(1 + tan q )dq
)
– 1/÷3
2 = Ú log ________ dq 1 + tan q 0
p /2
( )
__
p /4
( )
)
1/÷3
p /4
1 – tan q = Ú log 1 + ________ dq 1 + tan q 0
( )
p 2x 2x __ – sin– 1 _____ 2 + tan– 1 _____ 2 2 1+x 1–x = Ú __ ____________________________ dx x e + 1 – 1/÷3 __ p __ – 2tan– 1x + 2 tan– 1x 1/÷3 2 = Ú __ __________________ dx ex + 1 – 1/÷3 __
1/÷3
log(1 + x) m = Ú _________ dx 0 1 + x2
( )
( (
__
1/÷3
5p /4
...(i)
(
p x– __
)
1
cos x + sin x)e( 4 ) = Ú ________________ dx p x– __ – 3p /4 e( 4 )+ 1 5p /4
...(ii)
( x– __p ) )
(cos x + sin x) e( 4 ) + 1 2I = Ú _____________________ dx p ( x– __ ) – 3p /4 4 e + 1 5p /4
(
5p /4
= Ú (cos x + sin x)dx
– 3p /4
5p /4
fi 24. Given,
(
( (
1__ I = – ___ cos x + 2 ÷ 1
(
)
5p ___ p 4 __ = 3p 4 – ___ 4
))
0
)
( ) q = Ú ( _________ dq sin q cos q ) 2q = Ú ( __________ dq 2sin q cos q ) 2q = Ú ( _____ dq sin 2q )
0
t = Ú ____ dt, sin t 0
x = Ú ____ dt sin x 0
1 = __ I1 2
fi
(Let 2q = t)
( )
I1 Thus, __ = 2 I2
Now,
Thus,
(
)
p /2
x2 I2 = Ú _________ 3 dx 0 ex (2 – x3)
1
( ) Ú ( )
sin nx dx Ú ( _____ sin x )
)
dt 1 = __ Ú _________ 3 0 e1 – t(t + 1)
et dt 1 = ___ ______ 3e 0 (t + 1)
(
)
sin(n – 2)x = Ú2 cos(n – 1)x dx + Ú _________ dx sin x sin 5x dx Ú _____ sin x sin3x = Ú2 cos 4 x dx + Ú _____ dx sin x
( )
( )
( )
sin 5x Ú _____ dx sin x 0
1 (Let 1 – x3 = t fi x2 dx – __ dt) 3 0 dt 1 __ _________ = – Ú 1 – t dx 3 1 e (t + 1)
(
In + 1 = In = In – 1 = In – 2 = ... = I1 = I0
Hence, In = p, n ≥ 0. 27. We have, sin nx – sin(n – 2)x = 2 cos(n – 1)x sin x
0
( )
2 = _____ (sin(n + 1)x)p0 = 0 n+1
Therefore, I0 = Ú sin x dx = (x)p0 = p
p /4
1
( )
p
0
25. Given,
)
= 2 Ú cos(n + 1)x dx
fi
p /2
)
Thus, In + 1 – In
p /2
(
q = Ú ____ sec2 q dq tan q 0
p /4
)
1 1 sin n + 1 + __ x – sin n + __ x 2 2 ___________________________ = Ú dx x __ 0 sin 2
0
0
( ( ( )) ) ( (
1 sin n + __ x 2 __________ 26. Let In = Ú dx. x __ 0 sin 2 p
x
tan– 1x I2 = Ú ______ x dx 0
p /4
I1 Thus, __ = 3e I2
x
p /4
)
Now, In + 1 – In
__ p = ÷2 Ú sin x + __ d x 4 – 3p /4
(
ex dx 1 = ___ Ú ______ 3e 0 (x + 1) I1 = ___ 3e
Adding Eqs (i) and (ii), we get
2.101
Definite Integrals
p /2
p /2
( )
sin3x = 2Ú cos4 x dx + Ú _____ dx sin x 0 0
3sin x – 4sin3x = 0 + Ú ____________ dx sin x 0
= Ú (3 – 4sin2 x)dx
p /2
(
)
p /2 0 p /2
1
= Ú [3 – 2(1 – cos2 x)]dx 0
2.102 Integral Calculus, 3D Geometry & Vector Booster p
p /2
= Ú (1 + 2 cos 2 x)dx
sin y B = Ú _____ dy y +2 0
sin x = Ú _____ dx x +2 0
p cos x 1 = _____ Úsin x dx – Ú _______ dx 2 x+2 0 0 (x + 2)
0
p
= (x + sin 2x)p /2 0
( )
p = __ 2 28. Given
(
1
m = Ú x50(2 – x)50 dx 0
(Let x = 2t fi dx = 2dt) 1/2
m = 2 Ú 250 t 50 250(1 – t)50dt
fi
1/2
= =
1 2100 0
Ú
Ú x
50
(1 – x) dx = 2
•
0
(
)
(
)
•
( )
( )
3 3 sin3x sin x = __ Ú ____ x dx – __ Ú _____ dx 40 40 3x
siny 3 sin x 1 = __ Ú ____ x dx – __ Ú ____ y dy, (Let 3x = y) 40 40
( )
•
( )
•
)
3 p __ 1 p = __ × __ – × __ 4 2 4 2 p = __ 4 30. We have, p /2
(
)
sin2 x B = Ú ______ dx (x + 1) 0 y dy Let x = __ fi dx = ___ 2 2
(
( )
( )
3 sin x sinx 1 = __ Ú ____ x dx – __ Ú ____ x dx 40 40
(
)
)
p /2
3sin x _____ sin3x 1 = __ Ú _____ x – x dx 40
•
(
p /2
– sin3x 1 3sin x = __ Ú ___________ dx x 40
•
)
32. Let I = Ú log(a2 cos2 x + b 2 sin2x)dx
3
•
)
p p = ___________________ = ___ 2(2 + 3)(3 + 0)(0 + 2) 60
0
•
(
Let a = 2, b = 3 and c = 0, dx Ú ______________ 2 0 (x + 4)(x2 + 9)
( ) 4 sin x 1 = __ Ú ( ______ ) dx x 4 •
1 __ – A 2 1 1 A + B = _____ + __ p+2 2
(
sin3x 29. Let I = Ú _____ x dx 0
)
p = __________________ 2(x + a)(x + b)(x + c)
.n
•
(
)
1 = _____ + p+2
100
m 100 fi __ n = 2 Hence, the result.
•
p
(
x2 Ú ______________________ 2 dx 0 (x + a2)(x2 + b2)(x2 + c2)
– t)50dt
50
)
)
•
0
1 2100 t 50(1 0
(
p
– cos x p cos x = ______ – Ú _______ dx x + 2 0 0 (x + 2)2
31. We have,
= 2100◊2 Ú t 50(1 – t)50dt
( )
fi
0
( )
fi
= Ú log[a2 (cos2 x + k 2 sin2x)]dx 0
(Let k = b/a)
p /2
= p log a + Ú log (cos2 x + k 2 sin2x))dx
= p log a + I1,
0
p /2
Where
I1 = Ú log (cos2 x + k 2 sin2x))dx p /2
(
0
)
dI1 2 k sin2x fi ___ = Ú ______________ dx dk 0 cos2 x + k 2sin2x
( Ú (
)
p /2
tan2x ◊ sec2x = 2k Ú ____________________ dx 0 (1 + k 2 tan2x)(1 + tan2x)
)
p /2
t2 dt = 2k _______________ dx, 2 2 0 (1 + k t )(1 + t 2) (Let t = tan x) • 2k 1 1 = ______ 2 Ú _____ – _______ dt k – 1 0 1 + t 2 1 + k 2 t 2
(
(
)
)
• 2k 1 = ______ 2 tan– 1t – __ tan– 1(tk ) k 0 k – 1
Definite Integrals
(
•
p p 2k __ = ______ 2 – ___ k – 1 2 2k
p k – 1 2k = ______ 2 × __ _____ k – 1 2 k
)
( ) p k – 1 2k = ___ – 1 × __ ( _____ 2 k ) k
fi
2
When k = 1, I1 = 0, then c = – p log 2
| |
| | | | | |
1+k Therefore, I = p log a + p log ____ 2
•
| |
(
p ln2 2 p __ = ___ × ___ __ __ – 2 6 3 2 ÷
) )
(
•
(
)
0
1
ln(1 + x) = – Ú ________ dx x 0
)
(( ) ) ( )
1 ln __ y 1 1 __ ____________ __ = – Ú dy, Let t = y 2 • __ 1 1 1 2 + __ + 1 __ 2 y y y
– ln(y) 1 = – __ Ú __________ 2 dy 2 • (y + y + 1)
ln(y) 1 = __ Ú __________ 2 dy 2 • (y + y + 1)
0
0
0
(
•
(
(
)
)
)
ln(y) 1 = – __ Ú __________ 2 dy 2 0 (y + y + 1)
(
)
(
)
(
)
x2 x3 __ x4 x – __ + __ – + ... 2 3 4 = – Ú __________________ dx x 1
)
)
ln(1 + x) = 0 – Ú ________ dx x 0
0
ln t 1 Now, I2 = __ Ú ________ dt 2 0 (t2 + t + 1
(
(
ln(1 + x) = (ln x ln(1 + x))10 – Ú ________ dx x
1
= I1 + I2 (say)
(
)
( )
1
dt ln t dt ln2 1 = ___ Ú _________ + __ Ú _________ 2 0 (t2 + t + 1) 2 0 (t2 + t + 1)
•
(
1
ln2 + ln t = 2Ú ___________ dt 0 4(t2 + t + 1) •
( ( ) )
ln2 2 2 1 • = ___ × ___ __ tan– 1 ___ __ × t + __ 2 2 0 3 ÷3 ÷
1
)
(
)
)
(Let x = 2t fi dx = 2dt) •
( (
dt ln2 = ___ Ú __________ 2 0 3 1 2 __ __ t + + 2 4
ln x 34. Let I = Ú _____ dx. 1 +x 0
ln x 33. Let I = Ú __________ 2 dx 0 x + 2x + 4
)
a +b = p log a + p log _____ 2a a+b = p log _____ 2
(
p ln 2 = _____ __ 3÷3 p ln 2 p ln 2 Thus, I = 0 + _____ __ = _____ __ 3÷3 3÷3
1 + b/a = p log a + p log _______ 2
•
•
I1 = p log|k + 1| + c
I2 = 0
dt ln2 Also, I1 = ___ Ú __________ 2 0 (t2 + t + 1)
1+k Thus, I1 = p log _____ 2
)
fi 2 I2 = 0
p = _____ k+1 Integrating, we get
(
ln(t) 1 = – __ Ú _________ 2 dt 2 0 (t + t + 1) = I2
(
)
x x2 __ x3 = – Ú 1– __ + __ – + ... dx 2 3 4 0
(
)
1 x2 x3 __ x4 = – x – __2 + __ 2 – 2 + ... 2 3 4 0
(
)
1 1 1 = – 1 – __ 2 + __ 2 – __ 2 + ... 2 3 4
p 2 = – ___ 12 1 ln(1 + x4) 35. Let I = Ú _________ dx x 0
(
(
)
x8 x12 ___ x16 x20 x4 – __ + ___ – + ___ – ... 2 3 4 5 = Ú _________________________ dx x 1
)
0
2.103
2.104 Integral Calculus, 3D Geometry & Vector Booster 1
(
x7 = Ú x3 – __ + 2 0
(
4
8
x11 ___ – 3 12
)
x15 ___ + 4
x19 ___ – ... dx 5
In + 2 – 2In +1 – In
)
3 1 sin n + __ x – sin n + __ x 2 2 _______________________ = Ú dx x __ 0 sin 2
x 2 cos(n + 1)x ◊ sin __ 2 = Ú _________________ dx x 0 sin __ 2
= Ú 2 cos(n + 1)x dx
p
1
16
x x x x = __ – ___ + ___ – ___ + ... 4 16 36 64 0
(
)
1 1 1 1 = __ – ___ + ___ – ___ + ... 4 16 36 64
( (
p
) )
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
(
)
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
)
p 2 p 2 1 ___ = ___ – __ ◊ 24 8 6 p 2 = ___ 48 36. Given,
(
(
)
p
(
p
= In + 2 – In – 1 = I1 – I0.
Thus,
In – In – 1 = I1 – I0 = p – 0 = p
fi
In = p + In – 2 = p + p + In – 2 = 2p + In – 2 = 3p + In – 3 = ... = n p + I0 = n p + 0 = n p
)
)
) ( )
(
p
(
)
1–1 Now, I0 = Ú _______ dx = 0 1 – cos x 0 p
(
3 x 2 sin n + __ x sin __ 2 2 dx = Ú __________________ 1 – cos x 0
Subtracting, we get
3 x p sin n + __ x sin __ 2 2 = Ú ________________ dx x 2 __ 0 sin 2
( ( () ) ( )) ( ( ( ) ) ) ( ( ( ) ) )
3 sin n + __ x 2 __________ = Ú dx x __ 0 sin 2 p
Subtracting Eq. (ii) from Eq. (i), we get
I0 – I1 = p
fi
In = I0 + n p = n p
In = n p p /2
...(i)
( )
p /2
(
...(ii)
) )
1 – cos(2 nq ) = Ú ____________ dq 1 – cos(2q ) 0
1 – cos(n x) 1 = __ Ú ___________ dq = n p 2 0 1 – cos(x)
p
p
In + 2
sin2 nq Now, In = Ú ______ 2 dq 0 sin q
Similarly,
1 sin n + __ x 2 __________ – In = Ú dx x __ 0 sin 2
)
1 – cos x I1 = Ú ________ dx = p 1 – cos x 0
and
)
)
1 – x cos n x In = Ú __________ dx = n p. 1 – cos x 0
fi
cos(n + 1)x – cos(n + 2)x = ______________________ dx 1 – cos x 0
(
In + 2 – In +1 = In + 1 – In
p
)
(
In + 2 – 2In +1 – In = 0
1 – cos(n + 2)x 1 – cos(n + 1)x = Ú _____________ dx – Ú _____________ dx 1 – cos x 1 – cos x 0 0
p
)
=0
)
In + 2 – In + 1
( Ú (
( )
Similarly, In + 2 – In +1 = In + 1 – In
p
p
( )
)
( )
1 – x cos n x In = Ú __________ dx 1 – cos x 0
Now,
( )
)
2 = _____ sin(n + 1)x)p0 n +1
2 1 1 – __2 1 + __ 2 + ___ 2 + ... fi 4 2 12 fi
(
(
0
1 1 1 – 2 __ 2 – __ 2 + ___ 2 + ... 4 8 12
(
)
p
1 1 1 1 = __ 2 – __ 2 + __ 2 – __ 2 + ... 2 4 6 8
( (
(
37. We have
(
)
(12 + 22 + 33 + ... + n2)(13 + 23 + 33 + ... + n3) lim ______________________________________ n Æ • (16 + 26 + 36 + ... + n6)
Definite Integrals
( (
r = 1
)
( )
( )
n
n
S (r2) × S (r3) r = 1 r = 1 = lim _____________ n n Æ • (r6) S
)
1 n r 2 __ 1 n r 3 __ n S __ n × n S __ n r = 1 r = 1 = lim __________________ n Æ • 1 n r 6 __ n S __ n r = 1
( )
( ) ( ) ( Ú ) 1
1
Ú x dx × Ú x dx 0 0 ________________ = 2
3
(( () () ) ) 0
x x4 1 __ __ 3 0 4 0 = __________ x7 1 __ 7 0 1 1 __ × __ 3 4 ___ 7 _____ = = 1 12 __ 7 38. We have, n n – r 4r lim S _____ cos ___ n 2 n Æ • r = 1 n n n – r 1 4r = lim __ S _____ cos ___ n n n Æ • n r = 1
( ) )
( ( ) ( ) ) 1 r 4r = lim __ n S ( ( 1 – __ n )cos ( ___ n ) ) n
r = 1
1
= Ú (1 – x)cos(4x)dx
(
)
1
sin (4x) 1 __ 1 = (1 – x) ______ + Ú sin(4x)dx 4 4 0 0 1
1 = 0 + __ Ú sin(4x)dx 4 0
(
(
)
( Ú (
)
p /2
3 1
0
p
sinq Ú _______________ 2 2 dq , " a, b Œ R +, a > b 2 2 0 a cos q + b sin q p /2
x6 dx
n Æ •
)}
1 1 = lim __ – _____ n Æ • 2 n+2 1 = __ – 0 2 1 = __ 2 40. We have
sinq = 2 Ú _______________ 2 2 dq 0 a cos q + b2 sin2q
1
(
{ ( ( )
2.105
)
sinq = 2 ____________________ 2 2 dq 0 a cos q + b2 – b2 cos2q
(
p /2
)
sinq = 2 Ú _________________ 2 dq 0 b + (a2 – b2)cos2q 0
Let cosq = t
(
)
dt = – 2 Ú ____________ 2 1 b + (a2 – b2)t2 1
(
)
dt = 2 Ú ____________ 2 0 b + (a2 – b2)t2 1
(
)
dt 2 = _______ 2 2 Ú ___________ (a – b ) 0 _______ b2 2 2 + t2 (a – b )
(
______
)
1
÷a 2 – b2 2 1 = _______ 2 2 × ________ tan– 1 _______ t b b 0 (a – b ) ________ ______ 2 2 ÷ a – b
(
______
)
÷a 2 – b2 2 = __________ _______ × tan– 1 _______ 2 2 b b÷(a – b )
)
cos (4x) 1 = – _______ 16 0 1 = ___ [1 – cos (4x)] 16
b
39. We have, n
1
lim S Ú (xr – xr + 1)dx n Æ • r = 1 0 n
(
)
= b sin2q – a(1 – cos2q)
(
)
= b sin2q – a sin2q)
1 1 = lim S _____ – _____ n Æ • r = 1 r + 1 r+2
(
(x – a) = a cos2q + b sin2q – a
Now,
xr +1 xr + 2 1 = lim S _____ – _____ n Æ • r = 1 r + 1 r+2 0 n
÷
______
x–a Ú _____ d x b–x a Let x = a cos2q + b sin2q fi dx = (b – a)sin(2q )
1 = lim __ – n Æ • 2
) ( __ 13 – __ 14 ) + ( __ 14 – __ 15 ) + ... 1 1 + ( _____ – _____ n + 1 n + 2)
1 __ + 3
and
= (b – a)sin2q (b – x) = b – a cos2q – b sin2q = b (1 – sin2q) – a cos2q = (b – a)cos2q
2.106 Integral Calculus, 3D Geometry & Vector Booster a
The given integral reduces to p /2
÷
(b – a)sin2q Ú ___________ × (b – a)sin(2q )dq 0 (b – a)cos2q
0
sinq = Ú ____ × (b – a)sin(2q )dq cosq 0
= (b – a) Ú (2sin2q )dq
p /2
(
)
p /2
p /4
0
p aI + bJ = __ 2
Also,
b cos x – a sin x bI – aJ = Ú _____________ dx a cos x + b sin x 0
...(i)
(
)
= (log|a cos x + b sin x|)p /2 0
b = log __ a
( )
...(ii)
( ) ) ( ) )
a
( )
p Ú log(cot a + tan x)dx, a Œ 0, __ 2 0
(
a
(
)
)
= Ú log[cos(a – x)]dx – Ú log(sin a)dx
a
0
0
0
0
a
0
= Ú log(cos x)dx – Ú log(sin a)dx a
(
{ (
)}
)
p 4 – 3 sin2 2 __ – x 4 1 = __ Ú ______________________ dx p p 4 0 sin __ – x + cos __ – x 4 4 p /4
(
(
(
)
)
)
4 – 3 cos22x 1 = ____ __ Ú __________ dx cos x 4÷2 0
( Ú (
)
p /4
4 – 3(2cos22x – 1)2 1 = ____ __ Ú _________________ dx cos x 4÷2 0
)
1 = ____ __ Ú (sec x + 12cos x – 12cos3x)dx 4÷2 0
– Ú log(cos x)dx a
)
p /4
a
a
(
4 – 3(4cos4x – 4cos2x + 1) 1 = ____ __ ______________________ dx cos x 4÷2 0
cos(a – x) = Ú log _________ dx sin a cos x 0
)
p /4
sin x ____ cos x dx
)
4 – 3 sin22x 1 = __ Ú ___________ dx 4 0 sin x + cos x p /4
( (
cos a = Ú log _____ + sin a 0
( Ú (
sin6x + cos6x I = Ú ____________ dx sin x + cos x 0
p /4
3. We have,
)
p /4
ap b 1 ___ + b log __ a I = ________ (a2 + b2) 2 bp b 1 2 ___ – a log __ a . J = ________ 2 2 (a + b )
a
(
sin6x + cos6x = 2Ú ____________ dx sin x + cos x 0
1 – 3 sin2x cos2x = _____________ dx sin x + cos x 0
Solving Eqs (i) and (ii), we get
)
p /4
fi
p /2
(
...(ii)
sin6x + cos6x 2I = Ú ____________ dx sin x + cos x 0
2. Now, aI + bJ = Ú dx
and
)
Adding Eqs (i) and (ii), we get
p /2
(
...(i)
0
sin(2q ) p /2 = (b – a) q – _______ 2 0 p __ = (b – a) 2
)
p /2
= (b – a) Ú (1 – cos(2q ))dq
(
sin6x 4. Let Ú __________ dx. 0 sin x + cos x
cos6x = Ú __________ dx 0 sin x + cos x
0
= a log(cosec a) p /2
p /2
= – a log(sin a)
p /2
= – Ú log(sin a)dx
___________
– Ú log(cos x)dx 0
1__ = ____ [log|sec x + tan x| + 12sinx 4÷2
[
(
)]
p /4 1 – 12 sin x – __ sin3x 3 0
(
__ __ 1 1 = ____ __ log(÷2 + 1) + 6÷2 – 12 ___ __ – 4÷2 2 ÷ __ __ 1 = ____ __ [log(÷2 + 1) + ÷ 2 ] 4÷2
)]
1 ____ __ 6÷2
Definite Integrals
(
5. We have 1
(
)
dx I = Ú ______________________ 0 (5 + 2x – 2x2)(1 + e(2 – 4x))
...(i)
dx = Ú __________________________________ 2 2 – 4(1 – x) 0 {5 + 2(1 – x) – 2(1 – x) }(1 + e ) dx = Ú _____________________ 2 4x – 2 0 (5 + 2x – 2x )(1 + e )
...(ii)
(
p
{
}
{
}
dx 1 _____ 1 = Ú ____________ 2 , since _____ 2 – 4x + 4x – 2 = 1 0 (5 + 2x – 2x ) e e 1
dx 1 fi I = __ Ú _____________ 2 0 5 2 – 2 x – x – __ 2
(
)
= – p 2 + 2k
|
( ) ( ) ( ) ( ) ___
|)
|
___
1 + ÷11 1 ___ = ____ ___ log _______ 11 ÷10 ÷
(
)
(
4
)
( Ú ( Ú (
fi
p 2k = Ú _______ dx 0 1 + sin x
fi
dx 2k = 2p Ú _______ 1 + sin x 0
fi
dx k = p Ú _____________ p 0 1 + cos __ – x 2
p /2
fi
)
(
1
)
t4(t4 – 4t3 + 6t2 – 4t + 1) = _____________________ dt 0 t2 + 1
)
1
(
(
7
6
4t ___ + t5 6 1 2 = __ – __ + 1 – 7 3 t = __ – 7
(
( ) p p x k = __ Ú ( sec ( __ – __ dx 2 4 2 )) p x k = ( p tan ( __ – __ 4 2 ))
4
2
0
p __ 2 0
k=p
Thus, J – I = – p 2 + 2p = p (2 – p ) 8. We have, p /2
)
4 = Ú t – 4t + 5t – 4t + 4 – _____ 2 dt 0 t +1 5
)
fi
(t8 – 4t7 + 6t6 – 4t 5 + t4) = _____________________ dt 0 t2 + 1 6
)
dx k = p Ú ____________ p x 0 2cos2 __ – __ 4 2 p /2
fi
t4(t – 1)4 1 = Ú ________ 2 dt, Let t = __ x fi 0 t +1
1
(
p /2
(x – 1) Ú _______ dx 1 x8 + x10 1
(p – x) k = Ú _______ dx 0 1 + sin x
p
6. We have, •
fi
p
1
÷11 1 x – __ – ____ 2 2 1 1 ___ = – __ × ______ ___ log ______________ 4 11 ÷ 11 ÷ 1 2◊____ x – __ + ____ 2 2 2 0 (÷11 + 1) 1 1 = __ × ____ ___ log_________ ___ 4 ÷ 11 (÷11 – 1)
p
( ( ) ( ) )
___
)
p
)
|
(
– x2 x = _________ + 2 Ú _______ dx (1 + sin x) 0 0 1 + sin x
1
(
p
p
x where k = Ú _______ dx 1 + sin x 0
dx 1 = – __ Ú _________________ ___ 4 0 2 1 2 ÷11 x – __ – ____ 2 2
)
x2(1 + cos x) and J = Ú ___________ dx 0 (1 + x sin x)2 x2 cos x (J – I) = Ú _________ dx 2 0 (1 + sin x)
dx 1 1 2i Ú ___________ _________ + _________ 2 2 – 4x 0 (5 + 2x – 2x ) (1 + e ) (1 + e4x – 2 )
)
Now,
Adding Eqs (i) and (ii), we get
(
(
2 x 7. Given I = Ú ________ dx 1 + x sin x 0 p
1
1
p
1
1
)
22 = ___ – p 7
2
)
1
3
4t – ___ + 4t – 4 tan– 1t 3 0 4 – 1 __ + 4 – 4 tan (1) 3
)
____
_____
Ú (÷ sin x + ÷cos x ) – 4 dx 0
p /2
dx = Ú _______________ ____ 4 _____ 0 (÷sin x + ÷ cos x )
sec2 x dx = Ú ___________ ____ 0 (÷tan x + 1) 4
p /2
2.107
2.108 Integral Calculus, 3D Geometry & Vector Booster •
dt = Ú _______ 0 (t + 1) 4
• 1 1 = – __ × ______ 3 (t + 1) 0
1 = __ 3
(
10. Given, p /4
)
In = Ú (tan n x)dx 0
P /4
= Ú (tann – 2x◊ tan2x)dx
0
9. We have, p
(
P /4
)
0
p
(
0
))
dx = Ú ________________ 0 1 – tan2(x/2) ___________ a – 1 + tan2(x/2)
p
( (
(
sec2(x/2)dx = Ú _____________________ 0 (a – 1) + (a + 1)tan2(x/2)
(
)
•
(
)
÷ ÷
______
( ÷
_______ •
)
(a + 1) __ p 2 = ______ × _______ × (a + 1) (a – 1) 2 p = _______ _____ 2 ÷ a – 1
fi
1 In + In – 2 = _____ n –1
( )
1 _______ = 4 (I5 + I3 ) 1 _______ = 5 (I6 + I4 ) 1 and _______ = 6 (I7 + I5 ) Clearly, 3, 4, 5 and 6 are in AP. p
11. Let I = Ú q 3log(sinq )dq
...(i)
0 p
p
= Ú (p – q )3log(sin(p – q ))dq 0 p
p dx Thus, Ú _________ = _______ _____ ◊ a > 1 2 (a – cos x) 0 ÷a – 1
= Ú (p – q )3log(sinq )dq 0
Diffeerentiating w.r.t a, we get
Adding Eqs (i) and (ii), we get
p
t n – 1 1 In + In – 2 = _____ n –1 0
1 Thus, _______ = 3 (I2 + I4 )
(a + 1) (a + 1) 2 = ______ × ______ tan– 1 t _______ (a + 1) (a – 1) (a – 1) 0 _______
( )
fi
1 fi _________ = (n – 1) (In + In – 2 )
)
2
dt 2 = ______ Ú __________ (a + 1) 0 (a – 1) 2 ______ + t (a + 1)
0
0
sec (x/2)dx 1 = ______ Ú ________________ (a + 1) 0 (a – 1) ______ + tan2 (x/2) (a + 1)
0
= Ú (tann – 2x◊ sec2x)dx – In – 2
)
P /4
P /4
p
P /4
= Ú (tann – 2x◊ sec2x)dx – Ú (tann – 2x dx
sec2(x/2)dx = Ú __________________________ 2 2 0 a(1 + tan (x/2)) – (1 – tan (x/2)) p
= Ú (tann – 2x(sec2x – 1))dx
dx Ú a_______ – cos x
p a dx – Ú __________ 2 = – _________ 0 (a – cos x) (a2 – 1)3/2 p
p
2I = Ú {(p – q )3 + q 3}log(sinq )dq 0 p
p a dx fi Ú __________ 2 = _________ 2 3/2 0 (a – cos x) (a – 1)
= Ú {p 3– 3p 2q + 3p q 2}log(sinq )dq
= p 3 Ú log(sinq )dq – 3p 2 Ú (q )log(sinq )dq
(Let a = 2)
0
p
0
p
2p dx fi Ú __________ 2 = ____ __ 3÷3 0 (2 – cos x)
p
0 p
+ 3p Ú (q 2)log(sinq )dq 0
= – p 4log2 – 3p 2I1 + 3p I2
...(ii)
Definite Integrals p
(n + 1)In + 1 = In + 2 – G(n + 1)
0 p
(n + 1)In + 1 = In + 2 – G(n + 1)
Also,
n2In = nIn + 1 – nG(n + 1)
p
Thus,
In + 2 – (n + 1)In + 1 = nIn + 1 – n2In
0
fi 13. Given,
In + 2 – (n + 2)In + 1 + n2In = 0
Now, I1 = Ú q log(sinq )dq = Ú (p – q )log(sinq )dq
0
= p Ú log(sinq )dq – I1
p
fi 2I1 = p Ú log(sinq )dq
= – p 2log(2)
p fi I1 = – __ log(2) 2
(
(
)
2
p Now, 2I = – p log2 – 3p – ___ log(2) + 3p I2 2 2
(
)
(
) ( 1 1 f (x) = ( _______ sin ( x – __ x ) (a + b)x)
p
__ 3p p 3 3p = ___ ◊ ___ log(÷2 ) + ___ Ú [q 2log sin(q )]dq 2 6 2 0 p
I = Ú f (x)dx, (a π b) 2
3p p 3p = ___ ◊ ___ log2 + ___ Ú [q 2log sin(q )]dq 2 6 2 0
1 1 1 = ______ Ú __ sin x – __ x dx (a + b) 1/2 x
1 1 1 1 = ______ Ú __ sin y – __ y dy, x = __ y (a + b) 2 y 1/2
= – I
fi fi
2I = 0 I=0 2
0 •
{ (
) }
xn xn = e– xlog x ◊ __ n 0 – Ú e– x ◊ __ 1 – e– xlog x __ n dx x 0 •
1 = 0 – __ n Ú { (e– x ◊ xn – 1 – e– xlog x ◊ xn) }dx 0 •
= Ú f (x)dx = 0
= Ú {(e– xlog x) ◊x n – 1}dx
•
1 1 = __ n Ú e– xlog x ◊ xn dx – __ n Ú e– x◊ xn – 1dx 0 0 In – 1 ____ G(n) = ____ n – n Thus, nIn = In – 1 – G(n) Replacing n by (n + 1), we get
1/2
14. We have, p /6
(
___________
)
÷3cos(2x) – 1 Ú ____________ dx cos(x) 0 p /6
))
( ( ) ) 1 1 1 = – ______ Ú ( __ y sin ( y – __ y ) )dy (a + b) 1/2
In = Ú (e– xx n – 1log x)dx
)
( (
2
Hence, the result. 12. Given, •
•
) ( ) )
1 1 = Ú _______ sin x – __ x dx (a + b)x 1/2
2
p
__ 3p = ___ ◊ Ú q 2log(÷2 s in(q ))dq 2 0
•
( (
p
__ 3p 3p = ___ ◊ Ú q 2log(÷2 ) dq + ___ Ú [q 2 log sin(q )]dq 2 0 2 0
(
)
1/2
p
0
...(ii)
2
p 3p = ___ log2 + ___ Ú [q 2log sin(q )]dq 4 2 0 3
)
b 1 (a2 – b2) f (x) = a – __ x sin x – __ x
Let
p
4
(
fi
p 4 3p = ___ log2 + ___ I2 4 2
)
1 1 af (– x) + bf (x) = __ x sin x – __ x From Eqs (i) and (ii), we get fi
p 4 p 2 3p 3 fi I = – ___ log2 – __ p 2 – ___ log(2) + ___ I2 2 2 2 2
...(i)
1 1 af (– x) + bf (x) = – __ x sin __ x – x
4
( )
1 1 af (x) = bf (– x) = __ x sin x – __ 2 Replacing x by – x, we get
0
2.109
(
_________
)
2x – 4 ÷ 6cos dx = Ú __________ cosx 0 p /6
( ÷
_________
)
6cos2x – 4 = Ú _________ dx 0 cos2x
2 = Ú (÷ 6 – 4sec x) dx
p /6 0
__________
2.110 Integral Calculus, 3D Geometry & Vector Booster p /6
(
( )
)
6__________ – 4sec2x = Ú ___________ dx 2 0 x ÷ 6 – 4 cos
6 sec2x = Ú ___________ _________ dx + 4 Ú ___________ __________ dx 2 2 0 0 x ÷6 – 4sec x ÷ 6 – 4sec
6cos x sec2x = Ú ___________ __________ dx + 4 Ú ___________ _________ dx 2 0 0 ÷6cos 2x – 4 2 – 2tan x
6cos x sec2x = __________ _________ dx + 4 Ú ___________ _________ dx 2 0 0 2 – 6sin2x 2 – 2tan x
p /6
(
p /6
)
( Ú ( ÷
p /6
) )
p /6
p /6
p /6
( ( ÷ ( ÷
) ) )
Let sin x = t and tan x = v 1/2
(
)
__
1/÷3
(
( ÷ ) ( ÷ ) __
6p 1 2 = ____ __ – 4◊__ sin– 1 __ 2 3 3÷6 __ 2p 2 = ___ __ – 2sin– 1 __ 3 ÷6 15. We have
__
÷3
)
dv 6t = Ú ________ _______ dx _______ dx + 4 Ú ________ 2 0 0 ÷ 2 – 2v2 ÷ 2 – 6t
(
__
(
)
x4 + 2x2 + 1 = Ú __ ___________ 2 dx (x + 1)3 – ÷3 __
÷3
(
)
(x2 + 1)2 = Ú __ ________ 2 dx 3 – ÷3 (x + 1) __
÷3
(
)
dx = Ú __ _______ 2 (x + 1) – ÷3 __
÷3
dx = 2Ú _______ 2 0 (x + 1) __
( ) ( )
p /2 0
( )
9 G __ __ 2 = _____ ×÷ p 2G(5)
)
( )
7 5 3 1 __ ◊ __ ◊ __ ◊ G __ __ 2 2 2 2 ____________ = × ÷ p 2 ◊ (4!)
__
__ 105 × ÷p = ________ × ÷p 4 2 × 24
35 × p = _______ 7 2 So, L = 3, M = 5, P = 2 and Q = 7 Hence, the value of L+M+P–Q =3+5+2–7 =3 3. We have,
p I1 = Ú log(sin x)dx = __ log 2 8 0
__
= 2[tan– 1(÷3 ) – 0] 2p = ___ 3
Integer Type Questions 1. As we know that
( ) ( ) ( )
n+1 G _____ __ 2 Ú sinnx dx = _________ × ÷ p n+2 0 2.G _____ 2 11 p /2 G ___ __ 2 So, Ú sin10x dx = ______ ×÷ p 2G(6) 0 p /2
Thus, A = 2 and B = 8 Hence, the value of B – A is 6. 2. As we know that n+1 p /2 G _____ __ 2 Ú cosnx dx = __________ × ÷p n+2 0 2 . G _____ 2
p /2
= ( 2tan– 1(x) )÷0 3
__
__ 63 × 15 × ÷ p = ____________ 5 × ÷p 2 × 120 63 × p = _______ 5 2 ×8 63 × p = ______ 28
So, Ú cos8x dx
x4 + x3 + 2x2 + x tan2x + x2sin3x + tan– 1(x) + 1 Ú __ _______________________________________ dx x6 + 3x4 + 3x2 + 1 – ÷3 ÷3
9 7 5 3 1 __ ◊ __ ◊ __ ◊ __ G __ __ 2_____________ 2 2 2 2 = × ÷ p 2 ◊ (5!)
p /2
p and I2 = Ú log(cos x)dx = __ log 2 8 0 Hence, the value of I1 – I2 + 5 = 5 1
log(1 + x) 4. Let L = Ú _________ dx 0 (1 + x2) Let x = tanq fi d x = sec2q dq
p /4
log(1 + tanq ) 2 = Ú ___________ sec q dq 0 (1 + tan2q )
Definite Integrals •
p /4
= Ú log(1 + tanq )dq 0 p /4
(
...(i)
0
(
))
p = Ú log 1 + tan __ – q dq 4 0
1 + 1 – tan q = Ú log ___________ dq 1 + tan q 0
( ) 2 = Ú log ( ________ dq 1 + tan q )
p /4
0
p /4
p /4
0
0
= Ú log|2|dq – Ú log(1 + tan q )dq p /4
= Ú log|2|dq – I 0
p /4
fi 2I = Ú log 2 dq 0
p = __ log 2 4 p fi I = __ log 2 8
__ 3 = __ × ÷ p 2 3 = __ × (p)1/2 2 Hence, the value of L + M + N is 5 1 1 1 1 1 6. Clearly, M = __ × 1 × __ + __ × 1 × __ = __ 2 2 2 2 2 1 1 and N = 20 × __ × 1 × __ = 5 2 2 Hence, the value of 2 M + N + 2 = 1 + 5 + 2 = 8 7. We have, 2
M = Ú (– 1)[x] dx
p /4
and Let M = Ú log(1 + tan x)dx
(
(
))
p = Ú log 1 + tan __ – x dx 4 0
( ) 1 + tan x + 1 – tan x = Ú log ( _________________ ) dx 1 + tan x 2 = Ú log ( _______ dx 1 + tan x ) p /4
1 – tan x = Ú log 1 + ________ dx 1 + tan x 0
0
p /4
0
p /4
p /4
0
0
= Ú log 2 dx – Ú log(1 + tan x)dx
p = __ log 2 – I 4 p __ fi 2I = log 2 4 p __ fi I = log 2 8 p p p Now, L + M = __ log 2 + __ log 2 = __ log 2 8 8 4
So, A = 4 and B = 2 Hence, the value of A + B + 3 = 9. 5. We have, I = Ú e– xx3/2 dx 0
1
– 2
– 1
0
= Ú (– 1)
0
– 2
2
+ Ú (– 1)[x] dx 1
1
dx + Ú (– 1)
(– 2)
dx + Ú (– 1)(0) dx
(– 1)
– 1
0
2
+ Ú (– 1)(1) dx – 1
0
1
2
– 2
– 1
0
1
1
= Ú 1 ◊ dx + Ú (– 1) ◊ dx + Ú (1) dx + Ú (– 1)dx = (– 1 + 2) – (0 + 1) + (1 – 0) – (2 – 1) = 1 – 1 + 1 – 1 = 0 4
and N = Ú [x + 1]dx 0 4
= Ú ([x] + 1)dx 0
4
4
0
0
= Ú [x]dx + Ú 1◊dx
43 = ___ + 4 = 10 2 Hence, the value of M + N – 1 = 9. 8. We have,
3
L = Ú (|x – 1| + 1)dx – 2
•
0
– 1
p /4
– 2 – 1
= Ú (– 1)[x] dx + Ú (– 1)[x] dx + Ú (– 1)[x] dx
0
p /4
( ) 3 1 = __ G ( __ 2 2) 5 = G __ 2
p /4
= Ú e– xx5/2 – 1 dx
2.111
1 1 = __ (4 + 1) × 3 + __ (1 + 3) × 2 2 2
2.112 Integral Calculus, 3D Geometry & Vector Booster 23 = ___ 2
Thus, the value of 8
Now, M = Ú |x – 2|dx 0
1 = 2 × __ × 2 × 2 2 =4 2 2 23 Thus, __ (L – M) = __ ___ – 4 = 5 3 3 2
(
1
)
Ú x2 d(ln x) – 1 e
9. Let L = Ú [log|x|]dx 1
3
2p
1
__
10p
0
Hence, the value of B __ = 5. A 11. We have, 1 lim ___ __ + n Æ • ÷ n
(
(
1 ____ ___ + 2n ÷
1 ____ ___ + 3n ÷
)
1 1 ____ ___ + ... + ___ 2n 4n ÷
)
1 1 1 1 1 __ + ___ __ + ___ __ + ... + ___ = lim ___ __ 1 + ___ n Æ • ÷n 2n 2 ÷3 ÷ 4 ÷
1 4n 1_ = lim ___ __ S ___ n Æ • ÷ n r = 1 ÷r 1 4n 1__ = lim __ S ___ n Æ • n r = 1 __ r n 4 dx__ = Ú ___ x ÷ 0
÷
__
__
4 = (2÷x ) = (2÷4 – 0) = 4 0
( ) ( )
15.
( )
1 e4 – 1 = __ _____ 2 e2
1
dx Ú __________________ _____ _____ 1 + x + ÷ 1 – x + 2 0 ÷
Put fi
x = sin(2q ) dx = 2cos(2q )dq
p /4
2cos(2q )dq = Ú _____________________________ __________ __________ ÷1 + sin(2q ) + ÷ 1 – sin(2q ) + 2 0
2cos(2q )dq = Ú ___________________________ cosq + sinq + cosq – sinq + 2 0
2cos(2q )dq = Ú __________ 2(cosq + 1) 0
cos(2q ) = Ú _________ dq (cosq + 1) 0
n(n – 1) _______ Ú [x]dx 0 2 _______ n = _______ = (n – 1) n __ 2 Ú {x}dx
1 – 2sin2q ) = Ú __________ dq 0 (cosq + 1)
1 – 2(1 – cos2q ) = Ú _____________ dq (cosq + 1) 0
1 = Ú ________ – 2(1 – cosq ) dq 1 + cosq 0
1 = Ú _________ – 2(1 – cosq ) dq 2 0 2cos (q /2)
p /4
p /4
p /4
0
10
Ú [x]dx 0 Thus, ________ 10 = (10 – 1) = 9. Ú {x}dx
p /4
p /4
0
13. As we know that
)
f (x) b–a Ú ________________ dx = _____ 2 f (a + b + x) + f (x) a
p /4
p /4
n
(
x2 = __ 2 1/e 1 1 = __ e2 – __ 2 2 e
Clearly, m = 4 and n = 2. Hence, the value of (m + n) = 6. We have,
12. As we know that
b
( ) ( ) e
and B = Ú [sin x + cos x]dx = 5p
= Ú x dx 1/e
A = Ú [÷ 2 s in x]dx = p 0
1 = Ú x2 ◊ __ x dx 1/e e
and M = Ú [|log x|]dx and, then find the value L of __ + 5. M 10. We have,
)
8–2 = _____ = 3 2 14. We have
3
(
[x2] Ú ___________________ 2 dx 2 [x – 20x + 100] + [x2]
4
p /4
(
(
( ( )
)
)
)
q 1 = Ú __ sec2 __ – 2(1 – cosq ) dq 2 2 0
Definite Integrals
( ( )
p __
)
q = tan __ – 2(q – sinq ) 4 2 0
( ( ) (
p /2
= Ú 2((cos x + cos 3x + ... + cos(2k – 1)x)cos x dx 0 p /2
))
p p 1 = tan __ – 2 __ – ___ __ 4 ÷2 8 __ __ p = (÷2 – 1) – __ + ÷ 2 2 __ p = 2÷2 – __ – 1 2 Thus, a = 2, b = 2, c = – 1 Hence, the value of (a + b + c) = 3.
(
= Ú [2cos 2x + 2cos 3x cos x + ... 0
)
+ 2cos(2k – 1)x cos x]dx p /2
= Ú [(1 + cos 2x) + (cos 4x + cos 2x) + ... 0
Questions asked in Past Iit-Jee Examinations
+ cos(2k)x + cos(2k – 2)x]dx
(
(
)
1. Ú [ f (x) + f ( – x)](g(x)) – g( – x))dx
p = __ 2
= Ú h (x)dx = 0 – p /2
where h(x) = [ f (x) + f ( – x)](g(x) – g(– x)
0
x
the curve y + Ú f (t)dt = 2
p /2
2
2. Let I = Ú f (sin 2x)sin x dx
...(i)
0
= Ú f (sin 2x)cos x dx
...(ii)
0
Adding Eqs (i) and (ii), we get 2I = Ú f (sin 2x)sin x + cos x )dx
p /2
( (
__ x = ÷2 Ú f (sin 2x) sin x + __ dx 4 0
=
__ p /4 ÷2 f (cos2t)(cos t)dt
Ú
))
x Let __ – x = t 4
– p /4
)
__ p /4
= 2÷2 Ú f (cos2t)(cos t)dt
)
(
(
)
(
(
)
)
(
(
)
)
)
p (x – p)sin(2x)sin __ cos(x) 2 ______________________ = Ú dx (2x – p) 0 p
)
p (p – x)sin(2x)sin __ cos(x) 2 = Ú _______________________ dx p – 2x 0 p
(
(
p (p – x)sin(2p – 2x)sin __ cos(p – x 2 = Ú ______________________________ dx 2(p – x) – p 0 p
p /2 0
(
0
p x sin(2x)sin __ cos x 2 _________________ 5. Let I = Ú dx ...(i) 2x – p 0 p
p /2
x
4. If f is a continuous function with Ú f (t)dt Æ • as |x| Æ •, then show that every line y = mx intersects
fi h(x) is an odd function.
]
sin(2k)x ___________ cos(2k – 2)x p /2 + _______ + 2k (2k – 2) 0
– p /2 p /2
)
sin 2x p /2 sin 4x _____ sin 2x p /2 = 1 + _____ + _____ + 2 0 4 2 0
p /2
2.113
...(ii)
Adding Eqs (i) and (ii), we get
0
=
__ p /4 2÷2
Ú
p
f (cos2x)(cos x)dx
0
__ p /4
fi
2I = 2÷2 Ú f (cos2x)(cos x)dx
fi
I = 2÷2 Ú f (cos2x)(cos x)dx
__
0 p /4 0
p
(
)
sin2 kx = Ú ______ cos x dx sin x 0
))
(
(
))
p
(
(
))
fi
p = Ú sin(x)cos x sin __ cos(x) dx 2 0
fi
4 = – ___2 Ú (t sin t )dt, p p /2
3. Let I = Ú sin(2kx)cot x dx 0 p /2
(
p fi 2 = 2 Ú sin(x)cos x sin __ cos(x) dx 2 0
p /2
(
p 2 = Ú sin(2x)sin __ cos(x) dx 2 0
– p /2
(
p Let __ cos x = t 2
)
2.114 Integral Calculus, 3D Geometry & Vector Booster
( )
– p /2
fi
4 = – ___2 Ú (t sin t )dt p p /2
8 = – ___2 Ú (x sin x )dx p 0
8 = – ___2 (– x cos x + sin x )p /2 0 p
p /2
5
1 = ___ Ú |sin t |dt, 2p 0
1 = ___ Ú |sin x |dx 2p 0
1 2 = ___ × 4 = __ p 2p
)
4 = Ú 1 + _____ 2 dx 3 x –4
1
10. Given integral = Ú e x(x – 1)n dx
( | | ) = ( x + log _____ | xx +– 22 | ) = ( 5 + log __ | 37 | – 3 – log | __ 15 | ) = ( 2 + log ___ | 157 | ) x–2 5 1 = x + 4 ◊ ___ log _____ 2.2 x+2 3
Let I = Ú e x(x – 1)n dx 0
5
1
3
0
= – (– 1)n – n I n – 1
In = – (– 1)n – n I n – 1 1
= Ú f (x)dx
0
– p
1
= 0, since (1 – x2)sin x cos2x is an odd function.
= ( (x – 1)e x )10 – Ú exdx 0
= – (– 1) – (e x)10
= 1 – (e – 1) = 2 – e
– 1
From Eq. (i), we can write
1
= Ú |x – 1|dx
I2 = – 1 – 2I1
– 1
= – 1 – 2(2 – e) = 2e – 5
0
Again, from Eq. (i), we can write
= Ú |t|dx, – 2
(Let (x – 1) = t)
I3 = 1 – 3I2
= 1 – 3(2e – 5) = 16 – 6e
0
= Ú |x| – 2
p /2
= Ú (– x)dx – 2
(
)
1 11. Given integral is Ú ________ dx 0 1 + tan3x
0
...(i)
Now, I1 = Ú e x(x – 1)dx
1
= (– 1)n – n Ú (x – 1)n – 1e x dx 0
8. Given integral = Ú |1 – x|dx
= ( (x – 1)n e x )10 – Ú n(x – 1)n – 1e x dx 1
– p
0
1
7. Given, Ú (1 + x2)sin x cos2x dx
(Let t = 2p x)
2p
p
9. Given integral = Ú |sin(2p x)|dx 2p
p
=2
0
( )
(
1
x2 6. Given Ú _____ 2 dx 3 x –4
– x2 0 = ____ 2 – 2
8 8 = – ___2 (1 – 0) = ___ 2 p p 5
p /2
(
)
cos3x = Ú ___________ 3 dx 0 cos x + sin3x
...(i)
Definite Integrals p /2
(
)
sin3x = Ú ___________ 3 dx 0 sin x + cos3x Adding Eqs (i) and (ii), we get
p /2
(
3
...(ii)
sin x + cos x 2I = Ú ___________ dx 0 sin3x + cos3x
fi
3
(
)
( ( (
3
5
3
4
)
2
)
(2x3(x2 – 1)) + (x2 + 1)2 = Ú _____________________ dx 2 (x2 + 1)2(x2 – 1)
3
) ( ) 3
(2x3) dx = Ú _______ 2 dx + Ú _______ 2 2 2 2 (x + 1) (x – 1)
3
(
)
( | | )
( ) ( | |
p /4
= p (tan j – sec j )3p /4 p /4 __
( ) ( | | )
3 1 1 1 = Ú __ – __ dt + __ log __ t t2 2 2 5
3 1 10 1 = log|t| + __ + __ log __ t 5 2 2
__
= – 2p (÷2 + 1) __
fi I = – p (÷2 + 1) 1
(
| | ) ) ( | | )
) (
3p /4
(
( )
( ) Ú ( ) 1
x2 = – 2 _____ dx 3 – |x| 0
0 V
n p + V
0 V
V n p
0 V
0
= Ú |sin x|dx + Ú |sin x|dx = Ú sin x dx + Ú |sin x|dx p
= Ú sin x dx + n Ú |sin x|dx 0
V
= – (cos x) 0 + n × 2
= 1 – cos V + 2n
= (2n + 1 – cos V)
16. The given integral = 3
)
)
0
)
p – j = Ú _________ dj 1 + sin(j) p /4
1
1
p – j = Ú ____________ dj 1 + sin(p – j) p /4 3p /4
(
x2 = 0 – Ú _____ dx – 1 3 – |x|
)
(
1
sin x x2 = Ú _____ dx – Ú _____ dx – 1 3 – |x| – 1 3 – |x|
j 13. Let I = Ú ________ dj ...(i) 1 + sin j p /4
)
n p + V
( 1 1 = ( __ log|6| – ___ ) 2 10 3p /4
(
sin x – x2 14. Given integral = Ú ________ dx 3 – |x| – 1
15. Given integral = Ú |sin x|dx
3 1 1 = log|2| – ___ + __ log __ 10 2 2
__
= p [( – 1 – ÷ 2 ) – (1 + ÷ 2 ) ]
| | )
Let x 2 + 1 = t
)
3p /4
t–1 1 1 1 1 = Ú ____ 2 dt + __ log __ – __ log __ 2 2 2 3 t 5
)
= p Ú (sec2j – secj tan j)dj
(
( (
3p /4
x2.2x dx x–1 3 1 __ _____ = Ú _______ 2 + l og 2 x+1 2 2 (x + 1)2
10
)
(1 – sin j) = p Ú _________ dj p /4 cos2j
10
(
3p /4
(2x – 2x ) + (x + 2x + 1) = Ú ________________________ dx 2 (x2 + 1)(x4 – 1)
)
(1 – sin j )dj = p Ú ____________ p /4 1 – sin2(j )
2x5 + x4 – 2x3 + 2x2 + 1 12. Given integral = Ú ______________________ dx 2 (x2 + 1)(x4 – 1) 3
(
3p /4
p = Ú (1)dx = __ 2 0 p I = __ 4
3p /4
dj = p Ú _________ 1 + sin(j) p /4
p /2
Adding Eqs (i) and (ii), we get p 2I = Ú _________ dj 1 + sin(j) p /4
)
3
2.115
...(ii)
(
__
)
x ÷ Ú ___________ __ __ dx ÷5 – x + ÷ x 2
3– 2 = _____ 2
2.116 Integral Calculus, 3D Geometry & Vector Booster 1 = __ 2
1__ ___ 3 ÷
17. Given integral = Ú [2sinx]dx p/4
3p/4
p
0
p/4
3p/4
= Ú [2sinx]dx + Ú [2sinx]dx + Ú [2sinx]dx 5p/4
7p/4
2p
p
5p/4
7p/4
+ Ú [2sinx]dx + Ú [2sinx]dx + Ú [2sinx]dx
( ) 7p 5p 7p + ( ___ – ___ ) ◊ (– 2) + ( 2p – ___ ) ◊(–1) 4 4 4
)
3p p 5p = 0 + ___ – __ ◊ (1) + 0 + ___ – p ◊ (– 1) 4 4 4
(
p p p = __ – __ – p – __ 2 4 4 = – p
1__ ___ 3 ÷ x4 = _____ 4 p – 1__ 1 – x ___ – 3 ÷ 1 ___ __ 3 ÷ x4 = p _____ 4 dx 1__ 1 – x – ___ 3 ÷
=
( )
( )
__ p x 1 18. Given f (x) = Asin ___ + B, f ¢ __ = 2÷2 2 2
Ú
( (
Ú
( ( ) )
p x 2A fi Ú Asin ___ + B dx = ___ p 2 0
)
1 Asin(p x/2) 2A fi – __________ + B = ___ p (p/2) 0
(
)
2A 2A fi – ___ p + B = ___ p 4A fi B = ___ p __ 1 Also, f ¢ __ = 2÷2 2
Ú
( )
( )
p p 1 1 f ¢ __ = A __ cos __ × __ 2 2 2 2
(
)
( )
Ap p Ap = ___ cos __ = ____ __ 2 4 2÷2
__ Ap__ fi ____ = 2÷2 2÷2 8 fi A = __ p 32 4 8 ___ fi B = __ p × __ p = 2 p 1 ___ __ 3 ÷ x4 2x = _____ 4 cos–1 _____ 2 dx. 1__ 1 – x 1+x – ___ 3 ÷
(
( ) Ú ( ) Ú ( )
x4 = 2p Ú _____ 4 dx 0 1–x
1__ ___ 3 ÷
x4 I = p _____ 4 dx 0 1–x
fi
1 ___ __ 3 ÷
x4 = – p _____ 4 dx 0 x –1
(
(
))
)
1 = – p Ú 1 + _____ 4 dx 0 x –1
( )
fi
Ú
÷3
x4 2I = p Ú _____ 4 dx 1__ 1 – x – ___
fi
fi
19. Let I
( ( ( )))
x4 2x – Ú _____ 4 cos–1 _____ 2 dx 1 1 – x 1 +x __ – ___
1 ___ __ 3 ÷
p px f ¢(x) = A __ cos ___ 2 2
1 ___ __ 3 ÷
1 ___ __ 3 ÷ x4 4 dx – I p _____ 1__ 1 – x – ___ 3 ÷ 1 ___ __ 3 ÷
( ) ( )
)))
÷3 1 ___ __ 3 ÷
2A Also, Ú f (x) dx = ___ p 0
(
(
2x cos–1 _____ 2 dx 1+x
( )
1
1
))
÷3
0
(
(
x4 – 2x = Ú _____ 4 cos–1 _____ dx 1 1 – x 1 + x2 __ – ___
2p
= – p
1 ___ __ 3 ÷ 1 0
( Ú ( ( Ú
1 ___ __ 3 ÷ 1 0
)
1 + _____________ 2 dx (x – 1)(x2 + 1)
))
1 1 1 + __ _______ – _______ dx 2 (x2 – 1) (x2 + 1)
= – p
__ x–1 1 __ 1 = – p x + __ log _____ – tan–1x ÷ 3 2 2 x+1 0
1–÷ 3 __ p 1 1 1 __ = – p ___ __ + __ __ log ______ – 2 2 6 3 1+÷ 3 ÷
( ( | |
( |
(
20. We have, Un + 2 – Un + 1 p
(
)
p
(
1 ___
))
__
| )) )
1 – cos(n + 2)x 1 – cos(n + 1)x = Ú ______________ dx – Ú ______________ dx 1 – cosx 1 – cosx 0 0
Definite Integrals p
(
p
)
cos(n + 1)x – cos(n + 2)x = Ú ______________________ dx 1 – cosx 0
(
(
)
)
3 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
( )
21. Given
( (
) ( ) ( ) )
(
)
fi Un + 2 – Un + 1
1 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
( )
(
(
( ) x 2cos(n + 1)x sin ( __ ) 2 = Ú __________________ dx x sin ( __ ) 2
(
0
)
fi
))
)
p
= Ú (2cos(n + 1)x) dx 0
(
)
sin(n + 1)x 0 = 2 __________ (n + 1)x p =0
Thus,
Un + 1 + Un = 2Un + 1
[
fi Now,
Thus, fi fi
( ) 1 – cosx = Ú ( _______ dx = p 1 – cosx )
( (
)
( (
)
)
1 1 = _______ 2 2 Ú a __ x – 5 – b(x – 5) dx (a – b ) 1
x2 1 = _______ 2 2 a(log|x| – 5x) – b __ – 5x 3 1 (a – b )
22b 1 = _______ 2 a(log|2|) – ____ + 10b – 5a 2 3 (a – b )
22. Let
x sin2nx I = Ú ____________ 2n dx 0 sin x + cos2nx
2
( (
)
)
(
(
( (
)
(
)
(
)
(2p – x)sin2nx = Ú ____________ dx 0 sin2nx + cos2nx
sin2nx = 2p Ú ____________ 2n dx 0 sin x + cos2nx
( Ú ( Ú ( Ú ( Ú (
) ) )
sin2nx = 2p ____________ 2n dx – I 0 sin x + cos2nx
)
2p
)
fi
sin2nx 2I = 2p ____________ 2n dx 0 sin x + cos2nx
)
fi
sin2nx I = p ____________ 2n dx 0 sin x + cos2nx
(1 – cosnt) 1 = __ Ú _________ dt, 2 0 (1 – cost)
(
)
2p
(1 – cos2nq) = Ú __________ dq 0 (1 – cos2q) p
)
1 a __ x – 5 – b(x – 5) __________________ = Ú dx 1 (a2 – b2) 2
2sin (nq) = Ú ________ 2 dq 0 sin (q) p/2
)
x sin2nx – ____________ 2n dx 0 sin x + cos2nx
sin nq Also, Ú ______ 2 dq 0 sin q
(
2
2p
( )
(
(
2p
2
2
...(ii)
1 (a2 – b2) f (x) = a __ x – 5 – b(x – 5) 1 a __ x – 5 – b(x – 5) __________________ f (x) = (a2 – b2)
1
0
p/2
( )
2p
U1 – U0 = p U1 = U0 + p Un = U0 + np = np] p/2
...(i)
Ú f (x) dx
p
U1
1 __ x – 5
1 a f __ x + b f (x) = x – 5
2p
p
1–1 Now, U0 = Ú _______ dx = 0 1 – cosx 0
( )
Multiplying (i) by a and (ii) by b, and subtracting, we get 1 a2f (x) – b2f (x) = a __ x – 5 – b(x – 5)
3 1 2 sin n + __ x – sin n + __ x 2 2 = Ú __________________________ dx x 0 sin __ 2 p
( ( )
np 1 = __ × np = ___ 2 2
Replacing x by 1/x, we get
Therefore, (Un + 2 – Un + 1) – (Un + 1 – Un) p
)
(1 – cosnx) 1 = __ Ú __________ dx 2 0 (1 – cosx)
1 a f (x) + b f __ x =
3 2sin n + __ x 2 = Ú ___________ dx x __ 0 sin 2 p
(
2.117
2p
(Let 2q = t)
)
))
2
)
2.118 Integral Calculus, 3D Geometry & Vector Booster
( Ú (
) )
2p
fi
2p
37
sin2nx I = 4p Ú ____________ 2n dx 0 sin x + cos2nx sin2nx = 4p ____________ 2n dx 0 sin x + cos2nx
...(i)
1
p/2
(
)
sin2nx + cos2nx 2I = 4p Ú ____________ dx 0 sin2nx + cos2nx p/2
= 4p Ú dx
fi
p = 4pp __ = 4p 2 2 I = 2p 2 p
23. Let
(
)
2x sin x = 0 + Ú ________ dx – p 1 + cos2x x sin x = 4 Ú ________ dx 2 0 1 + cos x (p – x)sin(p – x) = 4 Ú ______________ dx 0 1 + cos2x (p – x)sinx = 4 Ú _________ dx 2 0 1 + cos x p
p sinx xsinx = 4 Ú ________ dx – 4 Ú ________ dx 2 2 0 1 + cos x 0 1 + cos x p sin x = 4 Ú ________ dx – I 2 0 1 + cos x p
p sinx 2I = 4 Ú ________ dx 2 0 1 + cos x p
psinx I = 2 Ú ________ dx 2 0 1 + cos x
sinx I = 4p Ú ________ dx 2 0 1 + cos x
= – 4p (tan–1 (cosx))2
= – (–1 – 1)
25. Given
=2
p __
(
0
p = – 4p 0 – __ 4 = p2
)
(
2
)
4
(
2
)
2x esin(x ) = Ú _______ dx 2 1
esin t = Ú ____ dt, x2 = t t 1
esin x = Ú ____ x dx 1
= (F(x))|16 1
= F(16) – F(1)
fi
k = 16
26. Given
a + b = 4.
Let
f (a) = Ú g(x) dx + Ú g(x) dx
16
( )
16
( )
b
0
0
a
4 – a
0
0
= Ú g(x) dx + Ú g(x) dx
df (a) fi _____ = g(a) – g(4 – a) da df (a) fi _______ = g(a) – g(4 – a) d(b – a) df (a) fi ________ = g(a) – g(4 – a) d(4 – 2a)
p/2
a
p
p
= – (cos(37 p) – 1)
p
2x 2x sin x = Ú ________ 2 dx + Ú ________ dx – p 1 + cos x – p 1 + cos2x
p
2esin(x ) Now, Ú ______ x dx 1
p
= – (cos t)|37p 0
4
2x(1 + sinx) I = Ú __________ dx – p 1 + cos2x
p
( )
p
(Let p ln x = t)
0
esinx fi Ú ____ x dx = F(x)
p
= Ú sin tdt,
d esinx ___ (F(x)) = ____ x , x > 0 dx
0
)
37p
...(ii)
Adding Eqs (i) and (ii), we get
(
e p sin(p lnx) 24. Given integral = Ú _________ dx x
fi
df (a) – _____ = g(a) – g(4 – a) 2da
df (a) _____________ g(a – 4) – g(a) fi _____ = >0 2 da Thus,
a
b
0
0
f (a) = Ú g(x) dx + Ú g(x) dx
It increases as (b – a) increases.
Definite Integrals
27. Given integral
x
g (x) = Ú cos4t dt 0
Now,
g(x + p)
= Ú cos t dt
0
x
= Ú cos t dt + Ú cos t dt 4
x
0
28. Given
= Ú cos t dt + Ú cos t dt 4
0
= g(x) + g(p)
x
fi
(1 + x) f (x) = 1
fi
1 f (x) = _____ 1+x 1 f (1) = __ 2
1
–1
= Ú (x – [x]) dx 1
1
(
31. Let
–1
= 0 – Ú ([x]) dx
p = __ – 2 Ú tan–1 x dx 2 0 1 p 1 __ = – 2 xtan–1 x – __ log|x2 + 1| 2 2 0 p p 1 = __ – 2 __ – __ log2 2 4 2 p p = __ – __ + log2 2 2 = log 2
( (
0
1
–1
0
0
1
dx I = Ú _______ ...(i) 1 + cosx p/4
–1
0
dx = Ú _______ p/4 1 – cosx 3p/4
3p/4
= Ú dx
(
( )
2 = Ú _____ 2 dx p/4 sin x
= Ú (2cosec2x) dx
3p/4
= (x)|0–1
30. Given,
= [0 – (– 1)] = 1
p/4
3p ___ 4 – (2cot x)|p __ 4
=
1 Ú tan–1 ________ dx 0 1 – x + x2
= – 2(–1 – 1)
1
=4
(
)
)
x – (x – 1) = Ú tan–1 ___________ dx 1 + x(x – 1) 0
...(ii)
fi
)
1 1 2I = Ú _______ + _______ dx 1 + cosx 1 – cosx p/4
–1
)
Adding Eqs (i) and (ii), we get
= – Ú (–1) dx – Ú (0) dx
(
)
3p/4
= – Ú ([x]) dx – Ú ([x]) dx
1
)
3p/4
= Ú (x) dx – Ú ([x]) dx
0
0
p 1 = __ – Ú tan–1 ________ dx 2 0 1 – x + x2
–1
= 2 Ú tan–1 x dx
1
0
1
– 1
1
= Ú tan–1 x dx – Ú tan–1 (x) dx
p = __ – Ú cot–1 (1 – x + x2) dx 2 0
–1
0
1
1
1
0
1
Ú f (x) dx
1
0
f (x) = 1 – x f (x)
0
1
1
1
fi
29. Given,
0
Also, Ú tan–1 (1 – x + x2)
Ú f (t) dt = x + Ú t f (t) dt
fi
1
1
x
0
1
0
4
0
0
= Ú tan–1 x dx – Ú tan–1 (– x) dx
p
x
0
= Ú tan–1 x dx – Ú tan–1 [(1 – x) – 1] dx
x + p
4
1
= Ú tan–1 x dx – Ú tan–1 (x – 1) dx
x + p
4
1
2.119
I=2
2.120 Integral Calculus, 3D Geometry & Vector Booster 32. Given,
2
fi
p/2
Adding Eqs. (i) and (ii), we get
Ú [2sinx]dx
5p/6
p
p/2
5p/6
= Ú [2sinx] dx + Ú [2sinx] dx 7p/6
7p/6
p
3p/2
(
)
(
)
p 7p __ ◊ 1 + (0) + ___ – p ◊ (–1) 2 6
5p = ___ – 6
(
)
3p 7p + ___ – ___ ◊ (– 2) 2 6 2p __ p ___ 4p ___ = – – 6 6 6 p __ = – 2
( Ú ( p
33. Let
...(i) ...(ii)
Adding Eqs (i) and (ii), we get
(
3
– 2
34. Given
)
2
ecosx sinx dx – 2 ____________
3
= + Ú 2dx odd function 2
= 0 + 2(3 – 2)
=2
| |
= Ú tdt, loge x = 1
( )
t2 2 = __ 2 –1
x
1 = 2 – __ 2
0
3 = __ 2
= p I = __ 2
g(x) = Ú f (t) dt
37. We have,
g(2) = Ú f (t) dt 1
2
0
1
lnt f (x) = Ú ____ dt 1 +t 1
Now,
lnt 1 f __ x = Ú ____ dt 1 +t 1
= Ú f (t) dt + Ú f (t) dt
1 Now, __ £ f (t) £ 1 2
( )
...(i)
1/x
1 1 Let t = __ y fi dt = – __2 dy y The given integral reduces to
1
1 fi Ú __ dt £ Ú 1 ◊ dt 02 0 1
1 fi __ £ Ú f (t) dt £ 1 2 0 1 Also, 0 £ f (t) £ __ 2 2
( ) x
0
2
– 2
–1
(x)p0
1
3
= Ú ecos x sinx dx + Ú 2dx
2
= Ú dx
2
2
2
fi
– 2
Ú
0
fi
3
loge x 36. Given Ú _____ x dx e–1
p
2
= Ú f (x) dx + Ú f (x) dx
e2
ecos x + e– cos x 2I = Ú ___________ dx e– cos x + ecos x 0
2
2
e– cos x = ___________ – cos x + ecos x 0 e p
1
35. Given Ú f (x) dx
) )
p
2
3 1 fi __ £ Ú f (t) dt + Ú f (t) dt £ __ 2 0 2 1
ecos x I = Ú ___________ cos x + e– cos x 0 e
1
...(ii)
1 1 __ £ Ú f (t) dt + Ú f (t) dt £ 1 + __ 2 0 2 1
+ Ú [2sinx]dx + Ú [2sinx] dx
1 0 £ Ú f (t) dt £ __ 2 1
3p/2
...(i) 2
1 fi Ú 0 dt £ Ú f (t) dt £ Ú __ dt 1 1 1 2
y
( ) ( )
1 ln __ y dy = Ú _______ × – ___2 1 1 1 + __ y y y
–ln(y) dy = Ú _____ × – ___ y 1 1 + y
Definite Integrals y
ln(y) = Ú _______ dy y(1 + y) 1 t
ln(t) = Ú ______ dt t(1 + t) 1
...(ii)
x
( )
3x2 3 F¢(x2) = ___ + 1 = ___ + 1 2x 2x
fi
3 f (x2) = ___ + 1 2x
Putting x = 2, we get
Adding Eqs (i) and (ii), we get
fi
x
ln(t) ln(t) 1 f (x) + f __ x = Ú ______ dt + Ú ______ dt 1 (1 + t) 1 t(1 + t)
fi
3 f (4) = __ ◊ 2 + 1 = 4 2
40. **
x
x
(1 + t)ln(t) = Ú _________ dt t (1 + t) 1
41. Given
ln(t) = Ú ____ dt t 1
(ln t)2 (ln x)2 = _____ = _____ 2 2 1
|
x
( ) ( )
lne 1 1 f (e) + f __ e = ___ = __ . 2 2 2
( ) Ú ( ) Ú ( ) p
38. Let
2
cos x I = Ú _____ dx x –p 1 + a
...(i)
x2 – f ¢(x) = 0
fi
x2 = ÷2 – x2
fi
x4 = 2 – x2
fi
x4 + x2 – 2 = 0
fi
(x2 + 2)(x2 – 1) = 0
fi
x = ± 1
...(ii)
(
)
p
= Ú (cos2x)dx
= 2 Ú (cos2x)dx
1 = __ Ú f (t)dt, 2 6
1 = __ Ú f (t)dt 2 0
1 = __ Ú f (x)dx 2 0
1 = __ × 6 Ú f (x)dx 2 0
= 3 I
6I
1
–p
p
0
fi
(
)
2
1 + cos x = 2 Ú ________ dx 2 0
( (
2
(Let 2x = t)
6I
(1 + ax) cos2x 2I = Ú ___________ dx (1 + ax) –p
p
_____
6 + 6I
ax cos2x = _______ dx 1 + ax –p p
Also,
3
Adding Eqs (i) and (ii), we get fi
f ¢(x) = ÷ 2 – x2
42. Given Ú f (2x)dx
cos2x = ______ dx – x –p 1 + a p
_____
fi
3 + 3T
p
_____
f (x) = Ú ÷2 – t2 dt 1
x
Thus,
2.121
))
p
sin x 1 I = __ x + _____ 2 2 0 p = __ 2
43. Given, 1/2
(
( ) )
1–x Ú [x] + ln _____ dx 1+x –1/2 1/2
1/2
( ( ))
1–x = Ú ([x])dx + Ú ln _____ dx 1+x –1/2 –1/2
= Ú ([x])dx + 0
x
39. Given
F(x) = Ú f (t)dt
fi
F ¢(x) = f (x)
Also,
F(x2) = x2 (x + 1) = x3 + x2
fi
F¢(x2) ◊ 2x = 3x2 + 2x
0
1/2
–1/2
0
1/2
–1/2
0
= Ú ([x])dx + Ú ([x])dx
2.122 Integral Calculus, 3D Geometry & Vector Booster
0
1/2
fi
f ¢(x) = 2x ( e– (x + 2x + 1) – e – x )
–1/2
0
fi
f ¢(x) = 2xe– x ( e– (2x + 1) – 1 )
= Ú (–1)dx + Ú (0)dx = (–1)(x)|0–1/2
(
)|
2 47. Given Ú x f (x)dx = __ t 5 5 0
I(m, n) = Ú t (1 + t) dt n
m + 1
)|
1
1
t n m + 1 = (1 + t)n _____ – _____ Ú t (1 + t)n – 1 dt m+1 0 m+10 n 2n = _____ – _____ I(m + 1, n – 1) m+1 m+1 I = Ú f (sin2x)sin x dx
...(i)
t 2 f (t 2)(2t) = 2t 4
1 f (t 2) = __ t 5 1 = __ ( ) ____ 2 2 ( __ ) 5
4 f ___ = 25
p/2
= Ú f (sin2x)cos x dx
...(ii)
0
Adding Eqs (i) and (ii), we get
2I =
p __ 2 0
Ú
p __ 2
1–x 48. Given Ú _____ dx 1+x 0
( (
)) p Let ( __ – x ) = t 4
p __
=
4 __ ÷2 f (cos2t)(cost)dt
=
4 __ 2÷2 f (cos2t)(cost)dt
Ú
p – __ 4
Ú
p __
fi
fi
=
Ú
__
0 p __ 4
2I = 2÷2 Ú f (cos2x)(cosx)dx 0
__
p __ 4
I = ÷2 Ú f (cos2x)(cosx)dx 0
x2 + 1
46. Given
f (x) = Ú e dt
fi
( (( ))) ( (( )))
Let x = cosq
q sin __ 2 q q _______ = Ú 2sin __ cos __ dq 2 2 q 0 __ cos 2
q = Ú 2sin2 __ dq 2 0
= Ú (1 – cosq)dq
p/2
p/2
( ) ( )
( )
p/2 0
p __
= (q – sinq)|02
p = __ – 1 2
| |
49. Let f (x) be a differentiable function defined as f : [0, 4] Æ R, show that (i) 8f ¢(a) f (b) = {f (4)}2 – {f (0)}2 when a, b Œ (0, 4)
(ii) Ú f (x)dx = 2[a f (a 2) + b f (b 2)] 0
x2
2
)
4
–t2
2
________
q p/2 sin __ 2 = Ú _______ sinq dq q 0 cos __ 2
0
4 __ 2÷2 f (cos2x)(cosx)dx
( ÷
1 – cosq = – Ú ________ sinq dq, 1 + cosq p/2
p __
÷
______
0
__ p = ÷2 Ú f (sin2x) sin x + __ dx 4 0
1
f (sin2x)(sinx + cosx)dx
2 Let t = __ , then 5
0
+
t2
p/2
45. Let
2
Clearly f (x) increases in (0, •).
m
(
4
4
O
1 = (–1) 0 + __ 2 1 __ = – 2 0
4
–
1
44. Given
4
4
f ¢(x) = e – (x + 1) ◊ 2x – e – x ◊ 2x
when 0 < a, b < 2 [IIT-JEE, 2004]
p/3
50. Let
(
)
2
p + 4x3 I = Ú _____________ dx p – p/3 2 – cos |x| + __ 3 p/3
Definite Integrals
(
(
)
(
)
)
(
p/3
Differentiating both sides w.r.t. x, we get x2
3
)
4x + Ú _____________ dx p – p/3 2 – cos |x| + __ 3 p/3
( (
(
)
)
p/3
)
)
)
(
)
(
)
(
( ( (
)
(
)
cosx + cosx ________ (2x) 1 + sin2x
( )
dy (–1) Now, ___ = 0 + (–1)(2p) ◊ _____ 1+0 dx x = p = 2p 52. The given integral can be written as 0
Ú ((x + 1)3 + 2 + (x + 1) cos (x + 1))dx – 2
Let (x + 1) = t 1
= Ú (t3 + 2 + t cost) dt –1 1
1
= Ú (t3 + t cos t) dt + Ú 2dt –1
–1
= 0 + Ú 2dt
53. Given,
= 2(1 – (–1)) = 4
–1
sinx
)
(Let t/2 = y)
fi
0 – sin2x f (sinx) ◊ cosx = – cosx
1 f (sinx) = _____ 2 sin x 1 fi f (x) = __ 2 x 1 1 __ = _____ fi f ___ 2 =3 1 3 ÷ ___ __ 3 ÷ 54. Given integral is = fi
)
sec2y dy = 4p Ú ____________ sec2y + 2tan2y p/6
sec2y = 4p Ú _________ dy 2 p/6 1 + 3tan y
sec2y 4p = ___ Ú _____________ dy __ 3 p/6 ( 1/÷3 )2 + tan2y
p/3
___ 16
1
p/3
__
Ú t2 f (t)dt = 1 – sinx
dy = 4p Ú _________ 2 , p/6 1 + 2sin y p/3
(
dy cos÷q __ ___ = – sinx Ú _________ dq dx 2 1 + sin2÷q p
1
dt = 2p Ú ____________ p/3 1 + 2sin2 (t/2) p/3
(
(
dt = 2p Ú ____________ 1 + (1 – cos t) p/3 2p/3
)
dt = 2p Ú _______ , (Let t = x + p/3) 2 – cos t p/3 2p/3
(
(
dx = 2p Ú _____________ p 0 2 – cos x + __ 3 2p/3
)
p = 2 Ú _____________ dx p 0 2 – cos |x| + __ 3 p/3
)
p = Ú _____________ dx + 0 p – p/3 2 – cos |x| + __ 3
(
)
cos÷q __ y = cosx Ú _________ dq 2 2 1 + sin ÷q p ___ 16
p = Ú _____________ dx p – p/3 2 – cos |x| + __ 3
(
__
2.123
)
)
__
÷ 3
( ) ( )
p
[ ( )
(
)]
1 1 Ú e|cosx| 2sin __ cosx + 3cos __ cosx sin x dx 2 2 0 p
[ ( )] Ú [ ( )] Ú [ ( ) ]
4p = ___ Ú __ __________ __ 2 3 1/÷3 ) + v2 (1/÷3
__ __ __ 4p __ = ___ × ÷ 3 ( tan–1 ( v÷3 )) ÷ 3 1/÷3 3
1 + e|cosx| 3cos __ cosx sinx dx 2 0
51. Given,
x2
(
__
)
cosx cos ( ÷q ) __ dq y(x) = Ú ___________ 2( 2 ) p /16 1 + sin ÷q
1 = Ú e|cosx| 2sin __ cosx sinx dx 2 0 p
p
1 = 0 + e|cosx| 3cos __ cosx sinx dx 2 0
2.124 Integral Calculus, 3D Geometry & Vector Booster
( Ú
)
2a
f (x) dx = 0, f (2a – x) = – f (x) 0
[ ( ) ] [ ( )] [ ( ) ] [ ( )]
p
1 = Ú e|cosx| 3 cos __ cosx sin x dx 2 0
1 = 2 Ú e|cos x| 3 cos __ cos x sin x dx 2 0
p/2
p/2
Ú (1 – x50)100 dx 0 5050 × _____________ = 5051 1 Ú (1 – x50)101 dx
1 = 2 Ú e|cos x| 3 cos __ cos x sin x dx 2 0
1 = 6 Ú ecos x cos __ cos x sin x dx 2 0
0
56. Let
sec2x
[ ( ) ] (Let t = cos x) t = 6 Ú e [ cos ( __ ) ] dt 2 e t t 1 = 6 _____ __ sin __ + cos ( __ ) ] 1[ 2 ( 2 ) 2 __ 1 + 0
1
p/2
( ) ( )
Thus, the value of
t , = – 6 Ú et cos __ 2 1
Ú f (t)dt 2 ________ L= lim p x Æ __ p2 4 x2 – ___ 16
( )
4
( 0
)
t
( ( ))
0
[ ( )
1
In = Ú (1 – x50)n dx
dx (A) Ú _____ 2 –1 1 + x 1
0
Integrating by parts, we get 1
In = (1 – x50)n ◊ x|10 + Ú x ◊ 50n ◊ (1 – x50)n – 1 ◊ x49 ◊ dx
dx = 2 Ú _____ 2 0 x + 1
= 2(tan–1x)10
p p = 2 __ – 0 = __ 4 4
0
1
= 0 + 50n Ú x ◊ (1 – x ) 50
50 n – 1
◊ dx
= – 50n Ú ((1 – x50) – 1) ◊ (1 – x50)n – 1 ◊ dx 0
(
1
0
0
)
= – 50n Ú (1 – x50)n ◊ dx – Ú (1 – x50)n – 1 ◊ dx = – 50n (In – In – 1) = – 50n In + 50n In – 1 fi
In = – 50n In + 50n In – 1
fi
(1 + 50n)In = 50n In – 1
In 50n fi ____ = ________ In – 1 (1 + 50n) Putting
(
)
1
0
1
1
( )
57.
( ) ]
24 1 1 1 = ___ e __ sin __ + cos __ – 1 2 2 5 2 1
55. Let
1
4
)
p p f sec2 __ . sec2 __ . 1 4 4 ___________________ = p __ 4 8f (2) = _____ p
1
t
(
2 f (sec x) 2sec2x tanx ________________ = lim p 2x x Æ __
dx ______ (B) Ú ________ _____ 0 1 ÷÷ – x2
p = (sin–1 x)10 = __ 2 3 dx (C) Ú _____ 2 2 1 – x
(
| | )
x–1 3 1 = – __ log _____ 2 x+1 2
1 1 1 1 = __ log __ – __ log __ 2 3 2 2
n = 101, then we get
( ) 1 2 = __ log ( __ ) 2 3 2
dx (D) Ú _______ _____ 1 x÷x 2 – 1
I101 50 × 101 5050 fi ___ = _____________ = _____ I100 (1 + 50 × 101) 5051
I100 5051 fi ___ = _____ I101 5050
= (sec–1 x)21
= (sec=12 – sec–11)
( )
(
Definite Integrals
58. Given
(
)
f (x) = f (1 – x)
Putting
x = 1/2, we get
1 1 1 f ¢ __ = – f ¢ 1 – __ = – f ¢ __ 2 2 2
( ) ( ) ( ) 1 fi 2f ¢ ( __ ) = 0 2 1 fi f ¢ ( __ ) = 0 2 1 Also, it is given that f ¢ ( __ ) = 0 4
fi
1 __ 2 1 Therefore, f x + __ sinx dx = 0. 2 1 – __ 2 1/2 Again, f (1 – t) esin(p t)dt 1 0 = – f (y) esin(p(1 – y)) dy, (Let 1/2
60.
sin(p(1 – py))
dy
0
= Ú f (y) e
)
(
)
p p 2 = ___ __ × __ = ____ __ 6 3 3÷3 ÷ p__ Tn > ____ 3÷3
( )
1
n
sin(p (1 – p y))
dy
0
Ú ÷
0
k = 1
Ú
__________ 2 1 – [f ¢(x)] =
fi ÷
0
f (x)
fi
1 – [f ¢(x)]2 = ( f (x))2
fi
dy 2 1 – ___ = y2 dx
( ) ( ) ÷
_____ dy fi ___ = ± 1 – y2 dx
dy _____ fi Ú ______ = ± dx 1 ÷ – y2
1 – t = y)
fi sin–1 y = c ± x fi y = sin(c ± x) When x = 0, y = 0, then c = 0 Thus, the equation of the curve is y = sin(± x) = ± sinx As
f (x) ≥ 0 for 0 £ x £ 1, we get
f (x) = sinx, for 0 £ x £ 1
Since
sin x < x for all x > 0, we get
f (x) < x for 0 < x £ 1
1/2
Thus,
0
and
= Ú f (t) esin(p y) dt, by Property I.
k = 0
x _________ 2 Given 1 – [f ¢(t)] dt = f (t) dt 0
( )
Ú
1/2
(
x
Ú
p 2 p __ = ___ __ __ – 3 3 6 ÷
(
( ) ( ) ( ) 1 Thus, f ( x + __ ) is an even function. 2 1 fi f ( x + __ ) is an odd function. 2
( )
( ) )
n – 1 k k h S f __ n > Ú f (x) dx > h S f __ n
1 Replacing x by x + __ , we get 2 1 1 fi f x + __ = f __ – x 2 2
= Ú f (y) e
))
__ 2 1 = ___ __ (tan–1 ( ÷3 ) – tan–1 ___ __ 3 ÷3 ÷
Now,
Thus, f ¢(x) = 0 vanishes at least twice. Again, f (x) = f (1 – x)
(
3 1 f ¢ __ = – f ¢ __ 4 4
1/2
(
( ) ( )
2x +__ 1 1 2 = ___ __ tan–1 ______ 3 ÷3 0 ÷
( ) ( ) 3 1 f ¢ ( __ ) = – f ¢ ( __ ) = 0 4 4
dx __ = Ú _______________ 3 2 ÷ 0 1 2 __ ___ x + + 2 2
1 x = __ in f ¢(x) = – f ¢(1 – x), we get 4
Ú
( )
1
f ¢(x) = – f ¢(1 – x)
)
1 n ___________ 1 __ 59. We have Sn < lim S = lim S 2 n Æ • n n Æ • n k = 1 k k __ 1 + n + __ n 1 dx = Ú ________ 2 0 1 + x + x
p = __ – 0 3 p = __ 3
fi
Putting
2.125
( ) ( )
1 f __ < 2 1 f __ < 3
1 __ 2 1 __ . 3
2.126 Integral Calculus, 3D Geometry & Vector Booster
( Ú ( p
61. Let
) )
sin n x In = Ú ___________ dx x –p (1 + p ) sinx p
sin n x = __________ dx x –p (1 + p )sinx
( Ú ( Ú ( p
...(i)
)
sin n (– x) = Ú ______________ dx – x –p (1 + p )sin(– x)
p
0
) )
sin n (x) = ____________ dx x –p (1 + p )sin(x)
p x sin n (x) = ____________ dx x –p (1 + p )sin(x)
...(ii)
Adding Eqs (i) and (ii), we get p
(
( p
p
fi
)
)
(
)
sin n (x) = 2 Ú _______ dx sin(x) 0
(
)
sin n (x) In = Ú _______ dx sin(x) 0
( Ú ( p
\
x = 0, y = 0, then A = 0 y = Ae x = 0 f (x) = 0 f (ln 5) = 0
When Thus, fi fi
) )
sin(n + 2)x – sin n x In + 2 – In = Ú _________________ dx sinx 0 p
2cos(n + 1)x × sin x = __________________ dx sinx 0
= Ú 2 cos(n + 1)x dx
(
)
p
sin(n + 1)x = 2 __________ = 0 n+1 0
fi
In + 2 = In
Now,
sinx I0 = 0 and I1 = Ú ____ dx = p sinx 0
Since and
In + 2 = In, so I1 = I2 = I5 = ... = I(2n – 1) = p I0 = I2 = I4 = ... = I2n = p
p
( )
Thus, S I2m + 1 = I3 + I5 + ... + I21 = 10p 10
fi
f ¢(0) – f (0) = ÷ 0 + 1 = 1
fi
f ¢(0) = f (0) + 1
fi
f ¢(0) = 2 + 1 = 3
Let
f –1 = g
y = f (x) ¤ x = g(y)
fi
f ¢(x) = f (x)
_____
when g is the inverse of f, then we shall use 1 g¢(y) = ____ f ¢(x) Putting
y = 2 and x = 0, then
1 1 g¢(2) = ____ = __ f ¢(0) 3
( Ú ( Ú ( Ú ( Ú ( Ú ( Ú ( 1
)
x4 (1 – x)4 64. Let I = Ú ________ dx 0 1 + x2
) )
1
x4 (x4 – 4x3 + 6x2 – 4x + 1) = _______________________ dx 0 x2 + 1
(x8 – 4x7 + 6x6 – 4x5 + x4 ) = _______________________ dx 0 x2 + 1
(x8 + 6x6 + x4 ) = _____________ dx 0 x2 + 1
(x6 (x2 + 1) + x4 (x2 + 1) + 4x6) = __________________________ dx 0 x2 + 1
(x2 (x2 + 1)(x2 + 1) + 4x6 ) = ______________________ dx 0 x2 + 1
4(x6 + 1 – 1) = (x4 + x2 ) + ___________ dx 0 x2 + 1
1
1
)
)
1
1
f (x) = Ú f (t) dt 0
f ¢(x) – e f (x) = Rx4 + 1
x = 0, we get
m = 1
x
–x
Putting
and S I2m = I2 + I4 + ... + I20 = 0 62. Given,
– x
e
10
m = 1
_____
fi
p
0
x
e– x f (x) = 2 + Ú ÷t 4 + 1 dt 0
sin n (x) = Ú _______ dx sin(x) –p
f (0) = Ú f (t)dt = 0
63. Given
(1 + p x)sin n (x) 2 In = Ú _____________ dx (1 + p x)sin(x) –p p
But
0
p
dy fi ___ = y dx dy fi Ú ___ y = Ú dx fi log|y| = x + C fi y = e x + C = Aex
1
) )
Definite Integrals
( Ú ( 1
)
2.127
p/2
8 2 2 = __ p Ú0 ((3 – 4 sin 3q) × (3 – 4 sin q))dq
4 = Ú (x + x ) – _____ + 4(x4 – x2 + 1) dx 2 0 x +1
4 = 5x4 – 3x2 + 4) – _____ 2 dx 0 x +1
8 = __ p Ú [{3 – 2(1 – cos 6q) × 3 – 2(1 – cos2q)}] dq 0
= (x5 – x3 + 4x – 4 tan–1 x)10
= (4 – 4 tan–1 (1)) = (4 – p)
8 = __ p Ú0 [(1 + 2 cos 6q) × (1 + 2 cos 2q)]dq
4
2
1
p/2
)
p/2
p/2
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q + 4 cos 2q cos 2q)dq 0
65. The given limit is
(
x
)
1 t ln(1 + t) lim __ Ú ________ dt x Æ 0 x3 0 t4 + 4
(
p/2
x
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q)dq 0
)
p/2
t ln(1 + t) Ú ________ dt 0 t4 + 4 = lim ___________ x Æ 0 x3 x ln(1 + x) _________ x2 + 4 _________ = lim x Æ 0 3x2
(
)
( (
8 + __ p Ú0 (4 cos 6q cos 2q)dq p/2
8 = __ p Ú0 (1 + 2 cos 2q + 2 cos 6q)dq p/2
16 + ___ p Ú0 (2 cos 6q cos 2q)dq
)
x ln(1 + x) = lim __________ x Æ 0 3x2 (x4 + 4)
ln(1 + x) = lim _________ x Æ 0 3x(x4 + 4)
p/2
8 = __ p Ú (1 + 2 cos 2q + 2 cos 6q)dq 0 p/2
)
16 + ___ p Ú0 (cos 8q + cos 4q)dq
( )
( (
( f (x) is an even function)
9x p sin ___ 2 4 = __ p Ú _______ dx x __ 0 sin 2
10
Thus, Ú f (x)cos(p x)dx –10
10
( ) sin 9q sin 3q 8 = __ p Ú ( _____ × _____ sin 3q sinq ) p/2
p/2
2
= 2 ◊ 5 Ú f (x)cos(p x)dx 0
2
)
sin 3 ◊ (3q) sin 3q 8 = __ p Ú ________ × _____ dq sin 3q sinq 0
(
= 2 Ú f (x)cos(p x)dx 0
0
p/2
)
and f is periodic with period 2. Also, since [– x] = 1 – [x] for non Integral values of x, so, f is an even function.
sin 9q 8 = __ p Ú _____ dq sin q 0
(
)
Ï1 – x : 0 £ x < 1 67. We have, f (x) = Ì Óx – 1 : 1 £ x < 2
p
p/2
p __
= 4
p
( (( )))
)
(
8 p 16 = __ p __ + 0 + ___ p (0 + 0) 2
2 __ Ú f (x)dx p – p 4 = __ p Ú f (x)dx, 0
p __
sin6q 2 ___ sin 4q 2 8 16 sin 8q _____ = __ p q + sin2q + _____ + p _____ + 3 0 8 12 0
ln(1 + x) ________ x = lim ________ x Æ 0 3(x4 + 4) 1 = ___ 12 66. The given integral is
= 10 Ú f (x)cos(p x)dx
where
= 10(I + J ),
I = Ú f (x)cos(p x)dx
0
)
(3 sin 3q – 4 sin33q) (3 sinq – 4 sin3q) 8 = __ p Ú ________________ × ______________ dq sin 3q sinq 0
1
0
2.128 Integral Calculus, 3D Geometry & Vector Booster 2
ln3
J = Ú f (x)cos(p x)dx
and
1
1
ln3
0
= Ú (1 – x)cos(p x)dx
1 = __ Ú dt 2 ln2 1 = __ (ln3 – ln2) 2
1
= Ú [1 – (1 – x))cos(p (1 – x)] dx 0
1
= – Ú x cos(p x)dx
2
J = Ú (x – 1)cos(p x)dx
fi
1 1
= Ú t cos(p (t + 1))dt,
0
(Let x – 1 = t)
1
6f (x) = 3f (x) + 3x f ¢(x) – 3x2 3f (x) = 3x f ¢(x) – 3x2 f (x) = x f ¢(x) – x2 dy fi y = x ___ – x2 dx y__ ___ dy fi x = – x dx dy y fi ___ – __x = x dx
= – Ú xcos(p x)dx 0
Thus,
=I I=J 1
= – Ú x cos(p x)dx
(
0
)
1
1 x 1 __ = – __ p sin(p x) 0 + p Ú0 sin(p x)dx 1 = – ___2 (cosp x)10 p 2 ___ = 2 p
Therefore,
which is a linear differential equation. dx ___ 1 So, IF = e – Ú x = e– logx = __ x Hence, the solution is y__ 1 x = Ú __2 dx + c x y 1 fi __x = – __ x + c when x = 1, y = 2, then c = 3 Thus, the equation of the curve is y 1 __x = – __ x + 3 Now, when x = 2, then y = –1 + 3x = –1 + 6 = 5
10
Ú f (x)cos(p x)dx –10
= 10 (I + J) = 20 I 2 = 20 × ___ 2 p 10
p2 Now, ___ × Ú f (x)cos(p x) 10 10 p 2 40 = ___ × ___ 10 p 2 =4 ___
(
÷ln3
5/6
p2 70. Given integral = ___ Ú sec(p x)dx ln 3 7/6
)
x sin(x2) 68. Let I = ___ dx Ú __________________ 2 2 ÷ln2 sin x + sin(ln(6 – x )) ln3
(
x
fi fi fi
0 1
69. Given, 6 Ú f (t)dt = 3x f (x) – x3
1
= – Ú t cos(p t)dt
( ( )) 3 1 I = __ (l n ( __ ) ) 4 2
3 1 = __ ln __ 2 2
0
Also,
)
ln3
0
(
sin(t) + sin(ln – 6) 1 2 I = __ Ú ________________ dt 2 ln2 sin(ln – 6) + sin(t)
1
)
Adding Eqs (i) and (ii), we get
I = Ú f (x)cos(p x)dx
Now,
(
sin(ln – 6) 1 = __ Ú _______________ dt 2 ln2 sin(ln – 6) + sin(t)
5p/6
p2 1 = ___ × __ Ú sec t dt ln 3 p 7p/6
p2 1 = ___ × __ Ú sec x dx ln 3 p 7p/6
Let x2 = t fi x dx = 1/2 dt
)
sin t 1 = __ Ú _______________ dt 2 ln2 sin t + sin(ln6 – t)
...(i)
5p/6
...(ii)
2.129
Definite Integrals 5p/6
1
p = ___ × Ú sec x dx ln3 7p/6
= 0 – 0 – 12[0 – 0] + 12 Ú 2x(1 – x2)5 dx
(1 – x2)6 1 1 = 12 – _______ = 12 0 + __ = 2 6 6 0 3x2 I = _____ x dx – 2 1 + e
( ) ( ) Ú ( ) Ú ( ) Ú ( ) 0
| (
5p/6 p = ___ (log|secx + tanx|) 7p/6 ln3
p 2__ ___ 1 2__ ___ 1 = ___ log – ___ – __ – – ___ + __ ln3 ÷3 ÷ 3 3 ÷3 ÷
74. Let
p 2 = ___ log ___ __ ln3 ÷3
3x2 = ______ dx – x – 2 1 + e
3ex x2 = _____ dx x – 2 1 + e
71. Given,
| | | )
( (
2
)|)
2
2
b
Ú ( f (x) – 3x)dx = a2 – b2 a
3 fi Ú f (x) dx = __ (b2 – a2) – (b2 – a2) 2 a
2
1 fi Ú f (x)dx = __ (b2 – a2) 2 a
fi
p p f __ = __ 6 6
2
–2
2
2
[
(
)]
p+x = Ú x2 + ln _____ cos x dx p – x –p/2
p __ 2 = x2 p – __ 2
[
(
)]
p+x + log _____ cos x dx p – x
= Ú (x2)cos x dx p – __ 2
=2
p __ 2 (x2)cos x 0
Ú
Ú
[ (
p2 = ___ – 4 sq.u. 2
0
( Ú (
)
2
x[x2] 77. I = Ú _________ dx –1 2 + [x + 1]
d2 73. Ú 4x ___ 2 (1 – x2)5 dx 0 dx 3
)
)
2
1
1 d d = 4x3 ___ (1 – x2)5 – Ú 12x2 ___ (1 – x2)5 dx dx dx 0 0
= (4x3 × 5(1 – x2)4 × (– 2x))10
[
f (x) = ax2 + bx
fi 2a + 3b = 6 fi (a, b) = (3, 0) and (0, 2) Thus, the number of polynomials is 2.
1
( Nr = Odd × Even = Odd function)
1
)
(
= 0,
Ú f (x)dx = 1
2sin x)p/2 0
)
p = 2 ___ – 2 4
Now, it is given that,
dx + 0
(
)]
+x dx Ú cos2x ◊ log 1_____ 1 –x –1/2 = _____________________ 1/2 1+x Ú cos2x ◊ log _____ dx 1–x 0
76. Let
= 2 (x sinx + 2x cos x – 2
( ) ( )
1/2
p __ 2 p+x + log p_____ cos x dx – x p – __ 2
2
(
0
= (x3)20 = 8
75. The given integral
p __ 2
0
I = Ú (3x2)dx
fi
= 2 Ú (3x2)dx
72. The given integral
Ú
)
= Ú (3x2)dx
( ) p/2
(
3(ex + 1)x2 2I = Ú _________ dx (1 + ex) – 2
b
f (x) = x
...(ii)
Adding Eqs (i) and (ii), we get
b
fi
...(i)
1
]
– 12 (x2 (1 – x2)5)10 – Ú 2x(1 – x2)5 dx 0
x[x2] = _________ dx – 1 2 + [x + 1] 0
(
)
1
(
)
__
(
÷2
)
x ◊ 0 x ◊ 0 x ◊ 1 = Ú ______ dx + Ú ______ dx + Ú ______ dx + 0 –1 2 + (0) 0 2 + (1) 1 2 + (0) __
÷2
( )
x = Ú __ dx 2 1
2.130 Integral Calculus, 3D Geometry & Vector Booster
( )
__
x2 ÷2 1 1 = __ = 2 – __ = __ 4 1 4 4
Thus, 4I – 1 = 0 78. We have
(
1
)
–1 12 + 9x2 a = Ú ( e9x + 3tan x ) _______ dx 0 1 + x2 Let 9x + 3tan–1 x = t
(
)
)
a = Ú et dt = ( e
3p 9 + ___ 4 –
fi
3p 9 + ___ 4
3p log(a + 1) = 9 + ___ 4 3p fi log|(a + 1)| – ___ = 9 4 79. We have f (x) = 7 tan8x + 7 tan6x – 3 tan2x fi
(
)
= 7 tan6x(tan2x + 1) – 3 tan2x(tan2x + 1)
= (7 tan6x – 3 tan2x)(tan2x + 1) 6
2
2
= (7 tan x – 3 tan x)sec x p/4
(i) Ú f (x)dx 0
Ú
= Ú (7t6 – 3t2)dt
6 1 1 2 1 = 0 – __ – __ = – 4 – ___ = ___ = ___ 24 24 12 6 4
0
7
7t = ___ – 7
fi
2 £ 2 + sin4(p x) £ 3
1 1 1 fi __ £ __________ £ __ 3 2 + sin4(p x) 2 192x3 __________ 192x3 192x3 _____ fi _____ £ £ 3 2 2 + sin4(p x) 192x3 192x3 fi _____ < f ¢ (x) < _____ 3 2 x
)
3 1
3t ___ = 0 3 0
0
=
x
x
x
x
1/2
1/2
192x3 192x3 fi Ú _____ dx £ Ú f ¢ (x)dx £ Ú _____ dx 2 1/2 3 1/2 1/2
)
( (
x
x4 x4 64 __ £ Ú f ¢ (x)dx £ 96 __ 4 1/2 1/2 4 1/2
fi
1 1 16 x – ___ £ Ú f ¢ (x)dx £ 24 x4 – ___ 16 16 1/2
(
(
x
)
4
)
(
1
1
)
1 fi Ú 16 x4 – ___ dx £ Ú f (x)dx 16 1/2 1/2
p __ 4
Ú {Ú
))
Ú
( )
fi
Integrating, we get
p __ 4 x f (x) 0 – f (x)dx dx 0 p __ 4 p __ 7 3 4 x tan x – tan x 0 – (tan7x 0 p __ 4 0 – (tan7x – tan3x)dx 0
( Ú
( )
x
x
– 3 tan x)sec x dx
(ii) Ú x f (x)dx
=
)
0 £ sin4(p x) £ 1
2
p/4
(
1
=
)
t 4 1 __ 4 0
t 6 = 0 – __ – 6
1/2
2
(
fi Ú (64x )dx £ Ú f ¢ (x)dx £ Ú (96x3)dx
=
= 0 – Ú (t2 – 1)t3dt
3
p __ 4 (7 tan6x 0
(
– 1)tan3x sec2x dx Let tan x = t
x
Ú
=0–
81. We have
1 )
a + 1 = e
p __ 4 (tan4x 0
0
3p 9 + ___ 4 0
– 1)tan3x dx
12 + 9x2 fi _______ dx = dt 1 + x2
Thus,
Ú
=0–
1
3 fi 9 + _____ 2 dx = dt 1+x
(
p __ 4 (tan4x 0
1
}
Ú
(
)
1 £ Ú 24 x4 – ___ dx 16 1/2 fi – tan3x)dx
(
)
1
x5 x 1 16 __ – ___ £ Ú f (x)dx 5 16 1/2 1/2
(
)
x5 x1 £ 24 __ – ___ 5 16 1/2 1
fi
2.6 < Ú f (x)dx < 3 × 9 1/2
Definite Integrals p __ 2
81. Ans. (a, c) 4p
Ú e (sin at + cos at)dt 0 L = ____________________ p 6
t
Given
4
83. Let
k p
Let
I(k) = Ú e (sin at + cos at)dt 6
t
4
I ¢(k) = e k p (sin6 (ak p ) + cos6 (ak p )) ◊ p k p
= p e
for a = 2 as well as a = 4
Integrating, we get I (k) = e k p + c
Since
I(0) = 0 , we get c = – 1
fi
I (k) = e
x
t2 Ú _____ 4 dt = 2x – 1 0 t + 1
Let
t2 f (x) = Ú __4 + 1dt – 2x + 1 0 t
fi
x2 f ¢(x) = __ 4 – 2 < 0 " x Œ [0, 1] x
fi
(
)
ex x2 cos x = Ú ________ dx 1 + e x p – __
...(ii)
1 2 t __4 + 0 t
Ú
t2 1 fi Ú __ 4 + 1dt < __ 2 0 t fi f (1) < 0 f (x) = 0 exactly one root in [0, 1]
p – __ 2
p __ 2 = (x2 cos x)dx p – __ 2
Ú
= 2 Ú (x2 cos x)dx 0
p __ 2
1dt
1 __ " t Π[0, 1] 2
1
2I = Ú ((ex + 1)x2 cos x/(1 + ex))dx
p __ 2
x
t2 0 £ _____ 4 < t +1
p __ 2
2
I(4) fi ____ = e 4 p – 1/e p – 1 I(1) 82. Given,
f (0) = 1 and f (t) =
)
x2 cos x = Ú ______ dx – x p 1 + e – __
p __ 2
–1
fi
p __ 2
Adding Eqs (i) and (ii), we get
k p
(
...(i)
2
0
fi
)
2
Ú e t(sin6 at + cos4 at)dt 0
(
x2 cos x I = Ú ______ dx 1 + ex p – __
2.131
fi
I = Ú (x2 cos x)dx 0
p p /2 __ 2 sin x) – (2 x sin x)dx 0 0
Ú
2
= (x
__ p 2 = ___ – 2 [x( – cos x)]2 – Ú ( – cos x)dx 0 4 0
__ p 2 = ___ – 2[ – (0 – 0) + (sin x)]2 0 4
p 2 = ___ – 2 4
[
p
]
p /2
p
(
)
Chapter
3
Area Bounded by the Curves
Concept Booster 3.1 Rules
to
Draw Different Types
of
y = x2
Curves
To find the area of the plane curves, first we need to draw the given curves. So my dear friends, you should remember the basic rules to trace the given curve. Find the domain and range of the given function y = f (x). II: Find the point of intersection on the co-ordinate axes. III: Find the intervals of the monotonocity of the function y = f (x). IV: Find the local maximum and the local minimum values of the function y = f (x). V: Find the concavity and the points of inflection of the function y = f (x). VI: Find out whether the function is periodic or not. VII: Find the symmetry as follows:
Rule I: Rule Rule Rule Rule Rule Rule
1 y = __ x
(a) A curve is symmetrical about x-axis, if the powers of y, which occur in its equation, are all even. y2 = x
(c) A curve is symmetrical about the line y = x, if the curve remains unchanged by interchanging x and y in the equation.
(b) A curve is symmetrical about y-axis, if the powers of x, which occur in its equation, are all even.
(d) A curve is symmetrical about the line y = – x if the curve remains unchanged when x and y are replaced by – x and – y, respectively.
3.2 Integral Calculus, 3D Geometry & Vector Booster
(e) A curve is symmetrical in opposite quadrants, if the equation of the curve remains unchanged when x and y are replaced by – x and – y, respectively.
The vertical asymptote to the given curve is x = 1 The oblique asymptote to the given curve is y = 2x + 3.
( ) ( ) 3 = lim ( _____ + 2 ) = 2 x–1
f (x) y where as m = lim ____ x = lim __ x Æ • x Æ • x
x Æ •
and
c= lim (y – m x) x Æ •
3.2 Area
(
)
3x = lim _____ + 2x – 2x = 3 x Æ • x – 1
of the
Cartesian Curve
Rule VIII: Find the asymptotes if any. Asymptote It is a straight line which touches the curve at infinity. There are three kinds of asymptotes.
(i) Vertical Asymptote: The straight line x = a is a vertical asymptote to the curve y = f (x) if at least one of the values of lim [ f (x)] or lim [ f (x)] tends x Æ a– x Æ a+ to + • or – •.
(ii) Horizontal Asymptote: The straight line y = b is a horizontal asymptote to the curve y = f (x), where lim [ f (x)] = b. lim [ f (x)] = b or x Æ + • x Æ – • (iii) Oblique Asymptote: The staright line y = m x + c is an oblique asymptote to the curve y = f (x), where
( )
If f (x) is a single valued and continuous function of x in the interval [a, b]. Let
f (x) ≥ 0
for every x in [a, b], then the area bounded by the curve t = f (x), the x-axis and the ordinates x = a and x = b, (a < b) is represented by
f (x) m= lim ____ x
and
c= lim (f (x) – m x)
Ú y dx = Ú f(x) dx
x y = _____ . x–1
3.3 area of the region bounded by a single curve and the co-ordinate axes
x Æ •
x Æ •
For examples,
1. Let
The vertical asymptote to the given curve is x = 1. The horizontal asymptote to the curve is
(
)
x y= lim _____ = 1 x Æ • x – 1
1 y = _____ . x2 – 1
2. Let
The vertical asymptotes to the given curve are
x = ± 1
The horizontal asymptote to the curve is 1 y= lim _____ 2 = 0 x Æ • x –1
3. Let
(
( ) )
3x y = _____ + 2x . x–1
b
b
a
a
(i) The area bounded by the curve y = f (x), x-axis and the straight lines x = a and x = b (a < b) is given by
b
b
a
a
A = Ú y dx = Ú f (x) dx
Area Bounded by the Curves
(ii) If f (x) £ 0 for all x in [a, b], the area bounded by the curve y = f (x), x-axis and the straight lines x = a and x = b, (a < b) is
b
b
a
a
d
d
c
c
A = Ú x dy = Ú f (y) dy
(v) When the curve x = f (y) crosses y-axis, i.e. changes its sign at y = c between y = a and y = b, its area is given by
A = Ú (– y) dx = |Ú f (x) dx|
(iii) If the graph of y = f (x) crosses x-axis between x = a and x = b at x = c, d, e, f respectively, its area (which lie below the x-axis is negative but the area is taken positive, so we take their modulus) is
c
b
a
c
A = Ú x dy + Ú (– x) dy
= Ú x dy + Ú (x) dy
| | | |
c
b
a
c
c
3.4 Area b
A = Ú y dx a
c
d
a
c
b
b
e
f
+ Ú f (x) dx + Ú f (x) dx + Ú f (x) dx d
c
| | | |
= Ú f (x) dx + Ú f (x) dx e
d
= Ú f (x) dx – Ú f (x) dx a
b
= Ú f (y) dy + Ú f (y) dy
3.3
a
between
c
two curves
(i) Suppose we are given two curves reprsented by y = f (x) and y = g (x), where f (x) ≥ g (x) in [a, b]. Here the points of intersection of these two curves are given by x = a and x = b is obtained by taking common values of y from the given equation of two curves. Hence the required area is b
A = Ú (f (x) – g (x)) dx a
c
e
b
b
d
e
f
+ Ú f (x) dx – Ú f (x) dx + Ú f (x) dx
(iv) The area enclosed by the curve x = f (y), y-axis and the abscissae y = c and y = d, (c < d) is given by
(ii) If the curves y = f (x) and y = g (x) intersect at x = c in [a, b], the required area is c
b
A = Ú ( f(x) – g (x)) dx + Ú (g (x) – f (x)) dx
a
c
3.4 Integral Calculus, 3D Geometry & Vector Booster
(iii) The area bounded by the curves x = f (y) and x = g (y) and the straight lines y = c and y = d where f (y) ≥ g (y) is given by d
A = Ú (f (y) – g (y)).
3.6 Least value
of a
variable area
Example 1. If the area is bounded by the curve x3 f (x) = __ – x2 + b 3
c
and the straight lines x = 0 and x = 2 and x-axis is minimum, find the value of b. Solution: Given fi
x3 f (x) = __ – x2 + b. 3 f ¢ (x) = x (x – 2)
Clearly f (x) is monotonic in (0, 2). Hence for minimum area, f (x) must cross the x-axis at
3.5 Area
of the
region bounded
by the
several curves
If y = f (x) is a strictly monotonic function in (a, b) such that f ¢ (x) = 0, the area bounded by the ordinates x = a and x = b, y = f (x) and y = f (c), where c Œ (a, b), is minimum a+b when c = _____ . 2
0+2 x = _____ = 1. 2
Thus,
1 f (1) = __ – 1 + b = 0. 3
fi
2 b = __ . 3
Exercises Area bounded by a function which changes sign
(Problems based on Fundamentals)
Area of a linear curve
1. Find the area of the region bounded by the curves y = x2 – 2x + 2, x = – 1 and x = 2. 2. Find the area bounded by the curves y = ln x + tan–1 x and the ordinates = 1 and x = 2. 3. Find the area bounded by y = x |sin x|, x-axis and the ordinates x = 0 and x = 2p. 4. Find the area bounded by y = 1 log__1 x and the x-axis between x = 1 and x = 2.
2
5. Find the area of the region bounded by the curve y = e2x – 3ex + 2 and the x-axis.
6. Find the area bounded by y = x3 and x-axis between the ordinates x = – 1 and x = 1.
7. Find the area bounded by the curve y = sin x and the x-axis, for 0 £ x £ 2p.
8. Find the area bounded by the curve y = cos x and the x-axis, for 0 £ x £ 2p.
9. Find the area bounded by the curve y = x (x – 1) (x – 2) and the x-axis. 10. Find the area bounded by the curve y = (x – 1)(x – 2) (x – 3) and the x-axis. Area of a region between two non-intersecting curves 11. Find the area of the region enclosed by the curves y = x2 and y = 2x – x2.
Area Bounded by the Curves
12. Find the area of the region bounded by the parabolas y2 = x and x2 = y. 13. Find the area of the region bounded by the parabolas 4y2 = 9x and 3x2 = 16y. 14. Find the area of the region bounded by the curves
15. 16. 17. 18. 19. 20. 21. 22.
23.
3x2 y = ___ and the line 3x – 2y + 12 = 0. 4 Find the area bounded by the curve x2 = 4y and the straight line x = 4y – 2. Find the area of the region included between the parabolas y2 = 4ax and x2 = 4ay. Find the area of the region enclosed by the parabola y2 = 4ax and the chord y = mx. Find the area of the region bounded by the curve y = x2 + 2, y = x, x = 0 and x = 3. Find the area bounded by the curve y = 2x – x2 and the straight line y = – x. Find the area bounded by the straight lines x = 0, x = 2 and the curves y = 2x – x2, y = 2x. Find the area bounded by the curves y = 6x – x2 and y = x2 – 2x. In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x? Find the area of the region between the circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.
24. Find the area of the region enclosed between the two circles x2 + y2 = 1 and (x – 1)2 + y2 = 1. 25. Find the area of the smaller region bounded by the 2 x2 y x y ellipse __ 2 + __2 = 1 and the straight line __ a + __ =1. b a b 26. 27. 28.
Find the area of the region {(x, y): x2 + y2 < 1 < x + y}. Find the area of the region {(x, y): y2 < 4x, 4x2 + 4y2 < 9}. Find the area of the region {(x, y): x2 + y2 < 2ax, y2 > ax, x > 0, y > 0}
29. Find the area of the region
{
}
2 x2 y x y (x, y): __ 2 + __ 2 £ 1 £ __ a + __ . b a b 30. Find the area of the region bounded by y = 1 + |x + 1|, x = – 2, x = 3, y = 0. 31. Find the area of the region bounded by y = 1 + |x + 1|, x = – 3, x = 3, y = 0. 32. Find the area enclosed by the curves y = |x – 1| and y = – |x – 1| + 1. 33. Find the area of the region bounded by the curves
3.5
f (x) = log x and g (x) = (log x)2. 34. Find the area enclosed by y2 = x2 – x4. 1 35. Find the area bounded by the curve |y| + __ £ e–|x|. 2 36. Find the area of the region f (x, y) = [(x, y): 0 £ y £ x2 + 1, 0 £ y £ x + 1, 0 £ x £ 2]. 37. Find the area bounded by the curves y = x and y = x3. 38. Find the area bounded by the curves y = x (x – 1)2 the y-axis and the line y = 2. 39. Find the area of the region bounded by y = x2 + 1, y = x, x = 0 and y = 2. 40. Find the area of the region bounded by the curves y = loge x and y = 2x, x = 1/2 and x = 2. 41. Find ther area enclosed by the parabola (y – 2)2 = (x – 1), the tangent to the parabola at (2, 3) and x-axis. 42. Find the area enclosed by the loop of the curve y2 = (x – 1) (x – 2)2. 43. Find the area bounded by the curves y = (x – 1)2, y = (x + 1)2 and y = 1/4. 44. Find the area bounded by the curves y = ex, y = e–x and the straight line x = 1. 45. Find the area bounded by the curve y = log x, y = log |x|, y = |log x| and y = |log |x||. 46. Find the area bounded by the curve |x – 2| + |y + 1| = 1. 47. Find the area of the region bounded by the curve [x] + [y] = 3 in first quadrant, where [,] = GIF. 48. Find the area of the region bounded by the curve |x| + |y| = 1 and |x – 1| + |y| = 1. 49. Find the area bounded by the curve
| |
p y = cos–1 (cos x) and |x – p | + y – __ = 2
p __ . 2
50. Find the area bounded by the curve |y| = – (1 – |x|)2 + 5.
__
51. Find the area bounded by the curves y = ÷ |x| and y = |x|. 52. Find the area of the region bounded by the curves |x + y| + |x – y| £ 2. 53. Find the area bounded by the curve y = x and y = x + sin x, where 0 £ x £ p. 54. Let f (x) be a non-negative continuous function such that the area bounded by the curve y = f (x), p p x-axis and the ordinates x = __ and x = b > __ is 4 4 __ p p b sin b + __ cos b + ÷ 2 b , find f __ . 4 2
(
)
( )
3.6 Integral Calculus, 3D Geometry & Vector Booster Area of a region between two intersecting curves
Least value of a variable area
55. Find the area bounded by the curves y = sin x and p y = cos x, x = 0 and x = __ . 2 56. Find the area enclosed by the curves y = 2 – |2 – x| 3 and y = __ . |x| 57. The line y = m x bisects the area enclosed by the lines 3 x = 0, y = 0, x = __ and the curve y = 1 + 4x – x2. 2 Find m. 1 58. Find the area bounded by the curve |y| + __ £ e–|x|. 2 Area of a region by a horizontal strip 59. Find the area bounded between y = sin–1 x and y-axis between y = 0 and y = p/2. 60. Find the area enclosed by the line y = x – 1 and the parabola y2 = 2x + 6. 61. Find the area enclosed by x = – 2y 2 and x = 1 – 3y2. 62. Find the area enclosed by the curves y = tan–1 x and y = cot–1 x and the y-axis. 63. Find the area bounded by the curve x = – y2 + y + 2 and the y-axis. Area of a region between several graphs
x3 71. If the area bounded by y = __ – x2 + a and the straight 3 lines x = 0 and x = 2 and the x-axis is minimum, find the value of a. 72. Find the value of a, for which the area bounded by the curve y = a2 x2 + ax + 1 and the straight lines y = 0, x = 0 and x = 1 the least. 73. For what value of k, the area enclosed by the curves y = x2 – 3 and y = kx + 2 is the least. Find also the least area. 74. If the area bounded by the curve y = x2 + 2x – 3 and the line y = mx + 1 is least, find the value of m. 75. Find the value of a for which the area bounded by x 1 the curve y = __ + __2 , and the lines y = 0, x = a and 6 x x = 2a is least.
(Mixed Problems) 1. The area bounded by the curve y = 4x – x2 and the x-axis is
30 31 (a) ___ sq.u. (b) ___ sq.u. 7 7
64. Find the area of the plane figure bounded by __ 2 y = ÷x , x Œ [0, 1], y = x , x Œ [1, 2]
32 (c) ___ sq.u. 3
and y = Рx2 + 2x + 4, x Π[0, 2].
65. Find the area of the region bounded by the curves 4|y| = |4 – x2 | and |y| (x2 + 4) = 12. 2
2
66. Find the area bounded by the ellipse x + 2y = 2 and the outside of the parabola y = 1 – x2. 67. Find the area common to the circle x2 + y2 = 4 and the ellipse x2 + 4y2 = 9. 68. Find the area enclosed by the curves y = ln (x + e) 1 and x = ln __ y , and x-axis.
( )
69. Find the area enclosed by the curves x2 + y2 = 4, parabola y = x 2 + x + 1 and the curve x x y = sin2 __ + cos __ and x-axis. 4 4 x2 70. Find the area enclosed by the curves __ + y2 = 1 4 x__2 and – y2 = 1. 2
[ ( )
( ) ]
34 (d) ___ sq.u. 3 ______
2. The area under the curve y = ÷3x + 4 between x = 0 and x = 4, is
56 (a) ___ sq.u. 9
64 (b) ___ sq.u. 9
(c) 8 sq.u.
(d) none of these
3. The area bounded by the curve y = x3, x-axis and two ordinates x = 1 to x = 2 equal to
15 (a) ___ sq.u. 2
15 (b) ___ sq.u. 4
17 (c) ___ sq.u. 2
17 (d) ___ sq.u. 4
4. The measurement of the area bounded by the coordinate axes and the curve y = loge x is (a) 1 (b) 2 (c) 3 (d) • 5. If the area bounded by the curves y2 = 4ax and a2 y = mx is __ , the value of m is 3
Area Bounded by the Curves
(a) 2 (c) 1/2
(b) – 2 (d) none of these
6. The area bounded by the parabola y = 4x2, y-axis and the lines y = 1, y = 4 is (a) 3 sq.u.
7 (b) __ sq.u. 5
7 (c) __ sq.u. (d) none of these 3 7. The area enclosed by the curves y = sin x, y = 0, p x = 0 and __ is 2 (a) p (b) 2p (c) 1 (d) 2 8. The area of the region bounded by the x-axis and the p p curve defined by y = tan x – __ £ x £ __ is 3 3
(
__
(a) log ÷2 (c) 2 log 2
)
__
(b) – log ÷2 (d) 0
9. The ratio of the areas bounded by the curves p y = cos x and y = cos 2 x between x = 0, __ and 3 x-axis is __
(a) ÷2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1 10. The area bounded by y = [x] and the lines x = 1 and x = 1.7 is
17 (a) ___ 10
(b) 1
17 (c) ___ 5
7 (d) ___ 10
11. The area bounded by the x-axis and the curve y = sin x and x = 0, x = p is (a) 1 (b) 2 (c) 3 (d) 4 12. The area bounded by the parabola y2 = 2x and the ordinates x = 1, x = 4 is __
__
4÷2 (a) ____ sq.u. 3
28÷2 (b) _____ sq.u. 3
56 (c) ___ sq.u. 3
(d) none of these 2
x2 y 13. The area of the ellipse __ 2 + __ 2 = 1 is a b 1 (a) p ab sq.u. (b) __ p ab sq.u. 2 1 (c) __ p ab sq.u. (d) none of these 4 14. The area of the smaller segment cut off from the circle x2 + y2 = 9 by x = 1 is
__ 1 (a) __ (9 sec–1 3 – ÷ 8 ) 2
(b) 9sec–1 (3) – ÷ 8
(c) ÷8 – 9 sec–1 3
(d) none of these
__
3.7
__
15. The area of the upper half of the circle whose equation is (x – 1)2 + y2 = 1 is given by 2
______
(a) Ú ÷2x – x2 dx 0
2
______
(c) Ú ÷2x – x2 dx 1
1
______
(b) Ú ÷2x – x2 dx 0
p (d) __ 4 __
16. The area bounded by the curves y = ÷ x , 2y + 3 = x and x-axis in the first quadrant is
(a) 9
27 (b) ___ 4
(c) 36
(d) 18
17. The area of the region {(x, y): x2 + y2 £ 1 £ x + y} is
p 2 (a) ___ 5
p 2 (b) ___ 2
p 2 (c) ___ 3
p 1 (d) __ – __ 4 2
(
)
18. The area bounded by the curves y = |x| – 1 and y = – |x| + 1 is (a) 1 __ (b) 2 (c) 2÷2 (d) 4 19. The area of the figure bounded by y = ex, y = e–x and the straight line x = 1 is
1 (a) e + __ e
1 (b) e – __ e
1 1 (c) e + __ e – 2 (d) e + __ e + 2 20. The area bounded by the curves y = loge x and y = (loge x)2 is
(a) 3 – e
(b) e – 3
1 1 (c) __ (3 – e) (d) __ (e – 3) 2 2 21. The area enclosed by the parabolas y = x2 – 1 and y = 1 – x2 is 1 2 (a) __ (b) __ 3 3 8 4 __ __ (c) (d) 3 3 22. The area bounded by the circle x2 + y2 = 4, line ___ x=÷ 3y and x-axis lying in the first quadrant, is
p (a) __ 2
p (b) __ 4
3.8 Integral Calculus, 3D Geometry & Vector Booster p (c) __ (d) p 3 23. The area formed by triangular-shaped region bounded by the curves y = sin x, y = cos x and x = 0 is
__
(a) ÷2 – 1
(c) ÷2
(b) 1
__
__
(d) 1 + ÷2
24. The area between the curve y = cos x and x-axis when 0 £ x £ 2p is (a) 2 (b) 4 (c) 3 (d) 0 25. The area of the greatest rectangle that can be inscribed 2 x2 y in the ellipse __ 2 + __ 2 = 1 is a b ___ (a) ÷ab (b) a/b (c) 2ab (d) ab 26. The area bounded by the curves y = |x – 2|, x = 1, x = 3 and the x-axis is (a) 4 (b) 2 (c) 3 (d) 1 27. The area bounded by lines y = 2 + x, y = 2 – x and x = 2 is (a) 3 (b) 4 (c) 8 (d) 16 28. The area enclosed by the parabola y2 = 4ax and the straight line y = 2ax, is
a2 (a) __ sq.u. 3
1 (b) ___ 2 sq.u. 3a
1 (c) ___ sq.u. 3a
2 (d) ___ sq.u. 3a
(a) 26a2
(b) 8a2
2
2
26a 104a (c) ____ (d) _____ 3 3 30. The area bounded by the curves y = In x, y = In |x|, y = |In x| and y = |log |x|| is (a) 4 sq.u. (b) 6 sq.u. (c) 10 sq.u. (d) none of these 31. The area bounded by the curves y = |x – 1| and y = 3 – |x|, is (a) 2 sq.u. (b) 3 sq.u. (c) 4 sq.u. (d) 1 sq.u. 32. The area of the region lying inside x2 + (y__ – 1)2 = 1 and outside c2 x2 + y2 = c2, where c = (÷2 – 1) is
__
p (a) (4 – ÷ 2 ) __ + 4
1 ___ __ 2 ÷
__
p (b) (4 + ÷2 ) __ – 4
( )
(a) 2
(b) 1
(c) 4
(d) none of these
34. The area of the region formed by x2 + y2 – 6x – 4y 5 + 12 £ 0, y £ x and x £ __ is 2 __ __ ______ 3 + 1 3 – 1 p ÷ p ÷______ __ __ (a) – (b) + 8 4 8 6 __
– 1 p ÷3 (c) __ – ______ (d) none of these 8 6 35. Let f (x) = maximum [x2, (1 – x)2, 2x (1 – x)] where 0 < x < 1. The area of the region bounded by the curves y = f (x), x-axis, x = 0 and x = 1 is
17 14 (a) ___ (b) ___ 27 27 19 (c) ___ (d) none of these 27 36. The area enclosed by the parabola ay = 3(a2 – x2) and x-axis is (a) 4a2 sq.u. (b) 12a2 sq.u. 3 (c) 4a sq.u. (d) none of these 37. The area bounded by the curve y = k sin x between x = p and x = 2p, is (a) 2k sq.u. (b) 0
29. The area bounded by the curves x = at2, y = 2at and the x-axis in 1 £ t £ 3, is
__ p 1 (c) (4 + ÷ 2 ) __ + ___ __ (d) none of these 4 ÷ 2 33. The area enclosed between the curves y = loge (x + e), 1 x = loge __ y and the x-axis, is
1 ___ __ 2 ÷
k 2 (c) ___ 2
(d) k sq.u.
38. Let f (x) be a non-negative continuous function such that the area bounded by the curve y = f (x), p p x-axis and the ordinates x = __ and x = b > __ is 4 4 __ p p b sin b + __ cos b + ÷2 b , then f __ is 4 2
( )
(
)
( ) p (c) ( __ + ÷ 2 – 1 ) 4
( ) p (d) ( __ – ÷ 2 + 1 ) 4
__ p (a) 1 – __ – ÷ 2 4 __
__ p (b) 1 – __ + ÷ 2 4 __
39. The area in the first quadrant between x2 + y2 = p 2 and y = sin x is
(p 3 – 8) (a) _______ 4 (p3 – 16) (c) ________ 4
p 3 (b) ___ 3 (p 3 – 8) (d) _______ 2
40. For 0 £ x £ p, the area bounded by y = x and y = x + sin x, is
Area Bounded by the Curves
(a) 2 (b) 4 (c) 2p (d) 4p 41. The area bounded by y = x sin x and x-axis between x = 0 and x = 2p, is (a) 0 (b) 2p sq.u. (c) p sq.u. (d) 4p sq.u. 42. The area bounded by the lines y = |x – 2|, |x| = 3 and y = 0 is (a) 13 unit2 (b) 5 unit 2 2 (c) 9 unit (d) 7 unit2 43. The area bounded by the curve x2 = ky, k > 0 and the line y = 3 is 12 unit2. Then k is
__
3 (c) __ (d) none of these 4 44. The area of the portion enclosed by the curve __ __ __ x + ÷ y = ÷a and the axes of reference is ÷
a2 (a) __ 6
e 2 (a) ___ (b) e2 2 (c) 2e2 (d) 1. 50. The area bounded by eln (x + 1), |x| £ 1, is (a) 1 (b) 2 (c) 4 (d) none __ 51. The area common to the region determined by y ≥ ÷x and x2 + y2 < 2 has the value
52.
(b) 3÷3
(a) 3
(b) a2
a2 a2 (c) __ (d) __ 2 4 45. The area bounded by the curve x = cos–1y and the lines |x| = 1 is
53. 54.
(a) sin 1
(b) cos 1
(c) 2 sin 1 (d) 2cos 1 46. The area enclosed between the curves |y| = 1– x2 and x2 + y2 = 1 is
3p – 8 (a) ______ 3 2p – 8 (c) ______ 3
)
1 (c) 2 e + __ e
(d) none
3 56. The area bounded by the curve y = __ and |x| y = 2 – |x – 2| is
(
y = |x – p|, is (a) p 2 (b) 2p 2 (c) p 2/2 (d) none 49. The area bounded by the curves ÏÔ x1/ln x y = Ì ÔÓe
)
p 1 (c) __ – __ (d) none 4 6 The area of the region bounded by ||x| – |y|| £ 1 and x2 + y2 in the xy-plane is (a) p (b) 2p (c) 3p (d) 1. The area bounded by 1 £ |x – 2| + |y + 1| £ 2 is (a) 2 (b) 4 (c) 6 (d) none The area bounded by the curve y = tan x + cot x| – |tan x – cot x| and between the lines x = 0 and p x = __ and the x-axis is 2 __ (a) ln 4 (b) ln__ ÷ 2 (c) 2 ln 2 (d) ÷ 2 ln 2 The area is enclosed by the graph of the curve y = ln2 x – 1 lying in the fourth quadrant is
p – 8 (b) _____ 3
( ) (
(
(b) (2p – 1)
2 (a) __ e
1__ B 1, ___ , C (2, 0) such that min {PA, PB, PC} = 1, 3 ÷ the area bounded by the curve traced by the point P is __ __ 3p p (a) 3÷ 3 – ___ (b) ÷ 3 + __ 2 2 __ __ p 3p __ (c) ÷ 3 – (d) 3÷ 3 + ___ . 2 2 48. The area bounded by the curves y = cos–1 (cos x) and
)
55.
(a) p
47. A point P moves inside a triangle formed by A(0, 0),
(
3.9
: x π1 and y = |x – e| is : x =1
)
(
4 (b) __ e
)
(
)
1 (d) 4 e – __ e .
4 – ln 27 (a) ________ (b) 2 – ln 3 3 (c) 2 + ln 3 (d) none 57. The area enclosed by the curve |x + y – 1| + |2x + y + 1| = 1 is (a) 2 (b) 1 (c) 4 (d) 3/2
58. The area of the region {f ( x, y) : x2 + y2 £ 1 |x| + |y|} is (a) p (b) p – 1 (c) p – 2 (d) p – 3 59. The area of the region of the xy-plane is bounded by |x| + |y| + |x + y| £ 1 is (a) 1/2 (b) 3/4 (c) 1 (d) 4 60. The area of the region bounded by the curves y = [x] and y = {x}, where [,] = GIF and {,} = LIF, is
3.10 Integral Calculus, 3D Geometry & Vector Booster (a) 2 (b) 1 (c) 1/2 (d) 4 61. The smaller area is bounded by the curves |x| + |y| = 1 __ __ and ÷ |x| + ÷ |y| = 1, is (a) 1/3 (b) 2/3 (c) 4/3 (d) 5/3. 62. The area bounded by the curves |x| + |y| = 1 and |x – 1| + |y| = 1 is (a) 1/2 (b) 1/4 (c) 3/2 (d) 3/4 63. The area of the region bounded by the curves x = – 2y2 and x = 1 – 3y2 is (a) 4/3 (b) 3/4 (c) 5/3 (d) 3/5 64. The area enclosed by the parabola (y – 2)2 = x – 1 and the tangent to it at (2, 3) and x-axis is (a) 10 (b) 7 (c) 9 (d) 5 65. The area enclosed by the curves (y – x)2 = x3 and the straight line x = 1 is (a) 5/4 (b) 4/5 (c) 2/3 (d) 3/2
(c) 2 (d) 4 70. A normal to the curve x2 + kx – y + 2 = 0 at the point, whose abscissa ‘1’ is paral let to the line y = x. Then (a) the value of k is – 3 (b) Equation of the normal is x – y = 1 (c) The area bounded by the curve the normal and the x-axis is 7/6 sq u. (d) Equation of tangent is x + y + 1 = 0 71. If the area bounded by the curve y = 3x3 + 2x and the lines x = a and y = 0 is unity, then the value of a is/are __ – 1 2 ___ (a) __ (b) __ 3 3 ÷
(More than one options are correct)
1__ (d) ___ 3 ÷ 72. Let the parabola by y2 = 4x and the solpe of the normal by – 1. Then (a) Equation of the normal is x + y = 3 (b) The points of intersections are (1, 2) and (9, – 6) (c) The area bounded by the curve and its normal is 64/3 sq. units (d) None of these
(Problems for JEE-Advanced)
1. Find the area bounded by the curve a2y2 = x2(a2 – x2) in xy-plane by the method of integration. [Roorkee, 1985]
2. Find the area bounded by the curve y = (x – 1)(x – 3) and x-axis lying between the ordinates x = 0 and x = 3. [Roorkee, 1986] 3. Find the area of the region bounded by the curves y = logex and y = sin 4(px) and x = 0. [Roorkee, 1987]
|x|
)
(b) n = 1 (d) m + n + p = 0 x 2 67. Let the curves y = 4ax, y = ax and y = __ a where 1 £ a £ 2. Then
(a) m = 1 (c) m + n + p = 3
(
)
8 1 (a) Area = __ a5 – __ a 5 (b) Max area = 84 (c) Max area = 75 a (d) Points of intersections are __ , 4 and 4 (4a3, 4a3)
(
{
}
)
[ ]
p 1__ 68. Let f(x) = min tan x, cot x, ___ " x Œ 0, __ . Then 2 3 ÷ the are bounded by the curve y = f (x) and the x-axis is p__ m ____ log __ + sq units, where m, n, p Œ N then n p÷3
(
( )
÷
__
2 (c) – __ 3
– x
66. The area bounded by the curve y = |e – e |, the 1 x-axis and x = 1 is em + __n + p sq unit, then e
(
÷
)
(a) m = 4 (b) n = 4 (c) p = 6 (d) m + n + p = 14 69. The area bounded by the curve y = x – x2 and the line y = m x is a/2 then the value of m is/are (a) 1 (b) – 2
4. Find the area between the curve y = 2x4 – x2, the x-axis and the ordinates of the two minima of the curve. [Roorkee, 1988] 5. Find the area of the portion of the circle x2 + y2 which is exterrior to the parabola y2 = 12x. [Roorkee, 1989] 6. Find the area enclosed between the curves 1 y = loge(x + e), x = loge __ y and the x-axis. [Roorkee, 1990] 7. The line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x – x2. Find the value of m. [Roorkee, 1991]
( )
Area Bounded by the Curves
8. Find the area bounded by the curve y = 2x – x2 and the line y = – x. [Roorkee, 1992] 9. Find the area bounded by the curve y = (x – 1)(x – 2) and the axis of x. [Roorkee, 1993] 10. Find the ratio in which the area bounded by the curves y2 = 12x and x2 = 12y is divided by the line x = 3. [Roorkee, 1994] 11. Find the area given by the curves x + y £ 6. x2 + y2 £ 6y and y2 £ 8x. [Roorkee, 1995] 12. Find the area of the region bounded by the curves x2 + y2 – 6x – 4y + 12 £ 0, y £ x, x £ 5/2. [Roorkee, 1996] 13. Indicate the region bounded by the curves x2 = y, y = x + 2 and x-axis and obtain the area enclosed by them. [Roorkee, 1997] 14. Indicate the region bounded by the curves y = x log x and y = 2x – x2 and obtain the area enclosed by them. [Roorkee, 1998] 15. Find the area of the region lying inside x__2 + (y + 1)2 and outside c2x2 + y2 = c2, where c = ÷ 2 – 1. [Roorkee, 1999] 16. Find the area enclosed by the parabola (y – 2)3 = x – 1, the tangent to the parabola at (2, 3) and the x-axis. [Roorkee, 2000] 17. A point P moves inside a triangle formed by A(0, 0), 1 B 1, ___ __ , C(2, 0) such that {PA, PB, PC} = 1, find 3 ÷ the area bounded by the curve traced by P. 18. Find the area bounded by the curve f(x) = cos–1(cos x), 0 £ x £ 2p with the tangent to the curve f(x) = |cos x| at x = p. 19. Find the area enclosed between the curves |y| = 1 – x2, x2 + y2 = 1. 1 20. Find the area bounded by the curves |y| = e–|x| – __ 2 and |x| + |y| = ln2. 21. Find the area enclosed between the parabolas y2 – 2y + 4x + 5 = 0 and x2 + 2x – y + 2 = 0. 22. Find the area enclosed between the smaller arc of the circle x2 + y2 – 2x + 4y – 11 = 0 and the parabola __ 2 y = – x + 2x + 1 – 2÷3 . 23. Find the smaller of the two areas enclosed between the ellipse 9x2 + 4y2 – 36x + 8y + 4 = 0 and the line 3x + 2y = 10. 24. Let f (x) = x2 + 3x + 2 and g(x) be its inverse. Find the area bounded by g(x), the x-axis and the ordinates x = 1 and x = 4. 25. Find the area bounded by the curve g (x), the x-axis, the ordinates x = – 1 and x = 4, where g (x) is the
( )
13 x3 x2 inverse of the function f(x) = ___ + __ + ____ + 1. 24 8 12x
3.11
26. Find the area bounded by the curves y = xex, y = xe–x and the line x = 1. 27. Find the area of the region bounded by |x + 2y| + |x + 2y| ≥ 8 and xy ≥ 2. 28. Find the area enclosed by |x| + |y| £ 3 and xy ≥ 2. 29. Find the area of the region bounded by |x – y| + |x + y| £ 8 and xy ≥ 2. 30. Find the area of the curve enclosed by |x + y| £ 2 2 + y2 ≥ 2.
(Tougher Problems for JEE-Advanced)
1. Find the area enclosed by 2x2 + 6xy + 5y2 = 1.
2. Find the area enclosed by the curve x2/3 + y2/3 = a2/3.
3. A polynomial function f(x) satisfies the condition f(x + 1) = f(x) + 2x + 1. Find f(x), if f(0) = 1. Find also the equations of the pair of tangents from the origin on the curve y = f(x) and compute the area enclosed by the curve and the pair of tangents. 2 4. Find the area bounded by the curve y = xe–x , the x-axis and the line x = c, where y(x) is maximum. 5. For what value of a, the area bounded by the curve y = a2x2 + ax + 1 and the straight lines x = 0, y = 0 and x = 1 is the least? 6. Find the possible value of a for which the parabola y = x2 + 1 bisects the area of the rectangle with vertices (0, 0), (a, 0), (0, a2 + 1), (a, a2 + 1). 7. Find the area bounded by the curve y = xe–x, xy = 0 and x = c where c is the x-coordinate of the curves inflection point. 8. Find the area of the region bounded by |x| + |y| ≥ 1 and x2 – 2x + 1 £ 1 – y2. 9. Find the area bounded by the curves |x + y| £ 1, |y – x| £ 1 and 3x2 + 3y2 ≥ 1. 10. Find the area bounded by the curves |x – 2| + |y – 2| £ 3, x2 – 4x + y + 3 £ 0 and x2 + y2 + 2x – 9 £ 0. 11. Find the area enclosed by the curves (x2 + y2) £ 4 £ 2(|x| + |y|) 4 – x2 12. Find the area bounded by the curves y = _____ and 4 y = 7 – |x|. 13. Find the area bounded by the curves y = x2 + x – 2, y = 2x for which |x2 + x – 2| + |2x| = |x2 + 3x – 2| is satisfied. 14. Find the area bounded by the curves y2 = 4a (x + a) and y2 = 4b (b – x). 15. Find the area bounded by the parabola y = x2 – 2x + 3, the line tangent to it at the point P(2, – 5) and the y-axis.
3.12 Integral Calculus, 3D Geometry & Vector Booster 16. Find the area enclosed between the circle x2 + y2 – 2x + 4y__– 11 = 0 and the parabola y = – x2 + 2x + (1 – 2÷3 ) . 17. Find the area enclosed by the curves a2x2 + b2y2 = 1 and b2x2 + a2 y2 = 1. 1 18. Find the area bounded by the curves |y| + __ £ e–|x| 2 and max {|x|, |y|} £ 2. 19. Find the area enclosed by the curves y = min {x3, |x – 2|, e3–x}, x-axis, y-axis and x = 4. 20. Find the area of the region bounded by the set S = S3 – S1 – S2, where
S1:{(x, y) : x2 + 2y2 £ 2}
S2:{(x, y) : 2x2 + y2 £ 2} 2
__
4÷2 (a) ____ 3 2 7 3
2
4÷2 (b) ____ 2 2 7 3
__
1. Find the area bounded by the curves x y = sin2x + cos __ , x = – 2 and x = 2 and x-axis. 4 2. Find the area bounded by the curves |x| + |y| = 1. 3. Find the area bounded by the curves |x – 2| + |y – 2| = 1. 4. Find the area bounded by the curve eln(x + 1) ≥ |y|, |x| £ 1. 5. Find the area bounded by the curves y = ln |x| and y = 1 – |x|. 6. Find the area bounded by the curves y = 3 – |x| and y = |x – 1|. 7. Find the area bounded by the curves |x| y = __ x , x π 0 and y = x (x – 1) (x – 3). 8. Find the area bounded by the curves y = ex, x = 0, y = e. 9. Find the area bounded by 1 £ |x – 2| + |y + 1| £ 2. 10. Find the area bounded by the curve |y| = sin (2x), where 0 < x < 2p. 11. Find the area enclosed by the curves
( ) ]
__________
|x + y| + |x – y|, £ 4, |x| £ 1 and y £ ÷x 2 – 2x + 1 . 12. Find the area bounded by the curves 4 £ x2 + y2 £ 2 (|x| + |y|). 13. If the area bounded by the curves 1 1 4 y = __ x , y = ______ , x = 2, x = a is ___ __ , find a. 2x – 1 ÷5
__
4÷2 4÷2 (c) ____ (d) ____ 723 733 2. The area of the region bounded by the curves y = f(x), the x-axis and the lines x = a and x = b, where – • < a < b < – 2, is
Integer type questions
[
assage I P Consider the functions defined implicity y3 – 3y + x = 0 on various interval in the real line. If x Œ (– •, 2) » (2, •), the equation implicity defines a unique real valued differentiable function y = f (x). If x Œ (– 2, 2), the equation implicity define a unique real valued differentiable function y = g (x) satisfying g(0) = 0. On the basis of the above information, answer the following questions. __ __ __ 1. If f (– 10÷2 ) = 2÷2 , then f ¢¢(–10÷2 ) is __
and S3:{(x, y) : x + y £ 2
Comprehensive Link Passages (For JEE-Advanced Examinations)
b
x (a) Ú ___________ dx + bf (b) – af (a) a 3((f (x))2 – 1)
x (b) – Ú ___________ dx + bf (b) – af (a) a 3((f (x))2 – 1)
x (c) Ú ___________ dx – bf (b) + af (a) a 3((f (x))2 – 1)
b
b
b
x (d) – Ú ___________ dx – bf (b) + af (a). a 3((f (x))2 – 1)
1
3. The value of Ú g¢(x) is
(a) 2g (–1) (c) – 2g (1)
–1
(b) 0 (d) 2g (1).
Passage II If x = f (y) and x = g (y) be two functional curves, the area bounded by the curves x = f (y), x = g(y), y = c and y = is given by d
Ú |f(y) – g(y)| dy, c
where d > c. d In case of g(y) always left of f(y), required area = Ú c [ f(y) – g(y)]. On the basis of the above information, answer the following questions. 1. The area bounded by y2 = 2x + 1 and x – y – 1 = 0, will be (a) 16/9 (b) 16/3 (c) 8/3 (d) none.
3.13
Area Bounded by the Curves
2. The area bounded by y = loge x, x-axis and y-axis is given by •
(a) Ú loge y dy
1
0
(b) Ú e–y dy –•
0
(c) Ú e y dy
– •
(d) none
3. The area bounded by y = tan–1 x, y = cot–1 x and y-axis is equal to __
(a) log ÷2
(b) log 4
(c) log 2
(d) none
Passage III Conseider the function 1 Ï : x œI Ô x - [ x] f (x) = Ì , 2 ÔÓ0 : x ŒI where [,] = GIF
and if g(x) = max {x2, f(x), [x]}, where – 2 £ x £ 2 On the basis of the above information, answer the following questions. 1. The area bounded by the curves y = f(x), x-axis and the ordinates x = 1/2 and x = 1 is (a) 1/4 (b) 1/8 (c) 1/2 (d) none. 2. The area bounded by y = g(x), x-axis and the straight line x = 1 is (a) 1/4 (b) 1/2 (c) 1/8 (d) none 3. The area bounded by y = g(x), where – 2 £ x £ 2, is (a) 175/48 (b) 275/48 (c) 175/24 (d) 275/24. Passage IV A normal to the curve x2 + kx – y + 2 = 0 at the point whose abscissa 1 is parallel to the line y = x. On the basis of the above information, answer the following questions. 1. The value of k is (a) –3 (b) 1 (c) 0 (d) 2 2. The equation of the normal is given by (a) y = x + 1 (b) y = x – 1 (c) y = – x + 1 (d) y = – x – 1. 3. The area in the first quadrant bounded by the curves, the normal and the x-axis is (a) 13/6 (b) 7/6 (c) 10/3 (d) 43/6.
Passage V
_____
Consider x 2 – 4, g(x) = |x – 2| and h(x) _____ the functions f (x) = ÷ = ÷x – 2 for x Œ R, a function is defined as F(x) = max or min {f (x), g(x), h(x)}. On the basis of the above information, answer the following questions. 1. The area of F (x) = min {f(x), g(x), h(x)} between the co-odinates axes for x < 0 is (a) 2p (b) p (c) 4p (d) none 2. The area enclosed by F(x) = min{ f(x), g(x), h(x)} and F(x) = max{ f(x), g(x), h(x)} for x Œ [0, 2] is (a) p (b) p + 2 (c) p – 2 (d) p – 1. 3. The area bounded by the curves F(x) = min{f (x), g(x), h(x)}, x = 2, y = 0 and x = 3, is (a) 3/2 (b) 1/2 (c) 1 (d) none
Matrix Match (For JEE-Advanced Examination Only)
1. Match the following Columns: Column I
Column II
(A) The area bounded by y = 3x and (P) y = x2 is
2/3
(B) The area bounded by x = 4 – y2 (Q) and the y-axis is
17/12
(C) The area in squares units of the2 (R) region bounded by the curve x = 4y and the line x = 2 and the x-axis is
32/3
(D) The area bounded by y = x3, (S) y = x2 and x = 1, x = 2 is (T)
32/9 9/2
2. Match the following Columns: Column I
Column II
(A) The area enclosed by the curve (y (P) – sin–1 x)2 = x – x2 is
0
(B) The area of the region represented (Q) by __ 2 £ |x + y| + |x – y| £ 2 is ÷
1
(C) If the area bounded by (R) y = x2 – 3 and the line y = ax + 2 attains its maximum value, the value of a is
p
(D) If k is a positive number and the area (S) of the region bounded by the curves y = x – kx2 and k y = x2 attains its (T) maximum value, the value of k is
6 p/4
3.14 Integral Calculus, 3D Geometry & Vector Booster
3. Match the following Columns: Column I
Column II
(A) The area of the figure (P) bounded by y = tan x, p p y = tan2 x, – __ £ x £ __ 2 2 is
4p
(B) The area bounded by (Q) the curve y = x sin x and the x-axis, x = 0 and x = 2p is
p 1 – __ 4
(C) The area bounded by (R) the curves x = tan–1 __ , x = 0, x = p/4 and ÷y y = 0 is
__
2(p + 1) + ÷ 2
(
)
(
Column II
(A) The area bounded by the (P) curves y = |sin x|, x-axis and the lines |x| = p.
2 s.u.
(B) The area bounded by the (Q) curves y = sin –1 x, y-axis, |y| = p/2
1 s.u.
(C) The area bounded by the (R) x2 curves y = ___ + 2 , 64 [,] = GIF, y = x – 1, x = 0 is
8 s.u.
(D) The area bounded by the (S) curves y = |x – 1| and y = 3 – |x|, is
4 s.u.
[
)
)
)
Questions asked in past IIT-JEE Examinations
4. Match the following Columns: Column I
(
(
p (D) The area bounded by (S) log 2 – 2 + __ e 2 the curves f(x) = max {1 + sin x, 1, 1 – cos x}, between the co-ordinates x = – p and x = p is
2. Assertion (A): The area bounded by the curve y = [x] n(n – 1) and the lines x = 0 and x = n is _______ s.u. 2 Reason (R): The area bounded by the curve y = [x] and the lines x = 0 and x = 10 is 45 s.u. 3. Assertion (A): The area bounded by the curves f(x, y) p 1 = {(x, y) : x2 + y2 £ a2, x + y ≥ a} is __ – __ s.u. 4 2 Reason (R): The area bounded by the curves f(x, y) = {(x, y) : x2 + y2 £ 1, x + y ≥ 1,} p 1 is __ – __ s.u. 4 2 4. Assertion (A): The area bouned by the curves y = min {|x + 2|, |x|, |x – 2|} is 2 s.u. Reason (R): The area bounded by the curves y = 1 – |x| and y = |x| – 1 is 2 s.u. 5. Assertion (A): The area of the region bounded by the curves y = |x – 1| and y = 3 – |x| is 4 s.u. Reason (R): The area of the region bounded by the curves y = ln x, y = ln |x|, y = |ln x| and y = | ln|x| | is 4 s.u.
]
Assertion and Reason (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true. 1. Assertion (A): The area bounded by the curve |x| + |y| = 1 is 1 s.u. Reason (R): The area bounded by the curve |x – 2| + |y – 2| = 1 is also 1 s.u.
1. Find the area bounded by the curve x2 = 4 and the straight line x = 4y – 2. [IIT-JEE, 1981] 2. Find the area enclosed by |x| + |y| = 1. [IIT-JEE, 1981] 3. Find the area bounded by the curve y = f(x), x-axis and the co-ordinates x = 1 and x = b is (b – 1) sin (3b + 4). Then f (x) is (a) (x – 1) cos (3x + 4) (b) sin (3x + 4) (c) sin (3x + 4) (x – 1) cos (3x + 4) (d) none. [IIT-JEE, 1982] 4. Find the area bounded by the x-axis, part of the curve 8 y = 1 + __ 2 and the ordinates at x = 2 and x = 4. x If the co-ordinates at x = a divides the area into two equal parts, find a. [IIT-JEE, 1983]
(
)
5. Find the area of the region bounded by the x-axis and the curves defned by
p p Ï ÔÔ y = tan x : - 3 £ x £ 3 Ì Ô y = cot x : p £ x £ p ÔÓ 6 2
[IIT-JEE, 1984] _____
6. Find the area bounded by the curves y = ÷ 5 – x2 and y = |x – 1| [IIT-JEE, 1985] 7. Find the area bounded by the curves x2 + y2 = 4, __
x2 = – ÷2 y and y = x.
[IIT-JEE, 1986]
Area Bounded by the Curves
8. Find the area bounded by the curves x2 + y2 = 25, 4y = |4 – x2| and x = 0 above the x-axis. [IIT-JEE, 1987]
9. Find the area of the region bounded by the curve p y = tan x, tangent drawn to the curve at x = __ and 4 the x-axis. [IIT-JEE, 1988]
10. Find all maxima and minima of the function y = x (x – 1)2 : 0 £ x £ 2. Also, determine the area bounded by the curve y = x (x – 1)2, the y-axis and the line y = 2. [IIT-JEE, 1989] 11. Find the area of the region bounded by the curves ln x y = e x ln x, y = ___ e x and x = 1. [IIT-JEE, 1990] 12. Sketch the curves and identify the region bounded 1 by y = ln x, y = 2x, x = __ and x = 2. Find the area 2 of the region. [IIT-JEE, 1991] 13. Find the area bounded by the curves y = x2 and 2 y = _____ 2 . [IIT-JEE, 1992] 1+x 14. The area of the region bounded by the curves y = |x – 1| and y = 1 is (a) 1 (b) 2 (c) 1/2 (d) none [IIT-JEE, 1993] 15. In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x? [IIT-JEE, 1994] 16. Consider a square with vertices at (1, 1), (–1, 1) (–1, –1) and (1, –1). Let S be the region consisting of all points inside the square which are near to the origin than to any edge. Sketch the region S and find its area. [IIT-JEE, 1995] 17. The slope of the tangent to the curve y = f(x) at a point (x, y) is (2x + 1) and the curve passes through (1, 2). The area of the region bounded by the curve, the x-axis and the line x = 1 is (a) 5/3 (b) 5/6 (c) 6/5 (d) 6 [IIT-JEE, 1996] 18. Let An be the area bounded by the curve y = tann x, p y = 0, x = 0, x = __ . If n ≥ 2, then (An + An+2) is 4 1 1 (a) ______ (b) __ n (n + 1) 1 (c) ____ (d) none n–1 [IIT-JEE, 1997] 19. Find all possible values of b > 0 so that the area of the region bounded between the parabolas y = x – bx2 x2 and y = __ is maximum. [IIT-JEE, 1997] b
3.15
( )
1__ 20. Let O (0, 0), A (2, 0), B 1, ___ be the vertices of a 3 ÷ triangle. Let R be the region consisting of all those points P inside D OAB, which satisfy d (P, OA) ≥ min {(P, OB), d (P, AB)} where d denotes the distance from the point to the corresponding line. Sketech the region R and find its area. [IIT-JEE, 1997] 2 2 21. Let f (x) = max {x , (1 – x) , 2x (1 – x)} where 0 £ x £ 1. Determine the area of the region bounded by the curves y = f (x), x-axis, x = 0, x = 1. [IIT-JEE, 1997] 22. Let C1 and 2 be the graphs of the functions y = x2, y = 2x, 0 £ x £ 1 respectively. Let C3 be the graph of a function y = f (x), 0 £ x £ 1, f (0) = 0.
For a point P on C1, let the lines through P parallel to the axes meet C2 and C3 at Q and R, respectively. If for every position of P on C1, the area of the shaded regions OPQ and ORP are equal, determine the function f (x). [IIT-JEE, 1998] 23. For what values of m, the area of the region bounded by the curves y = x – x2 and the line y = m x equals 9/2? (a) – 4 (b) – 2 (c) 2 (d) 4 [IIT-JEE, 1999] 24. Let f(x) be a continuous function given by : | x| £ 1 ÏÔ2 x f (x) = Ì 2 . ÔÓ x + ax + b : | x | > 1 Find the area of the region in the third quadrant bounded by the curves x = – 2y2 and y = f (x) lying in the left of the line 8x + 1 = 0. [IIT-JEE, 1999] No questions asked in 2000. 25. Let b π 0 and for j = 0, 1, 2, ..., n, let S1 be the area of the region bounded by the y-axis and the curve ( j + 1)p jp x exy = sin by, __ £ y £ _______ . Show that S0, S1, b b ..., Sn are in GP. Also, find their sum for a = – 1 and b = – 1 and b = p. [IIT-JEE, 2001]
3.16 Integral Calculus, 3D Geometry & Vector Booster 26. The area bounded by the curves y = |x| –1 and y = 1 – |x| is (a) 1 __ (b) 2 (c) 2÷2 (d) 4 [IIT-JEE, 2002] __ 27. The area bounded by the curves y = ÷ x , 2y + 3 = x and x-axis in the first quadrant is (a) 9 (b) 27/4 (c) 36 (d) 18 [IIT-JEE, 2003] 28. Find the curve passing through (2, 0) and having slope
(
)
(x + 1)2 + (y – 3) of tangent at any point P (x, y) as ______________ . (x + 1) Find also the area enclosed by the curve and x-axis in the fourth quadrant [IIT-JEE, 2004] 29. The area enclosed between the curves y = ax2 and x = ay2__ (a > 0) is 1 sq.u., the value of a is (a) 1/÷3 (b) 1/2 (c) 1 (d) 1/3 [IIT-JEE, 2004] 30. The area bounded by the curves y = (x + 1)2 and y = (x – 1)2 and the line y = 1/4 is (a) 4 s.u. (b) 1/6 s.u. (c) 4/3 s.u. (d) 1/3 s.u. [IIT-JEE, 2005] 31. Find the area bounded by the curves x2 = y, x2 = – y, and y2 = 4x – 3 [IIT-JEE, 2005] 32.
33.
È 4 a 2 4 a 1˘ È f (-1) ˘ È3a 2 + 3a ˘ Í ˙ Í ˙ If Í 4b2 4b 1˙ ÍÍ f (1) ˙˙ = Í3b2 + 3b ˙ Í 2 ˙ Í ˙ 4c 1˙˚ ÍÎ f (2) ˙˚ ÍÎ3c 2 + 3c ˙˚ ÍÎ 4c and f(x) is a quadratic function and its maximum value occurs at a point V. A is a point of intersection of y = f (x) with x-axis and point B is such that the chord AB subtends a right angle at V. Find the area enclosed by f (x) and chord AB. [IIT-JEE, 2005] No questions asked in between 2006 to 2007. The area of the region between the curves _______
÷
_______
÷
1 + sin x 1 – sin x y = _______ and y = _______ bounded by the cos x cos x p lines x = 0 and x = __ is __ 4 ÷2 – 1 t _____ (a) Ú _____________ dt 0 (1 + t2) ÷1 – t2
(
( Ú (
__
÷2 –1
) )
dt _____ (b) Ú _____________ 2 0 (1 + t )÷1 – t2 __
+ 1 ÷2
)
4t dt______ (c) _____________ 0 (1 + t2)÷1 – t2
__
÷2 + 1
(
)
t dt______ (d) Ú _____________ 2 0 (1 + t )÷1 – t2
[IIT-JEE, 2008]
Comprehension Consider the functions defined implicitly by the equation y2 – 3y + x = 0 on varoius intervals in the real line. If x Œ (– •, – 2) » (2, •), the equation implicitly defines a unique real-valued differentiable function y = f(x). If x Œ (– 2, 2), the equation uniquely implicity defines a unique real-valued differentiable function y = g(x) satisfying g(0) = 0 __ __ __ 34. If f (– 10÷2 ) = 2÷2 , then f ¢¢(– 10÷2 ) = __
__
4÷2 (a) ____ 73 32
4÷2 (b) – ____ 3 2 7 3
__
__
4÷2 4÷2 (c) ____ (d) – ____ 3 3 73 73 35. The area of the region bounded by the curve y = f (x), the x-axis and the lines x = a and x = b, where – < • < a < b < – 2, is
( Ú ( Ú ( Ú ( b
)
x (a) Ú _________ dx + bf (b) – af (a) a 3(f (x)2 – 1
x (b) – __________ dx + bf (b) – af (a) a 3(f (x)2 – 1)
b
) ) )
b
x (c) __________ dx – bf (b) + af (a) a 3(f (x)2 – 1)
x (d) – __________ dx – bf (b) + af (a) a 3(f (x)2 – 1)
b
1
36. Ú g¢(x) dx = –1
(a) 2 g (–1)
(b) 0
(c) – 2 g(1)
(d) 2 g (1) [IIT-JEE, 2008]
37. Area of the region bounded by y = ex and x = 0 and x = e is e
(b) Ú log (1 + e – y) dy
(a) e – 1 1
(c) e – Ú ex dx 0
1 e
(d) Ú ln y dy 1
[IIT-JEE, 2009]
Consider the polynomial f (x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f (x) and let t = |s|. 38. The real number s lies in the interval 3 1 (a) – __ , 0 (b) –1, __ 4 4
(
)
( )
Area Bounded by the Curves
(
( )
)
3 1 (c) – __ , – __ 4 2
1 (d) 0, __ 4
39. The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = , lies in the interval
( ) 21 (d) ( 0, ___ ) 64
( )
3 (a) __ , 3 4
(c) (9, 10)
21 11 (b) ___ , ___ 64 16
40. The function f ¢(x) is
( (
)
( ) ( )
1 1 (a) inc in – t, – __ and dec. in – __ , t 4 4 1 1 (b) dec. in – t, – __ and inc. in – __ , t 4 4 (c) inc in (– t, t) (d) dec in (– t, t) [IIT-JEE, 2010] 41. Let f : [– 1, 2] Æ [0, •) be a continuous function such that f (x) = (1 – x) for all x Œ [– 1, 2]. Let
)
2
R1 = Ú x f (x) dx and R2 be the area of the region –1
bounded by y = f (x), x = – 1 and x = 2 and the x-axis, Then (a) R1 = 2R2 (b) R1 = 3R2
(c) 2R1 = R2
(d) 3R1 = R2 [IIT-JEE, 2011]
42. Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts R1 (0 £ x £ b) and R2 (b £ x £ 1) such that R1 – R2 1 = __ , Then b is 4 (a) 3/4 (b) 1/2 (c) 1/3 (d) 1/4 [IIT-JEE, 2011] 2 43. Let S be the area of the region enclosed by y = e –x , y = 0, x = 0 and x = 1. Then 1 1 (a) S ≥ __ e (b) S ≥ 1 – __ e 1 1 (c) S ≥ __ 1 + ___ __ e ÷ 4 1 1 1__ __ + ___ (d) S ≥ ___ __ 1 – ___ [IIT-JEE, 2012] e ÷ 2 2 ÷ ÷ 44. The area of the region enclosed by the curves y = sin x + cos x and y = |cos x – sin x| over the p interval 0, __ is 2
(
)
(
)
[ ]
__
__
__
(a) 4(÷__ 2 – 1) (b) 2÷__ 2 ( ÷ __ 2 – 1) (c) 2(÷2 + 1) (d) 2÷2 ( ÷ 2 + 1) [IIT-JEE, 2013] No questions asked in 2014, 2015. 45. Area of the region ______ {(x, y) : y £ ÷|x + 3|, 5y £ x + 9 £ 15} is (a) 1/6 (b) 4/3 (c) 3/2 (d) 5/3 [IIT-JEE, 2016]
Answers Level-II
1. (c) 2. (d) 3. (b) 6. (c) 7. (c) 8. (c) 11. (b) 12. (b) 13. (a) 16. (a) 17. (d) 18. (b) 21. (d) 22. (c) 23. (a) 26. (d) 27. (b) 28. (c) 31. (c) 32. (a) 33. (a) 36. (a) 37. (a) 38. (b) 41. (d) 42. (a) 43. (a) 46. (a) 47. (c) 48. (c) 51. (c) 52. (a) 53. (c) 56. (a) 57. (a) 58. (c) 61. (b) 62. (a) 63. (a) 66. (a, b, d) 67. (a, b, d) 69. (b, d) 70. (a, b, c, d)
4. (d) 5. (a) 9. (d) 10. (d) 14. (b) 15. (a) 19. (c) 20. (a) 24. (b) 25. (c) 29. (d) 30. (a) 34. (c) 35. (a) 39. (a) 40. (a) 44. (a) 45. (c) 49. (b) 50. (c) 54. (a) 55. (b) 59. (b) 60. (b) 64. (c) 65. (d) 68. (a, b, c, d) 72. (b, c) 72. (a, b, c)
6. 4 11. 2
1. 4
2. 2
3. 2
7. 2 12. 8
8. 1 13. a = 8
9. 6
10. 4
Comprehensive Link Passages Passage Passage Passage Passage Passage
I : II : III : IV : V :
1. (b) 1. (b) 1. (b) 1. (a) 1. (b)
2. (a) 2. (c) 2. (a) 2. (b) 2. (c)
3. (d) 3. (c) 3. (b) 3. (b) 3. (b)
Matrix Match
1. 2. 3. 4.
(A)Æ(T), (A)Æ(T), (A)Æ(S), (A)Æ(S),
(B)Æ(R), (B)Æ(S), (B)Æ(P), (B)Æ(P),
(C)Æ(P), (C)Æ(P), (C)Æ(Q), (C)Æ(S),
(D)Æ(Q) (D)Æ(Q) (D)Æ(R) (D)Æ(S)
Assertion and Reason
INTEGER TYPE QUESTIONS
4. 4
5. 3
3.17
1. B
2. A
3. A
4. C
5. B
3.18 Integral Calculus, 3D Geometry & Vector Booster
Hints
and
2
= Ú y dx – 1
– 1
(
)|
=0 7. Hence, the required area
2
(
–1
x3 = __ – x2 + 2x 3 –1 8 1 = __ – 4 + 4 + __ + 1 – 2 3 3
2p
)
2. Hence, the required area 2
= Ú (ln x + tan x) dx
( (
)|
2 1 = (x log x – x tan x – __ log |1 + x2| 2 1 p 1 = (2 log 2 – 1) + 2tan–12 – __ log 5 – __ 2 4 p 1 = 2 (log 2 + tan–1 2) – __ log 5 – __ + 1 s.u. 2 4
x)|21 +
(
–1
)
(
))
3. Hence, the required area
0 p
2p
0
p
= Ú (x sin x)dx – Ú (x sin x)dx
= (– x cos x + sin x)|p0 – (– x cos x + sin x)|2p p = p – (– 2p – p) = 4p. sq.u. 4. Hence, the required area 2
= Ú (log1/2x) dx
1 = – ____ (x log x – x)|21 log2 1 = – ____ (2log 2 – 3) log2 1 = ____ (3 – 2 log2) sq.u. log2 5. Hence, the required area
log2
= Ú (e2x – 3ex + 2)dx 0
(
p
(– cos x)|p0 –
(– cos x)|2p p
= – (cos x)|p0 + (cos x)|2p p
= – (–1 – 1) + (1– (–1)) = 4 sq.u. 8. Hence, the required area 2p
= Ú cos x dx
0
p/2
3p/2
2p
0
p/2
3p/2
= Ú cos x dx – Ú cos x dx + Ú cos x dx (sin x)|p/2 – 0
(sin x)|3p/2 + p/2
)|
log2
e2x = ___ – 3ex + 2x 0 2
(sin x)|2p 3p/2
=
= (1 – 0) – (–1 – 1) + (0 – (–1)) = 4 sq.u.
9. Hence, the required area 2
= Ú {x (x – 1) (x – 2)}dx 0 1
2
= Ú {x(x – 1) (x – 2)}dx – Ú {x (x – 1) (x – 2)}dx 0 1
1
0
=
= Ú (x|sin x|) dx
2p
2p
0 p
= Ú sin x dx – Ú sin x dx
–1
= Ú sin x dx
= 3 + 1 – 2 = 2 sq.u.
1
)
= Ú x3dx
= Ú (x2 – 2x + 2) dx
)
1
2
( (
1 = 4 – 6 + 2 log 2 + 3 – __ 2 1 = __ + 2 log 2 sq.u. 2 6. Hence, the required area
1. Hence, the required area
solutions
1
2
= Ú (x3 – 3x2 + 2x)dx – Ú (x3 – 3x2 + 2x)dx 0
(
) | ( 1
4
1
)|
2
4
x x = __ – x3 + x2 – __ – x3 + x2 0 4 4 1 1 1 __ __ = – 1 + 1 – (4 – 8 + 4) – – 1 + 1 4 4 1 = 2 __ – 1 + 1 4 1 = __ sq.u. 2 10. Hence, the required area
(
) ( )
(
3
(
= Ú {(x – 1) (x – 2) (x – 3)}dx 0
))
Area Bounded by the Curves
3.19
1
= – Ú {(x – 1) (x – 2) (x – 3)}dx 0
2
+ Ú {(x – 1) (x – 2) (x – 3)}dx 1 3
– Ú {(x – 1) (x – 2) (x – 3)}dx
(
2
)|
1
x4 11 = – __ – 2x3 + ___ x2 – 6x 2 0 4
( (
) )
2 x4 11 + __ – 2x3 + ___ x2 – 6x 4 2 1 3 x__4 11 – – 2x3 + ___ x2 – 6x 4 2 2 1 11 = – __ – 2 + ___ – 6 + (4 – 8 + 22 – 12) 4 2 1 11 – __ – 2 + ___ – 6 4 2 99 81 ___ – – 54 + ___ – 18 + (4 – 8 + 22 – 12) 4 2 279 23 = – 2 ___ – 8 + 12 – ____ – 72 4 4 46 279 = ___ – ____ + 100 4 4 233 400 – 233 ____ 167 = 100 – ____ = _________ = sq.u. 4 4 4 11. Given curves are y = x2 and y = 2x – x2
( ( ( (
)
)
)
) (
)
Hence, the required area 1
= Ú (y2 – y1) dx 0 1
__
2 = Ú (÷x – x ) dx 0
(
)|
1
x3 2 = __ x3/2 – __ 3 0 3 2 1 1 = __ – __ = __ sq.u. 3 3 3
(
)
13. Do yourself. 3x2 14. Given curves are y = ___ and 3x – 2y + 12 = 0 4
Hence, the required area 4
= Ú (y2 – y1) dx – 2
Hence, the required area 1
= Ú (y2 – y1)dx 0
1
= Ú (2x – x2 – x2)dx 0
4
= Ú (2x – 2x2)dx 0
( )| 1 2 = ( 1 – __ ) = __ sq.u. 3 3 1 2 = x2 – __ x3 3 0
12. Given curves are y2 = x and x2 = y.
)
3x + 12 ___ 3x3 = Ú _______ – dx 2 4 – 2
3 = __ Ú (2x + 8 – x2) dx 4 – 2
3 x3 4 = __ x2 + 8x – __ 4 3 – 2
4
1
(
(
)
54 8 ___ ) – ( 4 – 16 + __ ) ) ( ( 3 3 ( ) 3 108 – 62 = __ ( ________ ) = ___ 464 = ___ 232 sq.u. 4 3
3 = __ 16 + 32 – 4 3 62 = __ 36 – ___ 4 3
15. Given curves are x2 = 4y and x = 4y – 2
3.20 Integral Calculus, 3D Geometry & Vector Booster
Hence, the required area
Hence, the required area
2
4a
= Ú (y2 – y1) dx
–1 2
0
(
)
x2 __ dx 4
4a
x+2 = Ú _____ – 4 –1
1 = __ Ú (x + 2 – x2) dx 4 –1
(
__ __
= Ú (2÷a ÷x – mx) dx 0
2
= Ú (y2 – y1) dx
4a
)
x3 2 1 x2 = __ __ + 2x – __ 4 2 3 –1 8 1 1 1 = __ 2 + 4 – __ – __ – 2 + __ 4 3 2 3
) ( )) ( ( 9 7 1 10 __ 1 27 = __ ( ___ + ) = __ ( ___ = __ sq.u. 4 3 4 6) 8 6 2
( ( ) ( ) ) ( ( ) ( ) )
2
__ __ x2 2 3/2 __ = 2÷a x – m 3 2 0
__ __ 16a2 2 3/2 ____ = 2÷a (4a) – m 3 2
32 = ___ – 8m a2 sq.u. 3
(
)
18. Given curves are y = x2 + 2, y = x, x = 0 and x = 3.
16. Given curves are y = 4 ax and x = 4 ay.
Hence, the required area
Hence, the required area
4a
3
= Ú (y2 – y1) dx 0
4a
= Ú (y2 – y1) dx 0
(
) ( ) ( ) ( ) ( )
__ __ x2 ___ = Ú 2÷a ÷x – dx 4a 0
x3 4a 2 3/2 4a __ ____ = 2÷a x – 3 12a 0 0
3 x3 = __ + 2x – x2 3 0
27 = ___ + 6 – 9 = 6 sq.u. 3
__
3
0
3
__ __ 64a 2 3/2 ____ = 2÷a (4a) – 3 12a 2
2
2
32a 16a 16a = ____ – ____ = ____ sq.u. 3 3 3
17. Given curves are y2 = 4ax and y = mx.
= Ú (x2 + 2 – 2x) dx
( (
)
)
19. Given curves are
y = 2x – x2 and y = – x.
Area Bounded by the Curves
Hence, the required area 3
= Ú (y2 – y1) dx 0
We have 6x – x2 = x2 – 2x. 2x2 = 8x x = 0, 4 Hence, the required area 4
3
= Ú (2x – x + x) dx 2
4
3
= Ú (3x – x2) dx
(
)
3x2 x3 3 = ___ – __ 2 3 0
(
)
27 = ___ – 9 = 2
= Ú (6x – x2 – x2 + 2x) dx 0 4
0
= Ú (y2 – y1) dx 0
0
= Ú (8x – 2x2) dx 0
9 __ sq.u. 2
20. Given curves are y = 2x – x2, y = 2x and x = 0, x = 2
( ) 128 64 = ( 64 – ____ = ___ sq.u. 3 ) 3 4 2 = 4x2 – __ x3 3 0
22. Let A1 be the area of the curve y = 4x – x2 with x-axis and A2 be the area of the curve y = x2 – x with x-axis 4
A1 = Ú y dx 0
4
= Ú (4x – x2) dx 0
Hence, the required area = Ú (y2 – y1) dx
64 32 = 32 – ___ = ___ 3 3
and
= Ú (2x – x – 2 ) dx 2
x
)
1
)
2x 2 ____ log 2 0
x3 = x – __ – 3
8 4 1 = 4 – __ – ____ + ____ 3 log 2 log 2
3 4 = __ – ____ sq.u. 3 log 2
)
2
( (
= Ú (x – x2) dx 0
)
(
x2 = __ – 2
(
)
x3 1 __ 3 0
)
1 1 = __ – __ = 2 3 Hence, the required ratio,
21. Given curves are 2
A2 = Ú y dx 0
0
( 1
2
(
)
x3 4 = 2x2 – __ 3 0
0
(
2
3.21
2
y = 6x – x and y = x – 2x.
A1 ___ = A2
32 ___ : 3
1 __ = 6
1 __ 6 32 ___ × 3
6 __ = 64 : 1 1
3.22 Integral Calculus, 3D Geometry & Vector Booster 26. Given curves are {(x, y) : x2 + y2 £ 1 £ x + y}.
23. Given curves are
x2 + y2 = 4 and (x – 2)2 + y2 = 4.
Hence, the required area 1
Hence, the required area
(
1
2
__________
1
(
( ) )
(x – 2) __________2 __ x–2 1 4 = 2 _____ ÷4 – (x – 2) + sin–1 ____ 2 2 2 0
(
_____
x + 2 __ ÷ 4 – x2 + 2
)
(
__
( ) )
x 2 4 __ sin–1 __ 2 2 1
( )
)
3 ÷ 1 = 2 – ___ + 2sin–1 – __ – 2sin–1(–1) 2 2
(
__
( ) )
3 ÷ 1 + 2 sin (1) – ___ – 2sin–1 __ 2 2
(
–1
__
__
_____
= Ú (÷1 – x2 – (1 – x)) dx 0
(
_____
( )
3 p 3 p ÷ p ÷ = 2 – ___ – __ + p + __ – ___ – __ 2 3 2 2 3
__ 3p 2p = 2 ___ – ___ – ÷ 3 2 3
__ 5p = ___ – 2÷3 sq.u. 3
( (
( ) ) ( ) )
(
)
)
(
)
24. Do yourself. x2 25. Given curves are __ 2 + a
y2 x __ 2 = 1 and __ a + b
y __ = 1. b
Solving y2 = 4x and 4x2 + 4y2 = 9 we get. 4x2 + 16x – 9 = 0 fi 4x2 – 2x + 18x – 9 = 0 fi 2x (2x – 1) + 9 (2x – 1) = 0 fi (2x + 9) (2x – 1) = 0 1 9 fi x = __ , – __ 2 2 Hence, the required area
Hence, the required area
pab = ____ – 4
(
1 __ ab 2
1 p = __ __ – 1 sq.u. 2 2
)
)
1
x÷1 – x2 __ x x2 1 + sin–1 __ – x + __ = _______ 2 2 2 2 0 1 –1 __ 1 1 __ __ = sin – 1 + 2 2 2 1 1 1 = __ sin–1 __ – __ 2 2 2 1 p = __ __ – 1 sq.u. 2 6 27. Given curves are {x : y2 £ 4x, 4x2 + 4y2 £ 9}.
)
(
= Ú (C – L) dx 0 1
= 2 Ú ÷4 – (x – 2)2 + Ú ÷4 – x2 dx 0
_____
1/2
3/2
0
1/2
= Ú P dx + Ú E dx 1/2
1/2
3/2
______
__ 1 __ = Ú 2÷x dx + Ú ÷9 – 4x2 dx 2 1/2 0 3/2
÷( )
________
3 2 1 2 __ __ = Ú 2÷x dx + Ú – x dx 2 1/2 2 0 __
Area Bounded by the Curves
(
( ÷( ) ) ________
)
1/2 3 2 2 1 x __ = 2 × __ × x2/3 + __ __ – x2 3 2 2 2 0
( )
9 2x 3/2 + __ sin–1 ___ 8 3 1/2
3.23
27 1 = __ (3 + 2 + 21) = ___ sq.u. 2 2 31. Given curves are y = 1 + |x + 1|, x = – 3, x = 3, y = 0.
9 1 4 1 = __ × ____ __ + __ × __ sin–1(1) 3 2÷2 8 2 __ 9 1 ÷2 1 – __ ___ + __ sin–1 __ 2 4 8 3 9p 9 2 1 1 = ____ __ + ___ – ____ __ – ___ sin–1 __ 3 3÷2 32 4÷2 16
(
(
( ) )
( )
( ) )
9p 9 5 1 = _____ __ + ___ – ___ sin–1 __ sq.u. 3 16 12÷2 32 28. Given curves are {x : x2 + y2 £ 2ax, y2 ≥ ax, x ≥ 0, y ≥ 0}
Hence, the required area 1 1 1 = __ × (3 + 1) × 2 + __ × (1 + 2) × 1 + __ × (2 + 5) × 3 2 2 2 28 1 = __ (4 + 3 + 21) = ___ = 14 2 2 32. Given curves are y = |x – 1| and y = – |x – 1| + 1.
Hence, the required area a
= Ú (y2 – y1) dx 0 a
(
__________
__
)
= Ú ÷a 2 – (x – a)2 – a÷x dx
Hence, the required area 1 1 1 __ __ __ a 2a – a = __ sin–1 x_____ – ___ x3/2 = 2 2 × 1 × 2 = 2 sq.u. ÷ a 2 2 3 0 33. Given curves are 2a __ __ a2 –1 = ___ a÷a – sin (–1) f (x) = log x and g (x) = (log x)2 3 2 0
(
__________ x–a 2 _____ a – (x – a)2 +
(
2
(
)
(
)
a
)
)
pa2 2a5/2 ____ = ____ + sq.u. 3 4 29. Do yourself.
Y
30. Given curves are y = 1 + |x + 1|, x = – 2, x = 3, y = 0.
X¢
O
X
Y¢
Hence, the required area e
= Ú (log x – (log x)2)dx 1
Hence, the required area 1 1 1 = __ × (2 + 1) × 1 + __ × (1 + 2) × 1 + __ × (2 + 5) × 3 2 2 2
34.
= (x log x – x – (x (log x)2 – 2 (x log x – x)))e1 = ((3 (x log x – x) – x (log x)2))e1 = (3 – e) sq.u. Given curve is y2 = x2 – x4.
3.24 Integral Calculus, 3D Geometry & Vector Booster Hence, the required area 1
2
= Ú (x2 + 1)dx + Ú (x + 1) dx 0
Hence, the required area 1
= 4 Ú (x – x ) dx 2
4
0
(
)
x5 1 __ 5 0
x3 = 4 __ – 3
8 1 1 = 4 __ – __ = ___ sq.u. 3 5 15
(
1
( ) ( ) 3
1
2 x __ + x 2 1 2
x = __ + x + 3 0
3 1 = __ + 1 + 4 – __ 3 2
(
) ( )
4 5 = __ + __ = 3 sq.u. 3 3 37. Given curves are
y = x and y = x3.
)
35. Given curve is
1 |y| + __ £ e– |x| 2
Hence, the required area 1
= 2 Ú (x – x3) dx 0
(
)
= 4 Ú e dx – x
0
log 2 = 4 ( – e– x )0
= 4 ( 1 – e– log 2 )
)
x2 x4 1 = 2 __ – __ 2 4 0
1 1 2 = 2 __ – __ = 2 × __ = 1 sq.u. 2 4 4
Hence, the required area log 2
(
(
)
38. Given curve is y = x (x – 1)
( )
1 1 = 4 1 – __ = 4 × __ = 2 sq.u. 2 2 36. Given curve is
f (x, y) = [(x, y) : 0 £ y £ x2 + 1,
0 £ y £ x + 1, 0 £ x £ 2] Hence, the required area 2
= Ú (2 – x (x – 1)2) dx 0
2
= Ú (2 – x (x2 – 2x + 1)) dx 0
Area Bounded by the Curves 2
Hence, the required area
0
= Ú (2 – (x3 – 2x2 + x)) dx
(
3
)
x4 2 3 __ x2 2 = 2x – __ + __ x – 4 3 2 0 16 = 4 – 4 + ___ – 2 3 10 ___ = sq.u. 3 39. Given curves are y = x2 + 1, y = x, x = 0 and y = 2.
(
)
= Ú (x2 – x1) dy 0
3
= Ú ((y – 2)2 + 1 – (2y – 4)) dy 0
3
= Ú (y2 – 4y + 4 + 1 – 2y + 4) dy 0
3
= Ú (y2 – 6y + 9)dy 0
(
)
3 y3 = __ – 3y2 + 9y 3 0
= 9 – 27 + 27
= 9 sq.u. 42. Given curve is y2 = (x – 1) (x – 2)2 40. Given curves are y = loge x and y = 2x, x = 1/2 and x = 2.
Hence, the required area 2
_____
= 2 Ú (x – 2)÷x – 1 dx 1
1
Hence, the required area 2
= Ú (2x – log x) dx
1/2
(
)
x
2
2 = ____ – (x log x – x) log 2 1/2
( (
__
( )
)
2 ÷ 4 1 1 1 = ____ – 21 log 2 + 2 – ____ + __ log __ – __ 2 2 log 2 log 2 2 __
)
4 – ÷ 2 __ 5 3 = ______ – log 2 + __ sq.u. 2 2 log 2 41. Given curve is (y – 2)2 = (x – 1) and the equation of the tangent to the curve at (2, 3) is x – 2y + 4 = 0.
= 4 Ú (t2 – 1)t2dt
|
0
1
|
= 4 Ú (t4 – t2)dt 0
| (
)|
t5 t3 1 = 4 __ – __ 5 3 0
1 1 = 4 __ – __ 5 3
| ( ) |
8 = ___ sq.u. 15 43. Given curves are
y = (x – 1)2, y = (x + 1)2 and y = 1/4.
3.25
3.26 Integral Calculus, 3D Geometry & Vector Booster Hence, the required shaded area 0
(
1/2
)
(
)
1 1 = Ú (x + 1)2 – __ dx + Ú (x – 1)2 – __ dx 4 4 –1/2 0
(
) (
46. Given curve is |x – 2| + |y + 1| = 1
)
0 1/2 (x + 1)3 __ (x – 1)3 __ 1 1 = _______ – x + _______ – x 3 4 –1/2 3 4 0
(
)
1 1 1 1 1 1 = __ – ___ + __ – ___ – __ + __ 3 24 8 24 8 3
(
)
1 1 1 1 = __ – 2 ___ + __ + __ 3 24 8 3 2 2 __ 2 1 __ 1 = __ – __ = – __ = sq.u. 3 6 3 3 3
Thus, the length of each side
44. Given curves are
y = ex, y = e– x and the straight line x = 1
________________
__
= ÷(2 – 1)2 + (0 + 1)2 = ÷ 2
Hence, the required area
__
= ( ÷2 )2 sq.u.
47. Given curve is [x] + [y] = 3
Hence, the required area 1
= Ú (ex – e– x ) 0
= (ex + e– x)10
1 = e + __ e – 2 sq.u.
(
)
45. Given curves are y = log x, y = log |x|, y = |log x| and y = |log|x||
If [x] = 0, 1, 2, 3 then [y] = 3, 2, 1, 0. Hence, the required area
= 4 (1 × 1 + 1 × 1 + 1 × 1 + 1 × 1)
=4×4
= 16 sq.u. 48. Given curves are |x| + |y| = 1 and |x + 1| + |y| = 1.
Hence, the required area 0
= 4 Ú x dy
= 4 Ú e ydy
= 4 (e y)0– •
= 4 (e0 – 0) = 4 (1 – 0) = 4 sq.u.
– • 0
– •
Hence, the required area
1 1 = 2 × __ × 1 × __ 2 2 1 = __ sq.u. 2
Area Bounded by the Curves
49. Given curves are
| |
p p y = cos–1 (cos x) and |x – p| + y – __ = __ 2 2
Hence, the required area 1
__
= 2 Ú (÷x – x) dx 0
(
)
x2 1 2 = 2 __ x2/3 – __ 3 2 0
2 1 = 2 __ – __ 3 2
( ) 4 – 3 1 1 = 2 ( _____ = 2 × __ = __ sq.u. 6 ) 6 3
52. Given curve is |x + y| + |x – y| £ 2
Hence, the required area
3.27
p 1 = 2 × __ × p × __ = 2 2
p2 __ sq.u. 2
Ï x £ 1, x ≥ –1 = Ì Ó y £ 1, y ≥ –1
50. Since the given curve is symmetrical about x-axis as well as y-axis.
Hence, the required area = (2)2 = 4 sq.u. 53. Given curves are y = x, y = x + sin x, 0 £ x £ p Hence, the required area
Hence, the required area __
÷5 + 1
= 4 Ú y dx 0
p
__
÷5 + 1
= 4 Ú (5 – (x – 1)2)dx
(
)
__
8 = __ ( 7 + 5÷5 ) sq.u. 3
51. Given curves are
__
+ 1 (x – 1)3 ÷5 = 4 5x – _______ 3 0
__
= Ú (x sin x – x) dx 0 p
0
= Ú (sin x) dx 0
= – (cos x)p0 = – (–1 – 1) = 2 sq.u.
54. Hence, the required area b
y=÷ |x| and y = |x|
(
__ p = Ú f (x)dx = b sinb + __ cosb + ÷ 2 b 4 p __
)
4
Applying, Newton and Leibnitz rule, we get __ p f (b) = b cos b + sin b – __ sin b + ÷ 2 4 __ p p f __ = 0 + 1 – __ + ÷ 2 2 4 __ p p f __ = 1 – __ + ÷2 2 4 55. Given curves are y = sin x and y = cos x p x = 0 and x = __ . 2
( ) ( ) (
)
3.28 Integral Calculus, 3D Geometry & Vector Booster
( ) ( ) ( 3 2 = ( 2 7 + __ log 2 – 5 ) + 3log ( ______ 2 2 + 7 )
)
__ 1 3 2 11 = __ + __ log 2 + 3log ______ __ + 2÷7 – ___ 2 2 2 2+÷ 7 __
÷
__
÷
(
)
__ __ 2 = (2÷7 – 5) + 3log ______ __ + 3log ( ÷2 ) 2+÷ 7
(
__
Hence, the required area p __ 4 (cos x 0
Ú
– sin x) dx + Ú (sin x – cos x) dx
fi
p __ 4
p __ cos x)04 –
= (sin x +
(
) ( ( )
)
57. Given curve is y = 4x – x2
p __ 2
__
2÷2 __ = ( 2÷7 – 5 ) + 3log ______ 2+÷ 7
(cos x +
y = 5 – (x – 2)2
p __ 2 sin x)__p 4
)
2 2__ = ___ __ – 1 – 1 – ___ 2 ÷2 ÷ 2 = 2 ___ __ – 1 2 ÷ __
= 2 ( ÷2 – 1 ) sq.u. 56. Given curves are
Now, the area OABO
3 y = 2 – |2 – x| and y = __ . |x|
3/2
= Ú (mx) dx 0
( ) ( )
9m mx2 3/2 = ____ = ___ 2 0 8
Hence, the area OABCDO 3/2
= Ú (1 + 4x – x2) dx 0
Hence, the required area 2
(
__
2 + ÷7
)
(
)
3 3 = Ú__ x – __ x dx + Ú 4 – x – __ x dx 2 ÷3
(
2
) ( 2
2
)
x x = __ – 3log x __ + 4x – __ – 3log x 2 2 ÷3 2
(
3 = 2 – 3log 2 – __ + 2
(
__
39 9m 1 ___ According to the question, ___ = __ ◊ 8 2 8 13 fi m = ___ 6
)
3 __ log 2 2 __
(________ 2 + ÷7 )2
+ 4 ( 2 + ÷7 ) –
__
2 + ÷7
( ) 9 27 3 = ( __ + 2 ◊ __ – ___ 4 24 ) 2 9 39 = ( 6 – __ ) = ___ sq.u 8 8 x3 3/2 = x + 2x2 – __ x 0
__
– 3 log ( 2 + ÷7 )
2
)
– 8 + 2 + 3 log 2
13 Hence, the value of m is ___ . 6 58. Do yourself. 59. Given curve is y = sin–1x and y-axis between y = 0 and y = p/2.
Area Bounded by the Curves
| (
)|
8 1 64 = __ ___ – 48 + __ + 4 – 16 2 3 3
1 = __ (24 – 48 + 4 – 16) 2
1 = __ (28 – 64) 2
36 1 = __ |(28 – 64)| = ___ = 18 sq.u. 2 2
| |
|
|
3.29
61. Given curves are Hence, the required area
p __ 2 (1 0
Ú
=
x = – 2y2 and x = 1 – 3y2.
– sin x) dx p __ 2 cos y)0
= (y +
p = __ – 1 sq.u. 2
(
)
60. Given curves are y2 = 2x + 6 and y = x – 1 On solving, we get,
– 2y2 = 1 – 3y2
y2 = 1
y = ± 1
Hence, the required area 1
= Ú (x2 – x1) dy –1
On solving, we get, (x – 1)2 = 2x + 6
y2 = 2 (y + 1) + 6
1
y – 2y – 8 = 0
(y – 4) (y + 2) = 0
1
= 2 Ú (y2 – 1) dy 0
4
= Ú (x2 – x1) dy 4
= Ú (y2 – 1) dy –1
y = – 2, 4 Hence, the required area
– 2
= Ú ((– 2y2) – (1 – 3y2)) dy –1
2
1
(
)
y2 – 6 = Ú _____ – (y + 1) dy 2 – 2
1 = __ Ú (y2 – 6 – 2y – 2) dy 2 – 2
4
( )
1 y3 = 2 __ – y 3 0
| ( ) |
1 4 = 2 __ – 1 = __ sq.u. 3 3 –1 62. Given curves are y = tan x and y = cot–1x and the y-axis. Hence, the required area
4
1 = __ Ú (y2 – 2y – 8) dy 2 –2
4 3 1 y = __ __ – y2 – 8y 2 3 – 2
(
)
p __ 4 tan y dy 0
Ú
p __ 2
+ Ú cot y dy
=
= (log (sec y))04 + (log (sin y))__p
p __ 4
p __
p __ 2
4
3.30 Integral Calculus, 3D Geometry & Vector Booster
(
( ) )
__ 1 = log ( ÷2 ) + 0 – log ___ __ 2 ÷
28 2 __ 7 = ___ – __ – 3 3 3
= 2log ( ÷2 )
= log (2) sq.u.
28 – 9 = ______ 3
19 = ___ sq.u. 3
__
63. Given curve is x = y2 + y + 2 and the y-axis.
65. Find the area of the region bounded by the curves 4|y| = |4 – x2| and |y| (x2 + 4) = 12. 65. Given curves are 4|y| = |4 – x2| and |y| (x2 + 4) = 12.
Hence, the required area 2
= Ú xdy
–1 2
= Ú (– y2 + y + 2)dy
Hence, the required area
–1
(
)
2 y2 __ + 2y 2 –1
y3 = – __ + 3
8 1 1 = – __ + 2 + 4 + __ – __ – 2 3 3 2
( ) 7 1 7 = ( 4 – __ – __ = __ sq.u. 3 2) 6
[
__
2÷2
__
2 64. Given curves are y = ÷ x , y = x
y = – (x2 – 2x – 4) = 5 – (x – 1)2
( ( ) ( Ú ( ) )
( )
2
4 – x2 12 = 4 Ú _____ 2 dx – Ú _____ dx 4 0 0 x +4 __
2÷2
x2 – 4 + _____ dx 4 2
[ (
( ) )
(
) ( ]
) ] __
2÷2 x ÷2 x3 2 x3 = 4 6 tan __ 2 – x – ___ – ___ – x 2 0 12 0 12 2 –1
__
[
__
__ 4 4 – 2÷2 = 4 6 tan–1 ( ÷2 ) – __ + _______ 3 3
__ __ 4 = __ ( 18 tan–1 ( ÷2 ) – 2÷2 ) sq.u. 3 66. Given curves are x2 + 2y2 = 2 and y = 1 – x2
Hence, the required area 2
1
0
0
2
__
= Ú (– x2 + 2x + 4)dx – Ú ÷ x dx – Ú x2dx
(
1
) ( ) ( )
2 1 x3 x3 2 2 = – __ + x2 + 4x – __ x3/2 – __ 3 3 3 1 0 0
(
) ( ) (
)
18 8 1 2 = – ___ + 12 – __ – __ – __ 3 3 3 3
Hence, the required area 1
(
______
_____
)
= 2 Ú ÷2 – 2y2 – ÷1 – y dy – 1/2
Area Bounded by the Curves
__ 1
1
_____
(
– 1/2
_____
)
__ y 1 1 = 2÷2 __ ÷1 – y2 + __ sin–1(y) 2 2 – 1/2
(
( ( ( ( (
)
1 2 – 2 ___ (1 – y)3/2 – 3 – 1/2
_____
= 2÷2 Ú ÷1 – y2 dy – 2 Ú ÷1 – y dy – 1/2
3.31
__
( ) ) ( )
__ p 1 1 ÷3 1 4 3 3/2 = 2÷2 __ + __ ◊ ___ + __ sin–1 __ – __ __ 4 4 2 2 2 3 2 __
) ( )
__
__
3 p 3 p ÷ 4 3÷__ = 2÷2 __ + ___ + ___ – __ ____ 4 8 12 3 2÷2 __
__
Hence, the required area
) ( ) ) ) __
0
•
1 – e
0
= Ú log (x + e) dx + Ú e– xdx
3 ÷3 p ÷ = 2÷2 __ + ___ – 2 ___ __ 3 8 2 ÷
= ((x + e) log (x + e) – (x + e))01 – e – (e– x)•0
__ __ p 3 ÷ = 2÷2 __ + ___ – ÷6 sq.u. 3 8
= (eloge – e) – (1log1 – (1)) – (e– x)•0
= (1 – (0 – 0))
__
67. Given curves are x2 + y2 = 4 and x2 + 4y2 = 9
= 2 sq.u. 69. Given curves are x2 + y2 = 4, y = x2 + x + 1 and
[ ( )
( ) ]
x x y = sin2 __ + cos __ and y = 0. 4 4
Solving the above equations, we get,
x2 + 4 (4 – x2) = 9
fi
3x2 = 7
fi
7 x = __ 3
÷
Hence, the required area
__
(
___
2
_____
_____
)
1 = 4 Ú __ ÷9 – x2 dx + ___ Ú ÷ 4 – x2 dx 0 2 ÷7/3
[ (
_____
( )
___ ÷7/3
9 x 1 x = 4 __ __ ÷9 – x2 + __ sin–1 __ 2 2 2 3 0 _____
( ÷ ) }
{
68. Given curves are fi
( ÷ )
1 7 1 7 = 4p + 9 sin __ __ – 8 sin–1 __ __ 3 3 2 3
__
( )
1 y = ln (x + e) and x = ln __ y y = ln (x + e), y = e– x, y = 0
2
_____
+ Ú (x2 + x + 1) dx + 2 Ú__ ÷4 – x2 dx –1
( ) ) ]
x x 2 + __ ÷4 – x2 + 2 sin–1 __ ___ 2 4 ÷7/3 __
–1
__
0
Hence, the required area ÷7/3
__
= ( ÷ 3 × 1 ) + ( ÷ 3 – 1 )
(
÷3
)
0 x x = ( 2÷3 – 1 ) + __ + __ + x 3 2 –1 _____ x x 2 2 –1 + 2 __ ÷ 4 – x + 2sin __ __ 2 2 ÷3
__
(
3
2
( ) )
( (
__ 1 = ( 2÷3 – 1 ) + 0 – – __ + 3
{
(
__
))
1 __ – 1 2
)}
3 2p ÷ + 2 (0 + p) – ___ + ___ 2 3 2p ___ – ÷ 3 } { 3 2p 1 = ( ___ + ÷ 3 – __ ) sq.u. 3 6 __ 5 = ( 2÷3 – 1 ) + __ + 6 __
__
3.32 Integral Calculus, 3D Geometry & Vector Booster
(
2
)]
__ 1 1 1 1 + 2 ___ __ + __ log ( ÷2 + 1 )– __ – __ log 3 sq.u. 2 3 4 2 ÷
70. Given curves are 2
x x __ + y2 = 1 and __ – y2 = 1 4 2
71. Let the area be A. 2
Then
2
x __ + 4
) )
(
2 x4 x3 = ___ – __ + ax 12 3 0
4 8 = __ – __ + 2a 3 3
Solving the above equations, we get
(
x3 A = Ú __ – x2 + a dx 3 0
) ( 4 = ( 2a – __ ) 3
dA ___ = 2 > 0 da
2
x __ = 2 2
So, the area is minimum. It will be minimum when A = 0.
fi
3x2 ___ = 2 4
fi
8 x2 = __ 3
fi
Now,
x2 y2 = 1 – __ 4
fi
8 2 1 y2 = 1 – _____ = 1 – __ = __ 4×3 3 3
2 Hence, the value of a is __ . 3 72. Hence, the required area
Thus,
1
1__ y = ___ 3 ÷ Hence, the required area __
1
1/÷3
) )
= 4 Ú (Hyperbola) dy + Ú __ (Ellipse) dy
=
0
1/÷3
__
1
1/÷3 __ _____ 4 ÷ 2 1 + y2 dy 0
Ú
( ÷
)
_____
+ Ú __ 2÷1 + y2 dy 1/÷3
[ (
_____ __ y _____ 1/÷3 1 = 4 ÷2 __ ÷1 + y2 + __ log y + ÷ y 2 + 1 2 2 0
|
(
|)
[ (
1 = 4 ÷2 ____ __ × 2÷3
|
2 ___ __ + 3 ÷
|
1 1 __ log ___ __ + 2 ÷3
(
|)
[ (
a2 a = __ + __ + 1 3 2
fi
dA 2a __ 1 ___ = ___ + da 3 2
fi
d 2A 2 ____2 = __ > 0 3 da
)
2a 1 For maximum or minimum, ___ + __ = 0 3 2
]
fi
|)
2 ___ __ 3 ÷
2a 1 ___ = – __ 3 2
3 a = – __ 4 Hence, the value of a is – 3/4 fi
)]
)
)
1 a2x3 ___ ax2 = ____ + + x 3 2 0
So, the area is minimum.
__ 1 1 1__ ___ 2 __ 1 + 2 ___ __ + __ log ( ÷2 + 1 ) – ____ × __ – log3 2 2÷3 ÷3 4 ÷2 __ 1 1 = 4 ÷2 __ + __ log (3) 3 4
( (
__
_____ 1 y _____ __ 1 + 2 __ ÷1 + y2 + log y + ÷ y 2 + 1 __ 2 1/÷3 2 __
= A = Ú (a2x2 + ax + 1) dx 0
fi
( (
4 2a – __ = 0 3 2 a = __ 3
73. Do yourself. 74. Given curves are y = x2 + 2x – 3 and y = mx + 1
Area Bounded by the Curves
3.33
75. Hence, the required area
( ) ( ) ( ) ( ) 2a
x = A = Ú __ + 6 a
x2 1 2a = ___ – __ 12 x a
Solving, we get x2 + (2 – m) x – 4 = 0 Let a, b be the roots such that a + b = m – 2, ab = – 4
Hence, the required area b
A = Ú (Line – Parabola) dx a
b
= Ú ((mx + 1) – (x2 + 2x – 3)) dx a
(
) ) (
b mx2 x3 = ____ + x – __ – x2 + 3x 2 3 a
( ( ( ( ( ( (
a2 1 1 ___ – ___ + __ 2a 12 a 1 ___ 2a
dA a ___ 1 ___ = __ – da 2 2a2 d 2A 1 1 ____ fi 2 = __ + __ 2 a3 da Now for maximum or minimum, dA ___ = 0 da a 1 __ fi – ___ = 0 2 2a2 a 1 __ fi = ___ 2 2a2 fi a3 = 1 fi a=1 Hence, the value of a is 1, when the area is least.
x2 + 2x – 3 = mx + 1
a2 = __ – 3 a2 = __ + 4
1 __ 2 dx x
))
mb 2 b3 ma2 a3 = ____ + 4b – __ – b 2 – ____ + 4a – ___ – a2 2 3 2 3 (Problems for Jee-Advanced)
)) )) )
(b3 – a3) m = __ (b2 – a2) + 4 (b – a) – ________ – (b2 – a2) 2 3
1. Given curve is a2 y2 = x2 (a2 – x2).
(b3 – a3) m = __ – 1 (b2 – a2) + 4 (b – a) – ________ 2 3
)
( ( ( (
(b2 + ab + a2) m = (b – a) __ – 1 (b + a) + 4 – ____________ 2 3
)
)
(m – 2)2 (m – 2)2 + 4 = (b – a) _______ + 4 – __________ 2 3
a
)
(m – 2)2 ___ 16 = (b – a) _______ + 3 6
(
)
_____________ (m – 2)2 ___ 16 (a + b)2 – 4ab _______ +
= ÷
(
3
6
)
____________ (m – 2)2 ___ 16 (m – 2)2 + 16 _______ +
= ÷
6
Hence, the required area
3
The area is minimum, if m = 2 and the least area is 64 = ___ . 3
= 4 Ú y dx 0 a
0 p/2
(
______
)
x = 4 Ú __ a ÷ a2 – x2 dx = 4 Ú a2 sin2q cosq dq, (Let x = a sinq) 0
1
= 4a2 Ú t2 dt, 0
4a2 = ___ sq.u. 3
(Let t = sinq)
3.34 Integral Calculus, 3D Geometry & Vector Booster
| |
Hence, the required area
2. Given curve is y = (x – 1)(x – 2)(x – 3).
1
1
0
0
= Ú logx dx + Ú sin4 x dx
( 1 – cos2p x = 1 + Ú ( __________ ) dx 2 1
)
1 – cos2p x 2 = |x(logx – 1)|10 + Ú __________ dx 2 0 1
2
0
1
Hence, the required area
1 = 1 + __ Ú (1 – 2cos(2p x) + cos2 (2p x)) dx 4 0
3
= Ú {(x – 1)(x – 2)(x – 3)}dx
1
1
= – Ú {(x –1)(x – 2)(x – 3)}dx 0
2
+ Ú {(x – 1)(x – 2)(x – 3)}dx
1 3
– Ú {(x – 1)(x – 2)(x – 3)}dx
(
2
)|
x 11 = – __ – 2x3 + ___ x2 – 6x 4 2 0
( (
))
(
)
( )
3 11 1 3 = 1 + __ __ – 0 = 1 + __ = ___ 4 2 8 8 4. Given curve is y = 2x4 – x2.
) )
2 x4 11 + __ – 2x3 + ___ x2 – 6x 4 2 1
3 x4 11 – __ – 2x3 + ___ x2 – 6x 4 2 2
(
(
2sin(2p x) ________ sin(4p x) 1 1 3x _________ = 1 + __ ___ – + 4 2 2p 8p 0
1
4
(
1 + cos(4p x) 1 = 1 + __ Ú 1 – 2cos(2p x) + ___________ dx 4 0 2
0
) 1 11 – ( __ – 2 + ___ – 6 ) 4 2 99 81 – ( ___ – 54 + ___ – 18 ) + (4 – 8 + 22 – 12) 4 2 279 23 = – 2( ___ – 8 ) + 12 – ( ____ – 72 ) 4 4 1 11 = – __ – 2 + ___ – 6 + (4 – 8 + 22 – 12) 4 2
46 279 = ___ – ____ + 100 4 4 233 _________ 400 – 233 ____ 167 = 100 – ____ = = sq.u. 4 4 4
3. Given curves are y = loge x, y = sin4 (p x) and x = 0
Hence, the required area
|
1/2
|
= 2 Ú (2x4 – x2) dx 0
| (
)
|
2x5 x3 1/2 = 2 ___ – __ 3 0 5
| (
1 = 2 ___ – 80
)|
1 ___ = 24
7 ____ sq.u. 120
5. Given curves are x2 + y2 = 64 and y2 = 12x.
Area Bounded by the Curves
Hence, the required area
(
4
8
)
= area of a semi-circle – 2 Ú P dx + Ú C dx 0
(
4
8
___ __
4
)
______
= 64p – 2 Ú ÷12 ÷ x dx + Ú ÷64 – x2 dx 0
4
(
(
6.
(
__
( ) )
( ) )
__ 32p 1 = 64p – 24÷3 – 2 ____ – 8÷3 – 32sin–1 __ 2 2 __ 64p = 32p – 8÷3 + ____ 3 __ 16p = ____ – 8÷3 sq.u. 3 Given curves are 1 y = ln(x + e) and x = ln __ y
(
= Ú (mx) dx 0
( ) ( )
9m mx2 3/2 = ____ = ___ 2 0 8
3/2
x x 8 – 2 __ ÷64 – x2 + 32sin–1 __ 2 2 4
3/2
Hence, the area OABCDO
)
__ 4 3 = 64p – 2 2÷3 × __ x3/2 2 0 ______
Now, the area OABO
)
( )
fi y = ln(x + e), y = e– x, y = 0
= Ú (1 + 4x – x2) dx 0
(
)
x3 3/2 = x + 2x2 – __ 3 0
9 27 3 = __ + 2 ◊ __ – ___ 2 4 24
( ) 9 39 = ( 6 – __ ) = ___ sq.u. 8 8
According to the question, 9m ___ = 8 fi
1 39 __ ◊ ___ 2 8
13 m = ___ 6
13 Hence, the value of m is ___ . 6 8. Given curve is
Hence, the required area 0
•
1 – e
0
= Ú log(x + e)dx + Ú e– x dx = ((x + e) log (x + e) – (x + e))01 – e – (e–x)•0 = (eloge – e) – (1log1 – (1)) – (e–x)•0 7.
= (1 – (0 – 1)) = 2 sq.u. Given curve is y = 1 + 4x – x2 fi y = 5 – (x – 2)2
Hence, the required area 3
= Ú (y2 – y1) dx 0
3
= Ú (2x – x2 + x) dx 0
3
= Ú (3x – x2) dx 0
(
)
3x2 x3 3 = ___ – __ 2 3 0
27 27 = ___ – ___ 2 3
( ) 9 27 – 18 = ( _______ = __ sq.u. 2 ) 2
3.35
3.36 Integral Calculus, 3D Geometry & Vector Booster
9. Given curve is
12
(
)
__ __ x2 Also, A2 = Ú 2÷3 ÷ x – ___ dx 12 3
( ( ) )
__ 2 x3 12 = 2÷3 __ x3/2 – ___ 3 36 3
(
) (
)
__ ___ 12 __ __ 27 2 2 = 2÷3 × __ × 12 ÷12 ___ – __ × 2÷3 × 3÷3 ___ 3 3 36 36
(
)
3 = (96 – 48) – 12 – __ 4
| |
3 3 147 = 48 – 12 + __ = 36 + __ = ____ 4 4 4
Hence, the required area 2
|
= Ú {(x – 1)(x – 2)}dx 1
2
|
Thus, A1 : A2 = 45 : 147 = 15 : 49
= Ú (x2 – 3x + 2) dx 1
| (
)|
11. Given curves are x + y £ 6. x2 + y2 £ 6y and y2 £ 8x
2 x3 3x2 = __ – ___ + 2x 3 2 1
| ( | ( | (
8 1 = __ – 6 + 4 – __ + 3 3
)|
3 __ – 2 2
)| 14 + 9 = ______ – 4 ) 6 | 23 1 = ( ___ – 4 ) = __ sq.u. 6 6 7 3 = __ + __ – 4 3 2
Hence, the required area
2
2
10. Given curves are y = 12x and x = 12y and x = 3.
2
______
___
= Ú {( 2÷2x ) – ( 3 – ÷9 – x2 ) } dx 0
3
______
{
}
+ Ú (6 – x) – ( 3 – ÷9 – x2 ) dx 2
(
3
)
__ 2 __ x2 3 2 = 2÷2 ◊ __ ( x3/2 )0 – Ú ( 3 – ÷ 9 – x2 )dx + 6x – __ 3 2 2 0
(
3
)
9 16 = ___ + 18 – __ – 12 + 2 3 2
(
)
__ __ x2 Here, A1 = Ú 2÷3 ÷ x – ___ dx 12 0
( ( ) )
__ 2 x3 3 = 2÷3 __ x3/2 – ___ 3 36 0
__ 2 __ = 2÷3 __ × 3÷3 – 3
3 45 = 12 – __ = ___ sq.u. 4 4
(
)
27 ___ 36
(
_____
( ))
3
x÷9 – x2 __ 9 x – 3x – ________ – sin–1 __ 2 2 3 0 9p 27 16 = ___ – 9 + ___ + ___ – 10 3 4 2
(
)
9p 16 27 = ___ + ___ + ___ – 19 4 3 2
(
)
9p 1 = ___ – __ sq.u. 4 6
Area Bounded by the Curves
12. Given curve is 2
Hence, the required area 2
2
x + y – 6x – 4y + 12 £ 0, y £ x, x £ 5/2
= Ú (x + 2 – x2)dx –1
fi (x – 3)2 + (y – 2)2 £ 1, y £ x, x £ 5/2
(
__________ 1 – (x – 3)2
( (
fi y = 2 ± ÷
__________ 1 – (x – 3)2 as
fi y = 2 – ÷
)
x2 x3 2 = __ + 2x – __ 2 3 –1
fi (y – 2)2 = {1 – (x – 3)2}
)
8 1 1 = 2 + 4 – __ – __ + 2 + __ 3 2 3 1 7 = 8 – __ – __ 2 3
5 x £ __ 2
)
48 – 3 – 14 ___ 31 = __________ = sq.u. 6 6 14. Given curve is y = x log x and y = 2x – x2
Hence, the required area
(
5/2
)
__________
(
)
5 1 1 = __ 2 + __ __ – Ú 2 – ÷1 – (x – 3)2 dx 2 2 2 2 5/2
__________
(
Hence, the required area
)
9 = __ – Ú 2 – ÷1 – (x – 3)2 dx 8 2
( { ( ) ( ( )
1
)
5/2 (x – 3) __________2 __ 9 1 = __ – 2x – _______ 1 – (x – 3) + sin–1 (x – 3) ÷ 8 2 2 2
__
3 ÷ 9 1 1 __ 1 = __ – 2 ◊ __ – __ – × ___ 8 2 2 2 2
)}
1 1 + __ sin–1 – __ – sin–1 (– 1) 2 2
( (
__
(
))
3 1 ÷ 9 p p = __ – 1 + ___ + __ – __ + __ 8 8 2 6 2 __
= Ú ((2x – 2x2) – xlogx) dx 0
( ) ( __ x2 logx – __ x4 ) 2 1 = ( 1 – __ ) + __ 3 4 1 2 = x2 – __ x3 – 3 0
1 = __ + 3 15. Given
2 1
2
0
7 1 __ = ___ sq.u. 4 12 curves are x2 + (y – 1)2 = 1 __
and c2 x2 + y2 = c2 where c = ÷ 2 – 1
)
3 p ÷ 9 = __ – 1 + ___ + __ 8 8 6
(
__
)
3 p 1 ÷ = __ – ___ – __ sq.u. 8 8 6 13. Given curves are x2 = y, y = x + 2 Hence, the required area __
1/÷2
______
______
) = p ◊ (1)2 – 2 Ú ( c ÷1 – x2 – 1 – ÷ 1 – x2 0
__
1/÷2
______
= p ◊ (1) – 2 Ú ( (c + 1) ÷1 – x2 – 1 ) 2
0
3.37
3.38 Integral Calculus, 3D Geometry & Vector Booster
} ) ( { x 1 = p ◊ (1) – 2( 2 { __ ÷1 – x + __ sin x } – x ) 2 2 = p – 2 2 { ____ ( 212 ◊ ___12 + __ 12 ◊ __p4 } – ___ 12 ) ______
__
1/÷2 x 1 = p ◊ (1)2 – 2 (c + 1) __ ÷ 1 – x2 + __ sin–1 x – x 2 2 0
______
__
2
2
÷
__
÷
–1
__
__
÷ ÷
__ 1/÷2
0
__
÷
__
(p – 2)÷2 = p – _________ 4 __ __ 1 = __ { ( 4 – ÷2 )p + 2÷2 } sq.u. 4 16. Given curve is (y – 2)2 = (x – 1) and the equation of the tangent to the curve at (2, 3) is x – 2y + 4 = 0.
Hence, the required area 3
= Ú (x2 – x1) dy 0
3
= Ú ((y – 2)2 + 1 – (2y – 4)) dy 0
3
= Ú (y2 – 4y + 4 + 1 – 2y + 4) dy 0
Hence, the required area
( (
)
1 1__ __ 1 = 3 2 ◊ __ ◊ ___ – p ◊ 1 2 ÷ 3 6
)
p 1 = 3 ___ __ – __ 3 6 ÷
(
__ p = ÷3 – __ sq.u. 2 18. The tangent at x = p to the curve f (x) = |cos x| will be parallel to x-axis and cuts the curve f (x) = cos–1 (cos x) at B and C.
)
Thus, AD = p – 1 Hence, the area of DABC 1 = __ × (2p – 2) × (p – 1) 2 1 = __ × 2(p – 1) × (p – 1) 2 = (p – 1)2 sq.u. 19. Given curves are |y| = 1 – x2, x2 + y2 = 1.
3
= Ú (y2 – 6y +9) dy 0
(
)
3 y3 = __ – 3y2 + 9y 3 0
= 9 – 27 + 27 = 9 sq.u. 17. Here D, E and F are the mid-points of the sides of the trangle ABC respectively.
Hence, the required area
(
1
)
= p – 4 Ú (1 – x2) dx 0
( )
x3 1 = p – 4 x – __ 3 0
( ) 8 = ( p – __ ) sq.u. 3
1 = p – 4 1 – __ 3
Area Bounded by the Curves
20. Given curves are 1 |y| = e–|x| – __ and |x| + |y| = ln 2 2
Hence, the required area
(
ln2
(
22. Given curves are
)
)
1 1 = 4 __ (ln2)2 – Ú e–x – __ dx 2 2 0
(
(
x2 + y2 – 2x + 4y – 11 = 0
and
y = – x2 + 2x + 1 – 2 ÷3
fi
(x – 1)2 + (y + 2)2 = 16
and
y + 2(÷3 – 1) = – (x – 1)2
__
__
( (
ln2
e 1 1 = 4 __ (ln 2)2 – ___ – __ 2 –1 2 0
)
2
= 2 Ú (Parabola – Circle) dx
( )
0
2 = 2 (ln 2) + 2 ln __ e . 21. Given curves are 2
Hence, the required area
) )
–x
3.39
y2 – 2y + 4x + 5 = 0 and x2 + 2x – y + 2 = 0 2
2
fi (y – 1) = – 4(x + 1) and (x + 1) = (y – 1).
2
__
= 2 Ú ((– x2 + 2x + 1 – 2 ÷3 ) 0
____________
(
)) )
– – (x – 1)2 dx – 2 + ÷16
[ (
)
__ 2 x3 = 2 – __ + x2 + x – 2÷3 x 3 0
(
( ) ) ] ÷15 8 1 = 2 ( – __ + 6 – 4 ÷ 3 – ( (– 4) + ____ + 8 sin ( __ ) ) 3 2 4 ÷15 1 + ( – ____ – 8 sin ( __ ) ) ) 2 4 8 = 2 ( 10 – __ – 4 ÷3 ) 3 22 = 2 ( ___ – 4 ÷3 ) sq.u. 3
(x – 1) ___________2 x–1 2 – – 2x + ______ 16 – (x – 1) + 8sin–1 _____ ÷ 2 4 0 ___
__
___
–1
–1
__
Hence, the required area –1
________
Ú __ ( 1 + ÷– 4(x + 1) ) – (1 + (x + 1)2) dx
=
–1 – 3÷4
( (
(
))
–1
(x + 1)3 2 = x + 2 ◊ __ (– (x + 1)3/2 ) – x + _______ 3 __ 3 3 –1 – ÷4
(
3
))
–1
(x + 1) 4 = __ (– (x + 1)3/2) – _______ 3 __ 3 3 –1 – ÷4
( ( (
( ) ) __
( 3 )3
– ÷4 4 = __ – ÷4 ) + _______ 3 3
| (
3 __ 3/2
)|
))
4 4 12 = __ (–2) – __ = ___ = 4 sq.u. 3 3 3
__
23. Given curve is 9x2 + 4y2 – 36x + 8y + 4 = 0 fi
(9x2 – 36x) + (4y2 + 8y) + 4 = 0
fi
9 (x2 – 4x) + 4(y2 + 2y) + 4 = 0
fi
9 (x – 2)2 + 4(y + 1)2 = 36 + 4 – 4 = 36
fi
9 (x – 2)2 + 4(y + 1)2 = 36
(y + 1)2 (x – 2)2 _______ fi _______ + = 1 4 9
3.40 Integral Calculus, 3D Geometry & Vector Booster 1 11 = 6 – __ = ___ sq.u. 2 2 25. The required area will be equal to area enclosed by y = f (x) and the y-axis between the abscissae y = –1 and y = 4 \ f (0) = 1, f (–2) = –1, f (2) = 4 Clearly, f (x) is monotonic in [–2, 2]. Hence, the required area Solving, we get x = 2 and x = 4 Hence, the required area 2
( ÷ ( ) ( ) ) Ú ( ) ÷ ( ) Ú ( ÷ ( ) ) _____________ 2
(x – 2) 10 – 3x = Ú –1 + 3 1 – _______ – _______ dx 4 2 2 ____________ 2
(x – 2) 3x = –1 + 3 1 – _______ – 5 + ___ dx 4 2 2 ____________
4
(x – 2)2 = – 6 + 3 1 – _______ + 4 2
(
3x ___ dx 2
( ÷ ( ) 3x x–2 ( _____ ) ) + ___ ) 2 4
___________
x–2 x – 2 = – 6x + 3 _____ 1 – _____ 4 2 2
1 + __ sin–1 2
(
\
–2
(
)
2
( )
0
(
)
x3 x2 13 + Ú ___ + __ + ___ x + 2 dx 8 12 –2 24
(
)
x4 x3 13x2 2 = 3x – ___ – ___ – ____ 24 0 96 24
(
)
0 13x2 x4 x3 + ___ + ___ + ____ + 2x 24 96 24 –2
(
) (
)
)
)
16 8 52 16 8 52 = 6 – ___ – ___ – ___ – ___ – ___ + ___ – 4 96 24 24 96 24 24
| | |
32 = 10 – ___ – 96
2 4
p = – 24 + 3 __ + 12 + 12 – 3 4 p = 3 __ – 1 sq.u. 4 24. The required area will be equal to area enclosed by y = f (x) and the y-axis between the abscissae y = – 2 and y = 6.
(
0
13 x3 x2 ___ = Ú 3 – ___ – __ – x dx 24 8 12 0
= Ú (E – L) dx
4
0
2
4
4
2
= Ú (4 – f (x)) dx + Ú ( f (x) – (–1)) dx
|
104 ____ 24
|
1 13 = 10 – __ – ___ 3 3 14 = 10 – ___ 3 16 = ___ sq.u. 3 26. Given curves are y = xe x, y = xe–x
)
f (0) = 2, f (–1) = – 2, f (1) = 6
Clearly, f (x) is monotonic in [–1, 1]. Hence, the required area 1
–1
0
0
= Ú (6 – f (x)) dx + Ú ( f (x) – (– 2)) dx 1
–1
0
0
= Ú (4 – x3 – x) dx + Ú (x3 + 3x + 4) dx
) (
(
x2 1 __ + 2 0
x4 = 4x – __ – 4
| ( |
) (
x4 __ + 4
)
0 3x2 ___ + 4x 2 –1
)|
1 1 1 3 = 4 – __ – __ – __ + __ – 4 4 2 4 2 2 = 8 – __ – 2 4
|
Hence, the required area 1
= Ú (xex – xe–x) dx 0
1
= (x (ex + e–x))10 – Ú (ex + e–x) dx 0
Area Bounded by the Curves
(
) (
3.41
)
1 1 __ = e + __ e – e – e
2 = __ e sq.u. 27. Given curves are
|x – 2y| + |x + 2y| £ 8 and xy ≥ 2.
Hence, the required area 4
Hence, the required area 4
(
)
(
)
2 = 2 Ú 4 – __ x dx 1/2
= 2 (4x – 2 log x)21/2
1 = 2 8 – 2log2 – 2 + 2 log __ 2
= 2 (6 – 4 log 2) sq.u.
(
( ) )
30. Given curves are
2 = 2 Ú 2 – __ x dx 1
= 2 (2x – 2 log x)41
= 2 (8 – 2 log 2 – 2)
= 2 (6 – 2 log 2) sq.u.
|x + y| £ 2 and x2 + y2 ≥ 2.
28. Given curves are |x| + |y| ≥ 3 and xy ≥ 2.
Hence, the required area
(
__
)
p ◊ (÷2 )2 1 = 2 __ × 2 × 2 – _______ 2 4
p = 4 – __ sq.u. 2
(
)
Hence, the required area 2
(
)
2 = Ú 3 – x – __ x dx 1
2 x2 = 3x – __ – 2 log x 2 1
1 = 6 – 2 – 2 log 2 – 3 + __ 2
(
)
( 3 = ( __ – 2 log 2 ) sq.u. 2
)
29. Given curves are
|x – y| + |x + y| £ 8 and xy ≥ 2.
(Tougher Problems for JEE-Advanced) 1. Given curve is
2x2 + 6xy + 5y2 = 1
Let x = r cos q, y = r sin q
1 fi r2 = __________________________ 2 2 cos q + 6 sinq cosq + 5 sin2q
Hence, the required area p/2
= Ú r2 dq –p/2
3.42 Integral Calculus, 3D Geometry & Vector Booster p/2
(
)
1 = Ú _________________________ dq 2 2 –p/2 2cos q + 6sinq cosq + 5sin q
sec2q = Ú _________________ dq 2 –p/2 2 + 6 tanq + 5 tan q
p/2
•
(
(
•
(
)
dt 1 = __ Ú __________ 6 5 – • 2 __ 2 t + t + __ 5 5 •
)
dt = Ú __________ 2 , – • 5t + 6t + 2
( (
–1
( ( ) )
= p sq.u.
y – 2 = ± 2 (x – 1)
(Let t = tan q)
y = ± 2x
)
•
– •
Hence, the required area 1
= 4 Ú (a2/3 – x2/3)3/2 dx 0
0
= 4 Ú (a sin3q) (– 3a cos2q sinq) dq p __ 2
(Let x = a cos2q) p __ 2
= 12 a2 Ú sin4q cos2q dq
3 1 1 p = 12a2 × __ ◊ __ ◊ __ ◊ __ 6 4 2 2
( )
Now,
m = ( f ¢(x))x = 0 = 0 y – b = m(x – a)
fi
y – b = 2a (x – a)
Since the point (a, b ) lies on the curve, so
2
1 – 2x2 = 0 1__ fi x = ± ___ 2 ÷ 1 By the sign scheme, it is maximum at x = ___ __ 2 ÷ 1 Thus, c = ___ __ 2 ÷ Hence, the required area
Equation of the tangent at (a, b) is
2
e–x – 2x2 e–x = 0
fi
3. Clearly, f (x) = x2 + 1 f ¢(x) = 2x
)
2 2 dy fi ___ = e–x – 2x2 e–x dx
fi
3pa2 = ____ sq.u. 8
fi
)
dy For maximum or minimum, ___ = 0 dx
0
)
( (
0 a
(
1 x3 x2 = 2 __ – __ + x 3 2 0 1 1 = 2 __ – __ + 1 3 2 4 1 = 2 __ – __ 3 2 5 = 2 × __ 6 5 = __ sq. u. 3 4. Given curve is 2 y = xe–x
= 4 Ú y dx
= 2 Ú (x2 + 1 – x) dx 0
a
a = ± 1, b = 2
Hence, the equation of the tangent is
Hence, the required area
From Eq. (i) and (ii), we get
)
2. Given curve is x2/3 + y2/3 = a2/3.
...(ii)
) ( ) 5t + 3 1 = __ × ( 5 tan ( ______ 1 )) 5
b = a2 + 1
dt = 1/5 Ú _____________ 3 2 1 – • __ t + + __ 5 5
p p = __ – – __ 2 2
c
= Ú x e–x dx 2
0
...(i)
– c2
1 = – __ Ú et dt, (Let – x2 = t) 2 0
Area Bounded by the Curves 2 1 = – __ (et)– c 0 2
2 1 = – __ (e– c – 1) 2
2 1 = __ (1 – e– c ) sq. u. 2 1 The area is maximum, when c = ___ __ 2 ÷ Hence, the maximum area
)
1 – __ e 2
1 = __ 1 – 2 1 1 = __ 1 – ___ __ ÷e 2 5. Let the required area be A.
(
)
0
)
1 a2 x3 ___ ax2 = ____ + + x 3 2 0
a2 a = __ + __ + 1 3 2
)
(
)
dA 2a 1 fi ___ = ___ + __ 3 2 da d 2A fi ____2 = da
fi
a3 = 3a
fi
a2 = 3
A = Ú (a2 x2 + ax + 1) dx
( (
fi
a3 2 __ + a = (a3 + a) 3 2a3 + 6a = 3a3 + 3a
2 __ > 0 3
So, the area is the least. For maximum or minimum,
7. Given curve is
fi x–2=0 fi x=2 So c=2 Hence, the required area is 2
2a fi ___ + 3
1 __ = 0 2 3 fi a = – __ 4 Hence, the value of a is –3/4. 6. Given curve is y = x2 + 1
a
= Ú (x2 + 1) dx 0
= Ú xe– x dx 0
2
= (– xe– x)20 + Ú e– x dx – x
= – (xe
+
0 – x 2 e )0
= (1 – 3e–2) sq.u. 8. Given curves are |x| + |y| ≥ 1 and x2 – 2x + 1 £ 1 – y2 |x| + |y| ≥ 1 and (x – 1)2 + y2 £ 1 fi
Hence, the required area
y = xe– x
dy fi ___ = e– x – xe– x dx d 2y fi ___2 = – e– x – e– x + xe– x dx For point of inflection, d 2y ___2 = 0 dx fi – e– x – e– x + xe– x = 0
dA ___ = 0 da
__
fi a = ÷3 __ Hence, the value of a is ÷ 3 .
1
Thus,
x3 = __ + x 3 0 a3 __ = + a 3 It is given that,
(
( ) ( ) ( )
a
3.43
Ï x + y ≥ 1, x + y £ - 1 and (x – 1)2 + y2 £ 1 Ì Ó x - y ≥ 1, x - y £ - 1
3.44 Integral Calculus, 3D Geometry & Vector Booster Hence, the required area
(
1
1 = 2 Ú ÷1 – (x – 1)2 dx + __ × 1 × 1 2 0 =
(
(x – 1) __________2 2 ______ 1 – (x – 1) +
÷
2
(
___________
| |
|
1
+ 2 Ú ÷10 – (x + 1)2 dx + Ú (1 – x)dx
)
__________
|
___
÷10 – 1
( ) )
1
x–1 1 __ sin–1 _____ + 1 2 2 0
( ) )
(
2
1
)
2
x3 = – __ + 2x2 – 3x 3 1
[
___________ (x + 1) 10 – (x + 1)2 ___________________
÷
( ) ] ( ) ___
÷10 – 1
x + 1 + 5sin–1 _____ ___ 2 ÷10 2
1 1 1 = 2 __ + __ sin–1 __ + 1 2 2 2
+ 2
p = 2 + __ sq.u. 6
x2 2 – x – __ 2 1 8 1 = – __ + 8 – 6 + __ – 2 + 3 3 3
(
)
9. Given curves are |x + y| £ 1, |y – x| £ 1 and 3x2 + 3y2 ≥ 1.
(
(
)
( ) )
3 3 1 + 2 0 + 5sin–1 (1) – __ – 5 sin–1 ____ ___ + __ 2 2 ÷10
( ) ( ) )
3 2 1 = __ + __ + 5p – 3 – 10 sin–1 ____ ___ 3 2 ÷10
(
3 11 = 5p – ___ – 10 sin–1 ____ ___ sq.u. 6 ÷10 11. Given curves are (x2 + y2) £ 4 £ 2(|x| + |y|) Hence, the required area = area of the square – area of the circle p = 2 – __ sq.u. 2 10. Given curves are
(
)
|x – 2| + |y – 2| £ 3, x2 – 4x + y + 3 £ 0
and
x2 + y2 + 2x – 9 £ 0
fi
|x – 2| + |y – 2| £ 3, (x – 2)2 £ – (y + 1)
and
(x + 1)2 + y2 £ 10
fi
(x – 2)2 £ – (y + 1), (x + 1)2 + y2 £ 10
Ï x + y £ 7, x + y ≥ 1 Ì Ó x - y £ 3, x - y ≥ - 3
Hence, the required area = area of a circle-area of a square = 4p – 4 = 4 (p – 1) sq.u. 12. Given curves are |4 – x2| y = _______ and y = 7 – |x| 4 |x2 – 4| fi y = _______ and y = 7 – |x| 4
Hence, the required area 2
= Ú (– x2 + 4x – 3) dx 1
Area Bounded by the Curves
Hence, the required area
[
( ) ( ) ( ) ] 2
)
4
]
4 – x2 x2 – 4 1 = 2 __ (7 + 3) × 4 – Ú ______ dx – Ú ______ dx 2 4 2 0 2
[ (
3 2
x = 2 40 – x – ___ – 12 0
[ ( [
On solving, we get,
____
[ ( ) ( ) ] Ú [ ( ) ] Ú [ ( )] Ú [ ( )] ___
x ___ – x 12 2
) (
____
B (b – a, ÷4ab ) , B ¢(b – a, – ÷4ab )
Hence, the required area
4
2
2÷ab
)]
3.45
y2 = Ú ___ b – ___ – 4b – 2÷ab
y2 ___ – a dy 4a
8 64 8 = 2 20 – 2 – ___ – ___ – 4 – ___ + 2 12 12 12 2 16 2 = 2 18 + __ – ___ + 4 + __ – 2 ] 3 3 3
2 1 y = ___ (a + b) – __ __ a + 4 – 2 ÷ab
= 32 sq.u.
2 1 y = 2(a + b) ◊ 2 ÷ab – __ __ a + 4 0
÷ab y2 y2 1 = 4 (a + b) ÷ab – __ __ a + __ dy 2 0 b
3 1 y = 4 (a + b) ÷ab – __ ___ + 2 3a
]
13. Given curves are y = x2 + x – 2, y = 2x where
___
2 ÷ab
___
x (x + x – 2) ≥ 0
fi
x (x – 1) (x + 1) ≥ 0 2
Solving, y = x + x – 2, y = 2x, we get,
x = – 1, x = 2
___
÷ab
y2 __ dy b
___
2
y2 __ dy b
___
___
( (
)
___
y3 2 ÷ab ___ 3b 0
___
___
)
___ _______ 8ab ÷ab 1 8ab ÷ab = 4 (a + b) ÷ab – __ _______ + 2 3a 3b ___ ___ 4 = 4 (a + b) ÷ab – __ (a + b) ÷ab 3 ___ 8 = __ (a + b) ÷ab sq.u. 3
15. Given curve is y = – x2 – 2x + 3 dy fi ___ = – 2x – 2 dx
( )
dy fi ___ = – 6 dx x = 2
Hence, the required area 0
2
–1
1
= Ú [2x – (x2 + x – 2)] dx + Ú [2x – (x2 + x – 2)]dx
[
] [
The equation of the tangent to the given curve at P (2, – 5) is 6x + y = 7
]
0 2 x2 x3 x2 x3 = __ – __ + 2x + __ – __ + 2x 2 3 2 3 –1 1
[ (
)]
10 13 7 = __ + ___ – ___ 3 6 6 14 7 = ___ = __ sq.u. 3 6 14. Given curves are y2 = 4a(x + a) and y2 = 4b(b – x)
| |
Hence, the required area
3
1 = __ ◊ 2 ◊ 12 – Ú x dy 2 – 5
= 12 – Ú (–1 + ÷ 4 – y ) dy
|
3
_____
– 5
| [
]
|
|
3 2 = 12 – – y – __ (4 – y)3/2 3 – 5
3.46 Integral Calculus, 3D Geometry & Vector Booster
[
]
2 = 12 – 10 – __ 3
Hence, the required area
and
y = – x + 2x + (1 – 2 ÷3 )
fi
(x – 1)2 + (y + 2)2 = 16
and
y = – (x – 1)2 + 2(1 – ÷ 3 )
]
_______
( Ú ( ÷
)
1 Let a = ________ _______ 2 ÷ a + b2
__
2
1 __ b
________
÷1 – a 2 x2 – b 2 x2 ÷ 1 = 4 Ú _________ + Ú ________ dx a b a 0
x2 + y2 – 2x + 4y – 11 = 0
[
a
8 = __ sq.u. 3 16. Given curves are
[
a
( ÷
_______
1 __ b
)
_______
)
]
a b 1 1 = 4 __ Ú __ 2 – x2 dx + __ a __ 2 – x2 dx b 0 a a b
__
(
Solving, we get x = –1, 3
( ) )
a 4 = ___ tan–1 __ sq.u. ab b 18. Given curves are 1 |y| + __ £ e–|x| and {|x|, |y|} £ 2 2
Hence, the required area 3
= Ú (yp – yc) dx –1
Hence, the required area
3
__
= Ú [(– x2 + 2x + 1 – 2 ÷3 ) – (– 2 – –1
___ ÷16 –
(x – 1)2)] dx
ln 2
[
__ (x – 1) ____________2 x3 = – __ + x2 + (3 – 2 ÷3 ) x + ______ ÷16 – (x – 1) 3 2
( ) ] )
16 x–1 3 + ___ sin–1 _____ 4 2 –1 __ 8p 32 = ___ – 4÷3 + ___ sq.u. 3 3 17. Given curves are
(
a2 x2 + b2 y2 = 1 and b2 x2 + a2 y2 = 1 On solving, we get, 1 _______ x = ± ________ =y 2 ÷ a + b2
(
)
1 = 4 Ú e–x – __ dx 2 0 ln 2 x = 4 – e– x – __ 2 0
(
)
( ) ( ) 1 ln2 = 4 ( __ – ___ ) 2 2
ln2 = 4 – e– ln 2 – ___ + 1 2 1 ln2 = 4 – __ – ___ + 1 2 2
= (2 – 2 ln 2) sq.u. 19. Given curves are y = min{x3, |x – 2|, e3 – x}, x-axis, y-axis and x = 4.
Area Bounded by the Curves
On solving, we get,
2
A = (1, 1), B = (3, 1), C = (4, e)
= Ú 1 dx = (x)2–2 = 4 –2
Hence, the required area
1
2
3
4
0
1
2
3
2. Given curve is |x| + |y| = 1
= Ú x3 dx + Ú (2 – x) dx + Ú (x – 2) dx + Ú (e3 – x) dx
( ) (
) (
)
3 x4 1 x2 2 x2 = __ + 2x – __ + __ – 2x + ( – e3 – x )43 4 0 2 1 2 2
( 9 1 = ( __ – __ sq.u. 4 e)
)
1 1 __ 1 1 = __ + __ + + 1 – __ e 4 2 2
Hence, the required area
20. Given curves are
S1:{(x, y): x2 + 2y2 £ 2}
= ar (ABCD)
S2: {(x, y): 2x2 + y2 £ 2}
= 4 (ar D OAB)
and
S3:{(x, y): x2 + y2 £ 2}
1 = 4 × __ × 1 × 1 2
= 2 sq.u.
3. Given curve is |x – 2| + |y – 2| = 1
The area of the first quadrant ___
÷2/3
__ ______
= Ú ( ÷2 ÷ 1 – x2 – x ) dx 0
( (
______
__
( ) )
)
___
Hence, the required area
x x x2 ÷2/3 1 = ÷2 __ ÷1 – x2 + __ sin–1 __ – __ 2 2 2 2 0
1 2 1 1 = ___ __ sin–1 __ + __ – __ 3 3 3 2 ÷
= 2 sq.u. 4. Given curve are
1 2 = ___ __ sin–1 __ 3 2 ÷ Hence, the required area
eln (x + 1) ≥ |y|, |x| £ 1
x + 1 ≥ |y|, |x| £ 1
|
( ÷ ) ( ÷ ) __
(
2p = 8 ___ – 8
(
|
__
1 = 4 × __ × 1 × 1 2
( ÷ ) ) ( ÷ ) ) __
1 2 ___ __ sin–1 __ 3 2 ÷ __
__ 2 = 2p – 4 ÷2 sin–1 __ sq.u. 3
Integer Type Questions
[
( ) ]
x x Œ [–2, 2], y = sin2x + cos __ = 1 4 Hence, the required area
1. For
Hence, the required area
3.47
3.48 Integral Calculus, 3D Geometry & Vector Booster
1 = 2 × __ × 2 × 2 2
= sq.u.
5. Given curves are
y = ln |x| and y = 1 – |x|.
Hence, the required area 1
= Ú (e – e x) dx 0
Hence, the required area
[
]
0
1 = 2 __ × 1 × 1 + Ú e x dx 2 – •
(
)
y = 3 – |x| and y = |x – 1|
Here,
AB = ÷1 2 + 12 =÷ 2
and
BC = ÷2 2 + 22 =÷ 8 = 2 ÷2
__
_______
__
= AB × BC
= ÷2 × 2 ÷2 = 4 sq.u.
__
= (e – e + 1)
= 1.
9. Given curves are
1 £ |x – 2| + |y + 1| £ 2
Putting
X = x – 2, Y = y + 1
Thus, the equation redices to 1 £ |X| + |Y| £ 2
Hence, the required area __
__
7. Given curves are
|x| y = __ x , x π 0 and y = x (x – 1)(x – 3)
Ï1 : x > 0 y= Ì and y = x (x – 1) (x – 3) Ó-1: x < 0
8. Given curves are y = e x, x = 0, y = e
Hence, the required area
= (e – e x )10
3 1 = 2 __ + 1 = 2 × __ = 3 sq.u. 2 2 6. Given curves are
_______
(
3 = 4 × __ = 6 sq.u. 2 10. Given curve is |y| = sin (2x), where 0 < x < 2p
)
1 1 = 4 __ × 2 × 2 – 2 × __ × 1 × 1 2 2
Area Bounded by the Curves
Hence, the required area p/2
= 4 Ú (sin (2x)) dx 0
(
)
cos(2x) p/2 = 4 – _______ 2 0
= – 2 (–1 – 1)
= 4 sq.u. 11. Given curve is
Hence, the required area,
( ) ( ( ) ) ( ) ( ( ) ( ) ) ( ) ( ) ( ) ( ) a
|x + y| + |x – y| £ 4, |x| £ 1, y ≥ |x – 1|
(
)
1 1 4 Ú __ x – ______ dx = log ___ __ 2x – 1 ÷5 2 a
fi fi fi fi
Hence, the required area
1 = __ × 2 × 2 2
= 2 sq.u.
4 £ x2 + y2 £ 2 (|x| + |y|)
(
64 a2 log ______ = log ___ 2a – 1 15
a2 ______ 2a – 1
| |
64 = ___ 15
1. Given curves are x2 = 4y and x = 4y – 2
Hence, the required area
Hence, the required area __
16 a2 4 log ______ – log __ = log ___ 2a – 1 3 5
fi a=8 Hence the value of a is 8.
12. Given curves are
x2 1 4 __ log ______ = log ___ __ 2 2x – 1 2 ÷5
)
2
= Ú (y2 – y1) dx
p (÷2 ) 2 = 4 ______ – (p – 2) 2
= 4 (p – (p – 2))
x + 2 = Ú _____ – 4 –1
= 8 sq.u.
1 = __ Ú (x + 2 – x2) dx 4 –1
x3 2 1 x2 = __ __ + 2x – __ 4 2 3 –1
13. Given curves are
1 1 ______ y = __ , x = 2, x = a, where a > 2 x , y = 2x – 1
–1 2
(
)
x2 __ dx 4
2
(
)
3.49
3.50 Integral Calculus, 3D Geometry & Vector Booster
) ( )) ( ( 9 7 1 10 __ 1 27 = __ ( ___ + ) = __ ( ___ = __ sq.u. 4 3 4 6) 8 6
8 1 1 1 = __ 2 + 4 – __ – __ – 2 + __ 4 3 2 3
2. The required area 1 = 4 × __ × 1 × 1 = 2 sq.u. 2
b
3. Given Ú f (x) dx = (b – 1) sin (3b + 4) 1
Differentiating both sides w.r.t. b, we get f (b) = sin (3b + 4) + 3 (b – 1) cos (3b + 4)
Hence, the required area
fi f (x) = sin (3x + 4) + 3 (x – 1) cos (3x + 4)
| | 0
4. Given curve is
= Ú tan x dx +
= (log secx)0 __p + (log sec)04
p – __ 3
|
It is given that
(
4
)
(
)
8 8 Ú 1 + __ 2 dx = Ú 1 + __ 2 dx a 1 x x
Ú
p __ 2
+ Ú cot x dx p __ 4
p __
|
– 3
p __
+ (log (sin x))p2
a
p __ 4 tan x dx 0
( )
( )
1 1 = log (2) + log ___ __ – log ___ __ 2 2 ÷ ÷
= log (2) sq.u. 6. Given curves are
( ) ( ) 8 8 fi ( a – __ a – 2 + 4 ) = ( 4 – 2 – a + __ a )
8 a 8 4 fi x – __ x 2 = x – __ x a
16 fi 2a – ___ a = 0 fi 2a2 = 16
Hence, the required area
fi a2 = 8
2
__
fi a = 2÷2
p p Ï ÔÔ y = tan x : - 3 £ x £ 3 Ì Ô y = cot x : p £ x £ p ÔÓ 6 2
2
_____
= Ú ÷5 – x2 dx – Ú |x – 1|dx –1
__
Hence, the value of a is 2÷2 . 5. Given curves are
__ 4
–1
( ) 3 5p = ( __ + ___ ) – 2 2 4 5p 1 = ( ___ – __ ) sq.u. 4 2 3 5p 5 = __ + ___ – __ 2 4 2
7. Given curves are
Area Bounded by the Curves __
Hence, the required area
(
__
0
( )
÷2
)
(
)
(
)
__
÷2 _____
x2 = Ú __ x dx + Ú – ___ dx – Ú __ ÷5 – x2 dx R2 0 – ÷2 – ÷2
( ) ( )
5 = – __ + (2 + 2p) 3 1 = 2p + __ sq.u. 3
Hence, the equation of the tangent is p y – 1 = 2 x – __ 4 p __ y – 1 = 2x – 2 p y = 2x + 1 – __ 2
x2 + y2 = 4, x2 = – ÷2 y and y = x
Hence, the required area p __ 4
x2 + y2 = 25, 4y = |4 – x2| and x = 0
( ( __ p4 – __ 12 ) ) × 1 p 1 1 p (log (sec x)) – __ ( __ – __ – __ 2 4 ( 4 2 ))
1 p Ú tan x dx – __ __ – 2 4 0
8. Given curves are
p __ 4 0
__ 1 log ( ÷2 ) – __ 4
1 1 __ log (2) – __ 2 4 10. Given curve is y = x (x – 1)2
Hence, the required area
( ( ) ( ) ( Ú ( ) Ú ( ) 2
4
5 ______
) )
p ◊ 52 x2 x2 = ____ – Ú 1 – __ dx + Ú __ – 1 dx + Ú ÷25 – x2 dx 4 4 0 2 4 4 2
4
5 _______
25p x2 x2 = ____ – 1 – __ dx + __ – 1 dx + Ú ÷25 – x2 d x 4 4 0 2 4 4
(
( ) )
25p 25p ___ 25 4 8 ____ 4 = ____ – __ + __ + – sin–1 __ + 6 4 3 3 4 2 5
Hence, the required area
25 4 = 2 + ___ sin–1 __ sq.u. 2 5
(
( ) )
9. Given curve is y = tan x dy fi ___ = sec2x dx
= Ú (2 – x (x2 – 2x + 1))dx 0
2
( )
dy p fi ___ x = __p = sec2 __ = 2 4 dx 4 p When x = __ , then y = 1 4
0
2
( )
2
= Ú (2 – x (x – 1)2)dx
= Ú (2 – (x3 – 2x2 + x))dx 0
(
)
x4 2 3 __ x2 2 = 2x – __ + __ x – 4 3 2 0
3.51
3.52 Integral Calculus, 3D Geometry & Vector Booster
(
)
16 = 4 – 4 + ___ – 3 3
10 = ___ sq.u. 3 11. Given curves are ln x y = ex ln x, y = ___ ex and x = 1
1
( 1
)
( )
4 – ÷2 __ 5 3 = ______ – log 2 + __ sq.u. 2 2 log 2
1
1 x2 x2 1 1 (log x) = __ e ______ + __ log x – __ 2 2 4 1/e 1/e 1
1 1 1 1 = – ___ – e ___ 2 – __ + ___ 2e 4 4e2 2e
e 1 1 1 = – ___ – ___ + __ – ___ 2e 2e 4 4e
) (
{
)
( ( 1
log x 1 = __ e Ú ____ x dx – e Ú x log x dx 1/e 1/e 2 1/e
__
2 13. Given curves are y = x2 and y = ______ 2 1 + x
(
( ) )
Hence, the required area
log x = Ú ____ ex – ex log x dx 1/e
) (
Hence, the required area
( (
__
2 ÷ 1 1 ____ – 2log 2 – 2 – __ log __ 2 2 log 2
4 = ____ – log 2
) ( ) }
)
2 = 2 Ú _____ 2 – x2 0 1 + x
)
x3 1 = 2 2tan–1x – __ 3 0 p 1 = 2 2 ◊ __ – __ 4 3 2 = p – __ sq.u. 3 14. Given curves are y = |x – 1| and y = 1
(
(
)
)
| |
e2 – 5 = _____ 4e 12. Given curves are
1 y = ln x, y = 2x, x = __ and x = 2 2 Hence, the required area 1 = __ × 2 × 1 = 1 sq.u. 2 15. Given curves are y = 4x – x2 and y = x2 –
Hence, the required area 2
= Ú (2x – log x) dx 1/2
( )
2x 2 = ____ – (x log x – x)21/2 log 2 1/2
Area Bounded by the Curves 5/2
Now,
A1 = Ú ((4x – x2) – (x2 – x)) dx
3.53
x2 = 1 + 2y, x2 = 1 – 2y
0
5/2
= Ú (5x – 2x2) dx
0
(
)
5x2 2x3 5/2 = ___ – ___ 2 3 0
125 ____ 250 125 = ____ – = ____ sq.u. 8 24 24
(
)
1
and
A2 = Ú (0 – (x2 – x)) dx 0
(
)
x2 x3 1 = __ – __ 2 3 0
1 = __ – 2
1 __ = 3
1 __ sq.u. 6
Now, Area of the above x-axis = A1 – A2
125 = ____ – 24
1 __ = 6
121 ____ 24
Hence, the required ratio
= (A1 – A2) : A2
121 1 = ____ : __ = 121 : 4 24 6
16.
Equation of the sides of a square are x = 1, x = –1, y = 1 and y = –1 Let (x, y) be any point inside the region S. Thus, according to the given conditions ______
÷x 2 + y2 < |1 + x|, |1 – x|, |1 + y|, |1 – y| fi
(x2 + y2) < (1 + x)2, (1 – x)2, (1 + y)2, (1 – y)2
y < 1 – 2x, y2 < 1 + 2x
and
2
2
x < 1 – 2y, x < 1 + 2y
The region S is the region lying inside the four parabolas y2 = 1 – 2x, y2 = 1 + 2x
The parabola y2 = 1 – 2x, x2 = 1 – 2y intersect the line y __= x and the point of intersection is (a, a), where a=÷ 2 – 1. Let A = Area of the region OPQO = Area of OPR + area RPQR 1/2
_______ 1 – 2x) dx = __ a2 + Ú ÷(1 2 a
1/2 1 2 1 × __ × (1 – 2x)3/2 = __ a2 – __ 2 3 2 a
1 1 (1 – 2a)3/2 = __ a2 + __ 2 3
(
)
a2 a3 = __ + __ 2 3 2 a = __ (3 + 2a) 6 __ __ 1 ) ( 3 + 2÷2 – 2 ) = __ ( 3 – 2÷2 6 __ 1 2 – 5 ) sq.u. = __ ( 4÷ 6 Hence, the required area
= Area of the shaded region
= 4(2A) = 8A
__ 8 = __ ( 4÷2 – 5 ) 6
__ 4 = __ ( 4÷2 – 5 ) sq.u. 3 17. It is given that, dy ___ = (2x + 1) dx
dy = (2x + 1)dx Integrating, we get
y = x2 + x + c
which is passing through the point (1, 2), so 2=1+1+c fi c=0
3.54 Integral Calculus, 3D Geometry & Vector Booster Hence, the equation of the curve is
y = x2 + x
Hence, the required area = A t
x2 = Ú x – __ dx t 0
x2 x3 = __ – __ 2 3t 01
t2 t2 = __ – __ 2 3
0
(
)
x3 x2 1 = __ + __ 3 2 0 1 = __ + 3
1 __ = 2
5 __ sq.u. 6
18. Given curves are
p y = tannx, y = 0, x = 0, x = __ . 4 p __ 4
Now,
An = Ú (tan x) dx n
0
Thus,
p __ 4
p __ 4
0
0
= Ú (tannx)dx + Ú (tann – 2x)dx p __ 4
= Ú (tann – 2x) (tan2x + 1)dx 0
p __ 4
= Ú (tann – 2x)sec2x dx 0
1
= Ú tn – 2dt, t = tan x 0
( )
t n – 1 1 = _____ n – 10
1 = _____ n–1
| |
19. Given curves are
(
)
( (
) )
x2 y = x – bx2 and y = __ b
t
3t2 – 2t2 = ________ 6 t2 = __ 6 2 b 1 = __ ______ 2 6 b +1 Taking logarith of both sides, we get
(
An + An – 2
)
t
= Ú (x2 + x) dx
) (
x2 b = Ú x – bx2 – __ dx, Let t = ______ 2 b 0 b +1
Hence, the required area 1
(
)
log A = 2log b – 2log (b2 + 1) – log 6
4b 1 dA 2 _______ fi __ ___ = __ – A db b (b2 + 1)
(
)
2 (b2 – 2b + 1) dA fi ___ = A _____________ db b (b2 + 1) dA For maximum or minimum, ___ = 0 gives db b = 1 Since b > 0, so b = 1 Thus A is maximum, when b = 1 20. Let the co-ordinates of P be (x, y) Equation of the line OA : y __= 0 Equation of the line OB : ÷__3 y = x Equation of the line AB : ÷3 y = 2 – x d (P, OA) = Distance of P from the line OA = y d (P, OB) = Distance of P from the line OB |R3y – x| = ________ 2 d (P, AB) = Distance of P from the line AB
|R3y – x – 2| = ___________ 2
Area Bounded by the Curves
It is given that, d (P, OA) £ min {d (P, OB), d (P,OA)}
3.55
From the graph, it is clear that
1 Ï 2 : 0£ x£ Ô(1 – x ) 3 Ô 1 Ô f (x) = Ì2 x(1 – x ) : £ x £ 1 3 Ô 1 Ô 2 : £ x £1 Ôx 3 Ó
Hence, the required area
{
__
}
|R3y – x| |___________ y – x + 2| ÷3 y £ min _______ , 2 2 |R3y – x| |R3y – x + 2| y £ ________ and y £ ___________ 2 2
|R3y – x| , then Case I: When y £ ________ 2 __ y ) ( x – ÷ 3 y £ ________ 2
( 2 + ÷3 ) y £ x
fi
y £ ( 2 – ÷3 ) x
fi
y £ (tan15°) x
__
__
|÷3 y – x + 2| , then Case II: When y £ ____________ 2 __ 2y £ 2 – x – ÷ 3 y __
fi
( 2 + ÷3 ) y £ 2 – x
fi
y £ ( 2 – ÷3 ) (2 – x)
fi
y £ – (tan (15°)) (x – 2)
__
2/3
1
0
1/3
2/3
(
) (
) ( )
(1 – x)3 1/3 2x3 2/3 = – _______ + x2 – ___ + 3 3 1/3 0
1 2 3 __ 1 = – __ __ + 3 3 3
__
fi
1/3
= Ú (1 – x)2dx + Ú 2x (1 – x)dx + Ú x2dx
x3 1 __ 3 2/3
( ( ) ) 2 2 2 1 2 1 + ( __ ) – __ ( __ – ( __ ) + __ ( __ 3 3 3) 3 3 3) 1 1 __ 2 + ( __ – __ 3 3(3) ) 2
3
2
3
3
17 = ___ sq.u. 27
22. Let C1 and C2 be the graphs of dunctions y = x2, y = 2x, 0 £ x £ 1 respectively. Let C3 be the graph of a function
y = f (x), 0 £ x £ 1, f (0) = 0.
Hence, the area of the shaded region
1 = __ × base × height 2
1 = __ × 2 × (1 ◊ tan15°) 2
= tan 15°
= ( 2 – ÷3 ) sq.u.
__
21. Given curve is
f (x) = max{x2, (1 – x)2, 2x (1 – x)}
Let the co-ordinates P be (x, x2), where 0 £ x £ 1. x
Area (ORPO)
= Ú t2 0
=
0
( ) ( ) ( ) ( ) x
Area (ORPO)
x
– Ú f (t)dt
_ ÷t dt
Ú 0
x2
t2 – Ú __ dt 0 2
x t2 x 2 = __ t3/2 – __ 3 4 0 0
2x3 x4 = ___ – __ 3 4
2
3.56 Integral Calculus, 3D Geometry & Vector Booster According to the question,
(
x
)
x3 2x3 x4 __ – Ú f (t)dt = ___ – __ 3 0 3 4 Differentiating both sides w.r.t. x, we get
x2 – f (x) = 2x2 – x3 3
(
)|
0
x2 x3 = (1 – m) __ – __ 2 3 1 – m
1 = – __ (1 – m)3 6
It is also given that
2
fi f (x) = x – x Hence, the required curve is f (x) = x3 – x2, where 0 £ x £ 1. 23. Case-I: When m < 0
9 1 – __ (1 – m)3 = __ 2 6
fi
(1 – m)3 = – 27
fi
1 – m = – 3
fi m = 4. 24. We have, Ï x 2 + ax + b : x > –1 Ô : –1 £ x £ 1 f (x) = Ì2 x Ô 2 Ó x + ax + b : x > 1 As f is continuous on R, f is continuous at –1 and 1,
Area of the region 1 – m
= Ú ((x – x2) – mx) dx 0
(
)|
1 – m
x2 x3 = (1 – m) __ – __ 2 3 0
1 1 = __ (1 – m)3 – __ (1 – m)3 2 3
1 = __ (1 – m) 6 Also, it is given that,
9 1 __ (1 – m)3 = __ 2 6 fi fi fi
(1 – m)3 = 27 (1 – m) = 3 m = – 2.
Case II: When m > 0
Area of the region
lim f (x) = lim f (x) = f (–1) – + and lim + f (x) = lim f (x) = f (1) –
= Ú (x – x2 – mx) dx 1 – m
x Æ 1
x Æ 1
Thus,
1 – a + b = – 2 and 1 + a + b = 2
fi
a – b = 3 and a + b = 1
fi
a = 2 and b = –1.
Hence,
Ï x 2 + 2 x – 1 : x > –1 Ô f (x) = Ì2 x : –1 £ x £ 1 Ô 2 Óx + 2x – 1 : x > 1
Let us find the point of intersection of
x = – 2y2 and y = f (x).
These two curves meet at (– 2, –1)
The required area
0
x Æ – 1
x Æ – 1
– 1/8
( ÷
___
)
– x = Ú ___ – f (x) dx 2 – 2
Area Bounded by the Curves – 1/8
( ÷ )
– 1/8
( ÷ )
– 1
___
Y
– x = Ú ___ dx – Ú f (x) dx 2 – 2 – 2 –1/8
___
P(9, 3)
M(0, 3)
– x = Ú ___ dx – Ú (x2 + 2x – 1) dx 2 – 2 – 2
– Ú f (x) dx
X¢
X
O
– 1/8
Y¢
– 1
) (
(
– 1/8 – 2 = ____ __ (– x)3/2 – 3÷2 – 2
Hence, the required area 3
( ( )
) (
)
2__ __ 1 3/2 1 = – ____ – 23/2 – – __ + 1 + 1 8 3 3÷2
(
) (
)
8 1 + – __ + 4 + 2 – ___ – 1 3 64 __
2 __ ÷ = ___ ( ÷2 – 2– 9/2 ) + 3
881 = ____ 192
5 __ + 3
= Ú (2y + 3 – y2) dy 0
– (– x2)– 1/8 – 1
)
– 1 x3 __ + x2 – x 3 – 2
63 ___ 64
25. Let b π 0 and for j = 0, 1, 2, ..., n, let S1 be the area of the region bounded by the y-axis and the curve jp (j + 1)p xe xy, __ £ y £ _______ . Show that S0, S1, ..., Sn are b b in G.P. Also, find their sum for a = –1 and b = p [IIT-JEE, 2001] 26. The given curves are y = |x| – 1 and y = 1 – |x|
(
)
y3 3 = y2 + 3y – __ 3 0 =9+9–9 = 9 sq.u.
dy (x + 1)2 + (y – 3) 28. Given ___ = _______________ (x + 1) dx dy y 3 fi ___ = (x + 1) + ______ – ______ (x + 1) (x + 1) dx dy y 3 fi ___ – ______ = (x + 1) – ______ (x + 1) dx (x + 1) which is a linear differential equation. Thus,
dx – Ú____ x + 1 =
IF = e
1 e– log(x + 1) = ______ (x + 1)
Therefore, the solution is y 3 _____ = Ú 1 – _______ dx x+1 (x + 1)2
(
fi
)
y 3 _____ = x + _____ + c x+1 x+1
Put x = 2 and y = 0, then c = – 3 Hence, the equation of the curve is y 3 _____ = x + _____ – 3 x+1 x+1 Hence, the required area
= ar (Quad ABCD)
= 4ar (DOAB)
1 = 4 × __ × 1 × 1 2
fi
y = x2 + x + 3 – 3x – 3 = x2 – 2x.
= 2 sq.u. 27. The given curves are
_
y = ÷x , 2y + 3 = x and y = 0 Hence, the required area
3.57
3.58 Integral Calculus, 3D Geometry & Vector Booster Hence, the required shaded area
2
= Ú (0 – (x2 – 2x))dx
0
0
(
)
x3 2 = x – __ 3 0
8 = 4 – __ = 3
2
( )
4 __ sq.u. 3
29. The given curves are
y = ax2 and x = ay2, where a > 0
(
)
1/2
(
)
1 1 = Ú (x + 1)2 – __ dx + Ú (x – 1)2 – __ dx 4 4 –1/2 0
0 (x + 1)3 __ 1 1 1/2 = _______ – x + (x – 1)3/3 – __ x 3 4 –1/2 4 0
1 1 1 1 1 1 = __ – ___ + __ – ___ – __ + __ 3 24 8 24 8 3
1 1 1 1 = __ – 2 ___ + __ + __ 3 24 8 3
2 2 __ 2 1 __ 1 = __ – __ = – __ = sq.u. 3 6 3 3 3
(
) (
(
)
)
(
)
31. The given curves are
x2 = y, x2 = – y and y2 = 4x – 3
Hence, the required area 1/a
__
(÷
)
x = Ú __ a – ax2 dx 0
1/a 1 2 1 = ___ __ ◊ __ x3/2 – __ ax3 ÷a 3 3 0
1 2 1 1 = ___ __ ◊ __ a3/2 – __ a ◊ __3 ÷a 3 3 a
(
)
(
)
1 = ___ 2 3a It is also given that, 1 ___ 2 = 1 3a 1 fi a2 = __ 3
÷
__
1 a = ± __ 3 1 fi a = ___ __ , since a > 0. 3 ÷ 30. The given curves are fi
y = (x + 1)2, y = (x – 1)2 and y = 1/4
Hence, the required area
(
1
1
0
3/4
______
)
= 2 Ú x2dx – Ú ÷4x – 3 dx
( ( ) (
))
3/2 1 x3 1 2 (4x – 3) = 2 __ – __ ◊ _________ 4 3 0 3 3/4
1 1 = 2 __ – __ – 0 3 6
1 1 = 2 ◊ __ = __ 6 3
( ( ) ( ) )
32.
The given system of equations 4x2f (–1) + 4x f (1) + f (2) = 3x2 + 3x is satisfied for 3 distinct real numbers a, b and c. Comparing the co-efficients of x2, x and constant terms, we get 4f (–1) = 3, 4f (1) = 3, f (2) = 0 Let
f (x) = ax2 + bx + c
Given
3 f (–1) = __ 4
fi
3 a – b + c = __ 4
3 f (1) = __ 4
fi
3 a + b + c = __ 4
b=0 f (2) = 0 4a + 2b + c = 0 c = – 4a we get
Thus, And fi fi Solving,
3.59
Area Bounded by the Curves
| |
|
130 = 80 + 8 – 3 – ____ 3 130 = 85 – ____ 3 125 = ____ 3
|
_______
÷
1 + sin x 33. We have C1 : y = _______ cos x
÷ ÷
______________
1 a = – __ , b = 0, c = 1 4
Thus,
1 f (1) = – __ x2 + 1 4
(
p 1 – cos __ + x 2 = _____________ p __ sin + x 2
(
)
)
______________________
(
p x 2sin2 __ + __ 4 2 = _____________________ p x p x 2 sin __ + __ cos __ + __ 4 2 4 2
(
) ) (
)
÷ ( ) p x 1 – sin x Similarly, C : y = ÷ _______ = tan ( __ cos x ÷ 4 – __2 ) _________
p __ x = tan __ + 4 2
_______
__________
2
Hence, the area of the shaded region
(
)
Let
1 – 0 1 m1 = m (AV) = _____ = – __ 0–2 2 t2 1 – __ –1 t 4 m2 = m (BV) = ________ = – __ t–0 4
m1 × m2 = –1
fi
t 1 – __ – __ = –1 2 4
p/4
( (
__
÷2 – 1
( ) ( )
__
( ( ) ) Ú ( ) ( )
____________ 2
____________ 2
(
)
)
)
(
) (
( ) )
4t dt_____ x = Ú ____________ , Let tan __ = t 2 2 2 0 (1 + t )÷1 – t
34. Given curve is
2
2
x2 3 = 4 – __ – __ x dx 4 2 – 8
x3 = 4x – ___ – 12
8 128 = 8 – ___ – 3 – – 32 + ____ – 48 12 3
3x2 2 ___ 4 – 8
) (
__________
2t (2dt) _____ = Ú ____________ 0 (1 + t2)÷1 – t2 ÷2 – 1
3 x2 = Ú 1 – __ – __ (x – 2) dx 4 2 – 8
( (
0
fi t = – 8 Therefore, the co-ordinates of B = (– 8, –15). Hence, the required area
0
p – AVB = __ , so 2
__________
p __ p __ x x = Ú tan __ + – tan __ – dx 4 2 4 2 0 p/4
( ÷ (
) ÷ ( ) ) x x ( 1 + tan ( __ ) ) – ( 1 – tan ( __ ) ) 2 2 = Ú ________________________ dx x ( 1 – tan ( __ ) ) 2 ÷ x 2tan ( __ ) 2 = Ú _____________ dx x ( 1 – tan ( __ ) ) 2 ÷ p/4
t2 Let the co-ordinates of B be B t, l – __ 4
))
y3 – 3y + x = 0
fi
3y2y ¢ – 3y ¢ + 1 = 0
fi
(3y2 – 3)y ¢ + 1 = 0
fi
1 1 y ¢ = – ________ = – ________ 2 2 (3y – 3) 3(y – 1)
...(i)
fi
2yy ¢ y ≤ = – ________ 2 2 3(y – 1)
...(ii)
3.60 Integral Calculus, 3D Geometry & Vector Booster From Eq. (i), we get
Also, the shaded area
From Eq. (ii), we get
1
__ 1 1 y ¢ (– 10 ÷2 ) = _______ = – ___ 3(1 – 8) 21 __
0
__ 4÷2 _______
2 (2÷2 ) (–1) y ≤(– 10÷2 ) = ___________ = – 2 3(1 – 8)2(21) (3) (7)3 __
35. Hence, the required area
1
1
= Ú e dx – Ú exdx
0
0
1
= e – Ú e xdx
0
b
38. Given f (x) = 1 + 2x + 3x2 + 4x3 f ¢(x) = 2 + 6x + 12x2 > 0 for all x in R Thus, f (x) is an increasing function on R. So, f (x) can have atmost one root. It is clear that f (x) cannot have a poistive real root. We have
= Ú f (x)dx
= (x f (x))ba – Ú (x f ¢(x))dx
x 1 = (bf (b) – af (a)) – __ Ú _____ 2 dx 3 a 1–y
x 1 = (bf (b) – af (a)) + __ _____ dx 3 a y2 – 1
x 1 = (bf (b) – af (a)) + __ _________ dx 3 a ( f (x))2 – 1
36. For
x Π(Р2, 2),
(g (x))3 – 3g (x) + x = 0
a
b
a
( ) Ú ( ) Ú ( ) b
b
3
(g (– x)) – 3g (– x) – x = 0
( ) 3 1 __ 1 1 and f ( Р__ ) = 1 Р1 + __ Р__ = > 0 2 4 2 4 3 1 Hence, f (x) has a root in ( Р__ , Р__ ). 4 2 3 1 39. Let a Π( Р__ , Р__ ) and t = |x| = |a| 4 2
b
= Ú (e – ex ) dx
...(i) ...(ii)
Adding Eqs. (i) and (ii), we get fi {(g (x))3 + (g (x))3} – 3{g (x) + g (– x)} = 0
fi {g (x) + g (– x)}{(g (x))2 + (g (– x))2 – g(x)g(– x)) – 3} = 0
fi {g (x) + g (– x)} = 0
As fi
1
3 3 27 ___ 27 1 f – __ = 1 – __ + ___ – = – __ < 0 4 2 16 16 2
3 1 – __ < a < – __ 4 2 3 1 – __ < t < – __ 4 2 t
1/2
Now, Ú g ¢(x)dx
fi
–1
= (g (x))1–1 = g (1) – g (–1) = g (1) + g (1) = 2g (1) 37. We have
3/4
Ú f (x)dx < Ú f (x)dx < Ú f (x) dx 0
0
2
3
0
x4)1/2 0
fi
(x + x + x +
< Ú f (x)dx < (x + x2 + x3 + x4)3/4 0
t
0
(
)
t
fi
1 1 __ 1 1 __ + __ + + ___ < Ú f (x)dx 2 4 8 16 0
9 3 81 27 < __ + ___ + ___ + ____ 4 16 64 256
fi
15 525 ___ < Ú f (x)dx < ____ 16 0 256
fi
3 __ < 4
(
)
t
Hence, the required area e
= Ú x dy 1 e
= Ú ln y dy
1 e
= Ú ln (1 + e – y) dy 0
t
15 525 ___ < Ú f (x)dx < ____ < 3 16 0 256
40. We have,
f (x) = 1 + 2x + 3x2 + 4x3
fi f ¢(x) = 2 + 6x + 12x2
Area Bounded by the Curves
fi f ≤(x) = 6 + 24x By the sign scheme, we get that
(
fi
(
fi 1 1 f ¢(x) decreases in – •, – __ and increases in – __ , • 4 4 fi 1 In particular f ¢(x) decreases in – t, – __ and increases 4 fi 1 in – __ , t . 4 43.
(
)
(
)
)
)
3.61
2 1 __ (1 – b)3 = ___ 3 12 1 (1 – b)3 = __ 8 1 (1 – b) = __ 2 1 b = __ 2
2
41. We have R1 = Ú x f (x) dx
– 1 2
= Ú (1 – x)f (1 – x) dx –1 2
= Ú (1 – x)f (x) dx –1
2
2
–1
–1
= Ú f (x)dx – Ú x f (x) dx
= Ú f (x)dx – R1 –1
2
fi
1 S ≥ __ e (Since area of a rectangle OCDS = 1/e),
2
2R1 = Ú f (x)dx = R2
2
Since
e– x ≥ e– x, " x Œ [0, 1]
fi
1 S ≥ Ú e– xdx = 1 – __ e 0
1
–1
42. The given curves are y = (1 – x)2, y = 0 and x = 0
(
)
Area of a rectangle OAPQ + Area of a rectangle QBRS > S 1__ 1 –__1 ___ 1 fi S £ ___ (1) + _____ __ e ÷ 2 ÷2 ÷
( ) ( )
(
)
1 1 1 Since __ 1 + ___ __ £ 1 – __ e ÷e 4 Thus, option (c) is incorrect. 44. The given curves are y = sin x + cos x and y = |cos x – sin x|
b
We have, R1 = Ú (1 – x)2dx 0
(
)
(1 – x)3 b = – _______ = 3 0
1 __ (1 – (1 – b)3) 3
1
Also,
R2 = Ú (1 – x2)dx b
Now, fi fi
(
)
(1 – x)3 1 __ 1 = – _______ = ((1 – b)3) 3 3 b 1 R1 – R2 = __ 4 1 1 1 __ (1 – (1 – b)3) – __ (1 – b)3 = __ 3 3 4 2 1 1 1 __ (1 – b)3 = __ – __ = ___ 3 3 4 12
Hence, the required area p/2
= Ú (sin x + cos x) dx 0
(
p/4
p/2
0
p/4
)
– Ú (cos x – sin x)dx + Ú (sin x – cos x)dx
3.62 Integral Calculus, 3D Geometry & Vector Booster
= – (cos x)p/2 + (sin x)p/2 0 0
__
(
– (sin x)p/4 + 0
(cos x)p/4 – 0
)
(sin x)p/2 p/4
(cos x)p/2 p/4
(
Area = { ( x, y) Œ R2 : y ≥ ÷|x| , 5y £ x + 6 £ 15 }
)
1 1 = – (0 – 1) + (1 – 0) – ___ __ + ___ __ – 1 2 ÷2 ÷
(
) (
)
1__ 1 – 0 – ___ – 1 – ___ __ 2 2 ÷ ÷
45. Shifting the origin to (– 3, 0), we get the
(
)
__ 1 1 = 2 – ÷2 – 1 + ___ __ – 1 + ___ __ 2 2 ÷ ÷ __
= 2 – ( 2÷2 – 2 )
= 4 – 2÷2
= 2÷2 ( ÷2 – 1 ). Sq.u.
__
__
__
Hence, the required area = Region (OPK) + Region (QLKR) + Region (OLQ) – Triangle (PQR)
8 5 3 1 5 = __ + 1 + __ – __ = 4 – __ = __ sq.u. 3 3 2 2 2
Chapter
4
Differential Equation
Concept Booster 1. Introduction A differential equation is a mathematical equation for an unknown function of one or more variables that relates the values of the function itself and its derivatives of various orders. It plays a prominent role in engineering, physics, economics, biology, and other disciplines. Differential equations arise in many areas of science and technology, specifically whenever a deterministic relation involving some continuously varying quantities (modelled by functions) and their rates of change in space and/or time (expressed as derivatives) is known or postulated. This is illustrated in classical mechanics, where the motion of a body is described by its position and the velocity as the time value varies. Newton’s laws allow us (given the position, velocity, acceleration and various forces acting on the body) to express these variables dynamically as a differential equation for the unknown position of the body as a function of time. In some cases, this differential equation (called an equation of motion) may be solved explicitly. An example of modelling a real world problem using differential equations is the determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The ball’s acceleration towards the ground is the acceleration due to gravity minus the acceleration due to air resistance. The gravity is considered constant, and the air resistance may be modelled as proportional to the ball’s velocity. It means that the ball’s acceleration, which is a derivative of its velocity, depends on the velocity (and the velocity depends on time). Finding the velocity as a function of time involves solving a differential equation. Differential equations are mathematically, studied from different perspectives, mostly concerned with their
solutions—the set of functions that satisfy the equation. Only the simplest differential equations admit solutions given by explicit formulae; however, some properties of solutions of a given differential equation may be determined without finding their exact form.
2. Definition An equation that involves the dependent variables, independent variables and the derivatives of dependent variables is called a differential equation, i.e.
{
{
}
}
dy dx x, y, ___ = 0 or x, y, ___ = 0 dx dy For examples, dy 1. ___ = sin x dx dy3 d 2y dy 2. ___3 + 4 ___2 + 5 ___ + 10 y = 0 dx dx dx ________
÷ ( ) ÷ ________
( )
dy 2 d2y 3. 1 + ___ = 2 + ___ 2 dx dx dw du dv ___ 4. ___ + ___ + = u + v + w dx dx dx
each are differential equations.
3. Ordinary Differential Equation A differential equation that involves the derivative with respect to a single independent variable is called an ordinary differential equation. It can also be expressed as a function of variables x and y, and the derivatives of y w.r.t. x, i.e.
(
)
dy f x, y, ___ = 0. dx
Thus, dy 1. ___ + 10 y = 0 dx
4.2 Integral Calculus, 3D Geometry & Vector Booster
dy 2. ___ + 2y = sin x dx
Example 6. The order of the differential equation whose
d2y dy 3. ___2 + 2 ___ + 5y = 0 dx dx
y = (c1 + c2) cos (x + c3) – c4 ex + c5,
where c1, c2, c3, c4 and c5 are arbitrary constants
d ny 4. ___n + ny = 0 dx
is 3.
are ordinary differential equations.
4. Partial differential equation A differential equation which contains two or more independent variables and partial derivatives w.r.t. them is called a partial differential equation. For examples,
∂z ∂ z 1. ___ + ___ = 0 ∂ x ∂ y
∂2 z ____ ∂2 z 2. ____2 + 2 = x ∂ x ∂ y
5. Order
of a
6. Degree
genral solution is given by
of a
differential equation
The index power of the highest order derivative of a differential equation is called the degree of a differential equation, and is free from any radical sign.
Example 1. The degree of the differential equation
( )
( )
d3y 2102 d 4y dy 2013 ___ 3 + ___4 + ___ + 5005y = 0 dx dx dx
is 1.
Example 2. The degree of the differential equation
÷(
) ÷(
_________
differential equation
The order of a differential equation is the highest order derivative of a differential equation. Example 1. The order of the differential equation
__________
)
5 d2y d2y 1 + ___ 2 = 10 + ___ dx dx2
4
is 5.
Example 3. The degree of the differential equation
dy dy ___2 + 2 ___ + 5y = 0 dx dx is 2.
dy ___ + dx
Example 2. The order of the differential equation
Example 4. The degree of the differential equation
d 2y 100 ___2 + dx
edx = x + 1
2
( )
d 4y ___ 4 + dx
dy ___ + 50y = 0 dx
is 4.
Example 3. The order of the differential equation
( )
dy sin ___ = x dx
is 1.
Example 4. The order of the differential equation of the
curve
y = (a + b) e3x + (c + d) e–2x
a, b, c, d, ΠR
is 2. Example 5. The order of the differential equation of the curve
y = (a + b) cos ((c + d) x + e),
where
a, b, c, d, e, ΠR
is 3.
dx ___ = y dy
is 2. dy ___
is not defined.
Example 5. The degree of the differential equation
(
)
dy log 1 + ___ = x dx is not defined.
Example 6. The order and the degree of the differential
equation
(
)
( )
dy 2/3 d3y 1 + 3 ___ = 4 ___ 3 dx dx
are 3 and 3, respesctively
Example 7. The order and the degree of the differential
equation
÷
_______
÷( ( ) )
__________
d2y 4 dy 5 1 + ___ 2 = y + ___ dx dx 5
are 2 and 4, respectively.
Differential Equation
7. Linear
and
non-linear differential equation
A differential equation in which the dependent variables and all its derivatives present occur in the first degree only and no products of dependent variables and/or derivatives occur is known as linear differential equation. A differential equation which is not linear is called a nonlinear differential equation. For examples,
dy 1. ___ + 5y = x is linear. dx
d2y dy 2. ___2 + 5 ___ + 6y = 0 is linear. dx dx
d 2y 3. ___2 + dx
d2013y dy 4. _____ + 5 ___ + 10y = 0 is linear. 2013 dx dx
8. Formation
first order
(ii) (iii) (iv) (v) (vi) (vii) (viii)
fi
dy = f (x) dx
Integrating, we get
Ú dy = Ú f (x) dx
dy ___ = f (x) ◊ f (y) dx fi
dy ____ = f (x) dx f (y)
Integrating, we get
and
first degree
All differential equations of the first order cannot be always solvable. However they can be solved by suitable methods if they belong to any of the following standard forms. (i)
dy ___ = f (x) dx
Let the differential equation be
A differential equation of the first order and the first degree can be written as the form of dy ___ = f (x, y) or M dx ± N dy = 0, where M and N are dx functions of x and y or constants.
9.2 Variable Separable Form
Let the equation of the curve be f (x, y, c1, c2 ..., cn) = 0, ...(i) where c1, c2, ..., cn are n arbitrary constants apart from x and y. To obtain the differential equation of the curve (i), we have to differentiate (i) up to n times w.r.t. x and eliminate those arbitrary constants from (n + 1) equations, which yield the required differential equation. of
9.1 differential equation of the form
Thus, y = j (x) + c is the required solution.
differential equation
9. Differential equation
(ix) Reducible to exact differential equation. (x) Clairauts differential equation.
fi
dy y ___ + 10x = 0 is not linear. dx
of a
4.3
dy A differential equation is of the form ___ = f (x). dx Variable separable form. Reducible to variable separable form. Homegeneous differential equation. Reducible to homegeneous differential equation. Linear differential equation. Bernoulli’s differential equation. Exact differential equation.
dy
= f (x) dx Ú ____ f (y) Ú
fi j (y) = y (x) + c, which is the required solution.
9.3 Reducible to variable separable form Let a differential equation is of the form dy ___ = f (ax + by + c) dx Let
(ax + by + c = v)
(
)
dy 1 ___ dv fi ___ = __ – a dx b dx The Eq. (i) reduces to
(
)
1 dv __ ___ – a = f (v) b dx dv fi ________ = dx b f (v) + a
Integrating, we get
dv fi Ú ________ = dx bf (v) + a Ú fi j (v) = x + c fi
j (ax + by + c) = x + c,
which is the required solution.
...(i)
4.4 Integral Calculus, 3D Geometry & Vector Booster
9.4 Homogeneous differential equation Homogeneous equation If the degree of each term throughout the equation is the same, it is called a homogeneous equation. For examples, 2
2
2
x3 + xy2 + x2y + y3 = 0 are the homogeneous equation of 3rd degree. A homogeneous differential equation is of the form f (x, y) dy ___ = ______ ...(i) dx g (x, y) where the degree of f (x, y) is the same as the degree of g (x, y). In this case, divide the numerator and the denominator by the highest power of x. Thus the given differential equation reduces to dy y ___ = f __ x ...(ii) dx
( )
dy dv ___ = v + x ___ ◊ dx dx y __ Let x = v fi y = vx. fi
dv
j (v) = log |x| + c
fi
y j __ x = log |x| + C
a1a + b1b + c1 = 0 and a2a + b2 b + c2 = 0.
Solving, we get
b1c2 – b2c1 a2c1 – a1c2 a = __________ and b = __________ . a1b2 – a2b1 a1b2 – a2b1 Then the Eq. (ii) reduces to ...(iii)
X V dY ___ = V + ___ dX dX
Then the Eq. (iii) reduces to dV a1 + b1V V + X ___ = ________ . dX a2 + b2V
dx
fi
Let us choose a and b in such a way that
fi
Integrating, we get = ___ Ú _______ f (v) – v Ú x
...(ii)
which is a homogeneous equation of the first degree. Y Let __ = V X
dv dx _______ = ___ x f (v) – v
(a1X + b1Y) + (a1a + b1b + c1) = ___________________________ (a2X + b2Y) + (a2a + b2b + c2)
dY a1X + b1Y ___ = _________ dX a2X + b2Y
The Eq. (ii) reduces to dv v + x ___ = f (v) dx
dx = dX, dy = dY
dY a1(X + a) + b1(Y + b) + c1 ___ = ________________________ dX a2(X + a) + b2 (Y + b) + c2
2. x3 + x2y + y3 = 0,
fi
fi
dy dY Thus, ___ = ___ . dx dX Now, Eq. (i) can be reduced to
2
1. x + y = 0, x + xy + y = 0 are the homogeneous equations of 2nd degree.
The Eq. (i) can be reduced into homogeneous equation form by means of suitable substitutions. x = X + a, y = Y + b, where a, b are constants.
Integrating, we get j (V) = log |X| + k
( )
fi
Y j __ = log (X) + c X
which is the required solution.
fi
y–b j x_____ – a = log (x – a) + c
9.5 Reducible to homogeneous differential equation
which is the required solution of the given differential equation.
( )
Let a differential equation is of the form dy a1x + b1y + c1 ___ = _____________ , dx a2x + b2y + c2 a1 where __ a2 π
b1 __ , b2
( )
9.6 Linear differential equation ...(i)
A differential equation is said to be linear differential eqation of the first order, when the dependent variable and its derivatives appear only in the first degree. Let a differential equation is of the form
Differential Equation
dy ___ + Py = Q ...(i) dx where P and Q are either functions of x or constants.
Ú pdx
Here IF = e = e . Multiplying both sides of Eq. (i) by IF, we get dy e px ___ + P e px y = Q ◊ e px dx d fi ___ (y e px) = Q ◊ epx dx Integrating, we get y ◊ e px = Ú Q ◊ epx dx + c
fi
y ◊ (I.F) = Ú Q ◊ (I.F) dx + c
fi
y ◊ (I.F) = j (x) + c
fi
y ◊ e px = j (x) + c
fi
y = (j (x) + c) e– px
Note: Sometimes it is convenient to express the differential equation in the form dx ___ + Px = Q dy where P and Q are either functions of y or constants. IF = e Ú Pdy = e Py
...(i)
where P and Q are either functions of x alone or constants.
dv Thus, ___ + P (1 – n) v = Q (1 – n) dx which is a linear differential equation. IF = e Ú P (1 – n) dx = e (1 – n) Ú pdx Thus the required solution is
v (IF) = Ú Q ◊ (IF) dx + c
(
)
fi y (1 – n) ◊ e (1 – n) Ú pdx = Ú Q e(1 – n) Úpdx dx + c
Its solution is given by u (x, y) = c, " c Œ R. For examples, 1. ex ◊ cos y dy + ex ◊ sin y dx = 0 is an exact differential equation. Since, it is derived by d (ex sin y) = 0 fi
ex sin y = c
which is the required solution. 2.
x dy + y dx = 0 is an exact differential equation. fi d (x y) = 0 fi xy = c which is the required solution.
Dear friends you should remember the follwoing exact differentials. These help us to find the integrating factors.
9.7 Bernouli’s differential equation
Dividing both sides of Eq. (i) by yn, we get dy y–n ___ + P y1–n = Q, dx dy dv 1 ___ Let y1– n = v fi y–n ___ = ______ . dx (1 – n) dx
du = Mdx + Ndy = 0
General form of variable seperation (By Inspection)
Thus the solution is x (IF) = Ú Q (I.F) dx + c
A differential equation is of the form dy ___ = Py = Qyn dx
9.8 Exact differential equation
which is the required solution of the given differential equation.
Then
is the required solution of the given differential equation.
A differential equation is said to be an exact differential equation, if it is formed by the equation, an exact differential to zero. Thus a differential equation is of the form M (x, y)dx + N (x, y) dy = 0 said to be exact, if the expression M (x, y) dx + N (x, y) dy is the exact differential of some function u (x, y) i.e.
px
4.5
1. d (x ± y) = dx ± dy
2. d (x y) = x dy + y dx
y dx – x dy x 3. d __ y = _________ y2
( ) y x dy – y dx 4. d ( __ x ) = _________ x 2
dx + dy 5. d [log (x + y)] = _______ x+y
x dy + y dx 6. d [log (xy)] = __________ xy
y dx – x dy x 7. d log __ y = _________ xy
y x dy – y dx 8. d log __ x = _________ xy
x dx + y dy 1 9. d __ log ( x2 + y2 ) = __________ 2 2 x + y2
( ( ) )
[ ( ) ]
[
]
4.6 Integral Calculus, 3D Geometry & Vector Booster
[ ( ) ]
x+y x dy – y dx 1 10. d __ log _____ = _________ 2 x – y 2 x – y2
( ) y dx – x dy x 12. d [ tan ( __ y ) ] = _________ x +y
x dy + y dx 1 _________ 11. d – __ xy = x2y2
fi
)
Æ (x2 + y2) x = r cos q, y = r sin q,
( )
y x2 + y2 = r2, q = tan–1 __ x x dx + y dy = r dr, x dy – y dx = r2 dq
16. when the expression is in the form of: Æ (x2 – y2) Let
x = r sec q, y = r tan q,
where
y x2 – y2 = r2, q = sin–1 __ x
fi
x dx – y dy = r dr, x dy – y dx = r2 sec q dq
( )
10. Orthogonal Tragectories If a curve intersects a family of curves at right angles, then previous one is the orthogonal trajectory to the later one. The differential equation of the orthogonal trajectories of dy the curves f x, y, ___ = 0 is the family of the curves, whose dx
(
)
(
pn + f1 pn–1 + f2 pn–2 + ...
15. when the expression is in the form of
fi
dy + fn–1 ___ + fn = 0 dx
______ x dx + y dy 14. d ÷x 2 + y2 = _________ ______ . ÷x 2 + y2
where
( )
dy + fn–1 (x, y) ___ + fn (x, y) = 0 ...(i) dx
( ) ( ) ( ) ( )
2
[ ( ) ]
Let
( )
dy n dy n – 1 dy n–2 fi ___ + f1 ___ + f2 ___ + ... dx dx dx
y x dy – y dx 13. tan–1 __ x = _________ 2 x + y2
(
( )
–1
2
( )
dy n dy n – 1 dy n – 2 ___ + f1 (x, y) ___ + f2 (x, y) ___ + ... dx dx dx
)
dx differential equation is f x, y, Р___ = 0. dy ule to find out the orthogonal trajectory: R (i) Let the equation of the family of curves is f (x, y, c) = 0, where c ΠR.
+ fn – 1 p + fn = 0 ...(i)
dy p = ___ dx and f1, f2, ..., fn are functions of x and y. The Eq. (i) can aslo be written as f (x, y, p) = 0 where
...(ii)
For examples, 1. The equation dy 5 dy y ___ + 2013 x ___ – y = 0 dx dx is a differential equation of 5th degree. 2. The equation dy 2014 dy ___ + x ___ – 2015y = 0 dx dx is a differential equation of 2014th degree. The differential equation f (x, y, p) = 0 can be solvable (i) for p (ii) for x (iii) for y (iv) for the first degree in x and y. Now, we shall discuss the varoius methods to solve the above types of equations.
( )
( )
( )
( )
(i) Equations solvable for p
Consider a differential equation of the form (p – f1 (x, y)) (p – f2 (x, y)) ... (p – fn (x, y)) = 0,
(ii) Form a differential equation of the given curve. dy dx (iii) Replace ___ by – ___ . dx dy dx (iv) Solve the equation f x, y, – ___ = 0, dy
which is solvable for p.
We will get the required orthogonal trajectories.
Each of these equations is of it first order and of first degree. Let the solutions of these equations are given by j (x, y, c1) = 0,
(
)
11. First Order Higher Degree Differential Equation The most general form of a differential equation of the first order and of higher degrees (say nth degree) is
fi [p – f1 (x, y)] = 0,
(p – f2 (x, y)) = 0
(p – fn (x, y)) = 0
j 2(x, y, c2) = 0
4.7
Differential Equation
j n (x, y, cn) = 0,
where c1, c2,
dy where P = ___ , dx
cn are arbitrary constants. Hence, the general solution is
y = Px + f (P),
is called the Clairaut’s differential equation.
j1(x, y, c1) ◊ j2(x, y, c2) ... qn (x, y, cn) = 0
Rule to find the Clairaut’s differential equation
which contains n arbitrary constants, whereas an equation of the first order and of first degree should contain only one arbitrary constant. So, we can take, without loss of generality, c1 = c2 = cn = c. Hence, the genral solution of Eq. (i) is
j 1(x, y, c) ◊ j 2 (x, y, c)
j n (x, y, c) ◊ j2 (x, y, c) = 0.
When the differential equation f (x, y, p) = 0 is solvable for x, it can be expressed in the form of x = f (y, p) ...(i) Differentiating w.r.t. y, we get dp dx ___ = f y, p, ___ dy dy
( ) ( )
dp 1 ___ fi __ ...(ii) p = f y, p, dy which is a differential equation of two variables y and p and hence it can be solved. Let us consider its solution be j (y, p, c) = 0 ...(iii) Eliminating p between (i) and (iii) gives the required solution. (iii) Equations solvable for y If the given equation is solvable for y, it can be expressed in the form y = f (x, p) ...(i) Differentiaing w.r.t. x, we get dy dp ___ = p = j x, p, ___ = 0 dx dx which is a differential equation in two variables x and p and hence its solution is possible. Let us consider its solution be
)
y (x, p, c) = 0
...(ii)
Eliminating p between (i) and (ii) we get the required solution of Eq. (i).
(iv) Clairaut’s Differential Equation
A differential equation of the first order in the form
y = Px + f (P)
...(i)
Differentiating w.r.t. x, we get dy dP dP ___ = P + x ___ + f ¢(P) ___ dx dx dx fi fi
(ii) Equations solvable for x
(
1. Given
fi
dP dP P = P + x ___ + f ¢(P) ___ dx dx dP (x + f ¢(P)) ___ = 0 dx dP ___ = 0 or (x + f ¢(P)) = 0 dx
fi
P = c
...(ii)
or,
(x + f ¢(P)) = 0
...(iii)
Eliminating P between Eq. (i) and (ii), we get,
y = cx + f (c)
where is the general solution of Eq. (i). Eliminating P between (i) and (iii), we get an equation, which has no constant. This is the singular solution of (i).
12. Higher Order Differential Equation
(i) Let a differential equation is of the form
d 2y ___2 = f (x) dx d dy ___ fi ___ = f (x) dx dx
( )
Integrating, we get dy ___ = f (x) + c1 dx Integrating again, we get
y = Ú f (x) dx + c1 x + c2
fi y = j (x) + c1 x + c2 which is the required solution. (ii) Let a differential equation is of the form
d 2y ___2 = f (y) dx
...(i)
dy Let ___ = p dx fi
d 2y ___ 2 = dx
dp ___ = dx
dp dp dy ___ ◊ ___ = p ◊ ___ dy dy dx
4.8 Integral Calculus, 3D Geometry & Vector Booster From Eq. (i), we get, dp p ___ = f (y) dy fi p dp = f (y) dy
Let the curve be y = f (x) and the point be P (x, y). dy Then tan (y) = ___ . dx (i) Length of the tangent (PT ) y siny = ___ PT fi PT = y cosecy
Integrating, we get
p2 __ = Ú f (y) dy + c1 2
fi
p2 = 2 Ú f (y) dy + 2c1
= j (y) + 2c1 fi
__________
__________ dy ___ = ± ÷ j (y) + 2c1 dx __________ fi dy = ± ÷ j (y) + 2c1 dx Integrating, we get
fi
y = y (x) + c2
which is the required solution. (iii) Let a differential equation is of the form
( )
dx 2 = y 1 + ___ dy
÷ ( )
( )
_________
y = Ú ± ÷j (y) + 2c1 dx + c2
d2y dy j ___ 2 , ___ = 0 dx dx
________
(ii) Length of the sub-tangent (TM): y tan y = ___ TM fi TM = y coty dx = y ◊ ___ dy (iii) Length of the normal (PN ): y cos y = ___ PN fi PN = y secy
fi
= y ÷1 + cot2y
p = ± ÷ j (y) + 2c1
_________
________
...(i)
( )
dy d2y dp d dy Let ___ = p fi ___ 2 = ___ ___ = ___ dx dx dx dx dx The Eq. (i) reduces to dp j ___ , p = 0 ...(ii) dx which is an equation of the first order. Then solve Eq. (ii) and get the required solution. Also, Eq. (i) can be written in the form dp j p ___ , p = 0 ...(iii) dy which is also an equation of the first order. Solve it and get the required solution.
( )
= y ÷1 + tan2y
dy 2 = y 1 + ___ dx
÷ ( )
________
(iv) Length of the sub-normal (MN): y tany = ____ MN
( )
dy fi MN = y ___ . dx (ii) Length of Intercepts of the
tangent by the axes:
( )
13. Applications
of
Differential Equation
Geometrical Application (i) Lengths of tangent, sub-tangent, normal and subnormal to the curve at a point
(i) The equation of the tangent to a curve at (x1, y1) is dy y – y1 = ___ (x – x1) dx at (x1, y1) (ii) The length of the x-intercept is
( )
( ) ( )
dx x1 – y1 ___ dy at (x1, y1) (iii) The length of the y-intercept is dx = x – y ___ dy at (x1, y1)
Differential Equation
4.9
Exercises (Problems based on Fundamentals) Find the order and the degree of each of the following differential equations: 2
d y 1. ___2 + 4y = 0 dx
dy 2. ___ + dx
d2y 3. ___2 + dx
( )
2
19. y = aex + be– x, where a and b are parameters.
20. y = sin (bx + c), b and c being parameters. 21. y = ae2x + be3x, where a and b are parameters.
dx ___ = 2 dy
( ) ÷ ( ) ÷ ÷ ( ) ÷ ( ) dy 2 ___ + xy = 0 dx
_____
_________
d2y dy 2 3 ___ 4. 1 + ___ = c ◊ 2 dx dx
dy dy 2 5. x + ___ = 1 + ___ dx dx
dy dy 2 6. y = x ___ + a 1 + ___ dx dx
d 4y d3y 7. ___4 + sin ___ 3 = 0 dx dx
d2y 2 8. ___ 2 + dx
9. edx = (x + 1)
_________
_________
( )
( ) ( )
( )
dy 6 d2y ___ = x sin ___ 2 dx dx
dy ___
( )
17. y = m x, where m is a parameter. 18. y = A cos x + B sin x, where A and B are parameters.
( )
dy dy 10. sin ___ + cos ___ = x dx dx Formation of a differential equation 11. Find the differential equation of all non-vertical lines in a plane. 12. Find the differential equation of the family of parabolas having vertex at the origin and the axis along positive x-axis. 13. Find the differential equation of the family of all circles, whose centre lies on x-axis and touches the y-axis at the origin. 14. Find the differential equation of all parabolas whose axes are parallel to the x-axis and have latus rectum 4a. 15. Find the differential equation corresponding to the family of curves y = c (x – c)2, where c is an arbitrary constant. 16. Find the differential equation of the system of y2 x2 ellipses __ 2 + __2 = 1, where a and b are arbitrary a b constants.
22. Find the differential equation of all circles touching the (a) x-axis at the origin. (b) y-axis at the origin. 23. Obtain the differential equation of all circles of the radius r. 24. Find the differential equation of all the circles in the first quadrant which touch the co-ordinate axes. 25. Find the differential equation of all the circles which pass through the origin and the centre lies on x-axis. 26. Find the differential equation of all non-vertical lines in a plane. 27. Find the differential equation of all the parabolas with the latus rectum 4a and whose axes are parallel to the x-axis. 28. Find the differential equation of all the ellipses having foci on the x-axis and the centre at the origin. 29. Find the differential equation of the family of parabolas having vertex at the origin and the axis along positive y-axis. 30. Find the differential equation of the family of curves __ y2 = 2c (x + ÷c ). 31. Find the order and the degree of the differential equation of all parabolas whose axis of symmetry is parallel to x-axis. Differential equation of first order and first degres dy 32. Solve: ___ = dx
x2 – 1 _____ . x2 + 1
dy 1 33. Solve: ___ = _____ . dx ex + 1 dy cos 2 x – cos x 34. Solve: ___ = ____________ . 1 – cos x dx dy 1 35. Solve: ___ = ________ . dx x (x4 + 1) ____
dy ÷sin x 36. Solve: ___ = ________ . dx sin x cos x dy x2 37. Solve: ___ = ________ . dx (3 + 2x)2
4.10 Integral Calculus, 3D Geometry & Vector Booster dy 38. Solve: ___ = dx dy 39. Solve: ___ = dx dy 40. Solve: ___ = dx
÷
_______ __ 1 – ÷x ________ . 1 + ÷x _____
61. Solve: (x2 – yx2) dy + (y2 + xy2) dx = 0. Reducible to variable separable Form
____
(÷ tan x + ÷cot x ) .
dy 62. Solve: ___ = (x + y + 1)2. dx
1 ____________ 3 . sin x + cos3 x
dy 63. Solve: ___ = sin (x + y) + cos (x + y). dx
dy 1 41. Solve: ___ = _______________ ____ 4 . _____ dx (÷sin x + ÷ cos x )
64. Solve: (x + y) (dx – dy) = (dx + dy).
Variable separable Form
dy 65. Solve: tan y ___ = sin (x + y) + sin (x – y). dx dy 66. Solve: ___ – x tan y ( y – x) = 1. dx
dy 42. Solve: ___ = 1 + x + y + xy. dx dy 43. Solve: ___ = ex–y + x2 ◊ e–y. dx x
2
x
44. Solve: 3 e tan y dx + (1 – e ) sec y dy = 0. ________________ dy 45. Solve: ÷ 1 + x2 + y2 + x2y2 + xy ___ = 0. dx
)
dy dy 47. Solve: x ___ = a y2 + ___ . dx dx dy 48. Solve: xy2 ___ = 1 – x2 + y2 – x2y2. dx dy 49. Solve: (x + 1) ___ = 2xy. dx 50. Solve: sec2x tan y dx + sec2y tan x dy = 0. 51. Solve: (1 + e2x) dy + (1 + y2) ex dx = 0. dy 52. Solve: ___ = ex+y + x2ey. dx
_____
54. Solve: ÷ 1 + x2 dy + ÷ 1 + y2 dx = 0. ________________ 1 + x2 + y2 + x2y2 +
55. Solve: ÷
xy = 0.
dy 56. Solve: xy ___ + 1 + x + y + xy = 0. dx 57. Solve: (1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0. _____
_____
58. Solve: x ÷1 – y2 dx + y ÷1 – x2 dy = 0.
(
) (
)
dy dy 59. Solve: y – x ___ = a y2 + ___ . dx dx dy 60. Solve: xy ___ = dx
dy 69. Solve: ___ = sec (x + y). dx dy 70. Solve: sin–1 ___ = x + y. dx dy 71. Solve: ___ = cos (x + y + 1). dx dy 72. Solve: (x2 + 2 xy + y2 + 1) ___ = 2 (x + y). dx dy 73. Solve: ___ = sin (10x + 6y). dx y dy ____________ 2 (x2 + y2) – 1 74. Solve: __ x ___ + 2 = 0. dx x + y2 + 1 Homogeneous differential equation 75. Solve: (x3 + y3) dy – x2 y dx = 0.
_____ _____ dy 53. Solve: y ÷1 + x2 + x÷1 + x2 ___ = 0. dx _____
68. Solve: (x + y) (dx – dy) = dx + dy.
( )
dy sin y + cos y 46. Solve: ___ = ___________ . dx x (2log x + 1)
(
dy (x – y) + 3 67. Solve: ___ = ___________ . dx 2 (x – y) + 5
1 + y2 _____2 (1 + x + x2). 1+x
______
76. Solve: x dy – y dx = ÷ x 2 + y2 dx. 77. Solve: (x2 – y2) dx = 2xy dy. 78. Solve: (1 + 2ex/y) dx + 2e x/y (1 – x/y) dy = 0.
(
) (
)
dy dy 79. Solve: x ___ + 1 = y 1 – ___ . dx dx dy 80. Solve: x ___ = x + y. dx dy y – x 81. Solve: ___ = _____ . dx y + x dy 82. Solve: 2 xy ___ = x2 + y2. dx 83. Solve: (y2 – 2xy) dx = (x2 – 2xy) dy.
Differential Equation
84. Solve: x2y dx – (x3 + y3) dy = 0. 85. Solve: (x3 – 3xy2) dx = (y3 – 3x2y) dy.
÷
______ 2
dy y y 86. Solve: ___ = __ – __ 2 – 1 . dx x x dy y y 87. Solve: ___ = __ x + sin __ x . dx dy y 88. Solve: x ___ = y – x tan __ x . dx
( ) ( ) x 89. Solve: ( 1 + e ) dx + e ( 1 – __ y ) dy = 0. x __ y
x __ y
Reducible to homogeneous differential equation dy 90. Solve: ___ = dx
x + 2y + 3 __________ . 2x + 3y + 4
dy (x + y)2 91. Solve: ___ = ____________ . (x + 2) (y – 2) dx 93. Solve: (x + 2y + 3) dx = (2x + 3y + 4) dy. dy x + 2y – 5 94. Solve: ___ = _________ . dx 2x + y – 4 dy 4x + 6y – 5 95. Solve: ___ = __________ . dx 6x + 9y + 7 dy 96. Solve: ___ = dx
6x – 2y – 7 __________ . 3x – y + 4
Linear differential equation 97. Solve: y dx – x dy + ln x dx = 0. dy 98. Solve: x2 ___ – 3xy = 4x4 + 2x2. dx dy 99. Solve: ___ = y tan x – sin x. dx dy 2 100. Solve: x log x ___ + y = __ x log x. dx x 101. Solve: x dx = __ 2 – y dy. y
( )
dy y 107. Solve: ___ + __ = x3. dx x dy 108. Solve: ___ = dx dy 109. Solve: ___ – dx
y tan x – 2 sin x. y __ x = 2x2.
dy 2 110. Solve: x log x ___ + y = __ x log x. dx
111. Solve: y dx – (x + 2y2) dy = 0. 112. Solve: y dx + (x – y3) dy = 0. dy 113. Solve: x ___ – ay = x + 1. dx dy 114. Solve: (x + 1) ___ – ny = ex (x + 1)n + 1. dx 115. Solve: (1 + y2) dx = (tan–1 y – x) dx. dy 116. Solve: (x + 3y + 2) ___ = 1. dx 117. Solve: (1 + y2) dx = (xy + y3 + y) dy. Bernoulli’s differential equation dy y y 118. Solve: ___ + __ x log x = __ 2 (log y)2. dx x dy 119. Solve: ___ + xy = x2 y6. dx dy sin2 y 120. Solve: ___ + _____ x = x3 cos2 y. dx dy 121. Solve: ___ – x3 y2 + xy = 0. dx dy 122. Solve: (1 – x2) ___ + xy = xy2. dx dy 123. Solve: x ___ + y = x3 y6. dx –1 dy 124. Solve: (1 + x2) ___ + y = e tan x. dx
dy 102. Solve: (x + 3y + 2) ___ = 1. dx
dy 125. Solve: ___ (x2 y3 + xy) = 1. dx
dy 103. Solve: ___ + 2y = e3x. dx
126. Solve: (y log x – 1) y dx = x dy
dy 104. Solve: x ___ = x + y. dx
dy 105. Solve: x ___ + y = xex. dx
dy 106. Solve: x ___ + y = x log x. dx
dy 127. Solve: ___ = x2 y3 – xy dx dy xy __ 128. Solve: ___ + _____ 2 = x÷y . dx 1 – x dy y y 129. Solve: ___ + __ x ◊ log y = __ 2 (log y)2 dx x
4.11
4.12 Integral Calculus, 3D Geometry & Vector Booster dy 1 130. Solve: ___ + __ ◊ sin 2y = x3 ◊ cos 2y. dx x
2
131. Solve: (xy2 – e 1/x ) dx – xy2 dy = 0. dy 132. Solve: ___ + x (x + y) = x3 (x + y)3 – 1. dx Exact differential equation 133. Solve: x dx + y dy = x dy – y dx. 134. Solve: x dy – y dx = x4dx.
dy 158. Solve: ___ = dx
= (2xy + sin x ◊ cosec2 y) dy
(
)
2x – 3y + 1 __________ . 3x – 2y – 2
159. Solve: x (dy + dx) = y (dx – dy).
(
137. Solve: x dy + y dx + xy2 dx – x2y dy = 0.
dy y – x ___ dx 1 1 162. Solve: ________ = __ 2 + __ 2 . dy x y y + x ___ dx
138. Solve: (4x – 3y) dx + (2y – 3x) dy = 0.
163. Solve: x2 y dy – xy2 dx = x4 dy + x3 y dx.
135. Solve: x dy + y dx = sin y dy. 136. Solve: x dy + y dx + y2(x dy – y dx) = 0.
(
)
164. Solve: dx + x (y dx + x dy) = e– ey dx.
1 139. Solve: sin y + y sin x + __ x dx.
(
)
1 + x cos y – cos x + __ y dy = 0.
x dx + y dy _________ y dx – x dy 140. Solve: _________ _______ = . 2 2 x2 ÷x + y
(
)
xy 160. Solve: x (tan–1 x – log x) dy + y dx = _____ 2 dx. 1+x 161. Solve: x (dy – x dx) + y dx = 0.
(
x – __
165. Solve: y (y2 dx + (x2 dy – xy dx)) = e y dy. 166. Solve: y dx – x dy + xy2 dx = 0.
(
)
(
)
sin 2 x sin2 x 141. Solve: _____ y + x dx + y – _____ 2 dy = 0. y dy x + y ___ y4 dx 142. Solve: _______ = x2 + 2y2 + __2 . dy x y – x ___ dx dy a2 x ___ – y dy dx 143. Solve: x + y ___ = ___________ 2 . dx x + y2
168. Solve: (x2 + y2 + a2) y dy = (x2 + y2 – a2) x dx.
144. Solve: x dx + y dy = (x2 + y2) y dy.
172. Solve: x dx + y dy = m (x dy – y dx).
145. Solve: (x + y) dx + (x – y) dy = 0.
dy x + y ___ x sin2 (x2 + y2) dx ____________ 173. Solve: ________ = . dy y3 ___ y – x dx dy a2 x ___ – y dy dx 174. Solve: x + y ___ = ___________ 2 . dx x + y2
)
(
)
146. Solve: y dx + x (x – 1) dy = 0. 147. Solve: y dx + x (1 – xy) dy = 0. 148. Solve: (x + y) (dx – dy) = (dx + dy). 149. Solve: dx + dy = x dy + y dx. 150. Solve: x dy – y dx = (x2 + y2) dx.
(
)
1 1 151. Solve: __ x + __ y (x dy + y dx) = dx + dy. ______
152. Solve: (x dy + y dx) ÷x 2 + y2 = x2 y dx + xy2 dy. 153. Solve: (x – x3) dy = y (dx + x2 dy). 154. Solve: (x + 2y) dy + y dx = 0.
169. Solve: (1 + xy) y dx + x (1 – xy) dy = 0.
(
)
x dy y 170. Solve: ______ 2 = ______ 2 2 – 1 dx. 2 x +y x +y dy x + y ___ y4 dx 171. Solve: _______ = x2 + 2y2 + __ 2 . dy x y – x ___ dx
(
)
÷ ÷
___________
x dx + y dy a2 – x2 – y2 175. Solve: _________ = ___________ 2 . x dy – y dx x + y2 __________
x dx – y dy 1 + x2 – y2 176. Solve: _________ = __________ . x dy – y dx x2 – y2
( )
155. Solve: (x + sin y) dy + y dx = 0.
x 177. Solve: sin __ y (y dx – x dy) = xy2 (x dy + y dx).
156. Solve: ey dx + (xe y – 2y) dy = 0.
178. Solve: x dy + y dx + y2 (x dy – y dx) = 0.
157. Solve: 2y dy + (cos x ◊ cot y – y2) dx.
)
dy dy 167. Solve: 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ . dx dx
179. Solve: (4x3 + ex sin y) dx + ex cos y dy = 0.
Differential Equation
Orthogonal Trajectories 180. Find the orthogonal trajectories of the family of the straight lines which are passing through the origin. 181. Determine the 45∞ trajectories of the family of concentric circles x2 + y2 = a2. 182. Find the orthogonal trajectories of the family of the curves ax2 + y2 = 1. 183. Find the orthogonal trajectories of the circles x2 + y2 – ay = 0, where a is a parameter. 184. Find the orthogonal trajectories of the family of parabolas y2 = 4ax, where a is a parameter. 185. Find the orthogonal trajectories of the family of rectangular hyperbola xy = c2. 186. Find the orthogonal trajectories of the family of curves x2 – y2 = c2. First Order Higher Degree Differential Equation. Equation Solvable for p 187. Solve: (x + y + p) (2x + p) = 0. 188. Solve: p2 – p (ex + e–x) + 1 = 0. 189. Solve: p2 + 2 py cot x = y2. 190. Solve: p2 – px – xy – y2 = 0. 2
191. Solve: p y + (x – y) p – x = 0. 192. Solve: (p2 – 1) xy = (x2 – y2) p. 193. Solve: x y p2 – (x2 + y2) p + xy = 0. Equation solvable for x 194. Solve: p x – y p2 = a p. 195. Solve: y = 2 p x + y2 p3. 196. Solve: y2 log y = x y p + p2. 197. Solve: p2 y + 2p x = y. 198. Solve: p3 – 4 x y p + 8p2 = 0. 199. Solve: y p = 2p2 x + y2 p4. 200. Solve: x p2 – y p – p + 1 = 0. Equations solvable for y 201. Solve: y = (1 + p) x + a p2. 202. Solve: y = y p2 + 2px. 203. y = 2p x + tan–1 (xp2). 204. x2 p4 + 2x p = y. 205. y + p x = x4 p2. 206. y = p sin + cos p. Clairaut Differential Equation 207. Solve: (y + 1) P – xP 2 + 2 = 0. 208. Solve: P3x – P2y – 1 = 0.
209. Solve: (y + 1) P – xP2 + 2 = 0. 210. Solve: sin y ◊ cos Px – cos y ◊ sin Px – P = 0. 211. Solve: (x – a) P2 + (x – y) P – y = 0. ______
212. Solve: y = px + a ÷1 + p2 . a 213. Solve: y = P (x – b) + __ . P Higher Order Differential Equation d2y Differential equation is the form ___ 2 = f (x) dx d 2y ___ 214. Solve: 2 = x + sin x. dx d 2y 215. Solve: ___ 2 = e2x + ex + 2014. dx d 2y 216. Solve: ___2 = sin2 x. dx d 2y 217. Solve: ___ 2 = cos3 x. dx d2y 1 218. Solve: ___ 2 = _________ 2 . dx sin x cos2 x d2y 219. Solve: ___ 2 = sin4 x + cos4 x. dx d2y 220. Solve: ___ 2 = xex. dx
d2y Differential equation of the form ___ 2 = f (y) dx d2y 221. Solve: ___ 2 + y = 0. dx d2y 1 222. Solve: ___ 2 = __ 3 . dx y d2y 1 223. Solve: ___ 2 = ____ __ . 4÷y dx d2y 224. Solve: a2 ___ – y = 0. dx2 d2y 225. Solve: ___ 2 = e2y. dx d2y 226. Solve: 2 ___2 = 3y2, dx
( )
dy y (– 2) = –1, ___ = – 1, dx x = – 2
4.13
4.14 Integral Calculus, 3D Geometry & Vector Booster
( )
d2y 227. Solve: y3 ___ 2 = –1, dx
242. A curve passes through (2, 0) and the slope of the
( )
dy y (1) = –1, ___ = 0. dx x = – 1
( )
d 2y ___ dy Differential equation is the form j ___ 2 , = 0 dx dx 2 d y dy 228. Solve: x ___2 + ___ + x = 0. dx dx d 2y 1 ___ dy 229. Solve: ___2 = __ x + x. dx dx d2y 230. Solve: y ___2 + dx
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dy 2 ___ = 1. dx
d 2y dy 2 dy 231. Solve: y ___2 – ___ = y2 ___ . dx dx dx d2y dy 2 232. Solve: (x + a) ___2 + x ___ = dx dx d 2y 1 ___ dy 2 233. Solve: x ___2 – __ = 4 dx dx
dy ___ . dx
dy ___ . dx
d2y dy 2 234. Solve: y ___2 = 1 – ___ . dx dx d2y dy 235. Solve: ___ 2 = 2y ___ . dx dx
d2y dy 2 dy 236. Solve: y ___2 – ___ = y2 ___ . dx dx dx d2y 1 ___ dy a 237. Solve: ___ 2 + __ x – __ 2 = 0. dx dx x
d2y dy dy 2 238. Solve: y ___2 – y ___ ln y = ___ . dx dx dx Geometrical Applications
239. The slope of the curve at any point is the reciprocal of twice the ordinate at the point. The curve passes through the point (4, 3). Find the equation of the curve. 240. Find the equation of the curve whose slope at any point is y + 2 x and which passes through the origin. 241. Find the equation of the curve which passes through the origin and the tangent to which at every point (x, y) has slope equal to 4
x + 2xy – 1 ___________ . x2 + 1
(x + 1)2 + y – 3 tangent at P (x, y) is ______________ . (x + 1) Find the equation of the curve. 243. A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve. 244. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve. 245. If the length of the tangent at any point on the curve y = f (x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve. 246. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve, given that it passes through (– 2, 1). 247. The ordinate and the normal at any point P on the curve meet the x-axis at points A and B respestively. Find the equation of the family of the curves satisfying the condition AB is the AM of abscissa and the ordinate of P. 248. Find the nature of the curve for which the length of the normal at any point P is equal to the radius vector.
(Mixed Problems) 1. The order and the degree of the differential equation
÷
___
dy dy ___ – 4 ___ – 7y = 0 dx dx are
(a) 1 and 1/2 (b) 2 and 1 (c) 1 and 1 (d) 1 and 2 2. The order of the differential equation whose general solution is given by
y = c1e2x + c 2 + c3 ex + c4 sin (x + c5) 3.
is (a) 5 (b) 4 (c) 3 (d) 2 The differential equation of all straight lines passing through the origin is
÷
____
dy (a) y = x ___ dx
dy (b) ___ = y + x dx
Differential Equation
dy y (c) ___ = __ dx x
(d) none of these
4. A differential equation associated by the promitive y = a + be5x + ce–7x is
(a) y3 + 2y2 + y1 = 0
(b) 4y3 + 5y2 – 20y1 = 0
(c) y3 + 2y2 – 35y1 = 0
(d) none of these 5. The differential equation of the family of curves y = a cos (x + b) is
d2y (a) ___2 – y = 0 dx
d2y (b) ___2 + y = 0 dx
d2y (c) ___2 + 2y = 0 (d) none of these dx 6. The differential equation, whose general solution is y = A sin x + B cos x, is
d2y (a) ___2 + y = 0 dx
d 2y (b) ___2 – y = 0 dx
dy (c) ___ + y = 0 dx
(d) none of these
7. The differential equation of all straight lines passing through the point (1, – 1) is
10. The solution of the differential equation dy 1 + x2 ___ + _____ = 0 x dx is 1 (a) y = – __ tan–1 x + c 2 x2 (b) y + ln x + __ + c = 0 2 1 (c) y = __ tan–1x + c 2 x2 (d) y – ln x – __ = c 2 11. The solution of the differential equation
x cos y dy = (x ex ln x + ex) dx is
1 (a) sin y = __ x ex + c
(b) sin y + ex ln x + c = 0
(c) sin y = ex ln x + c
(d) none of these
12. The solution of the differential equation x (e2y – 1) dy + (x2 – 1) ey dx = 0 is
x2 (a) ey + e–y = ln x – __ + c 2 x2 (b) ey – e–y = ln x – __ + c 2
dy (a) y = (x + 1) ___ + 1 dx
dy (b) y = (x + 1) ___ – 1 dx
dy (c) y = (x – 1) ___ + 1 dx
dy (d) y = (x – 1) ___ – 1 dx
8. The differential equation of the family of parabolas with focus at the origin and the x-axis as axis is
( ) ( ) ( ) ( )
dy dy (a) y ___ + 4x ___ = 4y dx dx 2
x2 (c) ey + e–y = ln x + __ + c 2 (d) none of these
13. Solution of the equation (1 – x2) dy + xy dx = xy2dx is
dy 2 dy (b) – y ___ = 2x ___ – y dx dx 2 dy dy (c) y ___ + y = 2xy ___ dx dx
(a) (y – 1)2 (1 – x2) = 0
(b) (y – 1)2 (1 – x2) = c2y2
(c) (y – 1)2 (1 + x2) = c2y2
(d) none of these
dy dy (d) y ___ + 2xy ___ + y = 0 dx dx
14.
The general solution of y dx + (1 + x2) tan–1 x dy is (a) y tan–1 x = c (c) y + tan–1x = c
2
9. The differential equation of all lines in the xy-plane is
dy (a) ___ – x = 0 dx
d2y (b) ___2 – dx
d 2y (c) ___2 = 0 dx
d2y (d) ___2 + x = 0 dx
dy x ___ = 0 dx
4.15
the differential equation = 0, (b) x tan–1y = c (d) x + tan–1 y = c
dy 15. For solving ___ = (4x + y + 1), suitable substitution dx is
(a) y = vx
(b) y = 4x + v
4.16 Integral Calculus, 3D Geometry & Vector Booster
(c) y = 4x
(d) y + 4x + 1 = v
16. The solution of the differential equation dy xy ___ = ______ dx x2 + y2 is 2
2
(a) ay2 = e x /y
(b) ay = ex/y
(c) y = ex2 + ey2 + c
(d) y = ex2 + y2 + c
17. (x2 + y2) dy = x y dx, if y (x0) = e, y (1) = 1, the value of x0 = ______ __ 1 (a) ÷3 e (b) e2 – __ 2 ______ ______
÷
÷
÷
e2 – 1 e2 + 1 (c) ______ (d) ______ 2 2 18. The solution of (1 + xy) y dx + (1 – xy) x dy = 0 is x x 1 1 (a) __y + __ xy = k (b) log __ y = __ xy + k
x (c) __y = exy + k
( ) x (d) log ( __ y ) = xy + k
19. The solution of the equation (x + log y) dy + y dx = 0 is
(a) xy + y log y = c
(b) xy + y log y – y = c (c) xy + log y – x = c (d) none of these
20. The solution of (x – y3) dx + 3xy2dy = 0 is
x (a) log x + __ 3 = k y y (b) log x + __ 3 = k x x (c) log x – __ 3 = k y (d) log xy – y3 = k
21. The solution of the differential equation dy 3x2 sin2 x ___ + _____ 3 y = _____ dx 1 + x 1 + x3 is
1 (a) y (1 + x3) = x + __ sin 2 x + c 2 1 (b) y (1 + x3) = cx + __ sin 2 x + c 2 1 3 (c) y (1 + x ) = cx – __ sin 2 x + c 2 x 1 (d) y (1 + x3) = __ – __ sin 2 x + c 2 2
22. The solution of the equation
dy x ___ + 3y = x dx is
x4 (a) x3y + __ + c = 0 4
x4 (b) x3y = __ + c 4
x4 (c) x3y + __ + c (d) none of these 4 23. The solution of the differential equation dy ___ + 2y cot x = 3x3 cosec2 x dx
is
(a) y sin2 x = x3 + c
(b) y sin x = c
2
(d) y sin x2 = c
(c) y cos x = c
24. The integrating factor of the differential equation x dy – y dx = xy2 dx is
1 (a) __2 x
1 (b) __2 y
1 (c) __ xy
1 (d) ____ x2y2
25. The equation of the curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx – xy dy = 0, is
(a) x2 + y2 = 1
(b) x2 – y2 = 1
(c) 2x2 + y2 = 2
(d) none of these
( )
7 26. The equation of a curve passing through 2, __ and 2 1 having gradient 1 – __2 at (x, y) is x (a) y = x2 + x + 1 (b) xy = x2 + x + 1 (c) xy = x + 1 (d) none of these 27. The differential equation of the family of circles passing through the fixed points (a, 0) and (– a, 0) is
( )
(a) y1(y2 – x2) + 2xy + a2 = 0
(b) y1y2 + xy + a2x2 = 0
(c) y1(y2 – x2 + a2) + 2xy = 0
(d) none of these
y+1 28. The number of solutions of y = _____ , y (1) = 2 x–1 is
(a) none (c) two
(b) one (d) infinite
Differential Equation
29. The solution of
(a) order 1
(b) order 2
dy ___ = 1 + x + y2 + xy2, y (0) = 0 dx is
(c) degree 3
(d) degree 4
2
(a) y =
(
2
)
x x + __ – 1 e 2
2
(
(b) y = 1 +
(c) y = tan (c + x + x2)
(
2
(b) y2 = ax + b
y = (c1 + c2) cos (x + c3) – c4 ex + c 5 where c1, c2, c3, c4, c5 are arbitrary constants, is (a) 5 (b) 4 (c) 3 (d) 2 A solution of the differential equation
( )
dy 2 dy ___ – x ___ + y = 0 dx dx is
(a) y = 2 (c) y = 2x – 4
(a) – 1/2 (c) e – 1/2
(b) y = 2x (d) y = 2x2 – 4
(b) e + 1/2 (d) 1/2
( )
dy 2 + sin x ___ 34. If y = y (x) and ________ = – cos x y (0) = 1, y + 1 dx
( )
p then y __ equals 2 1 (a) __ 3 1 (c) – __ 3 35. 36.
_____
39.
_____
)
(
_____
_____
)
is (a) 2 (b) 3 (c) 4 (d) none The degree of the differential equation
( )
d3y 2/3 d2y dy ___ 3 – 3 ___2 + 5 ___ + 4 = 0 dx dx dx is
(a) 1 (c) 3
2 (b) __ 3 (d) 1
If x dy = y (dx + y dy), y (1) and y (x) > 0, then y (– 3) = (a) 3 (b) 2 (c) 1 (d) 0 The differential equation representing the family __ of curves y2 = 2c (x + ÷c ), where c is a positive parameter, is of
(b) 2 (d) none.
(More than one options are correct) 40. If m and n be the order and the degree of the differential equation
( )
( ) ( ) ( )
d2y 3 ___ d2y d3y dx2 ___ 2 + 4 ______ + ___ 3 = x2 – 1, 3 dx dx dy ___ 3 dx 5
dy 33. If y (t) is a solution of (1 + t) ___ – ty = 1 and dt y (0) = – 1 then y (1) is equal to
(b) (x + 1) + ex
(a) x + 2
(
(c) y = log x (d) y = ex + C 31. The order of the differential equation whose general solution is given by 32.
÷1 + x2 + ÷ 1 + y2 = A x ÷1 + y2 – y÷1 + x2
)
x (d) y = tan x + __ 2 2 d y 30. If ___2 = 0, then dx (a) y = ax + b
37. If f ¢(x) + f (x) = x, when f (0) = 2, then f (x) is (c) (x – 1) + 3e–x (d) none. 38. The degree of the differential equation
)
x2 x + __ 2 c e
4.17
then
(a) m = 3 and n = 5 (c) m = 3 and n = 3
(b) m = 3 and n = 1 (d) m = 3 and n = 2
41. If y = e4x + 2e–x satisfies the relation d 2y dy ___2 + A ___ + By = 0, then dx dx
(a) A = – 3
(b) B = – 4
(c) A – B = 1
(d) A + B = – 7
dy x2 + y2 + 1 42. The solution of ___ = _________ satisfying 2xy dx y (1) = 1 is given by
(a) a system of hyperbola (b) a system of circles (c) y2 = x (1 + x) – 1 (d) (x – 2) + (y – 3)2 = 5.
4.18 Integral Calculus, 3D Geometry & Vector Booster 43. The function f (x) satisfying the relation 2
2
({f (x)} + 4 ◊ f (x) ◊ f ¢(x) + {f ¢(x)} ) = 0 is given by __
__
) x (a) ke (2 + ÷3
) x (b) ke (2 – ÷3
) x (c) ke (4 – ÷3
) x (d) ke(4 + ÷3
__
(a) m = 3 (b) n = 2 (c) m + n = 5 (d) m – n = 1 45. The value of k such that the family of parabolas y = cx2 + k is the orthogonal trajectory of the family of ellipses x2 + 2y2 – 2y = c is (a) 1/2 (b) 1/4 (c) 2 (d) 4. 46. A differential function f satisfies the relation x+y f ______ = f(x) f(y) " x, y Œ R –{– 1}, 1 + xy
)
where f(0) π 0 and f ¢(0) = 1. Then dy y (a) f(0) = 1 (b) ___ = _____ 2 dx 1 – x 1–x (c) f(x) = _____ (d) None of these 1+x d 47. Let ___ (x2y) = x – 1, where x π 0 and x = 1, y = 0. dx Then 1 1 ___ 1 (a) The solution of the curve is y = __ – __ + 2 x 2x2 (b) The set of values of x, where x is increasing is (– •, 0) (1, •) (c) The set of values of x where x is decreasing is (0, 1) (d) None of these. dy dy dy 2 48. Let y = x ___ + ___ – ___ . then dx dx dx
( ) ( ) ( )
(a) The genral solution is y = cx + c – c2 (b) The particular solution is y = 2x – 4 (c) The singular solution is 4y = (x + 1)2 (d) None of these dy 1 – 3x – 3y 49. Let ___ = _________ . dx 1 + x + y If the solution of the above differential equation is mx + y + n loge|1 – x – y| + C then (a) m = 2 (b) n = 3 (c) m + n = 5 (d) n – m = 1 dy ax + b 50. Let ___ = ______ . If the solution of the given differdx cy + d ential equation is a parabola, then
(a) a (b) c (c) a (d) c
= = = =
2, 0, 0, 0,
b d b a
= = = =
0 4 2, c = 4 2, b = 2, d = 4
__
44. If the order and degree of the differential equation y = c1 sin–1x + c2 cos–1x + c3 tan–1x + c4 cot–1x where c1, c2, c3 and c4 are arbitrary constants, are m and n respectively, then
(
(Problems for JEE-Advanced)
1. Find the equation of the curve which passes through the origin and the tangent to which at every point
x4 + 2xy – 1 (x, y) has slope equal to ___________ . 1 + x2 [Roorkee-JEE, 2001]
2. Solve the differential equation y y y cos __ x (x dy – y dx) + x sin __ x (x dy + y dx) = 0
( )
( )
p when y (1) = __ . [Roorkee-JEE, 1997] 2 3. Solve the differential equation dy cos2 x ___ – (tan 2 x) y = cos4 x dx
( )
__
3÷3 p p p where – __ < x < __ and y __ = ____ 4 4 8 6 [Roorkee-JEE, 1996]
d 4. If y + ___ (xy) = x (sin x + log x), find y (x). dx [Roorkee-JEE, 1995] 5. Solve the differential equation
x (1 – x2) dy + (2x2y – y – 5x3) dx = 0 [Roorkee-JEE,1994] 6. A tangent to the curve y = f (x) cuts the line y = x at a point which is at a distance of one unit from y-axis. Find the equation of the curve. ______
7. Solve: x dy – y dx = ÷x 2 + y2 dx. y y 8. Solve: xe y/x – y sin __ x dx + x sin __ x dy = 0. 9. Solve: (x + 2y) (dx – dy) = dx + dy.
(
( ) )
( )
10. Solve: (1 + y2) dx = (tan– 1 y – x) dy. dy 2xy 11. Solve: ___ = __________ . dx x2 – 2y – 1 12. Solve: x dy – {y + xy3(1 + log x)} dx = 0. 13. Solve: sec2x ◊ tan y dx + sec2y ◊ tan x dy = 0. ________ dy 14. Solve: ÷ x + y + 1 ◊ ___ = x + y – 1. dx
(
)
x 15. Solve: (1 + ex/y) dy + 1 – __ y ex/ydy = 0.
Differential Equation
16. Solve: (4x – y + 3) dy + (2x – 3y – 1) dx = 0. dy 17. Solve: (1 + y2) + (x – etan–1y) ___ = 0. dx 18. Solve: y dx – x (1 + xy) dy = 0. 19. Prove that the equation of a curve whose slope at – (x + y) (x, y) is _______ and which passes through the point x 20.
21. 22. 23. 24. 25.
26.
27.
(2, 1), is x2 + 2xy = 8. If the square of the intercept cut by any tangent on the y-axis is equal to the product of the coordinate of the point of contact, find the equation of such curves. Find the curves for which the length of the normal is equal to the radius vector. dy 2 dy Solve: ___ – (ex + e– x) ___ + 1 = 0. dx dx 2 dy dy Solve: ___ + 2x ___ = 3x2. dx dx 2 dy dy Solve: xy ___ – 1 = (x2 – y2) ___ . dx dx Find the equation of the curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve, given that it passing through (– 2, 1). Find the equation of the curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.
( ) ( ) { ( ) }
28. Find the equation of the curve which touches the line y = 1 and passes through the point (0, 1) and satisfies
( ) 2
dy the differential equation y3 ___ 2 – y = 1. dx 29. The ordinate and the normal at any point P on the curve meet the x-axis at points A and B respectively. Find the equation of the family of curves satisfying the condition AB = arithmetic mean of abscissa and ordinate of P. 30. A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. PQ is of the constant length k. Find the coordinates of Q. Also find the differential equation of the given curve. 31. Find the equation of the curve, the slope of whose 2y tangent at any point (x, y) is ___ x , for all x, y > 0 and which pass through the point (1, 1).
4.19
32. Let C be a curve such that the normal at any point P on it meets x-axis and y-axis at A and B, respectively. If PB : PA = 1 : 2 (internally) and the curve passes through the point (0, 4), prove that the curve is a ___
hyperbola and it passes through the point (÷10 , 6). 33. A normal is drawn at any point P (x, y) of a curve, it meets the x-axis and the y-axis in points A and B 1 1 respectively, such that ___ + ___ = 1, where O is OB OA the origin. Find the equation of such a curve passing through (5, 4).
(Tougher Problems for JEE-Advanced)
1. Find the order of the differential equation whose general solution is given by y = C1cos (2x + C2) – (C3 + C4)3x + C 5 + C6sin (x – C7) 2. Form the differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis. 3. The slope of the tangent to a curve at a point P (x, y) 2y is ___ x , x, y > 0 and which passes through the point (1, 1), find the equation of the curve. 4. A conic C passes through the point (2, 4) and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangentcy. Find the equation of the conic. 5. Solve the differential equation x2 + y2 dy x2 + y2 ______ 2y ___ = e x + ______ x – 2x . dx 6. Find the integral curve of the differential equation dy x (1 – x lny) ___ + y = 0, dx 1 which passes through the the point e, __ e . 7. Solve the differential equation dy dy 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ dx dx d 8. If y + ___ (x y) = x (sin x + log x), find y as a function dx of x. 9. Solve the differential equation (x dy + y dx) sin (xy) + (x2y dx + x y2dx) cos (x y) = 0.
(
)
(
)
( )
(
)
(
10. Solve the differential equation dy 2y2cos x + y sin 2x + 2cos x ◊ sin2x ___ = ____________________________ . dx sin2x 11. Solve the differential equation
)
4.20 Integral Calculus, 3D Geometry & Vector Booster
( )
( )
y y y cos __ x (x dy – yd x) + x sin __ x (x dy + y dx) = 0, p where y (1) = __ . 2 12. Solve the differential equation
÷
______________________________
dy x4y2 – x6 + 2x4y – x6y2 – 2x6y + x4 ___ = ______________________________ dx y2 – x2y2 + x3y2 – x5y2 x
log t 13. If f (x) = Ú ________ 2 dt, x ≥ 1, prove that 1 1 + t + t
( )
1 f (x) = f __ x . 14. Solve the differential equation dy x2 + y2 + 3x + 3y + 2xy + 1 ___ = _________________________ . dx x2 + y2 – 3x – 3y + 2xy + 2 15. A differentiable function f satisfies the relation f (x + y) + f (x) ◊ f (y) = f (xy + 1). Given f (0) = –1, f ¢(0) = f ¢(1) =1, find the function f. 16. A differentiable function f satisfies the relation f (xy) = x f (y) + yf (x) for every x, y in R. If f ¢(1) = 1, the find the function f. y y x 1 17. Solve: __ y sin __ y – __ 2 cos __ x + 1 dx x
(
( ) ) ( ) y x x 1 1 + __ x cos ( __ x ) – __ sin ( __ y ) + __ dy = 0. ( y y )
2
2
dy (1 + y2) 18. Solve: ___ = _________ . dx xy (1 + x2) y dy ________________ 1 19. Solve: __ x + ___ = 4 . dx sin (xy) + cos2 (xy) 20. Solve: (x2 + y2 + 1) dy + 2xy dx = 0.
(
)
2 2 2 dy 1 x + y 21. Solve: y ___ + x = __ ______ x . 2 dx y + sin x cos2(xy) x dy 22. Solve: _______________ dx + _______ 2 = 0. 2 cos (xy) cos (xy)
23. Find the general solution of the linear equation of dy the first order ___ + p (x) y = q (x), if one particular dx solution y1(x) is known. 24. Find the curve such that the square of the intercept cut by any tangent off the y-axis is equal to the product of the co-ordinates of the point of tangency. 25. If f (x) is a function such that x
x
0
0
x Ú (1 – t) f (t) dt = Ú t f (t) dt, f (1) = 1, find f (x).
26. Find the equation of the curve passing through (3, 4) and satisfying the differential equation dy 2 dy y ___ + (x – y) ___ – x = 0. dx dx y j (y/x) 27. Solve: y¢ = __ x + ______ . j¢(y/x) 28. A differentiable function f satisfies the relation x+y f ______ = f (x) ◊ f (y) for all x, y in R – {–1}, where 1 + xy
( )
(
)
f (0) π 0 and f ¢(0) = 1. Find f (x). A differentiable function f satisfies the relation f (xy) = xf (y) + yf (x) for all x, y in R+. If f (1) = 1, find f (x). Find the curve for which the portion of the tangent included between the coordinate axes is bisected at the point of contact. 31. Find the nature of the curve for which the length of the normal at a point P is equal to the radius vector of the point P. 32. Find the equation of the curve passing through (2, 2) such that the slope of the tangent at any point to the curve is reciprocal of the ordinate of the point. 33. A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is dy y2 – 2xy ___ – x2 = 0 dx and hence find the equation of the curve. 34. Find the orthogonal trajectories to the curve y2 = a (x + a). 35. Find the equation of the curve in which the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact. 29. 30.
Integer type questions
1. Find the degree of the differential equation
( ) ( ) ( )
d3y 2/3 d2y dy ___ 3 – 3 ___ 2 + 5 ___ + 4y = 0. dx dx dx 2. If the order of the differential equation of the curve y = (c1 + c2) sin (x + c3) – c4ex + c 5 is M, find the value of (M – 1). 3. If y = e4x + 2e–x satisfies the differential equation d3y dy ___3 + A ___ + By = 0, the value of A – B + 4. dx dx dy a x + b 4. The solution of ___ = ______ , where a < 0 and dx b y + k b ≥ 0, represents a parabola, find the value of a + b + 4.
Differential Equation
( )
dy 5. The solution of the differential equation ln ___ = dx 3x + 4y, y (0) = 0 is
Ae3x + Be– 4y + C = 0, where A + B + C + 5 is... 6. The polynomial f (x) satisfies the equation f (x + 1) = x2 + 4x. The area enclosed by y = f (x – 1) and the __ 16÷2 2 _____ curve x + y = 0 is , where k is ... k 7. If the number of straight lines which satisfy the dy dy 2 differential equation ___ + x ___ = y is m, the value dx dx of m + 2 is...
( )
8. If f (x) + f ¢(x) = x, where f (0) = 2, then f (x) is
(x – 1) + ke– x, where k + 2 is ... 9. The solution of the differential equation dz z z 1 1 ___ + __ ln z = __ 2 (ln z)2 is ____ = ____ 2 + C, dx x x ln z x mz
(
)
Comprehensive Link Passages
(b) –7 (d) –1
1. The degree of the differential equation
( )
( )
( )
( )
Passage I The differential equation corresponding to
y = c1em 1x + c2em 2x + c3em 3x,
where c1, c2 and c3 are arbitrary constants and m1, m2 and m3 are the roots of m3 – 7m + 6 = 0 is
( ) ( ) ( )
where a, b, g and d are arbitrary constants. On the basis of the above information, answer the following questions. 1. The order of the differential equation is (a) 1 (b) 2 (c) 3 (d) 4 2. The degree of the differential equation is (a) 1 (b) 2 (c) 3 (d) 4 3. The value of a and b, respectively, are
÷ ( ) ÷ ( ) _________
_________
4 d2y d3y 1 + ___ 2 = 1 + ___ 3 is dx dx
d 3y d2y dy a ___ 3 + f ___ 2 + g ___ + d = 0, dx dx dx
dy 3 ! xy ___ + ... to • is dx (a) 0 (b) 1 (c) 2 (d) not defined 2. The dgree of the differential equation
3
In these questions, a passage (paragraph) has been given followed by questions based on each of the passage. You have to answer the questions based on the passage given.
(b) –7 (d) –1
dy dy 2 dy 3 1 1 1 x = 1 + xy ___ + __ ! xy ___ + __ ! xy ___ + __ 2 3 4 dx dx dx
10. The solution of the differential equation e xy2 – __ 3 dx – x2 y dy = 0, x e 1 given y = 0 when x = 1, is y2 = __ __ – x2 , where k x4 k + 5 is ...
)
(b) 1, 0 (d) 0, –1
Passage II The order of the differential equation is the highest order derivative appearing in the equation and the degree of the differential equation is the index power of the highest order derivative and is free from any type of radical sign. On the basis of the above information, answer the following questions.
where m + 4 is
(
(a) 0, 1 (c) –1, 0 4. The value of g is (a) 6 (c) 2 5. The value of d is (a) 6 (c) 2
4.21
(a) 3 (b) 1 (c) 2 (d) not defined 3. The order and the degree of the differential equation of the curve y = mx + m2 + m3 + m4 is (a) 1, 3 (b) 1, 4 (c) 2, 3 (d) 2, 4 4. The order and the degree of the differential equations dy in ___ = x + 1 is dx (a) 1, 4 (b) 2, 4 (c) 1, not defined (d) 2, not defined
( )
Passage III Differential equations are solved by reducing them to get the exact differential of an expression x and y, i.e. they are reduced to the form d [ f (x, y)] = 0 For example, x dx + y dy _________ x dy – y dx _______ _________ = x2 ÷ x 2 + y2 + 2y dy _________ x dy – y dx 1 2x dx fi __ × ___________ ______ = 2 x2 ÷x 2 + y2
4.22 Integral Calculus, 3D Geometry & Vector Booster d (x2 + y2) y ______ fi _________ = – d __ x 2 2 ÷x + y _______ y fi d (÷x 2 + y2 ) = – d __x Thus the required solution is _______ y ÷x 2 + y2 + __ x = c. On the basis of the above information, answer the following questions. 1. The general solution of (2x3 – xy2) dx + (2y3 – x2y) dy = 0, is
( ) ( )
2
2 2
4
(a) x + x y – y = c
2
2 2
4
(b) x – x y + y = c
(c) x2 – x2y2 – y4 = c (d) x2 + x2y2 + y4 = c 2. The general solution of the differential equation x dy y ______ 2 + 1– ______ 2 2 dx = 0, 2 x +y x +y
(
is
)
( ) y ( __ x ) = c
( )
(1 – x)2 ______ (a) e 2
(1 – x)2 ______ (b) e 2 –
(c) ln (1 + x) – 1
(d) 1 + x.
Passage V A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. PQ is of constant length k. On the basis of the above information, answer the following questions. 1. The co-odinates of Q are
y (a) x + tan–1 __ x = c
x (b) x + tan–1 __ y = c
(c) x – tan–1
(d) none.
3. The general solution of the differential equation ey dy + (xey – 2y) dy = 0, is
(a) xey – y2 = c y
(c) ye + x = c
(b) xe y – x2 = c y
2
(d) xe – 1 = cy .
the function and get required result. On the basis of the above information, answer the following questions. 1. The equation of the curve to the point (1, 0) which satisfies the differential equation (1 + y2) dx – xy dy = 0, is
(a) x2 + y2 = 1
(b) x2 – y2 = 1
(c) x2 + y2 = 2
(d) x2 – y2 = 2
–1
–1
dx (d) x – y ___ , 0 . dy
2. The differential equation of the curve is
( ) ( )
( ) ( )
dy 2 dy 2 (a) y ___ = y2 – k2 (b) y ___ = k2 – y2 dx dx 2 dy dy 2 (c) ___ = y2 – k2 (d) ___ = k2 – y2 dx dx 3. The cartesian equation of the curve, if it passes through the point (0, k) is (a) x2 + y2 = k2
(b) x2 + y2 = 2k2 y (c) log y + ÷y 2 – k2 = 0 (d) sin–1 __ = x + c. k
(
______
( )
)
Given below are matching type questions, with two columns (each having some items) each. Each item of Column I has to be matched with the items of Column II, by encircling the correct match(es). Note: An item of Column I can be matched with more than one items of Column II. All the items of Column II have to be matched.
1. Observe the following Columns Column I
(
–1
(a) tan y + sin x = c (b) tan x + sin y = c
(c) tan–1 y ◊ sin–1 x = c
) )
dx (b) x + y ___ , 0 dy
Column II 3
(B) The degree of the differential (Q) Not dy defined equation log 1 + ___ = x is dx
1 + y2 _____ _______ = 0, is ÷ 1 + x2 –1
( (
) )
(A) The degree of the differential (P) equation of the curve __ y2 = 2c (x + ÷c ) is
2. The solution of the differential equation
dy ___ + dx
( (
dy (a) x + y ___ , 0 dx dy (c) x – y ___ , 0 dx
Matrix Match (For JEE-Advanced Examination only)
Passage IV
dy The differential equation ___ = f (x) ◊ g (y) can be solved by dx dy seperating the variable ____ = f (x) dx and then integrating g (y)
1
(d) tan–1 y ◊ – sin–1 x = c.
dy 3. If ___ = 1 + x + y + xy and y (–1) = 0, then y is dx
)
(C) The order (L) and the degree (R) (M) of the differential equation of the curve y = a sin (bx + c) where a, b and c are arbitrary constants, the value of L + M is
4
Differential Equation
(D) The order (L) and the degree (S) (M) of the family of parabolas having vertex at origin and axis along positive y-axis, then L + M + 3 is
5
(B)
(A) The integrating factor of the (P) – x2
e
(B) The integrating factor of the (Q) differential equation dy ___ + x (x + y) = x3 (x + y)3 – dx 1 is
3
1 __ 4 x 1 – __4 y
3. Observe the following Columns: Column II
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) R is true and A is false. (d) R is false and A is true. 1. Assertion (A): The order of all the non-vertical lines in a plane is 3. Reason (R): The order of all the non-vertical lines in a plane passing through the origin is 1. 2. Assertion (A): The order of the differential equation dy sin ___ = x is 1. dx Reason (R): The degree of the differential equation
( )
(B) x2 dy + (xy + y2) dx = 0, (Q) cos (y/x) = y = 1 when x = 1 log |cx| x dy y y ________ (C) ___ – __x + cosec __ x = 0, (R) y = 2 1 – log |x| dx (x π 0, x π e) y = 0, when x = 1
( )
dy (S) log (x2 + y2) + 2xy + y2 – 2x2 ___ = 0, dx 2 tan–1 (y/x) y = 2 when x = 1 p = __ + log 2 . 2
(
)
4. Observe the following Columns: Column I
Column II
(A) If m and n are the order and (P) the degree of the equation d 4y dy 4 ___4 = y + ___ , then (m dx dx + n) is
( )
1/3
Assertion and Reason
(A) (x + y) dy + (x – y) dy = (P) y + 2x = 3x2y 0, y = 1, when x = 1
(S)
(D) The integrating factor of the (S) differential equation 3 (xy2 – e1/x ) dx – x2 y dy = 0 is
(D)
(D) The solution of dy y ___ – dx ________ _______ = 2x2 is 2 ÷x + y2
Codes
3
Column I
(R) sin–1 (y/x) = x2 + c
3
x y dx – (x + y ) dy = 0 is
(C) The solution of dy y x ___ = y + x tan __x is dx
e –x
(C) The integrating factor of the (R) differential equation 2
5
( )
Column II
differential equation dy ___ = x3 y2 – xy is dx
(Q)
p y = 1 at x = 0, then y __ 2 is
2. Observe the following Columns: Column I
dy 2 + sin x ___ If ________ = – cos x, i + y dx
4.23
sin (y/x) = cx
dy ___
edx = (x + 1) is 2. 3. Assertion (A): The integrating factor of the differential dy equation ___ + xy = 10 is ex. dx
Reason (R): The integrating factor of the differential dy equation ___ + Py = Q is e Ú Pdx. dx 4. Assertion (A): The orthogonal trajectories of the curve xy = c is x2 – y2 = c. Reason (R): The solution of he differential equation x d y + y d x = 0 is x y = c 5. Assertion (A): The solution of the differential
2
3
3
equation x y dx = (x + y ) dy is y =
x3 ___ 3 3y ce .
Reason (R): The integrating factor of the differential 1 equation M dx + N dy = 0 is ________ . Mx + Ny
4.24 Integral Calculus, 3D Geometry & Vector Booster 6. Assertion (A): The order of the differential equation of the curve y = c1ex + c2e2x + c3ex + 5 is 2. Reason (R): The order of a differential equation is the highest order derivative of a differential equation.
Questions asked in Previous Years’ JEE-Advanced Examinations
1. A normal is drawn at a point P (x, y) of a curve, it meets the x-axis at Q. If PQ is of constant length k,
prove that the differential equation describing such ______ dy curve is y ___ = ÷k 2 – y2 . dx Find the equation of such a curve passing through (0, k) [IIT, 1994] Let y = f (x) be a curve passing through (1, 1) such that the triangle formed by the co-ordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves. [IIT, 1995] Determine the equation of the curve passing through the origin in the form y = f (x) which satisfies the dy differential equation ___ = sin (10x + 6y) dx [IIT, 1996] A curve y = f (x) passes through the point P(1, 1). The normal to the curve at P is a (y – 1) + (x – 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also, obtain the area bounded by the y-axis, the curve and the normal to the curve at P. [IIT, 1996] A spherical rain drop evaporates at a rate proportional to its surface area at any instant t. The differential equation giving the rate of change of the radius of the rain drop is ... [IIT, 1997]
2.
3.
4.
5.
6. Let u (x) and v (x) satisfy the differential equations
dv du ___ + p (x)u = f (x) and ___ + p (x)v = g (x), where dx dx p(x), f (x) and g (x) are continuous functions. If u (x1)
> v (x1) for some x1 and f (x) > g (x) for all x > x1, prove that any point (x, y) where x > x1 does not satisfy the equation y = u (x) and y = v (x). [IIT, 1997] 7. A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, then P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve. [IIT, 1998] 8. The order of the differential equation whose general solution is given by y = (c1 + c2) cos (x + c3) – c4ex + c 5,
where c1, c2, c3, c4, and c5 are arbitrary constants is ... (a) 5 (b) 4 (c) 3 (d) 2 [IIT, 1998] 9. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve. [IIT, 1999] 10. A solution of the differential equation dy 2 dy ___ – x ___ + y = 0 dx dx is (a) y = 2 (b) y = 2x (c) y = 2x – 4 (d) y = 2x2 – 4 [IIT, 1999] 11. The differential equation representing the family __ of curves y2 = 2c(x + ÷c ), where c is a positive parameter is of (a) order = 1 (b) order = 2 (c) degree = 3 (d) degree = 4. [IIT, 1999] 2 2 12. If x + y = 1, then (a) yy ≤ – 2(y ¢)2 + 1 = 0 (b) yy ≤ + (y ¢)2 + 1 = 0 (c) yy ≤ + (y ¢)2 – 1 = 0 (d) yy ≤ + 2(y ¢)2 + 1 = 0. [IIT, 2000] dy 13. If y (t) is a solution of (1 + t) ___ – t y = 1 and dt y (0) = –1, then y (1) is (a) –1/2 (b) e + 1/2 (c) e – 1/2 (d) 1/2 [IIT, 2003] 14. A right circular cone with radius R and height H contains a liquid which evaporates at a rate propeortional to its surface in contact with air (proportonality constant (k > 0). Find the after what time which the cone is empty. [IIT, 2003]
( ) ( )
( )
dy 2 + sin x ___ 15. If y = y (x) and ________ = – cos x and y + 1 dx
( )
p y (0) = 1, then y __ equals 2
(a) 1/3 (c) –1/3
(b) 2/3 (d) 1. [IIT, 2004]
Differential Equation
16. A curve C passes through (2, 0) and the slope at (x, (x + 1)2 + y – 3 y) as _____________ . Find the equation of the curve (x + 1)
(d) the fixed radius 1 and variable centres along the y-axis. [IIT, 2007]
and the area enclosed by the curve and the x-axis in the fourth quadrant. [IIT, 2004] 17. If y = y (x) and it follows the relation x cos y + y cos x = p, the value of y ≤ (0) is
24. Let f (x) be a differentiable on (0, •) such that
(a) 1 (c) p
(b) –1 (d) – p [IIT, 2005]
18. If the length of the tangent at any point on the curve y = f (x) intercepted between the point and the x-axis is of length 1, find the equation of the curve. [IIT, 2005] 19. The solution of the primitive integral equation 2
2
(x + y ) dy = xy dx is y = y (x). If y (1) = 1 and y (x0) = e, then the value of x0 is ________
(a) ÷2 (e – 1)
(c) ÷3 e
2
__
________ 2 (e2 + 1) ______
(b) ÷
÷
e2 + 1 (d) _____ 2
[IIT, 2005] 20. For the primitive integral equation y dx + y2dy = x dy, x ŒR, y > 0, y = y (x), y (1) = 1, then y (– 3) is
(a) 3 (c) 1
(b) 2 (d) 5
[IIT, 2005] 21. A curve y = f (x) passes through (1, 1) and the tangent at P (x, y) cuts the x-axis and y-axis at A and B, respectively such that BP : AP = 3 : 1, then (a) the equation of the curve is x y¢ – 3y = 0. (b) the normal at (1, 1) is x + 3y = 4. (c) the curve passes through (2, 1/8). (d) the equation of the curve is xy¢ + 3y = 0. [IIT, 2006] dy 2 22. If y satisfies the differential equation ___ = __ x + y, dx
y (1) = 1, then (x + y + 2)2 e – y equals ... ______ [IIT, 2006] dy 1 – y2 23. The differential equation ___ = _____ determines a y dx
÷
family of circles with (a) the variable radii and fixed centre at (0, 1) (b) the variable radii and fixed centre at (0, –1) (c) the fixed radius 1 and variable centre along the x-axis.
4.25
(
)
t2f (x) – x2f (t) f (1) = 1 and lim ____________ = 1 for each x > t Æ x t – x 0, then f (x) is
1 (a) ___ + 3 x 1 (c) – __ x +
2x2 ___ 3 2 __ 2 x
1 (b) – ___ + 3x 1 (d) __ x
4x2 ___ 3
[IIT, 2007] 25. Let a solution y = y (x) of the differential equation ______
______
2 x ÷x 2 – 1 dy – y ÷y 2 – 1 dx = 0 satisfies y (2) = ___ __ . 3 ÷ p –1 Assertion (A): y = (x) = sec sec x – __ 6
(
)
__
÷
_____
1 2÷3 1 Reason (R): y (x) is given by __ y = ____ x – 1 – __ 2 x
26.
27.
28.
29.
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explannation of A. (c) A is true and R is false. (d) A is false and R is true. [IIT, 2008] If y ¢ = y + 1 and y (0) = 1, the values of y (ln2) is ... [IIT, 2009] Find the domain of the definition of non-zero solution of the differential equation (x – 3)2 y ¢ + y = 0 [IIT, 2009] Let f be a real-valued differentiable function on R (the set of real numbers) such that f (1) = 1. If the y-intercept of the tangent at any point P (x, y) on the curve y = f (x) is equal to the cube of the abscissa of P, the value of f (– 3) is ... [IIT, 2010] Let y ¢(x) + y (x)g ¢(x) = g (x)g ¢(x), y(0) = 0, x ŒR,
df (x) where f ¢(x) denotes _____ and g (x) is a given dx non-constant differentiable function on R with g (0) = g (2) = 0. The value of y (2) is ... [IIT, 2011] 30. Let f : [1, •) Æ [2, •) be a differentiable function such that f (1) = 2.
4.26 Integral Calculus, 3D Geometry & Vector Booster __
x
If 6 Ú f (t) dt = 3x f (x) – x – 5 for all x ≥ 1, the value 3
0
of f (2) is ... [IIT, 2011] 31. If y (x) satisfies the differential equation y¢ – y tan x = 2x sec x and y (0) = 0, then
( ) p p (b) y¢ ( __ ) = ___ 4 18 p p (c) y ( __ ) = ___ 3 9 p 4p 2p (d) y¢ ( __ ) = ___ + ____ 3 3 3÷3 p p 2 (a) y __ = ____ __ 4 8÷2 2
2
2
__
[IIT, 2012] p 32. A curve passes through the point 1, __ . Let the 6 y y slope of the curve at each point (x, y) be __x + sec __ x ,
( )
( )
x > 0. The eaquation of the curve is y 1 (a) sin __ x = log x + __ 2 y (b) cosec __x = log x + 2 2y (c) sec ___ x = log x + 2 2y 1 (d) cos ___ x = log x + __ 2 [IIT-JEE, 2013]
( ) ( )
( ) ( )
33. The function y = f (x) is the solution of the differential dy xy x4______ + 2x equation ___ + _____ = _______ in (–1, 1) satisfying 2 dx x – 1 ÷ 1 – x2 __
3 ÷ ___ 2
f (0) = 0, then Ú __ f (x) = dx is
3 p ÷ (b) __ – ___ 3 4
p (c) __ – 6
p (d) __ – 6
__
3 ÷ ___ 4
Py ≤ + Qy ¢ + 1 = 0, where P, Q are functions of x, y and y dy dy Here y ¢ = ___ , y ≤ = ___ , which of the following dx dx
(
)
statements (is/are) true?
(a) P (b) P (c) P (d) P
= = + –
y+x y–x Q = 1 – x + y + y + (y ¢)2 Q = x + y – y ¢ – (y ¢)2
[IIT-JEE, 2015] 36. A solution curve of the differential equation dy (x2 + xy + 4x + 2y + 4) ___ – y2 = 0, x > 0 dx
passes through the point (1, 3). The solution curve (a) intersects y = x + 2 exactly at one point (b) intersects y = x + 2 exactly at two points (c) intersects y = (x + 2)2 (d) does not intersect y = (x + 3)2 [IIT-JEE, 2016]
Answers 198. c (4x – c)2 = 64 y
190. (y – cex) (y + x – ce– x) = 0
199. y2 = 2cx + c3
191. (x – y + c) (x2 + y2 + c) = 0
200. y = cx – 1 + 1/c
192. (xy – c) (x2 – y2 – c) = 0
–1 203. y = 2 ÷cx + tan (c)
193. (y – cx) (y2 – x2 – c) = 0
2 204. y = 2 ÷cx + c c 205. y = c2 – __ x 206. x = sin + c
196. ln y = cx + c2. 197. y2 = 2cx + c2
__
3 ÷ ___ 2 [IIT-JEE, 2014] 34. Let y (x) be a solution of the differential equation (1 + ex) y ¢ + yex = 1. If y (0) = 2, which of the following statements (is/are) true? (a) y (– 4) = 0 (b) y (2) = 0 (c) y(x) has a critical point in (–1, 0) (d) y(x) has no critical point in (–1, 0). [IIT-JEE, 2015] 35. Consider the family of all circles whose centres lie on the straight line y = x. If this family of circles is represented by the differential equation
3 ÷ – ___ 2
LEVEL I
__
3 p ÷ (a) __ – ___ 3 2
__ __
Differential Equation
24. (x y – c) (x2 – y2 – c) = 0
Level II
1. (d) 2. (b) 3. (c) 4. (c) 5. (b) 6. (a) 7. (d) 8. (b) 9. (c) 10. (b) 11. (c) 12. (a) 13. (b) 14. (a) 15. (d) 16. (a) 17. (a) 18. (b) 19. (b) 20. (b) 21. (d) 22. (b) 23. (a) 24. (b) 25. (d) 26. (b) 27. (c) 28. (a) 29. (d) 30. (a) 31. (a) 32. (b) 33. (c) 34. (d) 35. (b) 36. (c) 37. (c) 38. (a) 39. (b) 40. (a, c) 41. (a, b, c, d) 42. (a, c) 43. (a, b) 44. (a, b, c, d) 45. (b) 46. (a, b, c, d) 47. (a, b, c) 48. (a, b, c) 49. (a, b, c, d) 50. (b, c, d)
|
Level III
______
|)
_____ 1 + ÷1 – y2 ______ 1. ÷1 – y2 ln __________ = ± x + c 2 1–÷ 1 – y 3 3 1 4 –1 –1 3. __ tan __ tan 4x + tan __ – __ – 3 4 5 5 4 4. __ sq.u. 3 5. x2 + y2 = 2x
7. y + ÷ x 2 + y2 = cx2
[ (
]
–1
5x ___ 3
}
tan y
10. xe
]
–1
–1
tan y
+ c = (tan y –1) e
x2 1 __ 11. __ y + 2 ln y – y = c. 2
3
(
)
x 2x __ 2 12. – __2 = ___ + ln x + c 3 3 y 13. tan x . tan y = c ________ ________ 2 14. x + c = 2 ÷x + y + 1 + __ ln (÷x + y + 1 – 1) 3 _____ 8 – __ ln ÷x + y + 1 + 2 3 15. x + y ◊ ex/y = c
16. (x + y + z)5 = c (y – 2x – 1)2 –1
17. xetan y = c (y – 2x –1)2 1 18. ln y = __ xy + c ___
20. x = ce ± 2 ÷y/x 2
2
______
28. y2 + ÷y 4 – 1 = e± 2x 29. x2 – y2 = kx dy 30. Q x + y ___ , 0 , x2 + y2 = 2k2 dx
(
)
31. y = x2 32. y2 = 2x2 + 16 33. x2 + y2 – 8 (x + y) + 16 = 0
LEVEL IV
1. 5
( )
d 2y 2. 2a ___ 2 + dx
3. y = x2 4. xy = 8.
5. cx = e – e–
6. ((ln y + 1) + ey) x = 1
y 7. log |(x2 – y2)| + tan–1 __ x = C
8. y ◊ x2 = – x2cos x + 2x sin x + 2cos x
dy 2 ___ = 0 dx
( )
x2 + y2 _____ x
( ( ) )
x3 x3 + __ log x – __ + C 3 9
9. (xy) sin (xy) = C
(
)
4y + sin x 2 ___ 10. ____ tan–1 _________ ___ = log |sin x| + C ÷15 ÷15 sin x
( )
y p 11. y sin __x = ___ 2x 2 12. y – log |y + 1| = __ R1 + x3 + C 3 3 14. y – __ log |(x + y)2 + 1| = x + C 2 15. y = x – 1 16. y = x log |x|
2
21. x ± y = a x
2
27. y = Cex /2 – 1
35. y = cx3, x = cy3
y y 1 8. log x = c + __ e– y/x sin __x + cos __x 2 1 4 __ 9. x + c = x + 2y + __ ln (3x + 6y – 1) 3 3
[
26. (y + 3) = (x + 4)2
34. (x – 1)2 + (y – 1)2 = 25
______
{
25. x2 – y2 + 4 = 0
– x
22. (y – e + c) (y + e
+ c) = 0
23. (2y – x2 – c) (2y + 3x2 – c) = 0
( )
( )
y x 1 17. – cos __ y + sin __ x + x – __ y = c 18. (1 + y2) (1 + y2) = c2 x2
4.27
4.28 Integral Calculus, 3D Geometry & Vector Booster
(
)
35. x2 + y2 = cx
sin (4xy) 1 1 19. __ (xy – sin (2xy)) + __ xy + _______ 4 8 4
(
36. x2 + y2 = 25, y = x + 1
)
sin2 (xy) 1 + __ xy + _______ = log |x| + c 2 2
Integer Type
y3 20. __ + y + x2y = c 3
1 _______ 1 21. __ 2 2 = c x – (x +y)
23. y ◊ ep(x)x = Ú (ep(x)x ◊ q (x)) dx + c __
y 24. 2 __x + log |x| = c 1 – x ____
( )
31. x2 ± y2 = a2 32. y2 = 2x 33. x + y = cx
(
)
34. ÷x 2 + y2 = 2x + c
Hints
P-I : P-II : P-III : P-IV : P-V :
1. (c) 1. (b) 1. (b) 1. (b) 1. (a)
1. Clearly, the order is 2 and the degree is 1. 2. Clearly, the order is 1 and the degree is 3. 3. Clearly, the order is 2 and the degree is 1. 4. We have,
÷ ( ) ÷ ( ( ) ) ( ) ( ( ) ) ( ) _________
_____
d2y dy 2 3 ___ 1 + ___ = c ◊ 2 dx dx fi fi
2 1/2
2
1/3
dy dy 1 + ___ = c ___2 dx dx dy 2 3 1 + ___ = c dx
d2y 2 ___ dx
Thus, order is 2 and degree is also 2.
2. (a) 2. (a) 2. (a) 2. (a) 2. (b)
1. (A)Æ(P); 2. (A)Æ(P); 3. (A)Æ(S), 4. (A)Æ(Q),
and
1. (d) 6. (a)
3. (b) 3. (b) 3. (a) 3. (b) 3. (a)
(B)Æ(Q); (B)Æ(P); (B)Æ(P), (B)Æ(S),
2. (c)
5. (5) 10. (8)
4. (b) 4. (c)
5. (a)
(C)Æ(R); (C)Æ(S); (C)Æ(Q), (C)Æ(P),
(D)Æ(S) (D)Æ(R) (D)Æ(R) (D)Æ(R)
5. We have,
( ) ) ( ( ) )
3. (d)
4. (b)
5. (a)
solutions fi
4. (4) 9. (6)
Assertion and Reason
2
_______
3. (3) 8. (5)
Matrix Match
25. f (x) = x3e( x ) y 27. j __ x = cx 30. xy = c
2
2. (2) 7. (4)
Comprehensive Link Passages
22. y ◊ ep(x)x = Ú (ep(x)x ◊ q (x)) dx + c
÷
1. (2) 6. (3)
dy x2 + 2x ___ = 1 dx
So, the order is 1 and the degree is also 1.
6. We have,
÷ ( ) ) ( ( ) ) _________
dy dy 2 y = x ___ + a 1 + ___ dx dx
fi
dy 2 dy 2 y – x ___ = a2 1 + ___ dx dx
(
So, the order is 1 and the also degree is 1.
7. We have,
( ) ( ) ( ) ( )
d 4y d3y ___4 + sin ___ 3 = 0 dx dx
dy dy 2 x + ___ = R1 + ___ dx dx
d 4y ___4 + dx
fi
dy 2 dy 2 x + ___ = 1 + ___ dx dx
So, the order is 4 and the degree is not defined.
(
d3y ___ 3 – dx
3
3
dy 1 d3 5 1 ___ 3 __ ! + ___ 3 __ ! + ... 3 5 dx dx
Differential Equation
8. We have,
( ) ( ) 2
2
( ) 2
dy dy dy ___ 2 + ___ = x sin ___ 2 dx dx dx So, the order is 2 but the degree is not defined.
2
9. We have, dy ___
edx = (x + 1)
( )
dy x + y ___ = a. dx Thus, the required differential equation is fi
( )
(
)
dy 2 dy 2 y ___ + y2 = x + y ___ dx dx dy 2xy ___ = y2 – x2. dx 14. The equation of all parabolas whose axes parallel to x-axis is (y – k)2 = 4a (x – h) Here, h and k are two arbitrary constants. Differentiating both sides, we get dy (y – k) ___ = 2a ...(i) dx Differentiating again w.r.t. x, we get fi
( )
( )
dy 1 dy 2 __ 1 dy 3 1 + ___ + __ ! ___ + ! ___ + ... 2 dx 3 dx dx
So, the order is 1 and the degree is not defined. 10. We have,
( ) ( ( ) ( (
4.29
( ) ( ) ( ) ) ) ( ) ( ) )
dy dy sin ___ + cos ___ dx dx
dy 1 dy 3 __ 1 dy 5 ! ___ + ! ___ – ... = x ___ – __ 3 dx dx 5 dx
1 dy 2 __ 1 dy 4 __ 1 dy 6 + ! ___ – ! ___ + ... = x + 1 – __ ! ___ 2 dx 4 dx 6 dx So, the order is 1 and degree is not defined. 11. Let the equation of the family of non-vertical lines in a plane is y = mx + c, where m and c are arbitrary constants. dy d2y 2 = 0. Now ___ = m and ___ dx dx From the above three equations, we should eliminate m. Thus, the required differential equation is d2y ___2 = 0. dx 12. Let the equation of the parabola be y2 = 4ax. where a is an arbitrary constant. Differentiating w.r.t. x, we get dy 2y ___ = 4a dx y dy __ fi ___ = a 2 dx Thus, the required differential equation is dy y2 = 2xy ___ . dx 13. The equation of the circle be (x – a)2 + y2 = a2, where a is an arbitrary constant. fi x2 + y2 – 2ax = 0 dy fi 2x + 2y ___ – 2a = 0 dx
( )
d2y dy 2 (y – k) ___2 + ___ = 0 dx dx
...(ii)
Eliminating k from Eq. (i) and (ii), we get
( ) ( )
d2y dy 3 2a ___ 2 + ___ = 0 dx dx which is the required differential equation.
15. The given curve is y = c(x – c)2 Differentiating w.r.t. x, we get dy ___ = 2c (x – c) dx
...(i)
...(ii)
( )
1 dy 2 From Eq. (i) and (ii), we get, c = ___ ___ 4y dx Put the value of c in Eq. (ii), we get
( ) ( ( ) ) ( ( ) ) ( )
dy 1 dy 2 1 dy 2 ___ = 2 ◊ ___ ___ x – ___ ___ 4y dx 4y dx dx fi
dy 1 dy 2 2y = ___ x – ___ ___ 4y dx dx
fi
dy dy 3 8y2 = 4xy ___ – ___ dx dx
which is the required differential equation. 16. The given curve is 2 x2 y __2 + __ 2 = 1 a b Differentiating w.r.t. x, we get
2x 2 yy ≤ ___2 + _____ 2 = 0 a b fi
yy ¢ x __2 + ___ 2 = 0 a b
...(i)
4.30 Integral Calculus, 3D Geometry & Vector Booster Again differentiating w.r.t. x, we get 1 __2 + a fi
yy ≤ ____ 2 + b
(y ¢)2 ____ 2 = 0 b
x yy ≤ _____ x (y ¢)2 1 __2 + _____ 2 + 2 = 0 a b b 2
fi
yy ¢ – ___ + b2
x yy ≤ _____ x (y ¢) _____ 2 + 2 = 0 b b
fi
d2y dy 2 dy xy ___ 2 + x ___ – y ___ = 0 dx dx dx
( ) ( )
which is the required differential equation. 17. Given curve is y = mx dy fi ___ = m dx y = __ x which is the required differential equation. 18. Given curve is y = A cos x + B sin x
fi fi
dy ___ = – A sin x + B cos x dx d2y ___ 2 = – A cos x – B sin x = – y dx
Hence, the required differential equation is d 2y ___2 + y = 0 dx 19. We have, y = aex + be– x fi
dy ___ = aex – be– x dx 2
dy ___2 = aex + be– x = y dx Hence, the required differential equation is fi
d2y ___2 – y = 0 dx 20. We have, y = sin (bx + c) fi
y¢ = b cos (bx + c)
fi
y≤ = – b2sin (bx + c)
fi
y ≤ (1 – y2) + y (y ¢)2 = 0
( )
d2y dy 2 (1 – y2) ___ 2 + y ___ = 0 dx dx which is the required differential equation. fi
21. We have,
y = ae2x + be3x
fi
dy ___ = 2ae2x + 3be3x dx
fi
d 2y ___2 = 4ae2x + 9be3x dx
= 2 (2ae2x + 3be3x) + 3be3x
d2y dy ___ 2 = 2 ___ + 3be3x dx dx
dy dy = 2 ___ + 3 ___ – 2y dx dx
fi
dy 2 d2y 1 + ___ + (y – a) ___ 2 = 0 dx dx
fi
d2y dy ___ 2 = 5 ___ – 6y dx dx
fi
d 2y dy ___ 2 – 5 ___ + 6y = 0 dx dx
(
( )
( )
22. Let the equation of the circle be
x2 + (y – a)2 = a2
fi
x2 + y2 – 2ay = 0
fi
dy dy 2x + 2y ___ – 2a ___ = 0 dx dx
dy dy x + y – ___ = a ___ dx dx dy x + y ___ x + y y dx _______1 ________ fi a = = y dy 1 ___ dx Hence, the required differential equation is x + yy1 fi x2 + y2 = 2 _______ y y 1 dy ___ fi (x2 + y2 – 2y2) = 2xy dx fi
(
dy 2xy ___ = _______ dx (x2 – y2)
= – b2y
fi
2 y ¢ = – _______ ______ y ÷1 – y2
(b) Do yourself.
( )
)
)
Differential Equation
23. Let the equation of the circle be 2
2
2
(x – h) + (y – k) = r
where h and k are parameters but not r. dy fi 2 (x – h) + 2(y – a) ___ = 0 dx dy fi (x – h) + (y – k) ___ = 0 dx 2 dy d 2y fi 1 + ___ + (y – k) ___2 = 0 dx dx
( )
24. Let the equation of the circle be
(x – a)2 + (y – a)2 = a2
fi
x2 + y2 – 2a (x + y) + a2 = 0
fi
2x + 2y – 2a (1 + y1) = 0
fi
x + y – a (1 + y1) = 0 x+y fi a = _____ 1 + y1 Hence, the required differential equation is x+y x+y 2 (x + y) + _____ = 0 x2 + y2 – 2 _____ 1 + y1 1 + y1
( )
( )
fi (1 + y1)2 (x2 + y2) – 2 (x + y)2(1 + y1) + (x + y)2 = 0
25. Do yourself 26. Let the line be y = mx + c dy ___ fi = m dx
fi fi fi fi
fi
27.
fi
dy (y – k) ___ = 2a dx
( ) ( )
fi
d2y (y – k) ___2 + dx
fi
d2y 2a ___ 2 + dx
y d2y __ 2 ___ = 0 b dx2
( )
y ___ dy 1 dy 2 – ___ + __ 2 ___ + 2 dx bx b dx
y d2y __ 2 ___ = 0 b dx2
y dy – __x ___ + dx
( )
which is the required differential equation. 30. Given curve is
d2y ___ 2 = 0 dx which is the required differential equation. Let the equation of the parabola be (y – k)2 = 4a (x – h), where h and k are parameters. dy fi 2 (y – k) ___ = 4a dx
fi
( )
dy 2 d2y ___ + y ___ 2 = 0 dx dx which is the required differential equation. 29. Let the equation of the parabola be (x – h)2 = 4a (y – k) where h and k are parameters dy fi 2 (x – h) = 4a ___ dx dy fi (x – h) = 2a ___ dx 2 dy fi 1 = 2a ___2 dx 2 d y 1 ___ fi 2 = ___ 2a dx fi
fi
dy 2x 2y ___ ___2 + ___ 2 = 0 a b dx y dy x __ 2 + __ 2 ___ = 0 a b dx 1 1 dy 2 __ 2 + __ 2 ___ + a b dx
dy 2 ___ = 0 dx
fi
__
y2 = 2c (x + ÷c ) dy 2y ___ = 2c dx dy y ___ = c dx d 2y y ___2 + dx
( )
dy 2 ___ = 0 dx
which is the required differential equation. 31. Do yourself. 32. Given differential equation is x2 – 1 dy = ______ 2 dx. x +1 Integrating, we get
x2 – 1
dx Ú dy = Ú x______ 2 +1
which is the required differential equation 28. Let the equation of the ellipse be
fi
1–2 y = Ú _____ dx x2 + 1
y x2 __2 + __ 2 = 1 a b
dy 3 ___ = 0 dx
( )
= x – 2tan–1x + c
which is the required solution.
4.31
4.32 Integral Calculus, 3D Geometry & Vector Booster 33. Given differential equation is
1 dy = _____ dx ex + 1
which is the required solution.
Integrating, we get
36. The given differential equation is
1 x dx Ú dy = e_____ +1
fi
ex y = Ú ((ex + 1) – __ x + 1) dx e
fi
(
____
)
1 – ex y = Ú _____x dx 1+e = x – log |e + 1| + c,
(
)
cos 2x – cos x dx dy = ____________ 1 – cos x
Integrating, we get cos 2x – cos x
dx Ú dy = Ú ____________ 1 – cos x
fi
(2cos x + 1) (1 – cos x) y = – Ú ___________________ dx (1 – cos x)
= – Ú (2cos x + 1) dx
= – (2 sin x + x) + c
which is the required solution of the given differential equation. 35. The given differential equation is dy 1 4 ___ = __ (x + 1) dx x dx fi dy = ________ 4 x (x + 1) Integrating, we get
fi
____
fi
dx ________ Ú dy = Ú x (x 4 + 1) 3
x dx y = Ú ________ 4 4 x (x + 1)
dt 1 = __ Ú _______ 4 t (t + 1)
____
÷sin x dx Ú dy = Ú ________ sinx cos x
fi
sin x ÷ y = Ú ________ dx sin x cos x
2 sec x ____ = Ú _____ dx tan x ÷
____
(Dividing the numerator and the denuminator by cos2x) ____
fi y = 2 ÷tan x + C which is the general solution. 37. The given differential equation is dy ___ = x2 (3x + 2x)2 dx x2 fi dy = ________ dx (3 + 2x)2 Integrating, we get ________ dx Ú dy = Ú (3 + 2x)2
fi
x2dx y = Ú ________ (3 + 2x)2
Let
3 + 2x = t
1 1 t – 3 2 __ = __ Ú ____ ◊ 2 dt 2 2 t
1 t–3 2 = __ Ú _____ dt t 8
1 1 __ 1 = __ Ú __ – + 1 dt t t 4
t 1 = __ log ____ + c 4 t+1
| |
)
x2
(
sin x ÷ dy = ________ dx sin x cos x
Integrating, we get
34. The given differential equation is dy cos 2x – cos x ___ = ____________ 1 – cos x dx
dy ÷sin x ___ = ________ dx dx sin x cos x
x
which is the required solution.
fi
| |
x4 1 = __ log _____ 4 + c 4 x +1
( )
( ) 3 1 = __ Ú ( 1 – __ ) dt t 8 2
(
)
6 9 1 = __ Ú 1 – __ + __ 2 dt t 8 t 1 1 = __ 1 – 6log t – __ + c t 8
(
(
)
)
1 1 = __ (3 + 2x) – 6 log (3 + 2x) – _______ + c 8 (3 + 2x)
Differential Equation
which is the required differential equation. 38. The given differential equation is
Let
sin x – cos x = t
÷ 1 – x dy = ______ ÷1 + x dx
fi
(cos x + sin x) dx = dt
Also,
1 – sin 2x = t 2
fi
sin 2x = 1 – t2 __ dt _____ =÷ 2 Ú ______ ÷ 1 – t2
Ú dy = Ú ÷
= ÷2 sin–1(t) + c
= ÷2 sin–1 (sin x – cos x) + c
______ __
dy 1 – ÷x ___ = ______ __ 1+÷ x dx fi
______ __ ÷ __ ÷
Integrating, we get
______ __ 1 – ÷x ______ dx __ 1+÷ x
÷
fi Let
x = cos22q
fi
dx = – 2sin 4q dq
fi
Integrating, we get
÷
_________
fi
1– cos 2q y = Ú _________ × – 2sin 4q dq 1 + cos 2q = Ú tanq × – 2 sin 4q dq
= – 2 Ú cos 2q (1 – cos 2q ) dq
= – 2 Ú (2cos 2q – 1– cos 4q ) dq
sin 4q = – 2 sin 2q – q – _____ + c 4
= (– 2 sin 2q + 2q + sin 2q cos 2q ) + c
= – 2 ÷1 – x + cos–1 ÷x + ÷ x (1 – x) + c
(
)
_____
__
__
which is the required solution of the given differential equation. 39. The given differential equation is ____ ____ dy ___ = (÷tan x + ÷ cot x ) dx fi
____
____ ____ (÷tan x + ÷ cot x ) dx
Ú
fi
y=
sin x + cos x ________ = Ú __________ dx sin x cos x ÷
__ sinx + cos x _________ = ÷2 Ú __________ dx 2sin x cos x ÷
__ sin x + cos x =÷ 2 ___________ _____ dx sin 2x ÷
(
dx = Ú ____________ 3 sin x + cos3x
dx = Ú ________________________ (sin x + cos x) (1 – sinx cosx)
sin x + cos x 1 2 = __ Ú ___________ + _____________ dx 3 (sin x + cos x) (1 – sin x cos x)
dx 2 = __ Ú ___________ 3 (sin x + cos x)
____
+ ÷ cot x ) dx Ú dy = Ú (÷tan x
)
( Ú (
dx dy = ____________ 3 sin x + cos3x
dx Ú dy = Ú ____________ 3 sin x + cos3x
Integrating, we get
__
Integrating, we get
____ ____ (÷tan x + ÷ cot x ) dx
dy =
__
which is the required differential equation. 40. The given differential equation is dy 1 ___ = ___________ dx sin3x + cos3x
______ __
1–÷ x y = Ú ______ dx __ 1+÷ x
4.33
(
)
1 sin x + cos x + __ Ú _____________ dx 3 (1 – sin x cos x) dx 2 = ____ __ Ú _________ p 3÷2 sin x + __ 4
(
)
2 sin x + cos x + __ Ú __________ dx 3 (2 – sin 2x)
| |
x p 2__ = ____ log __ + __ 2 8 3÷2 2 + __ tan–1(sin x – cos x) + c 3
)
which is the required solution of the given differential equation. 41. The given differential equation is
)
dy 1 ___ = ______________ ____ 4 ____ dx (÷sin x + ÷ sin x )
4.34 Integral Calculus, 3D Geometry & Vector Booster fi
dx dy = _______________ 4 ____ _____ (÷sin x + ÷ cos x )
Integrating, we get fi
dx
____ 4 Ú dy = Ú _______________ _____ (÷sin x + ÷cos x )
sec x dx = Ú ___________ ____ (÷tan x + 1)4
2t dt = ______ , (Let tan x = t2) (t + 1)4
(t + 1) – 1 = 2 Ú _________ dt (t + 1)4
1 1 = 2 Ú ______ – ______ dt (t + 1)3 (t + 1)4
1 1 = 2 _________ 2 + ________ + c – 2(t + 1) 3 (t + 1)3
1 1 = 2 _____________ + ____________ + c ____ ____ – 2(÷tan x + 1)2 3(÷tan x + 1)3
)
fi
Ú e dy = Ú (e + x ) dx y
x
_____________ dy ÷(1 + x2)(1 + x2) + xy ___ = 0 dx
fi
_______ _______ dy ÷(1 + x2) + y2) = – xy ___ ÷ (1 dx
)
÷(1 + x2) y _______ dx = – ________ dx ________ x + y2) ÷ (1 Integrating, we get _______
+ x2) y ÷ (1 _______ dx = – Ú _________ dy Ú ________ x ÷(1 + y2) fi
x
44. The given differential equation is
______ ÷1 + x2 +
|
|
_____
______
1 + x2 – 1 ÷ _____ 1 __ + ÷1 + y2 =c log __________ 2 2 ÷1 + x + 1
which is the required solution of the given differential equation. 46. The given differential equation is dy sin y + cos y ___ = ___________ dx x (2log x + 1) fi
dy dx __________ = ___________ sin y + cos y x (2 log x + 1)
Integrating, we get dy dx __________ = ___________ Ú sin y + cos y Ú x (2log x + 1) fi fi
2
x3 fi e = e + __ + c, 3 which is the required solution. y
fi
_______
which is the required solution of the given differential equation. 42. Given differential equation is dy ______ = (1 + x) dx Ú (1 + y) Ú
dy = Ú (ex + x2) dx Ú ___ e– y
sec2y 3ex _______ _____ d x = dy Ú (1 – ex) Ú tan y
_______________ dy ÷1 + x2 + y2 + x2y2 + xy ___ = 0 dx
)
dy fi ___ = (ex + x2) dx e– y Integrating, we get
sec2y 3ex _______ _____ d x = dy tan y (1 – ex)
fi – 3 log |1 – e x| = log |tan y| + c which is the required solution of the given differential equation. 45. The given differential equation is
Integrating, we get dy ______ = (1 + x) dx. (1 + y) x2 fi log |1 + y| = x + __ + c, 2 which is the required solution. 43. Given differential equation is dy ___ = e– y(ex + x2) dx
fi
2
( (
3ex tan y dx + (1 – ex) sec2y dy = 0
On Integrating, we get
dx ____ y = Ú _______________ 4 _____ (÷sin x + ÷ cos x )
(
dy dx 1 ___ __ Ú _________ = Ú ___________ p x (2log x + 1) __ 2 sin y + ÷ 4 y p 1 1 ___ __ log tan __ + __ = __ log |2 log x + 1| 2 8 2 ÷2
(
|
(
)
)|
which is the required solution of the given differential equation. 47. The given differential equation is
(
)
dy dy x ___ = a y2 + ___ dx dx
Differential Equation
dy (x – a) ___ = ay2 dx a fi y – 2dy = ______ dx (x – a) Integrating, we get 1 fi – __ y = a log |x – a| + C
Integrating, we get
fi
)
fi
y2 1 – x2 _______ 2 dy = _____ dx x (1 + y )
fi
1 1 1 – _____ dy = __ x – x dx 1 + y2
(
)
(
)
Integrating, we get
x2 y – tan–1y = log |x| – __ + c 2
which is the required solution of the given differential equation. 49. Given differential equation is dy (x + 1) ___ = 2xy dx dy 2x ___ fi y = _____ dx x+1 dy 1 ___ fi y = 2 1 – _____ dx x+1 Integrating, we get
( ) ( )
tan–1(y) + tan–1(e x) = tan–1 c
fi
y + ex tan–1 _______ = tan–1c 1 – exy
(
log |y| = 2x – 2log |x + 1| + c
( )
dy ___ = e x + y + x2ey dx fi
dy ___ = (ex + x2)ey dx
fi
e– ydy = (ex + x2) dx
Integrating, we get x3 – e– y = ex + __ + c 3
which is the required differential equation. 53. Given differential equation is ______ ______ dy ___ y ÷1 + x2 + x ÷1 + x2 = 0 dx dy y + x ___ = 0 dx dy dx ___ x + ___ y = 0
fi fi
Integrating, we get
log |x| + log |y| = logc
fi xy = c which is the required solution. 54. Given differential equation is _____
_____
÷ 1 + x2 dy + ÷ 1 + y2 dx = 0
which is the required solution. 50. Given differential equation is
dy dx ______ _______ + _______ ______ = 0 2 ÷1 + y ÷1 + x2
sec2x tan y dx + sec2y tan x dy = 0
fi
fi
sec2y sec2x _____ dx + _____ dy = 0 tan x tan y
Integrating, we get
|
log |tan x| + log |tan y| = log c
fi
tan x tan y = c
which is the required solution. 51. Given differential equation is 2
_____
|
|
_____
|
log y + ÷y 2 + 1 + log x + ÷x 2 + 1 = logc
Integrating, we get
2x
)
y + ex ______ = c 1 – exy which is the required solution. 52. Given differential equation is fi
which is the required solution of the given differential equation. 48. The given differential equation is dy xy2 ___ = 1 – x2 + y2 – x2y2 dx dy fi xy2 ___ = (1 – x2) (1 + y2) dx
(
x
(1 + e ) dy + (1 + y ) e dx = 0
fi
dy ex _____ 2 + _______ dx = 0 1+y 1 + ex2x
(
_____
)
_____
) = c fi y + ÷ y 2 + 1 ( x + ÷x 2 + 1 which is the required solution. 55. Given differential equation is ________________ dy ÷1 + x2 + y2 + x2y2 + xy ___ = 0 dx
fi
______________ dy ÷(1 + x2) (1 + y2) + xy ___ = 0 dx
4.35
4.36 Integral Calculus, 3D Geometry & Vector Booster fi
______________ dy xy ___ = – ÷(1 + x2) (1 + y2) dx
Integrating, we get
______
fi
ydy ÷1 + x _______ ______ = – _______ dx x 2 ÷ 1 + y 2
|
Intergrating, wet get _____ 1 + y2 =
÷
_____ ÷x 2 + 1 +
fi
_____
|
x 2 + 1 – 1 ÷ _____ 1 __ log __________ + c 2 ÷x 2 + 1 + 1
which is the required solution. 56. Given differential equation is dy xy ___ + 1 + x + y + xy = 0 dx dy xy ___ + (1 + x)(1 + y) = 0 dx y 1+x fi _____ dy + _____ dx = 0 x 1+y
log |x + a| + logc = log |y| – log |1 – ay| y c (x + a) = ______ 1 – ay
which is the required solution. 60. We have,
dx xy ___ = dy
y dy (1 + x + x2) fi _____ 2 = ___________ dx 1+y (1 + x2) x
(1 + x + x2) = __________ d x (1 + x2) x
dx dx = ___ x + _____ 2 1+x
fi
( ) ( ) ( ) ( )
1 1 fi 1 – _____ dy + 1 + __ x dx = 0 1+y Integrating, we get y – log |1 + y| + x + log |x| = c fi
| |
x (x + y) – log _____ = c 1+y
57. Given differential equation is
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
) ( )
(
1+y 1+x fi _____ dx + _____ dy = 0 2 1+x 1 + y2 Integrating, we get 1 tan–1 x + tan–1 y + __ log |(x2 + 1) (y2 + 1)| = c 2 58. Given differential equation is _____
_____
x ÷1 – y2 dx + y ÷1 – x2 dy = 0 y x _____ fi ______ dx + ______ _____ dy = 0 2 ÷1 – y2 ÷ 1 – x Integrating, we get
______
_____
÷ 1 – x2 + ÷1 – y2 = c which is the required solution. 59. Given differential equation is dy dy y – x ___ = a y2 + ___ dx dx dy fi y(1 – ) = (x + a) ___ dx dy dx fi _____ = ________ x + a y (1 – ay)
(
) (
(
)
)
dx a 1 fi _____ = __ y + _____ x+a 1 – ay
1 + y2 _____2 (1 + x + x2) 1+x
1 fi __ log |1 + y2| = log |x| + tan–1 x + c 2 which is the required solution. 61. We have,
(x2 – yx2) dy + (y2 + xy2) dx = 0
fi
x2 (1 – y) dy + y2 (1 + x) dx = 0
(1 – y) (1 + x) fi ______ dy + ______ dx =0 2 y x2 Integrating, we get 1 1 __ – __ y – log |y| – x + log |x| = c
| |
(
)
x 1 1 log __ y = c + __ x + __ y which is the required solution. 62. The given differential equation is fi
Let
dy ___ = (x + y + 1)2 dx x+y+1=v
dy fi ___ = dx dv ___ – dx
dv ___ – 1 dx 1 = v2
dv fi _____ = dx 2 v +1 Integrating, we get
dv = Ú dx Ú _____ 2 v +1
fi fi
tan–1 (v) = x + c tan–1 (x + y + 1) = x + c
...(i)
Differential Equation
is the required solution. 63. The given differential equation is dy ___ = sin (x + y) + cos (x + y) dx
...(i)
dy dv Put x + y = v fi 1 + ___ = ___ dx dx dy dv fi ___ = ___ – 1 dx dx
dv ___ – 1 = sin v + cos v dx
Integrating, we get
(v + log |v|) = 2x + c
fi
(x + y + log |x + y|) = 2x + c
fi
(y + log |x + y|) = x + c
which is the required solution of the given differential equation. 65. The given differential equation is dy tan y ___ = sin (x + y) + sin (x – y) dx
tan y fi ____ dy = 2 sin x dx cos y
dv = x + c Ú ______________ 1 + sin v + cos v
dv fi Ú ________________________________ = x + c 2 1 – tan (v/2) 2tan(v/2) ___________ ___________ 1 + + 1 + tan2 (v/2) 1 + tan2 (v/2)
) (
fi
dy sin (x + y) + sin (x – y) fi ___ = ____________________ tan y dx 2 sin x cos y = _________ tan y
dv fi ______________ = dx 1 + sin v + cos v
(
Ú ( 1 + __ 1v ) dv = 2x + c
dv fi ___ = 1 + sin v + cos v dx
)
sin y fi _______ = 2 sin x dx cos2 y dy
sec2 (v/2)dv fi Ú ____________ = x + c 2(1 + tan(v/2))
Integrating, we get sin y = Ú 2 sin x dx Ú _______ cos2y dy
fi log |1 + tan(v/2)| = x + c
1 fi ____ = c – 2 cos x cos y
fi log |1 + tan(v/2)| = x + c
|
( ) |
x+y fi log 1 + tan _____ = x + c 2 which is the required solution of the given differential equation. 64. The given differential equation is
(x + y) (dx – dy) = (dx + dy)
fi
(x + y – 1) dx = (x + y + 1) dy
dy fi ___ = dx Let
(x + y – 1) __________ (x + y + 1)
dy dv x + y = v fi ___ = ___ – 1 dx dx
dv v–1 fi ___ – 1 = _____ v+1 dx dv v – 1 2v fi ___ = _____ + 1 = _____ v+1 dx v + 1
(
)
v+1 fi _____ dv = 2 dx v Integrating, we get
4.37
which is the required solution of the given differential equation. 66. The given differential equation is dy ___ – x tan (y – x) = 1 dx dy fi ___ = 1 + x tan (y – x) dx dy dv Let y – x = v fi ___ = ___ + 1 dx dx
...(i)
dv fi ___ + 1 = x tan v + 1 dx dv fi ___ = x tan v + 1 dx fi
cot v dv = x dx
Intergrating, we get
Ú cot v dv = x + c
x2 log |sin v| = __ + c 2 x2 fi log |sin(x – y)| = __ + c 2 which is the required solution of the given differential equation. fi
4.38 Integral Calculus, 3D Geometry & Vector Booster fi
67. We have, dy (x – y) + 3 ___ = ___________ dx 2(x – y) + 5
...(i)
Let x – y = v
dy dv fi 1 – ___ = ___ dx dx
dy dv fi ___ = 1 – ___ dx dx
(
)
– cosec (x + y) + cot (x + y) = 2x + y + c
which is the required solution. 70. We have,
( )
dy sin–1 ___ = x + y dx dy fi ___ = sin(x + y) dx dv fi ___ – 1 = sin v, (Let v = x + y) dx
dv v+3 fi 1 – ___ = ______ dx 2v + 5
dv fi ___ = 1 + sin v dx
dv v+3 fi ___ = 1 – ______ dx 2v + 5
dv fi _________ = dx (1 + sin v)
( (
2v + 5 – v – 3 ______ v+2 = _____________ = 2v + 5 2v + 5
)
2v + 5 fi ______ dv = dx v + 2 1 fi 2 + _____ dv = dx v+2
)
Integrating, we get (2v + log |v + 2|) = x + c fi
(2(x – y) + log |x – y + 2|) = x + c
fi
(x – 2y + log |x – y + 2|) = c
which is the required solution. 68. We have, (x + y) (dx – dy) = dx + dy
(1 – sin v)dv fi __________ = dx cos2v fi
(sec2v – sec v tan v) dv = dx
Integrating, we get
tan v – sec v = x + c
fi
tan (x + y) – sec (x + y) = x + c
which is the required solution. 71. We have dy ___ = cos (x + y + 1) dx dv fi ___ – 1 = cos v, (Let x + y + 1 = v) dx
d(x + y) fi _______ = d (x – y) (x + y) Integrating, we get log |x + y| = (x – y) + c
dv fi ________ = dx 1 + cos v
dy 69. We have ___ = sec (x + y) dx
fi
dv fi ___ – 1 = sec v (Let v = x + y) dx
= 1 + sec v
dv fi _______ = dx 1 + sec v cos v dv fi ________ = dx 1 + cos v cos v(1 – cos v) dv fi ________________ = dx sin2v fi [cosec v cot v – (cosec v – 1)] dv = dx Integrating, we get
– cosec v + cot v – v = x + c
1 – cos v fi _______ = dx sin2v dv (cosec2v – cosec v . cot v) dv = dx
Integrating, we get cosec v – cot v = x + c fi
cosec (x + y + 1) – cot (x + y + 1)
72. We have, dy 2(x + y) ___ = __________ dx 1 + (x + y)2 dv 2v fi ___ – 1 = _____ 2 ,v=x+y dx 1+v dv 2v fi ___ = 1 + _____ dx 1 + v2
2v = 1 + _____ = 1 + v2
(1 + v)2 _______2 1+v
Differential Equation
( (
)
)
2 2 fi 1 – _____ + _______ dv = dx 1 + v (1 + v)2 Integrating, we get 2 v – 2 log |1 + v| – ______ = x + c (1 + v) 2 fi y – 2 log |1 + x + y| – _________ = c (1 + x + y) which is the required solution. 73. We have, dy ___ = sin (10x + 6y) dx
)
1 dv fi __ ___ – 10 = sin v 6 dx dv fi ___ = 6 sin v + 10 dx dv fi __________ = dx 6 sin v + 10 dv fi __________________ = dx 2 tan(v/2) 6 ___________ + 10 1 + tan2(v/2)
(
)
5 tan (5x + 3y) + 3 5 fi __ tan–1 _______________ = 5x + c 4 4
2 fi 1 – _______ dv = dx (1 + v)2
(
)
5 tan(v/2) + 3 5 fi __ tan–1 ___________ = 5x + c 4 4
1 + v2 fi _______ dv = dx (1 + v)2
( (
)
which is the required solution. 74. We have, y dy 2(x2 + y2) – 1 __x ___ + ____________ = 0 dx x2 + y2 + 1 2v – 1 1 dv fi __ ___ – 1 + ______ = 0 2 dx v+1 2v – 1 1 dv fi __ ___ = 1 – ______ 2 dx v+1 2–v = _____ v+1
( ) v+1 fi ( _____ dv = – 2dx v – 1) v – 2 + 3 fi ( ________ dv = – 2dx v–2 ) 3 fi 1 + ( _____ dv = – 2dx v – 2) v + 1 fi _____ dv = 2dx 2–v
Integrating, we get
(x2 + y2) + log |(x2 + y2) – 2| = c – 2x
sec2(v/2)dv fi __________________________ = dx 10 (1 + tan2(v/2)) + 12 tan (v/2)
which is the required solution.
2 dt fi _____________ = dx, t = tan(v/2) 10(1 + t2) + 12t
dy x2y ___ = ______ 3 3 . dx x + y
dt fi ___________ = dx 5(1 + t2) + 6t
y __ x dy , fi ___ = _______ y 3 dx 1 + __ 3
dt fi __________ = dx 5t2 + 6t + 5 dt fi __________ = 5dx 6 2 __ t + t + 1 5 dt fi _____________ 2 = 5dx 3 2 4 __ t + + __ 5 5
( ) ( )
Integrating, we get
(
)
t + (3/5) 5 __ tan–1 _______ = 5x + c 4 4/5
(
)
4.39
5 5t + 3 fi __ tan–1 _____ = 5x + c 4 4
75. The given differential equation can be written as
( )
( )
y Let __x = v fi y = v x
dy dy fi ___ = v + x ___ . dx dx
fi fi
dv v v + x ___ = _____ dx 1 + v3 dv v x ___ = _____ – v dx 1 + v3
( )
fi
v3 + 1 dx _____ dv = – ___ x v4
fi
dx 1 __ v + v– 4 dv = – ___ x
(
)
4.40 Integral Calculus, 3D Geometry & Vector Booster
(
dy 1 __x fi ___ = __ – dx 2 y
Integrating, we get
1 log |v| – ___ 3 = c – log |x| 3v
fi
y x3 log __ x – ___ 3 = c – log |x| 3y
| |
1 log |v| – ___ 3 = c, 3v which is the required fi
y Let __x = v fi y = vx
dy dv fi ___ = v + x ___ dx dx
fi
x dy – y dx = ÷x 2 + y2 dx
fi
______ dy x ___ – y = ÷x 2 + y2 dx
______
x 2 + y2 dy y ÷_______ x = x fi ___ – __ dx
fi
_______
÷ ( )
y y 2 __ x = 1 + __ x
...(i)
y Let __x fi y = vx
dy dy fi ___ = v + x ___ dx dx _____
fi
dy v + x ___ – v = ÷ 1 + v2 dx
fi
_____ dy x ___ = ÷ 1 + v2 dx
(
)
dv 1 __ 1 v + x ___ = __ – v dx 2 v dv 3v 1 – 3v2 1 x ___ = ___ – ___ = ______ 2 2v dx 2v
2v dx fi _______ dv = ___ x 1 – 3v2 Integrating, we get
______
dy fi ___ – dx
fi
solution of the given differential equation. 76. The given differential equation is
)
y __ x
fi
2v dx 2 dv = Ú ___ Ú 1______ x – 3v 1 – __ log |1 – 3v2| = log c + log |x| 3 y 2 1 – __ log 1 – 3 __ x = log c + log|x| 3
|
( ) |
which is the required solution of the given differential equation. 78. The given differential equation is
(1 + 2e x/y) dx + 2e x/y (1 – x/y) dy = 0
dy fi ___ = dx
(1 + 2ex/y) __________ x 2e x/y __ y – 1
( ) x 2e ( __ y – 1 ) ___________ x/y
dv dx _____ fi _______ = ___ x 2 ÷ 1 + v
dy fi ___ = dx (1 + 2ex/y)
Integrating, we get
fi
dv dx _____ = Ú ___ Ú _______ x 2
|
_____
|
fi ( v + ÷1 + v ) = cx 2
( ÷ ( ) ) ______
)
dv v + y ___ = dy
fi
dv y ___ = dy
2vev – 2ev – v – 2vev = __________________ 1 + 2ev
2ev + v = ______v 1 + 2e
fi y + ÷x 2 + y2 = cx2 which is the required solution of the given differential equation. 77. The given differential equation is
(x2 – y2) dx = 2xy dy
fi
______ dy x ___ – y = ÷ x 2 + y2 dx
...(i)
2ev (v – 1) _________ (1 + 2ev)
fi
_______
y y 2 fi __ x + 1 + __ x = cx
(
dv dx fi ___ = v + y ___ dy dy
log v + ÷ 1 + v2 = log|x| + log c _____
x Let __y = v fi x = vy
÷1 + v
...(i)
2ev (v – 1) _________ –v (1 + 2ev)
dy 1 + 2ev fi _______v dv = – ___ y v + 2e
Differential Equation
fi fi
log |v + 2ev| + log|y| = log c
81. We have,
v
fi
y (v + 2e ) = c x y __ y + 2ex/y = c
fi
(x + 2yex/y) = c
(
dy ___ = dx
)
which is the required solution of the given differential equation. 79. The given differential equation can be written as dy ___ = dx
y–x _____ y+x y (Let __x = v fi y = vx)
dy dv fi ___ = v + x ___ dx dx
fi
dv v + x ___ = dx
v–1 _____ v+1
v–1 = _____ – v v + 1
v – 1 – v2 – v = _____________ v+1
1 + v2 = _____ v+1
fi
dv v – 1 x ___ = _____ – v v+1 dx v–1 = _____ – v v+1
1 fi __ log |v2 + 1| + tan–1 (v) = c – log |x| 2 y 1 fi __ log |x2 + y2| + tan–1 __x = c 2 which is the required solution of the given differential equation.
( )
80. We have,
dy x + y y fi ___ = _____ x = 1 + __ x dx dv fi v + x ___ = 1 + v, v = dx dv fi x ___ = 1 dx dx fi dv = ___ x Integrating, we get
y __ x
log |v – 1| + log |x| = log c
y fi __x = log |x| + c which is the required solution.
1+v dx fi ______2 dv = – ___ x 1+v Integrating, we get 1 tan–1 (v) + __ log |1 + v2| = c – log |x| 2 y 1 tan–1 __ x + __ log |x2 + y2| = c 2 which is the required solution. 82. We have, dy 2xy ___ = x2 + y2 dx
( ) ( )
y 2 2 + __x dy x______ + y2 1_______ ___ fi = = x 2xy dx 2 __ y fi
dv 1 + v2 v + x ___ = _____ 2v dx
fi
dv 1 + v2 1 – v2 x ___ = _____ – v = _____ 2v 2v dx
2v dv dx fi _____ = – ___ x v2 – 1 Integrating, we get
dy x ___ = x + y dx
2
1 + v2 = – ______ 1+v
v+1 dx fi _____ dv = – ___ x v2 + 1
dv v – 1 v + x ___ = _____ dx v + 1
( )
(y/x) – 1 _______ (y/x) + 1
fi
...(i)
y–x _____ = y+x
log |v2 – 1| + log |x| = log c
fi
(v2 – 1) x = c
fi (y2 – x2) = cx which is the required solution. 83. We have, dy ___ = dx
y2 – 2xy _______ x2 – 2xy
( )
( ) ( )
y 2 y __ x – 2 __ x dy fi ___ = __________ y dx 1 – 2 __ x fi
dv v2 – 2v v + x ___ = _______ 1 – 2v dx
4.41
4.42 Integral Calculus, 3D Geometry & Vector Booster fi
dv 3v2 – 3v x ___ = _______ 1 – 2v dx
1 – 2v fi ________ dv = 3v(v – 1)
dx ___ x
( (
)
dx 1 1 2 fi __ _______ – ______ dv = ___ x 3 v(v – 1) (v – 1)
)
dx 1 1 1 2 fi __ _____ – __ – ______ dv = ___ x 3 v – 1 v (v – 1) fi
(
)
dx 1 1 1 – __ _____ + __ dv = ___ x 3 v–1 v
Integrating, we get fi
1 – __ log |v(v – 1)| = log |cx| 3 y y 1 – __ log __ x __ x – 1 = log |cx| 3
| (
fi fi
x2y dx – (x3 + y3) dy = 0
1 – 3v2 – v4 + 3v2 = _______________ v3 – 3v
1 – v4 = _______ v3 – 3v
( ) ( )
Integrating, we get
| |
v2 – 1 3 1 – __ log |1 – v4 | + __ log _____ = log |x| + log c 4 4 v2 + 1
(
)
y dv v v + x ___ = _____ , Let v = __ x dx 1 + v3 dv v x ___ = _____ – v dx 1 + v3 v – v – v4 = _________ 1 + v3 4
v = – _____ 1 + v3
( )
x3 – ___3 + log |y| = c 3y which is the required solution 85. Given differential equation is fi
(x3 – 3xy2) dx = (y3 – 3x2y) dy
dy x – 3xy fi ___ = ________ dx y3 – 3x2y
dv 1 – 3v2 x ___ = ______ – v dx v3 – 3v
v3 3v dx fi _____ 4 dv – _____ 4 dv = ___ x 1–v 1–v
( )
2
fi
Integrating, we get
y __ x = _______ y 3 1 + __ x
3
)
v3 3v dx fi _____ 4 dv – _____ 4 dv = ___ x 1–v 1–v
1 + v3 dx fi _____ dv = – ___ x 4 v Integrating, we get 1 – ___ 3 + log |v| = c – log |x| 3v
dv v + x ___ = dx
( ) ( ) ( )
)|
dy x2y fi ___ = ______ dx x3 + y3
(
y 1 – 3v2 _______ 3 , Let v = __ x v – 3v
fi
v3 – 3v dx fi _______ dv = ___ x 1 – v4
which is the required solution. 84. Given differential equation is
( ) ( ) ( )
y 2 1 – 3 __ x __________ = y 3 y __ – 3 __ x 3
| | ( ) ( )
y 2 __ x – 1 y 4 3 1 __ __ __ _______ fi – log 1 – x + log y 2 = log |xc| 4 4 __ x + 1
|
( ) |
which is the required solution. 86. The given differential equation is
÷
_____
dy ___ = dx
y __ x –
y2 __ 2 – 1 x
fi
_____ y dv v + x ___ = v – ÷v 2 – 1 Let __x = u dx
fi
dv x ___ = – ÷v 2 – 1 dx
(
_____
dv dx _____ fi ______ = – ___ x 2 ÷ v – 1 Integrating, we get
_____
| _____ | fi log | v + ÷v – 1 | = log | __ cx | _____ ) = ( __ cx ) fi ( v + ÷ v – 1 _______ y y fi ( __ x + ÷ ( __ x ) – 1 ) = ( __ cx )
log v + ÷v 2 – 1 = log c – log c 2
2
2
)
Differential Equation ______
fi y + ÷ y 2 – x2 = c
which is the required solution. 87. The given differential equation is
vev – ev – v – vev = ________________ 1 + ev
– ev – v = _______v 1+e
(
dy fi ___ = dx fi fi
)
( )
y y __ x + sin __ x
(
)
y dv v + x ___ = v + sin (v) Let __x = u dx dv x ___ = sin (v) dx
dv dx fi _____ = ___ x sin (v)
fi
|
)
dv 1 + ev fi _____v dv = – ___ y v+e Integrating, we get
log |v + ev| = log c – log|y|
(
) ( )
x c fi __y + ex/y = __ y fi (x + y ex/y ) = c which is the required solution. 90. The given differential equation is
Integrating, we get fi
(
( ) |
v log tan __ = log c + log |x| 2 y tan ___ = cx 2x
( )
dy x + 2y + 3 ___ = __________ . 2x + 3y + 4 dx Let
which is the required solution. 88. The given differential equation is dy y x ___ = y – x tan __ x dx
x = X + a and y = Y + b
dY (X + a) + 2(Y + b) + 3 Thus, ___ = _____________________ dX 2(X + a) + 3(Y + b) + 4
( ) dy y y fi ___ = ( __ x ) – tan ( __ x ) dx
(X + 2Y) + (a + 2b + 3) dY ___ = _______________________ dX (2X + 3Y) + (2a + 3b + 4)
(
)
y dv v + x ___ = v – tan (v) Let __x = u dx dv fi x ___ = – tan (v) dx dv dx fi _____ = – ___ x tan(v)
Let us choose a and b in such a way that
Integrating, we get
dY X + 2Y ___ = ________ 2X + 3Y dX which is a homogeneous equation.
fi
fi fi
| |
( ) | p v c log tan ( __ + __ = log | __ x | 4 2 )|
p v log tan __ + __ = log c – log|x| 4 2
(
)
( )
(
x x __ __ x ( 1 + ey ) dx + e y 1 – __y dy = 0
(
)
x __ 1 ey
)
x __ y – dx ___ ________ fi = dy ( 1 + e__ xy ) fi
v dv (v – 1) e v + x ___ = ________ , v = v dy 1+e
fi
v dv (v – 1) e x ___ = ________ – v v dy 1+e
Solving, we get a = 1, b = – 2. Equation (i) reduces to
Equation (ii) reduces to dV 1 + 2V V + X ___ = _______ dX 2 + 3V 2 + 3V dX fi _______2 dV = ___ X 1 – 3V Integrating, we get
x __ y
...(i)
a + 2b + 3 = 0 and 2a + 3b + 4 = 0
Y Put __ = V X dV dY fi ___ = V + X ___ . dX dX
y p c tan __ + ___ = __ x 4 2x which is the required solution. 89. The given differential equation is fi
4.43
2 + 3V dX dV = Ú ___ Ú _______ 2 X 1 – 3V
|
__
|
1 + ÷3 V 1__ 1 __ fi ___ log ________ – __ log |(1 – 3V 2)| 2 3 1 – ÷3 V ÷
= log C + log |X|
...(ii)
4.44 Integral Calculus, 3D Geometry & Vector Booster
| | __
fi
X + ÷3 Y 1 1 ___ __ __ log _______ – __ log |(X2 – 3Y2)| = log c 2 3 X – ÷3 Y ÷
fi
(x – 1) + ÷ 3 (y + 2) 1 ___ __ __ log _________________ 3 (x – 1) – ÷3 ( y + 2) ÷
|
__
|
1 – __ log|(x2 – 3y2 – 2x – 12y – 11)| = log c 2 which is the required solution of the given differential equation. 91. The given differential equation is
dy (x + y)2 ___ = ____________ (x + 2)(y – 2) dx
Let
X = x + 2 and Y = y + 2
...(i)
Equation (i) reduces to 2 (X + Y)2 dY (X – 2 + Y + 2) ___ = ______________ = _______ ...(ii) XY XY dX Let
dV dY Y = V X fi ___ = V + X ___ dX dX
Equation (ii) reduces to 2 dV (1 + V) V + X ___ = _______ V dX
fi
dy y – x ___ + ln x = 0 dx
fi
dy x ___ – y = ln x dx
dy y ___ ln x fi ___ – __ = x dx x
...(i)
which is a linear differential equation. dx ___ 1 IF = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating, we get
y ◊ (IF) = Ú Q . (IF) dx + c
ln x 1 ___ y ◊ __ x = Ú x2 dx + c log x __ y 1 fi __x = – ____ x + x + c fi
(
fi
)
y = – (log x + 1) + c
which is the required solution of the given differential equation. 98. The differential equation is
fi
2 dV (1 + V) X ___ = _______ – V V dX
fi
dV (1 + 2V) X ___ = _______ V dX
dy y fi ___ – 3 __x = 4x2 + 2 dx
Integrating, we get
)
dX 1 1 fi __ Ú 1 – _______ = Ú ___ X 2 (2V + 1) 1 fi __ (V – log (2V + 1)) = log |X| + c 2
(
|
|)
1 Y Y fi __ __ – log 2 __ + 1 = log |X| + c X 2 X
( ( ) | ( ) | )
y–2 1 y–2 fi __ _____ – log 2 _____ + 1 2 x+2 x+2
= log |x + 2| + c
which is the required solution of the given differential equation. 97. The given differential equation is
dx ___ 1 IF = e – 3 Ú x = e – 3 log x = __ 3 x
Multiplying both sides of Eq. (i) by IF and integrating, we get
VdV dX _______ = ___ Ú (2V + 1) Ú X
(
...(i)
which is a linear differential equation.
VdV dX fi _______ = ___ X (2V + 1)
dy x2 ___ – 3xy = 4x4 + 2x2 dx
y dx – x dy + ln xdx = 0
y ◊ (IF) = Ú Q.(IF) dx + c
fi
4x2 + 2 1 y ◊ __3 = Ú ______ dx + c x x3
(
(
)
) )
y 4 2 fi __3 = Ú __ x + __ 3 dx + c x x
(
1 = 4 log x – __ 2 + c x
which is the required solution of the given differential equation. 99. The given differential equation is dy ___ = y tan x – sin x dx dy fi ___ + (– tan x) y = – sin x dx
...(i)
Differential Equation
which is a linear differential equation.
Multiplying both sides of Eq. (i) by IF and integrating, we get
IF = e– Ú tan xdx = elog(cos x) = cos x
Multiplying both sides of Eq. (i) by I.F and integrating, we get
x ◊ (IF) = Ú Q.(IF) dy + c x ◊ e– y = Ú (3y + 2) e– y dy + c
y ◊ (I.F.) = Ú Q.(I.F) dx + c
fi
fi
y (cos x) = – Ú (sin x cos x) dx + c
fi
1 = – __ Ú sin 2 x dx + c 2
cos 2 x = ______ + c 4 which is the required solution of the given differential equation. 100. The given differential equation is
dy y 2 ___ + ______ = __ dx x log x x2
...(i)
which is a linear differential equation.
dx Ú ______
IF = e x log x = elog(log x) = log x
Multiplying both sides of Eq. (i) by IF and integrating, we get y (IF) = Ú Q (IF) dx + c 2 fi y ◊ log x = Ú __2 log x dx + c x 2 = – __ ...(i) x (1 + log x) + c which is the required solution of the given differential equation. 101. The given differential equation is x x dx = __ 2 – y dy y
( )
fi
dx x ___ + y = xy2 dy
dy y fi ___ + __ = y2 dx x
...(i)
which is a linear differential equation.
IF = e– Ú dy = e– y
x = – 3 (y + 2) + cey
which is the required solution of the given differential equation. 103. The given differential equation is dy ___ + 2y = e3x ...(i) dx which is a linear differential equation. IF = e Ú 2 dx = e2x Multiplying both sides of Eq. (i) by IF and integrating, we get
y ◊ (IF) = Ú Q.(IF) dx + C
fi
y (ex) = Ú (e2x ◊ ex) dx + c
fi
= Ú (e3x) dx + c
( ) ( )
e3x = ___ + c 3
e3x y = ___ + ce– x 3
which is the required solution. 104. The given differential equation is dy x ___ = x + y dx dy y fi ___ = 1 + __ x dx
dy y fi ___ – __ = 1 dx x
...(i)
which is a linear differential equation. dx ___ 1 IF = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating we get
dy (x + 3y + 2) ___ = 1 dx
dx fi ___ – x = 3y + 2 dy
= 3 (– ye– y + 2e– y) + c
Thus,
which is a linear differential equation. 102. The given differential equation is
4.45
...(i)
y ◊ (IF) = Ú Q ◊ (IF) dx + c
y dx fi __x = Ú ___ x + c y fi __x = log |x| + c which is the required solution.
4.46 Integral Calculus, 3D Geometry & Vector Booster
105. The given differential equation is
dy x ___ + y = xe x dx
dy fi ___ + dx
which is the required solution. 108. The given differential equation is dy ___ = y tan x – 2 sin x dx
y __ x = ex
dy fi ___ + (– tan x) y = – 2 sin x dx
which is a linear differential equation.
Thus,
which is a linear differential equation
dx
___ IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú Q.(IF) dx + c
fi
y ◊ x = Ú xex dx + c
fi (xy) = ex (x – 1) + c which is the required solution. 106. The given differential equation is dy x ___ + y = x log x dx dy fi ___ + dx
y __ x = log x
...(i)
which is a linear differential equation. Thus,
dx
___ IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú x log x dx + c
fi
y ◊ x = Ú x log x dx + c
– IF = e Ú tan x dx = e– log(sec x) = x
Multiplying both sides of Eq. (i) by IF and integrating we get
y ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
y ◊ cos x = Ú (– sin x ◊ tan ◊ x) dx + c
1 – cos2x = – Ú _________ cos x dx + c
= – Ú (sec x – cos x) dx + c
(
)
= – log |sec x + tan x| + sin x + c
which is the required solution. 109. The given differential equation is dy y ___ – __ = 2x2 dx x
...(i)
which is a linear differential equation. dx ___ 1 I.F. = e – Ú x = e– log x = __ x Multiplying both sides of Eq. (i) by IF and integrating we get
( )
Thus,
Thus,
Thus,
1 = Ú x log x dx – __ Ú x dx + c 2
x2 x2 = __ log x – __ + c 2 4 which is the required solution. 107. The given differential equation is dy y ___ + __ = x3 dx x which is a linear differential equation.
...(i)
y ◊ (I.F) = Ú Q ◊ (IF) dx + c
1 y ◊ __ x = 2 Ú x dx + c y fi __x = x2 + c which is the required solution. 110. The given differential equation is dy 2 x log x ___ + y = __ x log x dx fi
...(i)
dx ___
IF = e Ú x = elog x = x
Multiplying both sides of Eq. (i) by IF and integrating we get
dy y 2 fi ___ + ______ = __ dx x log x x2
...(i)
which is a linear differential equation. dx – Ú ______ x log x =
y ◊ (IF) = Ú Q.(I.F.) dx + c
Thus,
fi
xy = Ú x4 dx + c
Multiplying both sides of Eq. (i) by IF and integrating we get
x5 = __ + c 5
IF = e
elog (log x) = x
y ◊ (IF) = Ú Q ◊ (IF) dx + c
Differential Equation
fi
log x y ◊ log x = 2 Ú ____ dx +c x2
which is the required solution. 113. The given differential equation is
= 2 Ú (te– t) dt + c, t = log x
= – 2 (te– t – e– t) + c
log x – 1 = – 2 _______ + c x
dy a 1 __ fi ___ + – __ x y = x + x dx
(
y dx – (x + 2y2) dy = 0
fi
dx y ___ + (x – y2) = 0 dy
fi
dx y ___ + x = y2 dy
dx x fi ___ + __ = y dy y
Thus,
Thus,
IF =
elog y = y
Multiplying both sides of Eq. (i) by IF and integrating we get x ◊ (I.F) = Ú y2dy + c
fi
x ◊ y = Ú y2dy + c
y = __ + c 3 which is the required solution. 112. The given differential equation is fi
dx y ___ + (x – y3) = 0 dy dx y ___ + x = y3 dy
fi
dx fi ___ + dy
x __ y = y2
...(i)
Multiplying both sides of Eq. (i) by IF and integrating we get
fi
x ◊ y = Ú y3 dy + c
y4 = __ + c 4
)
...(i)
dx
– n _____ I.F. = e Ú x + 1 = e– n log (x + 1)
(
)
1 log _______ (x + 1)n =
= e
1 _______ (x + 1)n
y ◊ (IF) = Ú Q ◊ (IF) dx + c
y fi _______ = Ú ex dx + c (x + 1)n
dy
x ◊ (IF) = Ú Q ◊ (IF) dy + c
(
Multiplying both sides of Eq. (i) by IF and integrating we get
___ IF = e Ú y = elog y = y
)
dy n fi ___ – _____ y = ex (x + 1) n x+1 dx
which is a linear differential equation Thus,
(
x– a + 1 x– a = _____ – a + ___ – a + c 1
which is the required solution. 114. The given differential equation is dy (x + 1) ___ – ny = ex (x + 1) n + 1 dx
Thus,
y dx + (x – y3) dy = 0
which is a linear differential equation
3
y ◊ (IF) = Ú Q.(IF) dx + c
y fi __a = Ú (x– a + x– a –1) dx + c x ...(i)
dx – a ___ 1 ) IF = e – a Ú x = e– a log x = e log(x = __ a x
Multiplying both sides of Eq. (i) by IF and integrating we get
which is a linear differential equation. dy ___ e Ú y =
...(i)
which is a linear differential equation
which is the required solution. 111. The given differential equation is
dx x ___ – ay = x + 1 dy
( )
)
4.47
= ex + c
which is the required solution. 115. The given differential equation is
(1 + y2) dx = (tan–1 y – x) dy
–1 dx (tan y – x) fi ___ = __________ dy 1 + y2
dx x fi ___ + _____ = dy 1 + y2
tan–1 y ______ 1 +y2
which is a linear differential equation
...(i)
4.48 Integral Calculus, 3D Geometry & Vector Booster dy
Thus,
2 Ú _____ –1 IF = e 1 + y = e tan y
Multiplying both sides of Eq. (i) by IF and integrating we get
x ◊ (IF) = Ú Q ◊ (IF) dy + c
fi
–1 tan–1y tan–1 y x ◊ etan y = Ú _____ e dy 1 + y2
fi
x ◊ etan y = Ú tet dt + c, t = tan–1y
fi
x ◊ etan y = et (t – 1) + c
fi
x ◊ etan y = e tan y (tan–1y – 1) + c
–1
DIviding both the sides by y (log y)2, we get dy 1 1 __ 1 1 _______ 2 ◊ ___ + ____ ◊ x = __ 2 dx log y y(log y) x
–1
which is the required solution. 116. The given differential equation is dx (x + 3y + 2) ___ = 1 dy dx fi ___ = (x + 3y + 2) dy dx fi ___ – x = (3y + 2) dy which is a linear differential equation
...(i)
Thus, IF = e Ú dy = e– y Multiplying both sides of Eq. (i) by IF and integrating we get –
x ◊ (IF) = Ú Q.(IF) dy + c
fi
x ◊ e– y = Ú {3(ye– y) + 2(e– y)} dy + c – y
– y
= 3 (– ye
= – 3 (y + 1) e– y – 2e– y + c
fi
– e ) – 2e
+c
ydy – Ú _____ y2 +1 =
1 – __ log |y2 + 1|
we get
e 2
dv 1 __ 1 – ___ + v ◊ __ x = x2 dx dv v 1 fi ___ – __ = – __2 dx x x which is a linear differential equation. fi
1 –1 __ 1 IF = e – Ú x dx = e– log x = e log x = __ x . Hence the solution is
Thus,
v ◊ (IF) = Ú Q.(IF) dx + c
fi
1 1 1 v ◊ __ x = Ú – __2 ◊ __ dx + c x x
...(i)
1 = _______ _____ 2 ÷ y + 1 Multiplying both sides of Eq. (i) by IF and integrating IF = e
dy dv 1 fi _______ 2 ◊ ___ = – ___ dx dx y(log y)
which is a linear differential equation Thus,
( )
( )
x 1 dy fi __6 ___ + __ 5 = x3 dx y y
(1 + y2) dx = (xy + y3 + y) dy xy dx fi ___ = _____ 2 + y dy y + 1
)
1 Let ____ = v log y
119. The given differential equation is dy ___ + xy = x3y6 dx
which is the required solution. 117. The given differential equation is
(
1 1 fi ______ – ___ = c x log y 2x2
x = – 3(y + 1) – 2 + cey
y dx fi ___ + – _____ x = y 2 dy y +1
_____
which is the required solution. 118. The given differential equation is dy y y ___ + __x log x = __ 2 (log y)2 dx x
–1
– y
( ) ( ÷ )
ydy x _____ fi _______ _____ = Ú _______ + c ÷ y 2 + 1 ÷y 2 + 1 x fi _______ _____ = ÷ y 2 + 1 + c 2 y + 1
( )
–1
x ◊ (IF) = Ú Q ◊ (IF) dy + c
...(i)
1 1 dy 1 dv Let __5 = v fi __ 6 ___ = – __ ___ dx 5 dx y y
1 dv – __ ___ + vx = x3 5 dx dv fi ___ – 5vx = 5x3 dx which is a linear differential equation. fi
5x2 – ___ 2
IF = e– 5 Ú xdx = e
...(ii)
Differential Equation
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + C
fi
v . e
5x2 – ___ 2 = 2
5x – ___
5x2 – ___ 2 dx
– 5 Ú x3 ◊ e
(
)
+C
2
5x – ___ 5x2 e 2 __ 2 ___ 2 + C fi ____ = + 1 e 5 2 y5
which is the required solution of the given differential equation. 120. The given differential equation is dy sin 2 y ___ + _____ x = x3cos2y dx fi
dy 2 tan y sec2y ___ + ______ x = x3 dx
...(i)
Let tan y = v
dy dv fi sec2y ___ = ___ dx dx
dv 2v fi ___ + ___ x = x3 dx
...(ii)
which is a linear differential equation
dx ___
IF = e 2 Ú x = e2 log x = x2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
v ◊ x2 = Ú x5 dx + c
fi
which is a linear differential equation. IF = e– 2 Ú xdx = e – x Multiplying both sides of Eq. (ii) by IF and integrating, we get 2
v ◊ (IF) = Ú Q ◊ (IF) dx + c
fi
v ◊ e– x = Ú x3 e– x dx + c
fi
2 1 v ◊ e– x = – __ (x2 + 1) + c 2
6
x x2 tan y = __ + c 6
which is the required solution of the given differential equation. 121. The given differential equation is dy ___ – x3y3 + xy = 0 dx dy fi ___ + xy = x3y3 dx 1 dy fi __3 ___ + xy– 2 = x3 ...(i) y dx
2
2
2
2 e– x 1 fi ____ = – __ (x2 + 1) e– x + c 2 2 y
which is the required solution of the given differential equation. 122. The given differential equation is
4.49
dy (1 – x2) ___ + xy = xy2 dx
dy xy xy2 fi ___ + _____ 2 = _____ dx 1 – x 1 – x2 x 1 dy x fi __2 ___ + __ (1 – x2) = _____ y dx y 1 – x2
dy dv 1 1 ___ __ ___ Let __ y = v fi y2 dx = – dx
fi
...(i)
dv vx x ___ – _____ = _____ dx 1 – x2 1 – x2
...(ii)
which is a linear differential equation.
x – Ú ______ dx 1 – x2 2 =
IF = e
1 __ log |1 – x2|
e 2
_____
= ÷1 – x2
Multiplying both sides of Eq. (ii) by IF and integrating, we get, fi
v ◊ (IF) = Ú Q ◊ (IF) dx + C _____
_____
x ÷1 – x2 v ÷1 – x = – Ú ________ dx 1 – x2 2
x _____ = – Ú ______ dx ÷ 1 – x2
= ÷1 – x2 +C
_____
_____ 1 dy 1 dv Let y –2 = v fi __ 3 ___ = – __ ___ = ÷1 – x2 + C 2 dx y dx which is the required solution of the given differential dv 1 ___ 3 __ equation. fi – + vx = x 2 dx 123. The given differential equation is dy dv fi ___ – 2vx = – 2x3 ...(ii) x ___ + y = x3y6. dx dx
4.50 Integral Calculus, 3D Geometry & Vector Booster dy fi ___ + dx
y __ x = x2y6
–1
fi
– 5 1 dy y fi __6 ___ + ___ x = x2 y dx
put
...(i)
y– 5 = v
dy – 5 ___ fi ___ = 6 dx y
1 dv v – __ ___ + __ = x2 5 dx x
1 dx fi __2 ___ – x dy ...(ii)
which is a linear differential equation dx – 5Ú ___ x
IF = e
= e– 5 log x
1 = __ 5 x
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v ◊ (IF) = Ú Q ◊ (IF) dx + c
...(i)
which is a linear differential equation dx –1 Ú _____ I.F = e x2 + 1 = e tan x
Multiplying both sides of Eq. (i) by IF and integrating, we get
( e ) –1 y ◊ etan x = Ú _______ dx + c (x2 + 1) tan x 2 –1
= Ú e2t dt + c, t = tan–1 x
e2t = Ú ___ + c 2
fi
1 Let v = – __ x
dv fi ___ = dy
dv fi ___ + vy = y3 dx which is a linear differential equation
y2 __
v ◊ e2 = Ú y3 e2 dy + c
fi
y y2 __ e 2 __ 2 dy + c – ___ = 2 y e Ú x 2
( )
...(ii)
( )
2
y2
__ e 2 2 fi – ___ = (y – 2) e 2 dy + c x which is the required solution.
126. The given differential equation is (y log – 1) y dx = x dy
y2 log x dx – y dx = x dy
dy x ___ + y = y2 log x dx
dy ___ + dx
y y2 log x __ x = ______ x
log x 1 dy 1 __2 ___ + __ xy = ____ x y dx
x log |x2 + c1| + 2a tan–1 ___ + y = c2 C1
1 dx __ 2 ___ x dy
y2 __
y2 __
2
Thus,
y2 __
which is the required solution. 124. The given differential equation is
...(i)
y2
5 ___ 2 + c 2x
–1 dy (1 + x ) ___ + y = etan x dx –1 dy y etan x fi ___ + _______ 2 = _______ 2 dx (x + 1) (x + 1)
y __ x = y3
__ Thus, IF = e Ú y dy = e2 Multiplying both sides of Eq. (ii) by IF and integrating, we get
dx 1 fi ____ 5 = – 5 Ú ___3 + c x (xy) 1 fi ____ = (xy)5
which is the required solution of the given differential equation. 125. The given differential equation is
dx fi ___ = x2y3 + xy dy dx fi ___ – xy = x2y3 dy
dv 5v fi ___ – ___ x = – 5x2 dx
Thus,
dy ___ (x2y3 + xy) = 1 dx
dv ___ dx
1 dy 1 dv fi __6 ___ = – __ ___ dx 5 dx y fi
–1 e2tan x y ◊ etan x = ______ + c 2
dv – ___ + dx
(
)
v log x 1 __ x = ____ x , Let v = __ y
log x dv v ___ – __ x = – ____ x dx
...(i)
Differential Equation
which is a linear differential equation 1 __ 1 I.F. = e – Ú x dx = e– log x = __ x Multing both sides of Eq. (i) IF and integrating, we get
Thus,
log x 1 ____ v ◊ __ x = – Ú x2 dx v fi __x = – Ú te– t dt, (Let t = log x)
)
log x +1 1 ________ fi __ + c xy = x
1 Let __2 = v y 2 dy dv fi – __3 ___ = ___ y dx dx
fi
1 dv – __ ___ + vx = x2 2 dx
dv fi ___ – 2vx = – 2x3 dx which is a linear differential equation Thus,
...(i) __
Let ÷y = v
dy dv 1 ___ fi ____ __ = ___ 2 ÷y dx dx
dy dv 1__ ___ fi ___ = 2 ___ ÷y dx dx
fi
dv vx 2 ___ + _____ = x dx 1 – x2
(
)
dv x x fi ___ + ________ v = __ 2 dx 2(1 – x2)
which is the required solution of the given differential equation. 127. The given differential equation is dy ___ = x3y3 = xy dx dy fi ___ + xy = x3y3 dx x 1 dy fi __3 ___ + __ = x3y3 ...(i) y dx y2
__
x ÷y _____ = x 1 – x2
= (t + 1) e– t + c
(
dy 1__ ___ fi ___ + ÷y dx
...(ii)
IF = e– 2 Ú x dx = e – x
4.51
...(ii)
which is a linear differential equation. Thus,
xdx 2 Ú ________
1 – __ log |1 – x2|
IF = e 2(1 – x ) = e 4
1 = _______ 4 _____ 1 – x2 ÷
Multiplying both sides of Eq. (ii) by IF and integrating, we get
1 _____ v ◊ _______ = 4 ÷1 – x2
x dx 1 __ _____ Ú _______ + c 2 4÷1 – x2
__
÷y (1 – x2)3/4 _____ fi 4_______ = ________ +c 3 ÷ 1 – x2 which is the required solution. 129. The given differential equation is dy y y ___ + __ ◊ log y = __ 2 (log y)2 dx x x dy 1 1 1 ___ + _______ fi ________ = __ y (log y)2 dx x (log y) x2
...(i)
2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
v (IF) = Ú Q (IF) dx + c
fi
v ◊ e– x = Ú 2x3 e – x dx + c
= (1 + x2) e– x + c
fi
v = (1 + x2) + cex
2
1 Let ______ =v (log y) fi
dy 1 ___ – ________ 2 = y (log y) dx
dv ___ dx
2
2
2
2 1 fi __2 = (1 + x2) + cex y which is the required solution of the given differential equation. 128. The given differential equation is
dy xy __ ___ + _____ = x ÷y . dx 1 – x2
dy dv 1 ___ fi _______ 2 = – ___ dx y(log y) dx fi
dv v __ 1 – ___ + __ = dx x x2
dv v 1 fi ___ – __ = – __2 dx x x which is a linear differential equation. Thus,
dx ___ 1 IF = e – Ú x = e– log x = __ x
...(ii)
4.52 Integral Calculus, 3D Geometry & Vector Booster Multiplying both sides of Eq. (ii) by IF and integrating, we get dx v __x = – Ú ___ 3 + c x
3
e1/x 1 dv v fi __ ___ – __ x = – ____ 2 dx x2 3
dv 2v 2e1/x fi ___ – ___ x = – _____ dx x2
1 = ___ 2 + c 2x
...(ii)
which is a linear differential equation dx ___ 1 IF = e – 2 Ú x = e– 2 log x = __ 2 x
1 1 fi _______ = ___ + c x (log y) 2x2
Thus,
which is the required solution. 130. The given differential equation is
Multiplying both sides of Eq. (ii) by IF and integrating, we get
dy 1 ___ + __ ◊ sin2y = x3 ◊ cos2y dx x
e1/x v __2 = – 2 Ú ____ 4 dx + c x x
3
fi
dy 2 tan y x = x3 sec2y ___ + ______ dx
...(i)
Let tan y = v dy dv fi sec2y ___ = ___ dx dx
dv 2v fi ___ + ___ x = x3 dx
...(ii)
which is a linear differential equation Thus,
IF = e 2 Ú x = e2 log x = x2 dx ___
Multiplying both sides of Eq. (ii) by IF and integrating, we get fi
v ◊ x2 = Ú x5 dx + c
x6 (tan y) x2 = __ + c 6 which is the required solution of the given differential equation. 131. The given differential equation is fi
3
(xy2 – e 1/x ) dx – x2 ydy = 0
fi
3 dy x2y ___ = xy2 – e 1/x dx
dy y ___ = dx
y e1/x __ x – ____ 2 x
y2 – __ x =
e1/x ____ 2 x
dy ___ + x(x + y) = x3 (x + y)3 – 1 dx
( (
) )
dy + dx x 1 _______ 3 _______ + _______ = x3 dx (x + y) (x + y)2 d(x + ) x 1 _______ 3 _______ + _______ = x3 ...(i) dx (x + y) (x + y)2 1 put _______ = v (x + y)2 d(x + ) ___ dv 2 fi – _______ 3 _______ = dx dx (x + y)
d(x + ) 1 1 dv fi _______ 3 _______ = – __ ___ 2 dx dx (x + y)
fi
1 dv – __ ___ + vx = x3 2 dx
dv fi ___ – 2x v = – 2x3 dx which is a linear differential equation Thus,
3
fi
which is the required solution. 132. The given differential equation is
x6 = __ + c 6
2
( )
y 2 2 1/x3 fi __x = __ e + c 3
IF = e– 2Ú x dx = e – x 2
Multiplying both sides of Eq. (ii) by IF and integrating, we get
3
...(i) Let y2 = v dy 1 ___ dv fi y ___ = __ dx 2 dx
fi
v ◊ e– x = – 2 Ú x3 e– x dx + c 2
2
2
= (x2 + 1) e– x + c 2
v = (x2 + 1) + cex
Differential Equation 2 1 fi _______ = (x2 + 1) + cex 2 (x + 1)
fi
( )
( )
1 __x d – __ xy = – d y
which is the required solution of the given differential equation. 133. The given differential equation is
Integrating, we get
1 fi __ xy =
x dx + y dy = x dy – y dx
1 fi __ d (x2 + y2) = x dy – y dx 2 fi
d (x2 + y2) = 2(x dy – y dx)
d (x2 + y2) ____________ 2(x dy – y dx) fi _________ = 2 2 (x + y ) (x2 + y2)
( ) ( )
y d __ x = 2 _______ 2 y 1 + __ x
Integrating, we get
( )
y log |x2 + y2| = 2 tan–1 __ x + c
which is the required solution of the given differential equation. 134. The given differential equation is
x dy – y dx = x4 dx
fi
y d __ x = x2 dx
( )
fi
4.53
1 __y – __ xy = – x – c y __ x + c
which is the required solution of the given differential equation. 137. The given differential equation is
x dy + y dx + xy2 dx – x2y dy = 0.
fi
d (xy) = xy (x dx – y dx)
d (xy) (x dy – y dx) = __________ fi _____ xy 2 2 xy fi
( ) (
dy 1 ___ d – __ xy = y –
)
dx ___ x
Integrating, we get
( ) 1 __y fi __ xy + log ( x ) + c = 0 fi
1 __y – __ xy = log x + c
which is the required solution of the given differential equation. 138. The given differential equation is
(4x – 3y) dx + (2y – 3x) dy = 0
y x3 fi __x = __ + c 3
fi
4x dx + 2y dy – 3 (x dy + y dx) = 0
fi
4x dx + 2y dy – 3d (xy) = 0
which is the required solution of the given differential equation. 135. The given differential equation is
Integrating, we get
x dy + y dx = sin y dy
fi
d(xy) = sin y dy
which is the required solution of the given differential equation. 139. The given differential equation is
Integrating, we get
Integrating, we get fi
xy = c – cos y
which is the required solution of the given differential equation. 136. The given differential equation is
x dy + y dx + y2 (x dy – y dx) = 0.
fi
d(xy) = y2 (y dx – x dy)
d(xy) ____________ y2 (y dx – x dy) fi _____ = x2y2 x2y2
fi
2x2 + y2 – 3xy = c
(
)
1 sin y + y sin x + __ x dx
(
)
1 + x cos y – cos x + __ y dy = 0 fi (sin y dx + x cos y dy) – (cos x dy – y sin x dx)
(
dx + ___ x +
)
dy ___ y = 0
(
dx fi d (x sin y) – d (y cos x) + ___ x + Integrating, we get
)
dy ___ y = 0
4.54 Integral Calculus, 3D Geometry & Vector Booster fi x sin y – y cos x + log (xy) = c
Integrating, we get
which is the required solution of the given differential equation. 140. The given differential equation is
which is the required solution of the given differential equation. 143. The given differential equation is
d(x2 + y2) y ______ fi _________ = 2d __ x 2 2 ÷x + y
( )
______ y d ÷x 2 + y2 = 2d __ x
(
( )
)
Integrating, we get ______ y fi ÷x 2 + y2 = 2 __ x + c
(
( )
)
which is the required solution of the given differential equation. 141. The given differential equation is
(
)
(
2
)
sin 2 x sin x _____ y + x dx + y – _____ 2 dy = 0 y
(
)
fi
sin 2 x sin2x (x dy + y dy) + _____ y dx – _____ 2 dy = 0 y
fi
sin2x d (xy) + d _____ = 0 y
( )
which is the required solution of the given differential equation. 142. The given differential equation is
2 2
x dx + y dy _____________ x + 2x y + y fi _________ = y dx – x dy x2 2 2
(x + y ) = ________ x2
x dy + y dx _________ y dx – x dy fi _________ = . 2 2 2 (x + y ) x2 2
2
( )
fi
and x dx + y dy = r dr, x dy – y dx = r2 dq
fi
y x2 + y2 = r2, tan q = __ x
a2 ◊ r2 dq r dr = _______ 2 . r
Integrating, we get
Ú r dr = Ú a2 ◊ dq
d(x + y ) y fi _________ = – 2d __ x , 2 2 2 (x + y )
(x2 + y2) y fi _______ = a2 tan–1 __ x + c 2 fi
4
...(i)
Let x = r cos q and y = r sin q,
(x2
( ) y + y ) = 2 a tan ( __ x ) + k 2
2
–1
which is the required solution. 144. The given differential equation is
dy x + y ___ y4 dx _______ = x2 + 2y2 + __ 2 . dy x y – x ___ dx
2
)
r2 fi __ = a2 q + c 2
sin2x xy + _____ y = c
4
(
dy a2 x ___ – y dy dx x + y ___ = __________ 2 dx x + y2
Integrating, we get fi
y 1 – _______ 2 = c – __ x 2 (x + y )
y 1 fi __x – _______ 2 2 = c (x + y )
x dx + y dy _________ y dx – x dy ______ __________ = x2 ÷ x 2 + y2
fi
x dx + y dy = (x2 + y2) y dy
x dx + y dy fi _________ = y dy x2 + y2 1 fi __ d (log(x2 + y2)) = y dy 2 Integrating, we get y2 c 1 __ (log (x2 + y2)) = __ + __ 2 2 2 fi
log(x2 + y2) = y2 + c
which is the required solution. 145. The given differential equation is
Differential Equation
(x + y) dx + (x – y) dy = 0
fi
fi
(x dx – y dy) + (x dy – y dx) = 0
Integrating, we get
(x dx – y dy) + d (xy) = 0
d (x + y) = d (xy) (x + y) = xy + c
Integrating, we get
which is the required solution. 150. The given differential equation is
2 x2 y __ – __ + xy = c 2 2
which is the required solution. 146. The given differential equation is
y dx + x (x – 1) dy = 0
fi
y dx – xdy = – x2 dy
fi
xdy – ydx = x2 dy
xdy – ydx fi _________ = dy x2 y fi d __ x = dy
( )
Integrating, we get
( )
xdy – ydx = (x2 + y2) dx
xdy – ydx fi _________ = dx (x2 + y2) ydx – xdy fi _________ = – dx (x2 + y2) ydx – xdy ________ x2 _________ fi = – dx y 2 1 + __ x
( ( ) ) x d ( __ y ) fi _________ = – dx y ( 1 + ( __ x ) ) 2
y __ x = y + c
Integrating, we get
which is the required solution. 147. The given differential equation is
which is the required solution. 151. The given differential equation is
ydx – x (1 – xy) dy = 0
ydx – xdy fi _________ = ydy x2 fi
( )
y d __ x = ydy
Integrating, we get
( )
2
y y __ x = __ + c 2 which is the required solution. 148. The given differential equation is
(x + y) (dx – dy) = (dx + dy)
fi
(x + y) d (x – y) = d (x + y)
fi
d(x + y) d (x – y) = _______ (x + y)
Integrating, we get
(x – y) = log |(x + y)|
which is the required solution. 149. The given differential equation is dx + dy = xdy + ydx
( )
y tan–1 __ x = c – x
( ) x+y fi ( _____ ) (xdy + ydx) = d(x + y) xy
1 1 __ x + __ y (xdy + ydx) = dx + dy
( )
x+y fi _____ d (xy) = d (x + y) xy d (xy) fi _____ = xy
d (x + y) _______ (x + y)
Integrating, we get
log |xy| = log |x + y| + log c
fi
(xy) = c(x + y)
which is the required solution. 152. The given differential equation is ______
(xdy + ydx) ÷x 2 + y2 = x2ydx + xy2dy
xdy + ydx xdx + ydy ______ fi _________ = _________ xy ÷ x 2 + y2
4.55
4.56 Integral Calculus, 3D Geometry & Vector Booster d(xy) __ d(x2 + y2) 1 _________ ______ fi _____ = xy 2 x 2 + y2
÷
(
)
d ÷ x + y ______ = __________ 2 2 ÷x + y
Integrating, we get
______ 2
2
______
log |xy| = ÷ x 2 + y2 + c
which is the required solution. 153. The given differential equation is
(x – y3) dy = y (dx + x2dy)
fi
xdy – y3dy = ydx + x2ydy
xdy – ydx fi ________ = y2dy + x2dy y xdy – ydx fi _________ = y dy (x2 + y2)
( ) ( ( ) )
x d __ y _________ fi = y dy y 2 1 + __ x Integrating, we get
y y2 tan–1 __ x = __ + c 2
( )
which is the required solution. 154. The given differential equation is
(x + 2y) dy + ydx = 0
fi
(eydx + xe ydy) = 2y dy
fi
d (xey) = 2y dy
Integrating, we get
y2 (xey) = __ + c 2
which is the required solution. 157. The given differential equation is 2ydy + (cos x ◊ cot y – y2) dx = (2xy + sin x ◊ cosec2y) dy
fi
2ydy – d (xy2)
– (sin x cosec2y dy – cos x cot y dx) = 0
fi
2ydy – d (xy2) + d (sin x cot dy) = 0
Integrating, we get
y2 – xy2 + (sin x cot dy) = c
which is the required solution. 158. The given differential equation is dy ___ = dx
2x – 3y + 1 __________ 3 – 2y – 2
fi
3xdy – 2(y + 1) dy = 2xdx – 3ydx + dx
fi
3 (xdy + ydx) – 2(y + 1) dy = (2x + 1) dx
fi
3d (xy) – 2(y + 1) dy = (2x + 1) dx
Integrating, we get
( ) ( )
y2 x2 3(xy) – 2 __ + y = __ + x + c 2 2
fi
(xdy + ydx) + 2y dy = 0
fi
d (xy) + 2y dy = 0
which is the required solution. 159. The given differential equation is
Integrating, we get
(xy) + y2 = c
which is the required solution. 155. The given differential equation is
(x + sin y) dy + ydx = 0
fi
(xdy + ydx) + sin y dy = 0
fi
d (xy) + sin y dy = 0
Integrating, we get
(xy) – cos y = c
which is the required solution. 156. The given differential equation is
eydx + (xe y – 2y) dy = 0
x (dy + dx) = y (dx – dy)
xdy – ydx xdx + ydy fi _________ = _________ 2 2 2 (x + y ) (x + y2) xdy – ydx _________ 2 y d(x2 + y2) 1 _________ fi _________ = __ 2 2 x 2 1 + __ y 2 (x + y )
( ( ) ) x d ( __ y ) 1 d(x + y ) _______ fi = __ ________ x__ 1 + ( y ) 2 (x + y ) 2
Integrating, we get
2
2
2
2
Differential Equation
( )
ydx – xdy ________ y2 d(xy) _________ fi = _____ 2 x2 __ 1 + y (xy)
x 1 tan–1 __ y = __ log |(x2 + y2)| + c 2
which is the required solution. 160. The given differential equation is
(
)
xy x (log x – tan–1 x) dy + ydx = _____ 2 dx 1+x
fi
dy xy x (log x – tan–1 x) ___ + y = _____ 2 dx 1+x
fi
dy xy x(log x – tan x) ___ = _____ – y dx 1 + x2
(
fi
dy x x(log x – tan–1 x) ___ = y _____ 2 – 1 dx 1+x
fi
dy x 1 (log x – tan–1 x) ___ = y _____ 2 – __ x dx 1+x
d fi ___ {y(log x – tan–1 x)} = 0 dx Integrating, we get
y (tan x – log x) = C
x (dy – x dx) + ydx = 0
fi
xdy + ydx = x2dx 2
d (xy) = x dx
Integrating, we get
)
)
2
x xy = __ + c 3
which is the required solution. 162. The given differential equation is dy y – x ___ dx 1 fi ________ = __ 2 + dy x y + x ___ dx
1 __ 2 y
ydx – xdy _______ x2 + y2 fi _________ = ydx + xdy (xy)2 ydx – xdy ydx + xdy fi _________ = _________ 2 2 (x + y ) (xy)2
2
( )
x 1 tan–1 __ y + __ xy = c
which is the required solution. 163. The given differential equation is
x3 xdy – x2y2dx = x4dy + x3ydx
x2y(xdy – ydx) fi ____________ = (xdy + ydx) x3
(
)
fi
xdy – ydx xy ________ = d (xy) x2
fi
y d(xy) d – __x = _____ (xy)
–1
which is the required solution. 161. The given differential equation is
fi
2
Integrating, we get
–1
(
( ( ) ) x d ( __ y ) d(xy) _______ fi = _____ x__ 1 + ( y ) (xy)
( )
Integrating, we get
y log |xy| – __x = c
which is the required solution. 164. The given differential equation is
dx + x (ydx + xdy) = e– xy dx
fi
dx + x d (xy) = e– xy dx
fi
(1 – e– xy)dx + x d(xy) = 0
d(xy) dx ________ fi ___ = 0 x + (1 – e– xy ) Integrating, we get
log |x| + log |e xy – 1| = log c
fi
x (exy – 1) = c
which is the required solution. 165. The given differential equation is x – __
y (y2 dx + (x2dy – xydx)) = e y dy
fi
y3dx + xy(xdy – ydx) = e y dy
fi
x xdy – ydx – __y y3dx + xy3 ________ = e dy y2
x – __
(
)
4.57
4.58 Integral Calculus, 3D Geometry & Vector Booster fi
x – __
Integrating, we get
y3 dx + xy3 d (– xy) = e y dy
(xy)2 ____ + 2
166. The given differential equation is
ydx – xdy + xy2dx = 0
fi
xdy – ydx = xy2dx
( )
x d __ y = xdx
Integrating, we get
( )
(1 + xy) ydx + x (1 – xy) dy = 0
fi
ydx + xdy + xy (ydx – xdy) = 0
fi
d (xy) + xy (ydx – xdy) = 0
d(xy) fi _____ = (xdy – ydx) (xy)
which is the required solution. 167. The given differential equation is
( (
d(xy) fi _____ + (ydx – xdy) = 0 (xy)
x x2 __ y = __ + c 2
a2y2 __ x4 a2x2 ____ = – ____ + c 2 4 2
which is the required solution. 169. The given differential equation is
xdy – ydx fi _________ = xdx x2 fi
y4 __ + 2
(
)
)
dy dy 2 x – y ___ (x2 + y2) = (x2 – y2) y – x ___ dx dx
) (
)
(
)
d(xy) xdy – ydx dy dx = ___ y – ___ x fi _____2 = ________ xy (xy) Integrating, we get
| |
1 __y – __ xy = log x + c
dy dy 2 x – y ___ y – x ___ dx dx fi _________ = ________ 2 2 2 (x – y ) (x + y2)
2(xdx – ydy) (ydx – xdy) fi ___________ = __________ 2 2 2 (x – y ) (x + y2)
xdy y 2 = ______ 2 2 – 1 dx ______ 2 x +y x +y
(
)
which is the required solution. 170. The given differential equation is
(
)
ydx – xdy ________ 2 2 y2 d(x – y ) _________ __________ fi 2 = x2 (x – y2) 1 + __ y
fi
xdy = ydx – (x2 + y2) dx
fi
xdy – ydx = – (x2 + y2) dx
xdy – ydx ________ x2 ________ fi = – dx y2 1 + __ x
( ( ) ) x d ( __ y ) ________ = x ( 1 + ( __ y ) ) 2
Integrating, we get
( )
x log |x2 – y2| = tan–1 __ y + c
(x2 + y2 + a2)y dy = (x2 – y2 – a2)x dx
fi x2ydy + y2xdx + (y3 + a2y) dy = (x3 – a2x) dx fi xy (xdy + ydx) + (y3 + a2y)dy = (x3 – a2x) dx 3
2
3
( ( ) ) x d ( __ y ) fi ________ = – dx y __ ( 1 + ( x ) ) 2
which is the required solution. 168. The given differential equation is
xdy – ydx = – dx fi ________ (x2 + y2)
2
fi xyd (xy) + (y + a y) dy = (x – a x) dx
Integrating, we get
( )
y tan–1 __ x = c – x
which is the required solution. 171. The given differential equation is
Differential Equation
dy x + y ___ y4 dx ________ = x2 + 2y2 + __ 2 dy x y – x ___ dx 2
2 2
xdx + ydy (x + y ) fi _________ = ________ ydx – xdy x2 2 2 ydx – xdy 1 d(x + y ) _________ fi __ ________ = 2 2 2 2 (x + y ) x2 2
2
Integrating, we get y 1 __x – ________ 2 2 = c 2(x + y )
x dx + y dy = m (x dy – y dx)
xdx + ydy ____________ m (xdy – ydx) fi ________ = 2 2 (x + y ) (x2 + y2)
( ) ( ( ) )
y m d __ x d (x2 + y2) 1 _________ __ _________ fi 2 = y 2 2 (x + y2) 1 + __ x Integrating, we get y 1 tan– 1 __ x + _________ 2 2 = c 2 (x + y ) which is the required solution. 173. The given differential equation is
( )
dy x + y ___ x sin2 (x2 + y2) dx ________ = ____________ dy y3 ___ y – x dx xdx + ydy ____________ x sin2(x2 + y2) fi _________ = ydx – xdy y3 xdx + ydy x (ydx + xdy) fi __________ = ____________ 2 2 2 sin (x + y ) y3 2 2 y x 1 d (x + y ) fi __ ___________ + __ d __ x 2 sin2(x2 + y2) y
( )
Integrating, we get
( )
1 x2 __ __ + 2 y
a2(x dy – y dx) x dx + ydy = ____________ (x2 + y2)
fi
(
)
x dy – y dx a2 _________ y2 _____________ = x 2 1 + __ y
( ( ) ) x a d ( __ y ) = ________ x ( 1 + ( __ y ) ) x a d ( __ y ) + y ) = ________ x ( 1 + ( __ y ) ) 2
1 fi __ d (x2 2
2
2
which is the required solution. 172. The given differential equation is
)
dy a2 x ___ – y dy dx x + y ___ = ___________ 2 dx x + y2
( )
y 1 d (x + y ) fi __ _________ + d __ x = 0 2 (x2 + y2)2
(
4.59
1 __ cot (x2 + y2) = c 2
which is the required solution. 174. The given differential equation is
2
2
Integrating, we get
( )
x 1 fi __ (x2 + y2) = tan– 1 __ y + c 2 which is the required solution. 175. The given differential equation is
÷
___________
x dx + y dy a2 – x2 – y2 __________ = __________ x dy – y dx x2 + y2
Let x = r cosq, y = r sinq
y fi x2 + y2 = r2, __ x = tanq
...(i)
x dy – y dx x dx + y dy = r dr, _________ = sec2 q dq x2
÷ ÷
_____
r dr 1 – r2 fi ____ = _____ 2 r dq r2 _____
dr 1 – r2 fi ___ = _____ rdq r2
dr fi ___ = ÷ 1 – r2 dq
dr _____ fi ______ = dq ÷ 1 – r2
_____
Integrating, we get
sin– 1(r) = q + c
fi
______ y sin– 1 ÷x 2 + y2 = tan– 1 __ x + c
(
)
( )
which is the required solution. 176. The given differential equation is
4.60 Integral Calculus, 3D Geometry & Vector Booster
÷
__________
x dx – y dy 1 + x2 – y2 _________ = __________ x dy – y dx x2 + y2
...(i)
Let x = r secq , y = r tanq
fi
x2 – y2 = r2
fi
x dx – y dy = r dr, x dy – y dx = r2 secq dq ______
÷
x4 + exsin y = c
which is the required solution. 180. Let the equation be y = m x, where m is an arbitrary constant Differentiating w.r.t. x, we get,
...(i)
r dr 1 + r2 ______ fi ________ = r r2sec q dq
dy ___ = m dx
dr ______ fi _______ = secq dq ÷ 1 + r2
Eliminating m, between Eqs (i) and (ii), we get dy ___ = dx
Integrating, we get
_____
log |r + ÷ r 2 + 1 | = log |secq + tanq | + log c
_____ fi r + ÷r 2 + 1 = c(secq + tanq) ______ ___________ x+y ______ fi x2 – y2 + (x2 – y2) + 1 = c _______ x2 – y2
(
)
( ÷
÷
)
( ÷ )
...(ii)
y __ x
...(iii)
whcih is the differential equation of a family of lines. dy dx Now replacing ___ by – ___ in Eq. (iii), we get, dx dy
dx x – ___ = __ dy y
which is the required solution. 177. The given differential equation is x sin __ y ( y dx – x dy) = xy3(x dy + y dx)
dy dx ___ fi ___ y = – x
fi
dx Ú ___y = – Ú ___ x
fi
fi
c log |y| = log |c| – log |x| = log __ x xy = c
( ) x y dx – x dy sin ( __ y ) _________ ( y ) = x y d (x y) x x sin ( __ y ) d ( __ y ) = x y d (x y) 2
Integrating, we get (xy)2 x – cos __ y = ____ + c 2 which is the required solution. 178. The given differential equation is
( )
x dy – y dx + y2 = y2(x dy + y dx) = 0 2
fi
y d (xy) = y dx – x dy
fi
y dx – x dy d (x y) = _________ y2
( )
x = d __y
Integrating, we get
(
)
x fi xy – __ y = c
which is the required solution. 179. The given differential equation is
(4x3 + exsin y) dx + excos y dy = 0
(4x3) dx + (exsin y dx + excos y dy) = 0
fi
fi (4x3) dx + d (exsin y ) = 0 Integrating, we get
Integrating, we get
fi
dy
| |
which is required orthogonal trajectories. 181. The given family of corves are
x2 + y2 = a2
Differentiaint w.r.t. x, we get
dy 2x + 2y ___ = 0 dx
dy x + y ___ = 0 dx dy ___ – tan 45° dy dx Replacing ___ by ____________ = dx dy ___ 1 + tan 45° dx dy ___ – 1 dx ______ in Eq. (i), we get dy ___ 1 + dx fi
( )
dy ___ – 1 dx x + y ______ = 0 dy ___ 1 + dx
...(i)
Differential Equation
) ( )
(
y2 __ + log |y| = c 2
dy dy x 1 + ___ + y ___ – 1 = 0 dx dx
x2 fi __ – 2
fi
dy (x + y) ___ = (y – x) dx
fi (x2 – y2) + 2 log |y| = c which is the required orthogonal trajectories.
dy fi ___ = dx
y–x _____ y+x
...(ii)
183. Given curve is
which is a homeogeneous differential equation. fi dy dv Put y = v x fi ___ = v + x ___ dx dx dv _____ v–1 ___ fi v + x = dx v + 1 fi dv x ___ = dx
v–1 v – 1 – v2 – v _____ – v = _____________ v+1 v+1
fi
v+1 dx fi _____ dv = – ___ x v2 + 1
( )
1 __ log |v2 + 1| + tan– 1v = c – log x 2 y fi log |x2 + y2| + tan– 1 __ x = c which is the required trajectories of the given family of curves. 182. The given curve is
( )
x2 + y2 = 1
fi
dy 2a x + 2y ___ = 0 dx
fi
dy a x + y ___ = 0 dx
...(i)
...(ii)
fi fi fi
fi
dy xy ___ = 1 – y2 dx
dy 1 – y2 fi ___ = _____ xy dx dy dx Replace ___ by – ___ , we get dx dy fi fi fi
2 dx 1 – y – ___ = _____ xy dy 1 x dx = y – __ y dy
(
)
dy x dx – y dy + ___ y = 0
Integrating, we get
dy 2x + 2y ___ dx a = __________ dy ___ dx
( )
dy 2x + 2y ___ dx (x + y ) – _________ y = 0 dy ___ dx 2
2
(
)
dy dy (x2 + y2) ___ – 2x + 2y ___ y = 0 dx dx dy (x2 + y2 – 2y2) ___ = 2xy dx dy (x2 – y2) ___ = 2xy dx
dy 2xy fi ___ = _______ 2 2 dx (x – y ) Replace dy/dx by – dx/dy, we get
Solving Eqs (i) and (ii), we get dy xy ___ + y2 = 1 dx
dy dy 2x + 2y ___ – a ___ = 0 dx dx
Eliminating a between (i) and (ii), we get
Integrating, we get
x2 + y2 – ay = 0
2xy dx – ___ = _______ 2 2 dy (x – y )
dy y2 – x2 fi ___ = ______ 2xy dx y2 __ x – 1 _______ = y 2 __ x
( )
( )
fi
y dv v2 – 1 v + x ___ = ______ , v = __ x 2v dx
fi
dv v2 – 1 x ___ = _____ – v 2v dx
v2 – 1 = – ______ 2v
2v dv dx fi _______ = – ___ x 2 (v + 1) Integrating, we get
log |v2 + 1| + log |x| = log c
4.61
4.62 Integral Calculus, 3D Geometry & Vector Booster fi
x (v2 + 1) = c 2
Integrating, we get log |x| + log |y| = log (c2)
2
fi x + y – c x = 0 which is the required orthogonal trajectories. 184. The given curve is fi fi
y2 = 4ax dy 2y ___ = 4a dx y dy a = __ ___ 2 dx
Eliminating a from Eqs (i) and (ii), we get y dy y2 = 4x __ ___ 2 dx dy fi y = 2x ___ dx
( )
Replacing dy/dx by – dx/dy, we get dy y = 2x × – ___ dx fi y dy + 2x dx = 0 Integrating, we get
...(i)
...(ii)
fi xy = c2 which is the required orthogonal trajectories. 187. The given equation is (x + y + p)(2x + p) = 0 fi
(x + y + p) = 0
...(i)
and
(2x + p) = 0
...(ii)
From Eq. (i), we get, dy ___ + y + x = 0 dx dy fi ___ + y = – x dx which is a linear differential equation.
fi
fi
y . ex = – Ú x ex dx + c1 = – ex(x – 1) + c1 y = – (x – 1) + c1e– x
y2 __ + x2 = c 2 which is the required orthogonal trajectory. 185. Given curve is xy = c2 dy fi x ___ + y = 0 dx dy y fi ___ = – __x dx
Also, from Eq. (ii), we get,
Replacing dy/dx by – dx/dy, we get y dx – ___ = – __x dy fi x dx – y dy = 0
p2 – p (ex + e– x) + 1 = 0
fi fi
1 p2 – p ex + ___ – x + 1 = 0 e p2 ex– p (e2x + 1) + ex = 0
fi
p2 ex – pe2x – p + ex = 0
fi
pex (p – ex) – 1(p – ex) = 0
fi
(p – ex) (pex – 1) = 0
fi
(p – ex) = 0
fi 2x dx – 2y dy = 0 Integrating, we get x2 – y2 = a2 which is the required orthogonal trajectories. 186. The given curve is x2 – y2 = c2. dy fi 2x – 2y ___ = 0 dx dy x fi ___ = __ dx y Replacing dy/dx by – dx/ , we get dx x – ___ = __ dy y dx dy ___ fi x + ___ y = 0
...(iii)
dy ___ + 2x = 0 dx fi
dy = – 2x dx
y = c2 – x2
fi
...(iv)
Hence, the required solution is
(y + (x – 1) – ce– x) (y + x2 – c) = 0.
188. The given equation is
and
(
x
( pe – 1) = 0
...(i)
)
...(ii) ...(iii)
From Eq. (i), we get, dy ___ = ex dx fi dy = ex dx Integrating, we get fi
y = ex + c1
From Eq. (iii), we get,
...(iv)
Differential Equation
fi fi When
dy ___ ◊ ex – 1 = 0 dx fi ex dy = dx fi dy = e– xdx Integrating, we get fi y = c2 – e– x Hence, the general solution of Eq. (i) is (x – e– x – c) (y – e– x – c) = 0. 189. The given equation is fi
p2 + 2py cot x = y2 2
2
2
2
...(i)
2
( p + y cot x) = y (1 + cot x) 2
( p + y cot x) = y cosec x
fi
( p + y cot x) = ± y cosec x
p = y (cosec x – cot x)
...(ii)
p = – y (cosec x – cot x)
...(iii)
and
From Eq. (ii), we get, dy ___ = y (cosec x – cot x) dx dy fi ___ y = (cosec x – cot x) dx
(
)
dy 1_______ – cos x fi ___ dx y = sin x dy x __ fi ___ y = tan 2 dx Integrating, we get x log |y| = 2 log |sec __ | + 2log (c1) 2 x fi y = c21 sec2 __ 2 From Eq. (iii), we get
( )
( ) ( ( ) )
dy ___ = – y (cosec x + cot x) dx dy fi ___ y = – (cosec x + cot x) dx
( )
( ( ) )
x log y = 2 logc2 – 2 log sin __ 2 x fi y sin2 __ = c2 2 Hence, the required solution is x x y – c sec2 __ y sin2 __ – c = 0. 2 2
fi
( ( ) )
( ( ( ) ) ) ( ( ) )
190. The given differential equation is fi
When
log |y| = x + c1 ( p + x + y) = 0,
dy fi ___ + (x + y) = 0 dx dy fi ___ + y = – x dx which is a linear differential equation Thus,
IF = eÚdx = ex
Thus, the solution is
y . e x = – Ú x ex dx + c
fi
y . e x = – (x – 1)ex + c2
Hence, the required solution is
(log |y| – x c1) ( y . ex + (x – 1) ex – c2) = 0
191. The given differential equation is p2y + (x – y) p – x = 0 fi p2y + px – py – x = 0 fi py( p – 1) + x( p – 1) = 0 fi ( py + x)( p – 1) = 0 fi ( py + x) = 0, ( p – 1) = 0 When
( py + x) = 0
y dy fi ____ + x = 0 dx fi y dy + x dx = 0 Integrating, we get x2 fi __ + 2
Integrating, we get dy x __ fi ___ y = – cot 2 dx
( p – y) ( p + x + y) = 0 ( p – y) = 0, ( p + x + y) = 0 (p – y) = 0
dy fi ___ = y dx dy fi ___ y = dx Integrating, we get
fi fi
p2 + px – xy – y2 = 0 ( p2 – y2) + x( p – y) = 0
4.63
fi When
y2 c2 __ = __ 2 2
x2 + y2 = c2 (p – 1) = 0
p=1 dy fi ___ = 1 dx fi dy = dx Integrating, we get y = x + c1 Hence, the solution is (x2 + y2 – c2) (y – x – c1) = 0
4.64 Integral Calculus, 3D Geometry & Vector Booster 192. The given differential equation is ( p2 – 1) xy = (x2 – y2) p
fi
p2xy – xy – px2 + py2 = 0
fi
p2xy – px2 + py2 – xy = 0
fi
px ( py – x) + y ( py – x) = 0
fi
( px + y) ( py – x) = 0
fi When
( px + y) = 0, ( py – x) = 0 ( px + y) = 0
(
dy fi ___ x dx dx fi ___ x +
)
+ y = 0 dy ___ y = 0
log x + log y = log c
fi
x y = c
When
(py – x) = 0
(
Integrating, we get
fi
x2 – y2 = c 21
193. The given differential equation is x y p2 – (x2 + y2) p + x y = 0 2
2
2
fi
x y p – x p – y p + x y = 0
fi
px( py – x) – y( py – x) = 0
fi
( px – y)( py – x) = 0
fi
( px – y) = 0, ( py – x) = 0
When
( px – y) = 0 dy fi ___ x – y = 0 dx dy dx ___ fi ___ x – y = 0 Integrating, we get log x – log y = log c fi x = c y
(
)
( )
log |1 – p2| = log c – 2 log |y| c fi log |1 – p2| = log __ 2 y c fi (1 – p2) = __ 2 y c 2 fi p = 1 – __2 ...(ii) y Eliminating p between Eqs (i) and (ii), we get _____ c x = a ± y 1 – __ 2 y
| |
(
Hence, the required solution is (xy – c) (x2 – y2 – c21) = 0
)
( )
xdx – ydy = 0 2 c 21 x2 y __ – __ = __ 2 2 2
...(i)
– 2dy – 2p fi _____ dp = _____ y 2 1–p Integrating, we get
)
(xy – c) (x2 – y2 – a2) = 0 194. The given equation can be written as x = y p + a Differentiaing w.r.t. y, we get dp dx ___ = p + y ___ dy dy dp 1 ___ fi __ p = p + y dy dp 1 fi __ p – p = y ___ dy dy p fi _____ 2 dp = ___ y 1–p
dy fi y ___ – x = 0 dx fi
x2 – y2 = a2 Hence, the required solution is
(
Integrating, we get
Integrating, we get
)
÷
which is the required solution of Eq. (i) 195. The given equation can be written as 1 y x = __ __ – y2p2 2 p
(
)
...(i)
Differentiaing w.r.t. y, we get
dx ___ = dy
( ( (
)
y dp dp 1 __ 1 __ – __ ___ – 2yp2 – 2py2 ___ 2 p p2 dy dy
)
y ___ dp dp 1 __ 1 __ 1 __ 2 2 ___ fi __ p = 2 p – p2 dy – 2yp – 2py dy
fi
dp dp 2 p = p – y ___ – 2yp4 – 2p3y2 ___ dy dy
fi
( py – x) = 0 dy fi y ___ – x = 0 dx
dp dp p + 2 yp4 = – y ___ – 2p3y2 ___ dy dy
fi
dp p (1 + 2yp3) = – y ___ (1 + 2p3y) dy
fi
fi
dp (1 + 2yp3) p + y ___ = 0 dy
When
(
)
xdx – ydy = 0
(
(
)
)
)
Differential Equation
(
)
dp fi p + y ___ = 0 dy
fi
4.65
Differentiating, w.r.t. y, we get
(
)
dp dp 2p 1 – p2 – 2py ___ – 2 (y – p2y) ___ dy dy dx ___ = ______________________________ 2 dy 4p
d ( py) = 0
Integrating, we get p y = c y fi p = __ c
...(ii)
(
)
dp dp p 1 – p2 – 2py ___ – (y – p2y) ___ dy dy 1 __ p = _____________________________ 2 2p
fi
Eliminating p between Eq. (i) and (ii), we get
c c3 y = 2 __y . x + __ 3 . y2 y
fi
dp dp dp 2p = p – p3 – 2p2y ___ – y ___ + p2y ___ dy dy dy
y2 = 2c x + c3
fi
dp dp p = – p3 – p2y ___ – y ___ dy dy
fi
which is the required solution. 196. The given differential equation is
y2 log y = x yp + p2
fi
y2 log y – p2 x = ___________ y p
...(i)
fi
dp fi ___ p +
Differentiaing w.r.t. y, we get
(
dp p ( p2 + 1) = – ( p2 + 1) y ___ dy dp p = – y ___ dy
fi
(
)
(
)
dy ___ p = 0
fi p y = c ...(ii) dp dp yp 2 y logy + y – 2p ___ – (y2 log y – p2) p + y ___ Eliminating p between Eqs (i) and (ii), we get dy dy dy ___ = ___________________________________________ 2 dx y2 + 2c x = c2 (yp) which is the required solution. dp 2 ___ fi y p = yp 2y log y + y – 2p 198. c (4x – c)2 = 64y dy
)
(
)
199. The given differential equation is dp – (y2 log y – p2) p + y ___ dy y p = 2p2x + y2p4
fi
dp y ___ ( y2log y – p2) = p ( y2 log y – p2) dy
fi
dp y ___ = p dy
dy fi ___ p =
y – y2p3 x = ________ 2p
fi
...(i)
Differentiating, w.r.t. y, we get
(
)
dp dp p 1 – 2yp3 – 3y2p2 ___ – (y – y2p3) ___ dy dy dx 1 ________________________________ fi ___ = __ 2 dy 2 p
dp ___ p
Integrating, we get
dp dp fi p + 2yp4 + 2y2p3 ___ + y ___ = 0 dy dy
Eliminating p between Eqs (i) and (ii), we get log y = c x + c2
dp fi p (1 + 2yp3) + y ___ (1 + 2 yp3) = 0 dy
which is the required solution of (i) 197. The given differential equation is p2y + 2px = y
dp fi p + y ___ (1 + 2yp3) = 0 dy
dp fi p + y ___ = 0 dy
dy dp ___ fi ___ p + y = 0
fi
log p = log y + log c
fi
p = c y
fi
2p x = y – p2y
fi
y – p2y x = _______ 2p
...(ii)
...(i)
( (
) )
4.66 Integral Calculus, 3D Geometry & Vector Booster
fi
p y = c
...(ii)
Eliminating p between Eqs (i) and (ii), we get
Hence, the required solution of Eq. (ii) is
which is the required solution. 200. The given differential equation is 2
xp – yp – p + 1 = 0
fi
yp + p – 1 x = _________ p2
...(i)
(
x.e p = – 2a Ú pe p dp = – 2ae p (p – 1) + c
fi
x = – 2a (p – 1) + ce– p
...(iii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i). 202. The given equation is y = yp2 + 2px
Differentiating, w.r.t. y, we get
IF = e Údp = e p
y2 = 2c x + c3
which is a linear differential equation.
2p x y = _____ 1 – p2 Differentiaing w.r.t. x, we get
fi
)
dp dp dp p2 p + y ___ – ___ – 2 (yp + p – 1) p ___ dy dy dy dx ___ = _________________________________ 4 dy p
2 dp dx ________ = ___ x p ( p2 – 1)
dp dp fi p3 = p3 + p2y ___ – p2 ___ dy dy
(
1 1 fi _____ + _____ – p–1 p+1
)
dx 2 ___ __ p dp = x
dp dp dp – 2p2y ___ – 2p2 ___ + 2p ___ dy dy Integrating, we get dy dp dp dp log | p – 1| + log | p + 1| – log | p2| = log |x| + log c fi p2y ___ – p2 ___ – 2p2y ___ dy dy dy p2 – 1 dp dp fi log _____ = log |c x| 2 ___ ___ – 2p + 2p = 0 p2 dy dy p2 – 1 dp fi ______ = cx fi ___ = 0 dy p2
| |
fi dp = 0
fi p = c
...(ii)
Eliminating p between Eqs (i) and (ii) we get
p2 – 1 = p2c x
fi
p2 (1 – c x) = 1
fi
1 p2 = _______ (1 – c x)
(
)
÷
_______
2
201. The given equation is y = (1 + p) x + ap
...(i)
Differentiaing w.r.t. x, we get
fi
fi
dp (x + 2 ap) ___ = – 1 dx
dx fi ___ = – (x + 2ap) dp
dx fi ___ + x = – 2ap dp
1 p = _______ (1 – c x)
fi
...(ii)
Eliminating p between Eqs (i) and (ii), we get ______
2x ÷1 – c x + c x y = 0.
which is the required solution.
dy dp dp fi ___ = (1 + p) ◊ 1 + x ___ + 2ap ___ dx dx dx dp dp p = (1 + p) ◊ 1 + x ___ + 2ap ___ dx dx
fi
yc = c2x – c + 1
which is the required solution.
203. The given differential equation is y = 2 p x + tan– 1(xp2) Differentiating w.r.t. x, we get
...(ii)
dy dp ___ = 2p + 2x ___ + dx dx
(
...(i)
dp p2 + 2px ___ dx __________ 1 + (xp2)2
)
dp dp fi – p – 2x ___ (1 + p4x2) = p2 + 2px ___ dx dx
dp dp dp fi – p – 2x ___ – p5x2 – 2x3p4 ___ = p2 + 2px ___ dx dx dx
Differential Equation
dp dp dp fi – p – p5x2 – p2 = + 2x ___ + 2x ___ + 2x3p4 ___ dx dx dx dp fi – p (1 + p4x2 – p) = + 2x (p + 1 + x2p4) ___ dx dp fi – p = 2x ___ dx
2dp ___ dx fi ____ p + x = 0
fi 2 log p + log x = log c
fi p2x = c
dy dp dp ___ + p + x ___ = 2px4 ___ + 4x3p2 dx dx dx
fi
dp dp p + p + x ___ = 2px4 ___ + 4x3p2 dx dx
fi
fi
fi
dp x (1 – 2px3) ___ = – 2p (1 – 2x3p) dx
fi
dp x ___ = – 2p dx
dp ____ 2dx fi ___ = 0 p + x
...(ii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i). __
( ÷ )
c y = 2 __ x x + tan– 1(c)
fi
– 1 y = 2÷c x + tan (c)
___
fi
2 4
x p + 2xp = y 4 2
y = p x + 2px
dy dp dp fi ___ = 2x p4 + 4x2p3 ___ + 2p + 2x ___ dx dx
fi
fi
fi fi fi
dp (x – 2px4) ___ = 4x3p2 – 2p dx dp x (1 – 2px3) ___ = 2p (2x3p – 1) dx
Integrating, we get
which is the required solution. 204. The given differential equation is
...(i)
dp dp p = 2x p4 + 4x2p3 ___ + 2p + 2x ___ dx dp – 2x p4 – p = (4x2p3 + 2x) ___
log p + 2log x = log c
fi
p x2 = c
dp 2x (2xp3 + 1) ___ = – p (2xp3 + 1) dx dp 2x ___ = – p dx dp dx ___ 2 ___ p = – x
...(ii)
Eliminating p between Eqs (i) and (iii), we get the required solution of Eq. (i) c y + __ x + c2 which is the required solution. 206. The given differential equation is
y = p sin p + cos p
...(i)
dy dp dp dp fi ___ = sin p ___ + p cos p ___ – sin p ___ dx dx dx dx dp fi p = p cos p ___ dx
fi
dp cos p ___ = 1 dx
fi
cos p dp = dx
Integrating, we get
Integrating, we get
2 log |p| = log c – log |x|
fi
c 2 log |p| = log __x
Eliminating p from Eqs (i) and (ii), we get
| |
___
2 y = 2÷c x + c
y + px = x4p2
Differentiating w.r.t. x, we get
...(ii)
y = (x + c) sin– 1(x + c) + ÷ 1 – (x + c)2
which is the required solution. 207. The given differential equation can be written as
which is the required solution. 205. The given differential equation is
sin p = x + c
__________
x p2 = __ c ...(ii) Eliminating p between Eqs (i) and (ii), we get
4.67
2 y = Px – __ – 1 P
...(i)
Differentiating w.r.t. x, we get ...(i)
dP dP ___ = P + x ___ + dx dx
2 dP ___ 2 ___ P dx
4.68 Integral Calculus, 3D Geometry & Vector Booster fi
dP 2 dP P = P + x ___ + __ ___ dx p2 dx
(
)
211. The given differential equation is
2 dP fi x + ___ 2 ___ = 0 P dx
(
)
dP 2 fi ___ = 0 or x + ___ 2 = 0 dx P fi
P = c
...(ii)
or
2 x = – __2 p
...(iii)
(x – a) P2 + (x – y) P – y = 0
fi
P2x + Px – aP2 = ( P + 1) y
fi
( P + 1) y = P (P + 1) x – aP2
aP2 y = Px – _______ ( P + 1)
fi
which is a Clairaut differential equation. Hence, the solution is ac2 y = cx – ______ (c + 1)
Eliminating p between Eqs (i) and (ii), we get
2 y = cx – __ c – 1 which is the genral solution of Eq. (i).
212. The given differential equation is
Eliminating p between Eqs (i) and (iii), we get
which is a Clairaut differential equation. Hence, the solution is _____
(y + 1)2 + 8x = 0 which is the singular solution of Eq. (i). 208. The given differential equation is 3
2
Px–Py–1=0
fi
P2y = P3x – 1
1 y = P x – ___ 2 P
fi
( y + 1) P – x P2 + 2 = 0
fi
( y + 1) P = p2x – 2
fi
2 ( y + 1) = Px – __ p
fi
2 y = P x – __ + 1 P
(
)
which is a Clairaut differential equation. Hence, the solution is 2 y = c x – __ c + 1 210. The given differential equation is
)
sin y cos P x – cos y sin Px – P = 0
fi
sin (y – Px) = P
fi
(y – Px) = sin– 1 (P)
fi
y = Px + a ÷1 + P2
y = cx + a ÷1 + c2
213. The given differential equation is a y = p (x – b) + __ p a fi y = px + __ p – bp
(
which is a Clairaut differential equation. Hence, the solution is 1 y = c x – __ 2 c 209. The given differential equation is
(
______
– 1
y = Px + sin (P)
which is a Clairaut differential equation. Hence, the solution is y = c x + sin– 1(c)
)
which is a Clairaut differential equation. Hence, the solution is a y = cx + __ c – bc 214. The given differential equation is
(
)
d 2y ____2 = x + sin x
( )
d dy fi ___ ___ = x + sin x dx dx Integrating, we get dy x2 ___ = __ – cos x + c1 2 dx Again integrating, we get x3 y = __ – sin x + c1x + c2 6 which is the required solution. 215. The given differential equation is fi
d2y ___ 2 = e2x + ex + 2014 dx
d dy fi ___ ___ = e2x + ex + 2014 dx dx
( )
Integrating, we get
dy ___ = dx
e2x ___ + ex + 2014 + c1 2
Differential Equation
Again integrating, we get 2x
e y = ___ + ex + 1007x2 + c1x + c2 4 which is the required solution. 216. The given differential equation is
d 2y ___2 = sin2x dx d dy fi ___ ___ = sin2x dx dx
( )
Integrating, we get
( ) Ú ( ) ( dy ___ = dx
1 __ 2 sin 2x dx + c1 2
dy 2sin2x 1 fi ___ = __ x – ______ + c1 2 2 dx
)
Again Integrating, we get
(
)
cos2x 1 x2 _____ y = __ __ + + c1x + c2 2 2 4
which is the required solution. 217. The given differential equation is d 2y ___2 = cos3x dx d dy fi ___ ___ = cos3x dx dx
( )
( ) ( )
dy ___ = Ú cos3x dx dx dy fi ___ = Ú cos3x dx dx
(
)
)
x2 1 cos3x y = __ + __ x + _____ + c1x + c2 2 3 3
(
)
which is the required solution. 218. The given differential equation is d 2y 1 ___2 = _________ 2 2 dx sin x cos x
( ) ( )
d dy sin2x + cos2x fi ___ ___ = ____________ dx dx sin2x cos2x
d dy fi ___ ___ = sec2x + cosec2x dx dx
Integrating, we get
y = log (cos x) – log (sin x) + c1x + c2
fi
y = log (cot x) + c1x + c2
which is the required solution. 219. The given differential equation is d2y ___2 = sin4x + cos4x dx = 1 – 2 sin2x cos2x 1 = 1 – __ sin2x 2 1 = 1 – __ (1 – cos4x) 4 3 __ 1 __ = + cos4x 4 4 3 1 d dy fi ___ ___ = __ + __ cos4x 4 4 dx dx
( )
Integrating, we get dy 3x 1 ___ = ___ + ___ sin4x + c1 4 dx 16 Again Integrating, we get
( )
3x2 cos (4x) y = ___ + _______ + c1x + c2 8 64
d 2y ___2 = x ex dx
sin3x y = Ú x – _____ dx + c1x + c2 3
fi
fi
which is the required solution. 220. The given differential equation is
sin3x = x – _____ + c1 3 Again Integrating, we get
(
Again Integrating, we get
Integrating, we get
( )
dy ___ = tan x – cot x + c1 dx
( )
d dy fi ___ ___ = x ex dy dx Integrating, we get
( )
dy fi ___ = (x – 1)ex + c1 dx Again integrating, we get fi
y = (x – 2)ex + c1x + c2
which is the required solution. 221. The given differential equation is d 2y ___2 + y = 0 dx dp fi p ___ + y = 0 dy
fi
dp p ___ = – y dy
fi
2p dp = – 2y dy
On integrating, we get
4.69
4.70 Integral Calculus, 3D Geometry & Vector Booster p2 = c 21 – y2
fi
dy 2 fi ___ = c 21 – y2 dx
_______ dy fi ___ = ± ÷( c21 – y2) dx
( )
( )
fi
dy 4p dp = ___ __ y ÷
1 ____ __ 4÷y
__
2p2 = 2÷y + 2c1
p2 = ÷y + c1
which is the required solution. 222. The given differential equation is 1 __ 3 y
dp p ___ = dy
Integrating, we get c1 2 sin– 1 __ y = c ± x
dy ___2 = dx
fi
Integrating, we get
dy _______ fi ________ = ± dx ÷( c21 – y2)
2
fi
__
( ) ( )
dy 2 __ fi ___ = ÷y + c1 dx _______ dy __ fi ___ = + ÷ ÷ y + c1 dx dy ___ fi ________ = ± dx __ ÷÷ y + c1
Again integrating, we get dy ___ c2 ± x = Ú ________ dy dp __ __ 1 ___ ___ fi p = y , Let p = + c1 ÷ ÷ y dy dx 2t dt dy _____ = Ú ______ , where y = t2 fi p dp = – ___3 ÷t + c1 y 2{(t + c1) – c1}dt Integrating, we get _____ = Ú _______________ ÷t + c1 2 c p 1 1 _____ __ = ___ 2 + __ = 2 Ú (÷t + c1 – c1(t + c1)– 1/2) dt 2 2 2y fi
(
c1y2 + 1 p2 = _______ y2
( )
dy 2 c1y2 + 1 fi ___ = _______ dx y2
÷
________ 2
c1y + 1 dy fi ___ = ± _______ dx y2
y ____ fi _____ + 1dy = ± dx ÷ c 1y2
)
_______
which is the required solution. 223. The given differential equation is d2y ___2 = dx
1 ____ __ 4÷y
( ) ( )
d dy fi ___ ___ = dx dx
dy d dy ___ fi ___ ___ = dy dx dx
1 ____ __ 4÷y 1 ____ __ 4÷y
__
÷
1
3/2
which is the required solution. 224. The given differential equation is d2y a2 ___ – y = 0 dx2
Again integrating, we get ÷c 1y2 + 1 ________ = c2 ± x c1
[ 1 = 4 [ __ ( y + c ) 3
d2y y fi ___2 = __ 2 dx a y dp fi p ___ = __ dy a2
y dy p dp = ____ 2 a Integrating, we get
fi
p2a2 = y2 + c 21
fi
y2 + c21 p2 = _______ 2 a
] – c ÷ y + c ]+ c
_____ 1 = 4 __ (t + c1)3/2 – c1÷t + c1 + c2 3
( )
y2 + c21 dy 2 _______ ___ fi = 2 dx a
1
_______ __ ÷ 1 2
Differential Equation
( ) ÷
______
y2 + c21 dy fi ___ = ± ______ dx a2
dy dx ______ fi _______ = ± ___ 2 2 2 a ÷y + c 1
|
_______
|
fi
2p dp = 3y2 dy
p2 = y3 + c1
( ) ( )
______ dy fi ___ = ± ÷y 3 + c1 dx
When
\
( )
dy x = – 2, y = – 1 and ___ = – 1 dx x = – 2 c1 = 2
( )
_____ dy Thus, ___ = ± ÷y 3 + 2 dx
dy _____ fi _______ = ± dx ÷ y 3 + 2
c1 p2 e2y __ __ = ___ + 2 2 2
227. The given differential equation is
p2 = e2y c1
( ) ( )
dy 2 fi ___ = e2y + c1 dx _______ dy fi ___ = ± ÷e 2y + c1 dx
( )
d2y y3 ___ 2 = – 1 dx
d2y 1 fi ___2 = – __3 dx y
dy _______ fi ________ = ± dx 2y ÷e + c1
Again integrating, we get
dy 2 fi ___ = y3 + c1 dx
d2y ___2 = e2y dx dp fi p ___ = e2y dy fi p dp = e2y dy Integrating, we get
dp 2p ___ = 3y2 dy
x log y + ÷y 2 + c21 = c2 ± __ 2 a which is the required solution. 225. The given differential equation is
fi
fi
Integrating, we get
Again integrating, we get
fi
dp 1 2p ___ = – __3 dy y
dy 2p dp = – ___3 y Integrating, we get 1 p2 = – ___ 2 + c1 2y
dy _______ c2 ± x = Ú ________ ÷e 2y + c1
fi
e– ydy _________ = Ú __________ ÷1 + c1e– 2y
dt _______ = – Ú ________ , t = e– y 2 ÷1 + c1t
dy 2 1 fi ___ = – ___ 2 + c1 dx 2y
dt 1__ __________ = – ___ Ú _________ c 1 2 ÷ 1__ 2 ___ t + ÷c 1
dy 2 2c1y2 – 1 fi ___ = _________ dx 2y2
÷ ( ) | ÷ | ÷
_______
|
When
1__ 1 = – ___ log t + t2 + ___ __ ÷c 1 ÷c 1
1__ 1 = – ___ log e– y + e– 2y + ___ __ ÷c 1 ÷c 1
_________
226. The given differential equation is
d 2y 2 ___2 = 3y2 dx
( ) ( )
\
|
( )
dy x = – 1, y = – 1, ___ = 0 dx x = –1 1 c1 = __ 2
( )
÷
______
dy y2 – 1 \ ___ = ± ______ 2 dx 2y
y dy dx _____ fi _______ = ± ___ 2 ÷ y – 1 2
4.71
4.72 Integral Calculus, 3D Geometry & Vector Booster
_____
x fi ÷y 2 – 1 = c2 ± __ 2
Again integrating, we get
which is the required solution. 228. The given differential equation is
d2y x ___2 + dx
dy ___ + x = 0 dx
dy d2y Let ___ = p fi ___ 2 = dx dx
fi
dp x ___ + p + x = 0 dx x dp + p dx + x dx = 0
fi
d (xp) + xdx = 0
fi
Integrating, we get
)
x2 y = – __ + c1 log |x| + c2 4 which is the required solution. 229. The given differential equation is
2
d y 1 dy x ___ + x ___2 = __ dx dx dp p + x fi ___ = __ dx x
fi
x dp = p dx + x2dx
x dp – p dx fi __________ = dx x2 p fi d __ x = dx Integrating, we get p __ x2 __ = + c1 x 2
( )
3
x p = __ + c1x 2 dy x3 fi ___ = __ + c1x 2 dx fi
fi
dp y p ___ + p2 = 1 dy
(
dy 2 ___ = 1 dx
dp y p ___ = 1 – p2 dy dy p fi _____ 2 dp = ___ y 1–p fi
1 – __ log |1 – p2| = log y + log c1 2
1 fi _____ = (c1y)2 1 – p2
fi
1 1 – p2 = ____ 2 2 c1 y
fi
1 p2 – 1 = ____ 2 2 c1 y
dy 1 fi ___ = 1 – ____ 2 2 dx c 1 y
c1y _______ fi ________ dy = dx 2 2 ÷c 1 y – 1
÷
_______
Again integrating, we get _______
fi
d2y y ___2 + dx
fi
( )
Integrating, we get
x x p = – __ + c1 2
(
dp ___ dx
)
c1x2 ____ + c2 2
which is the required solution. 230. The given differential equation is
2
dy x2 x ___ = – __ + c1 2 dx dy x c1 x fi ___ = – __ + __ 2 dx x c1 x dx fi dy = – __ + __ 2 Again Integrating, we get
...(i)
(
x4 y = __ + 6
÷c 21 y2 – 1 _________ = x + c2 c21 which is the required solution. 231. The given differential equation is
fi
fi
dp y ___ – p = y2 dy
fi
y dp – p dy = y2dy
)
x3 dy = __ + c1x dx 2
( )
( )
d2y dy 2 dy y ___2 – ___ = y2 ___ dx dx dx dp py ___ – p2 = y2p dy
ydp – pdy fi _________ = dy y2 p fi d __ y = dy
( )
Differential Equation
Integrating, we get p __ y = y + c1 fi p = y2 + c1y
dy fi ___ = y2 + c1y dx
dy fi _______ = dx 2 y + c1y
dy fi ________ = dx y(y + c1)
(
Integrating, we get
( )
x log |x2 + c1| + 2a tan– 1 __ c + y = c2 1
which is the reqired solution. 233. Given differential equation is
)
1 1 fi __ y – _____ dy = dx y + c1
Again integrating, we get y log _____ = x + c2 y + c1
| |
which is the required solution. 232. Given differential equation is
( ) ( )
( ) ( )
d2y 1 ___ dy dy x ___2 – __ 2 = ___ 4 dx dx dx
fi
dp 1 x ___ – __ p2 = p dx 4
fi
dp p2 p x ___ = __ + p = p __ + 1 4 4 dx
4 dp dx fi _______ = ___ x p(p + 4)
dx 1 1 fi __ p – _____ dp = ___ x p+4
(
(
)
Integrating, we get
| |
d 2y dy 2 dy (x + a) ___2 + x ___ = ___ dx dx dx
fi
dp (x + a) ___ + xp2 = p dx
p fi _____ = c1x p+4
p log _____ = log |x| + log c1 p+4
p dp xp2 fi ___ + ______ = _____ dx (x + a) x + a
p + 4 ___ 1 fi _____ = c x p 1
dp p xp2 fi ___ – _____ = – ______ (x + a) dx x + a
fi
x 1 dp ________ 1 fi __2 ___ – = – ______ (x + a) p dx p/(x + a)
4 ______ 1 fi __ p = c1x – 1
– c1x 4 1______ fi __ p = c1x
c1x p fi __ = ______ 4 1 – c1x
fi
Put
1 – __ p = v
dv v x fi ___ + ______ = ______ dx (x + a) (x + a) which is a linear differential equation. \
dx _____ IF = eÚ x + a = elog |x + a| = (x + a)
fi
v (x + a) = Ú x dx
fi
x2 v(x + a) = _____ 2+c
fi
x2 1 __ – __ p (x + a) = 2 + c
2 (x + a) fi ________ dx + dy = 0 x2 + C1 2x 2a fi ______ dx + ______ 2 dx + dy = 0 x2 + C1 x + C1
)
4 ___ 1 1 + __ p = c1x
4c1x dy = ______ dx 1 – c1x
Again integrating, we get 1 y = – 4x + __ c log |c1x – 1| + c2 1 which is the required solution. 234. Given differential equation is fi
( )
d2y dy 2 ___ y ____ = 1 – dx dx2 dp py ___ = 1 – p2 dx
dy 2p dp fi _______ = – 2 ___ y 2 (p – 1)
4.73
4.74 Integral Calculus, 3D Geometry & Vector Booster Integrating, we get
fi
fi
fi
2
log |p – 1| = log C1 – log |y |
| |
C1 log |p – 1| = log ___ 2 y C 1 (p2 – 1) = ___ 2 y C1 p2 = ___ 2 + 1 y 2
C1 + y2 = ______ y2
÷ ÷
_______
C1 + y2 p = ± _______ 2 y
fi
C1 + y2 dy fi ___ = ± ______ dx y2
y dy ______ fi ________ = ± dx 2 ÷y + C1
fi ÷y 2 + C1 = C2 ± x
_______
______
which is the required solution. 235. Given differential equation is
( )
d 2y dy ___2 = 2y ___ dx dx
2
fi
dp y p ___ – p2 = y2p dy
fi
dp y ___ – p = y2 dy
dp fi ___ – dy
p __ y = y
...(i)
which is a linear differential equation. \
dy
– ___ 1 IF = e Ú y = e– log y = __ y
Multiplying both sides of Eq. (i) by IF and integrating, we get p __ y = Ú dy + c1 p __ y = y + c1 p = y2 + c1y dy ___ = y2 + c1y dx dy ________ = dx y( y + c1) Integrating, we get
| |
y 1 _____ __ | = x + c2 c1 log | y + c1 which is the required solution. 237. Given differential equation is
( ) ( )
d2y 1 ___ dy a ___2 + __ x – __ 2 = 0 dx dx x
dp fi ___ = 2y p dx dp fi p ___ = 2y p dy
dp fi ___ = 2y dy
fi
fi
fi
dy d2y x2 ___ + x ___ = a 2 dx dx dp x2 ___ + xp = a dx a dx x dp + p dx = ____ x
fi
a dx d (xp) = ____ x
fi
dp = 2y dy
Integrating, we get p = y2 + C1
dy fi ___ = y2 + C1 dx
Integrating, we get xp = a log |x| + c1
dy fi ________ = dx 2 (y + C1)
fi
dy x ___ = a log |x| + c1 dx
y 1 – 1 ____ ___ fi ____ = x + C2 tan ___ ÷C 1 ÷C 1
fi
c1 a log |x| ___ dy = ______ x + x dx
( )
which is the required solution. 236. Given differential equation is fi
d2y y ___2 – dx
( )
( )
dy 2 dy ___ = y2 ___ dx dx
(
)
Again integrating, we get
(log |x|)2 y = a _______ + C1 log |x| + C2 2
which is the required solution.
Differential Equation
238. Given differential equation is
( )
2
dy dy y ___2 – y ___ ln y = dx dx dp fi y ___ – yp ln y = p2 dx lny 1 1 dp ___ fi __2 ___ + – p = __ y p dx
fi
240. Given
( )
dy ___ = y + 2x dx dy fi ___ + (– y) = 2x dx
dy ___ dx 2
which is a linear differential equation.
fi
dp fi ___ – dy
p __ y = ln y
...(i)
dy
___ 1 IF = e – Ú y = e– log|y = __ y
Multiplying both sides of Eq. (i) by IF and integrating, we get log y p ____ __ dy + c1 y = Ú y
(log y)2 p ______ fi __ = + c1 y 2 dy fi ______________ = dx 2 (log y) + 2c1 y ____________ 2
(
Integrating, we get
÷
(Let log y = t)
( )
( )
log y 2 fi ___ tan– 1 _____ ____ = x + c2 c1 ÷2c 1
which is the required solution. 239. The slope of the curve = The slope of the tangent
dy = tany = ___ = dx
fi
2y dy = dx
y2 = x + c
fi
fi
y ◊ e– x = – 2e– x (x + 1) + c
241. Given
y ◊ e– x = – 2e– x (x + 1)
dy x4 + 2x y – 1 ___ = __________ dx x2 + 1 2xy = x2 – 1 + _____ 2 x +1 dy 2x fi ___ – _____ y = x2 – 1 dx x2 + 1
t 2 – 1 _____ ___ fi ____ = x + c2 tan ____ 1 ÷2c 1 ÷ 2c ___
y ◊ e– x = 2 Ú xe– x dx + c
which is a linear differential equation.
)
2 dt fi ________ = dx, (t2 + 2c1)
which is passing through (0, 0) then c = 0. Hence, the required equation of the curve is
which is a linear differential equation. \
IF = e – Údx = e– x
Hence, the solution is
dp y p ___ – y p ln y = p2 dy
dp y ___ – y ln y = p dy dp p fi ___ – ln y = __ y dy
4.75
1 ___ 2y
which is passing through (4, 3), so c = 5 Hence the equation of the curve is y2 = x + 5.
2x – Ú _____ dx x2 + 1 =
IF = e
2
e – log |x
1 = _____ 2 x +1
+ 1|
Hence, the solution is 1 y ◊ _____ = x – 2 tan– 1x + c 2 x +1 which passes through (0, 0), then c = 0. Hence, the equation of the curve is
y = (x2 + 1) (x – 2 tan– 1x)
242. Given dy (x + 1)2 + y – 3 ___ = _____________ (x + 1) dx
Let
dy fi ___ = dx
fi
...(i)
x + 1 = X, y – 3 = Y dY ___ dX
dY Y ___ = X + __ X dX
( )
dY 1 fi ___ + – __ Y = X X dX which is a linear differential equation. \
1 __ 1 IF = e – Ú x dX = e– log X = __ X
4.76 Integral Calculus, 3D Geometry & Vector Booster Hence, the solution is
( )
1 1 Y ◊ __ = Ú X ◊ __ dX + c = X + c X X
Y fi __ = X + c X y–3 fi _____ = (x + 1) + c x+1 which is passing through (2, 0) then c = – 4 Hence, the required equation of the curve is y–3 _____ = (x + 1) – 4 x+1 fi
y – 3 = (x + 1)2 – 4(x + 1)
fi
y = x2 – 2x
243. The equation of the tangent at any point (x, y) on the curve is dy Y – y = ___ (X – x) dx dy It meets x-axis at A x – y ___ , 0 dx
(
(
)
)
dy and y-axis at B 0, y – x ___ dx \ Mid-point of AB
( (
) (
))
dy 1 dy 1 = __ x – y ___ , __ y – x ___ 2 dx 2 dx
It is given that,
( (
) )
dy 1 and __ x – y ___ = x, 2 dx dy 1 __ y – x ___ = y 2 dx dy Thus, x ___ = – y dx dy dx ___ fi ___ y = – x dy ___ dx fi ___ y + x = 0 fi log y + log x = log c fi
log (y x) log c
fi x y = c which is passing through (1, 1), then c = 1. Hence, the equation of the curve is
x y = 1.
244. The equation of the normal at the point (x, y) is dx Y – y = – ___ (X – x) ...(i) dy The distance of perpendicular from the origin to the normal (i) is
|
|
dy y + x ___ dx ________ = _________ dy 2 ___ 1 + dx
÷ ( )
Also the distance between P and x-axis is |y|.
|
|
dy y + x ___ dx ________ Thus, _________ = |y| dy 2 ___ 1 + dx
÷ ( ) ( ) ( ) ( ( ) ( )
( ( ) )
fi
dy dy 2 dx 2 y2 + ___ x2 + 2xy ___ = y2 1 + ___ dy dx dx
fi
dx 2 dx (x2 – y2) ___ + 2xy ___ = 0 dy dy
)
dx dx fi ___ (x2 – y2) ___ + 2xy = 0 dy dy y2 – x2 dx dx fi ___ = 0, ___ = ______ 2xy dy dy dx Now, ___ = 0 gives x = k dy which is passing through (1, 1), so k = 1. Thus, the equation of the curve is x = 1 y2 – x2 dx Also, ___ = 0 = ______ 2xy dy which is a homogeneous differential equation. dy dv Let y = v x fi ___ = v + x ___ dx dx fi
dv v2 – 1 v + x ___ = _____ 2v dx
fi
dv v2 – 1 v2 + 1 x ___ = _____ – v = – _____ 2v 2v dx
2v dx fi ___2 + 1dv = – ___ x v fi log |v2 + 1| = log c – log x c fi |v2 + 1| = __ x fi
x2 + y2 = c x
which also passes through (1, 1), so c = 2. Hence, the equation of the curve is x2 + y2 = 2x. 245. The Equation of the tangent at any point (x, y) is dy Y – y = ___ (X – x) dx dy It meets the x-axis at x – y ___ , 0 dx
(
Given
AP = 1
fi
AP 2 = 1
)
Differential Equation
( ) ( )
dx 2 fi – y ___ + y2 = 1 dy
2v dx fi _________ dv = ___ x 2 1 + v – 2v
2 dx 2 1 – y fi ___ = _____ dy y2
2v dx fi _________ dv = – ___ x 2v2 – v – 1
dy y _____ fi ___ = ± ______ dx ÷1 – y2 ______
÷ 1 – y2 fi _______ dy = ± dx y
|
Integrating, we get
|
______ y _____ log __________ + ÷ 1 – y2 = c ± x 1 + ÷1 – y2
which is the required equation of the curve. 246. The slope of the line segment joining the points (x, y) and (– 4, – 3) is y+3 _____ x+4
( )
dy y+3 Thus, ___ = 2 _____ x+4 dx
(4v – 1)dv dv 2dx fi ___________ + ___________ = – ____ x (2v2 – v – 1) (2v2 – v – 1)
|
|
2v – 2 1 fi log |2v2 – v – 1| + __ log ______ = c – log x 3 2v – 1
|
| | |
2y2 fi log ___ 2 – x
2y – 2x 1 fi log |2y2 – xy – x2| + __ log _______ – log x = c 3 2y – x
log |y + 3| = log c + 2 log (x + 4)
fi
( y + 3) = c (x + 4)2
which is passing through (– 2, 1), so c = 1 Hence, the required equation of the curve is 2
( y + 3) = (x + 4)
247. The equation of the normal at P is dx Y – y = ___ (X – x) dy Thus, the points A and B are (x, 0) and
(
(4v – 1) + 1 2dx fi ___________ dv = – ____ x 2v2 – v – 1
y 2y – 2x 1 __ x – 1 + __ log ______ = c – log x 3 2y – x
| |
which is the required equation of the curve. 248. The equation of the normal at P is
dy 2 ◊ dx fi _____ = _____ y+3 x+4 Integrating, we get
4.77
)
dy x + y ___ , 0 dx Given condition is
dy x+y x + y ___ – x = _____ 2 dx
fi
dv 1 + v 1 + v – 2v2 x ___ = _____ – v = __________ 2v 2v dx
dx Y – y = – ___ (X – x) dy
(
)
dy It meets the x-axis at A x + y ___ , 0 dx Given condition is
OP2 = OA2
( )
dy 2 x2 + y2 = y2 ___ + y2 dx dy x fi ___ = ± __y dx fi
Taking positive sign, dy x ___ = __ dx y fi
y dy = x dx
x2 fi __ – 2
y2 c2 __ = __ 2 2
fi x2 – y2 = c2, which represents a rectangular hyperbola. dy x + y Taking negative sign, ___ = _____ 2y dx dy x ___ = – __y which is a homogeneous equation. dx dy dv y dy = – x dx Put y = vx fi ___ = v + x ___ fi dx dx y2 __ x2 __ a2 __ dv 1_____ +v fi + = ___ fi v + x = 2 2 2 2v dx fi
x2 + y2 = a2.
which represents a circle.
4.78 Integral Calculus, 3D Geometry & Vector Booster
(Problems for JEE-Advanced)
when
1. It is given that
2
e – log (x
1 = _______ 2 (x + 1)
+ 1)
Hence, the sloution is
which is a linear differential equation. 4
1 x – 1 y ◊ _______ = Ú ________ dx + c 2 (x + 1) (x2 + 1)2
dy cos2 x ___ – (tan2x) y = cos4 x dx dy (tan2x) fi ___ – ______ y = cos2 x dx cos2 x
which is a linear differential equation. – Ú_____ 2x dx 1 + x2 =
p p x = 1, y = __ then c = __ 2 2
( )
dy 2x x4 – 1 fi ___ – _____ 2 y = ______ dx 1 + x 1 + x2
IF = e
( )
y (x y) sin __ x = c
Hence, the solution is y p (x y) sin __x = __ 2 3. The given differential equation is
dy x4 + 2xy – 1 ___ = ___________ dx 1 + x2
\
fi
\
(
(
)
fi
– 1
) (
)
cos 2x cos 2x y ◊ _________ = Ú _________ cos 2x dx + c 1 + cos 2x 1 + cos 2x
2 = Ú 1 – _____ 2 dx + c x +1
cos 2x _________ 1 + cos 2x
Hence, the solution is
y x2 – 1 fi _______ = Ú _______ dx + c 2 (x + 1) (x2 + 1)
tan(2x) – Ú ______ dx cos2 x =
IF = e
= (x – 2 tan x) + c
which is the required solution. 2. The given differential equation is y y y cos __ x (x dy – y dx) + x sin __ x (x dy + y dx) = 0
(
)
cos 2x 1 y ◊ _________ = __ Ú cos 2x dx + c 1 + cos 2x 2 sin 2x = _____ 4+c
__
3÷3 p when x = __ , y = ____ then c = 0 8 6 Hence, the required curve is
( )
( ) y y dy __y y dy __y fi __ x cos ( __ x ) ( ___ – x ) + sin ( __x ) ( ___ + x ) = 0 dx dx
1 y = __ ◊ tan (2x) ◊ cos 2x 2 dv dv ___ ___ 4. The given differential equation is fi v cos v v + x – v + sin v v + x + v = 0 dx dx d y + ___ (x y) = x (sin x + log x) y dx __ Let x = v dy dv dv fi x ___ + 2y = x sin x + x log x fi xv cos v ___ + sin v 2v + x ___ = 0 dx dx dx dy 2y dv fi ___ + ___ x = sin x + log x fi x (v cos v + sin v) ___ = – 2v sin v dx dx
(
)
(
( )
)
(
dv 2v sin v fi x ___ = – _____________ dx (v cos v + sin v)
(v cos v + sin v) 2dx fi _____________ dv = – ____ x v sin v
2dx 1 fi cot v + __ v dv = – ____ x
(
)
Integrating, we get
log |sin v| + log |sin |v| = log c – 2 log x
c fi v sin v = __ 2 x y__ y__ c fi x sin x = __ 2 x
( ) ( )
)
which is a linear differential equation. \
2dx ____ IF = e Ú x = x2
Hence, the solution is
y ◊ x2 = Ú (x2 sin x + x2log x) dx + c
fi
2sin x cos x y = – cos x + _____ x + 2 _____ x2 x x c + __ log x – __ + __ . 3 9 x2
5. The given differential equation is
x (1 – x2) dy + (2 x 2 y – y – 5 x3) dx = 0
Differential Equation
dy (2x 2 – 1) y _______ 5 x3 fi ___ + __________ = 2 dx x (1 – x ) x (1 – x2)
dy fi ___ – dx
2
fi
dy (2x – 1) y 5 x2 fi ___ + __________ = _______ 2 dx x(1 – x ) (1 – x2)
which is a linear differential equation \
IF =
(2 x2 – 1) Ú ________ dx 2 e x(1 – x ) =
Hence, the solution is
1 _______ _____ x÷1 – x2
5x 1 _____ y ◊ _______ = Ú ________ dx + c 2 3/2 2 (1 – x ) x÷1 – x
y 5 _____ fi _______ = ______ _____ + c 2 x÷1 – x ÷1 – x2
6. The equation of the tangent at P (x, y) is dy Y – y = ___ (X – x) dx
(
It cuts the line y = x, so the point of intersection
)
dy dy y – x ___ y – x ___ dx dx = _______ , ________ dy dy ___ ___ 1 – 1 – dx dx
fi
dy dy y – x ___ = 1 – ___ dx dx
fi
dy (1 – x) ___ = (1 – y) dx
dy dx fi ______ = ______ (y – 1) (x – 1)
Integrating, we get
log |x – 1| = log |y – 1| + log c
fi
x–1 log _____ = log c y–1
x–1 fi _____ = c y–1
| |
( )
which is the required equation of the curve. 7. The given differential equation is
fi
_______
÷ ( )
y y2 __ x = 1 + __ x
_____ y dv v + x ___ – v = ÷1 + v2 , Let v = __ x dx _____ dv fi x ___ = ÷ 1 + v2 dx dv dx ______ fi _______ = ___ x 2 + v ÷ 1
fi
|
_____
(
_____
|
fi ( v + ÷ 1 + v2 ) = c x
fi y + ÷ y 2 + x2 = c x2
(
______
)
which is the required solution. 8. The given differential equation is y y x e y/x – y sin __ x dx + x sin __ x dx = 0
(
( ) ) ( ) y y __ x sin ( __ x ) – e dy fi ___ = ___________ y dx __ sin ( x ) dv v sin – e fi v + ___ = _______ , ( Let dx sin v y __x
v
fi
dv v sin – ev ___ = _______ –v dx sin v
fi
dv v sin – ev – v sin v ___ = _______________ dx sin v
( ) )
y v = __x
fi e– vsin v dv = – dv Integrating, we get e– vsin v _______ (– sin v – cos v) = – x – c (1 + 1)
e– vsin v fi _______ (sin v + cos v) = x + c (1 + 1)
( ) ( ()
y __y e – x sin __x y y _________ fi sin __ x + cos __x = x + c 2 which is the required solution. 9. The given differential equation is
( ) )
(x + 2y) (dx – dy) = dx + dy
fi
(x + 2y – 1) dx = (x + 2y + 1) dy
dy fi ___ = dx
(x + 2y – 1) ___________ (x + 2y + 1)
...(i)
Let
x + 2y = v
fi
dy dv 1 + 2 ___ = ___ dx dx
______
dy x ___ = – y = ÷x 2 + y2 dx
)
log v + ÷1 + v2 = log c + log |x|
It is given that dy y – x ___ dx ________ =1 dy 1 – ___ dx
4.79
4.80 Integral Calculus, 3D Geometry & Vector Booster
(
(
)
v+1 1 dv __ ___ – 1 = _____ 2 dx v–1
dv fi ___ = dx
2v – 2 ______ +1 v+1
dv 3v – 1 fi ___ = ______ v+1 dx
v + 1 fi ______ dv = dx 3v – 1
fi
fi
(x + 2y) + 4 log |3x + 6y – 1| = 3x + c
fi
2 (y – x) + 4 log |3x + 6y – 1| = c
which is the required solution. 10. The given differential equation is
x2 = v dx dv 2x ___ = ___ dy dy
(
)
dy
1 – ___ IF = e Ú y = e– log y = __ y
( )
1 1 __ v ◊ __ y = – Ú 1 + y2 dy + c
( )
v 1 fi __y = y – __ y + c
x2 1 fi __ y = – y – y + c
x2 + 1 fi ______ + y = c y
(
)
which is the required solution. 12. The given differential equation is
– 1
(1 + y ) dx = (tan y – x) dy.
tan– 1y dx x fi ___ + _____ 2 = ______ dy 1 + y 1 + y2 dy
Ú _____ 2 IF = e 1 + y = etan– 1y
x dy – y dx = x y3 (1 + log x) dx
x dy – y dx fi _________ = x y (1 + log x) dx y2
fi
which is a linear differential equation.
( )
x __x 2 2 – __ y d y = x dx + x log x dx
Integrating, we get
Hence, the solution is tan– 1y etan– 1y y = Ú ___________ dy + c 1 + y2
tan– 1
x ◊ e
)
Hence, the solution is
v + 4 log |3v – 1| = 3x + c
(
x2 1 __ y = – y + __ y Let
\
fi
\
dx 2x ___ – dy
dv v 1 fi ___ – __ = – y + __ y dy y which is a linear differential equation.
2
fi
( ) 3v – 1 + 4 fi ( _________ dv = 3 dx 3v – 1 ) 3 fi ( 1 + ______ dv = 3 dx 3v – 1 )
)
dy dv fi ___ = 1/2 ___ – 1 dx dx
= Ú t e dt + c, t = tan y
= (t – 1) e t + c
= (tan– 1 y – 1) etan– 1y + c
t
– 1
fi
( ) __ x3 + __ x3 log x – __ x9 + C 2x x 1 x – __ ( __ = ___ + __ log x + C 2 y) 3 3 1 x2 – __ __ = 2 y 2
3
3
3
3
3
which is the required solution. 13. The given differential equation is
sec2x ◊ tan y dx + sec2y ◊ tan x dy = 0
which is the required solution. 11. The given differential equation is
sec2 y sec2 x fi _____ dx + _____ dy = 0 tan x tan y
dy 2x y ___ = _________ 2 dx x – 2y – 1
Integrating, we get log tan x + log tan y = log c
2 2 dx x – y – 1 fi ___ = __________ 2x y dy
fi
dx 2x y ___ = x2 – y2 – 1 dy
fi
dx 2x y ___ – x2 = – (y2 + 1) dy
fi (tan x tan y) = c which is the required solution. 14. The given differential equation is dy x+y–1 ________ ___ = _________ dx ÷x + y + 1 Let
x + y + 1 = v2
...(i)
Differential Equation
( )
c dy dv log |v + ev| = log __ y 1 + ___ = 2v ___ fi dx dx c fi (v + ev) = __ y dy dv ___ ___ fi = 2v – 1 dx dx x c fi __ y + ex/y = __ y fi
(
2
dv v –2 fi 2v ___ –1 = _____ v dx
dv v2 – 2 + v fi ___ = _________ dx 2v2
fi
2v2 fi _________ 2 dv = dx v + v – 2
– 1
( (
)
which is a linear differential equation. dy
)
5 dx 1 (2v + 1) 1 fi 1 – __ ________ + __ ◊ ___________ dv = ___ 2 v2 + v – 2 2 (v + 2) (v – 1) 2
| |
v–1 15 x 1 fi v – __ log|v2 + v – 2| + ___ log _____ = __ + C 2 2 v+2 2 ________
_________
1 fi ÷x + y + 1 – __ log |(x + y + 1) + ÷ (x + y + 1) – 2| 2
|
_________
÷(x + y + 1) – 1 15 x _________ = __ fi + ___ log ______________ + C 2 2 ÷(x + y + 1) + 2 which is the required solution. 15. The given differential equation is
(
)
x (1 + ex/y) dx + 1 – __ y ex/y dy = 0
(
)
x __ y – 1 ex/y dx fi ___ = __________ dy (1 + ex/y) fi fi
dv (v – 1) e x v + y ___ = ________ , v = __ y dy 1 + ev v
dv (v – 1) e y ___ = ________ – v dy 1 + ev v
v
ve – e – v – ve = _______________ 1 + ev ev + v = – _____v 1+e dy 1 + ev fi _____v dy = – ___ y v+e
Integrating, we get
\
log |v + ev| = log c – log y
2 Ú _______
– 1
IF = e (1 + y ) = e tan y
Hence, the solution is
( etan y ) = Ú _______ dy +c 1 + y2 2
– 1
tan– 1y
x ◊ e
( etan y ) fi x ◊ e = _______ + C 2 which is the required solution. 18. The given differential equation is – 1
tan– 1y
2
y dx – x (1 + xy) dy = 0
fi
y dx – x dy = x2y dy
y dx – x dy fi _________ = y dy x2 y fi – d __ x = y dy fi
v
v
– 1 dx (1 + y2) ___ + (x – e tan y) = 0 dy
etan y dx x fi ___ + _______ 2 = _______ dy (1 + y ) (1 + y2)
dx 1 (2v + 1) – 5 fi 1 – __ ___________ dv = ___ 2 v2 + v – 2 2
)
which is the required solution. 16. Do yourself. 17. The given differential equation is – 1 dy (1 + y2) + (x – etan y) ___ = 0 dx
dv v2 – 2 v2 – 2 + v fi 2v ___ = ______ + 1 = _________ v v dx
|
4.81
( ) x d ( __ y ) + y dy = 0
Integrating, we get y y2 __ x + __ = c 2
( )
which is the required solution. 19. It is given that dy x+y y ___ = – _____ = – 1 – __ x x dx
(
fi fi
)
dv v + x ___ = – 1 – v, dx dv x ___ = – 1 – 2v dx
dv dx fi ______ = – ___ x 2v + 1
(Let v = y/x)
4.82 Integral Calculus, 3D Geometry & Vector Booster
( ) ( )
dy 2 fi ___ = dx
Integrating, we get
log |2v + 1| + log (x2) = log c
fi
(2v + 1) x2 = c
(
x2 __ 2 y
dy x fi ___ = ± __y dx
)
2y 2 fi ___ x + 1 x = c fi x2 + 2xy = c which is passing thorugh the point (2, 1), so, c = 8 Hence, the equation of the curve is
fi
y dy
fi
x dx ± y dy = 0
x dx = 0
Integrating, we get
20. The equation of tangent at (x, y) is dy Y – y = ___ (X – x) dx
x2 ± y2 = a2 which is the required solution. 22. The given differential equation is dy 2 dy ___ – (ex + e– x) ___ + 1 = 0 dx dx
Putting
fi
p2 – (ex + e– x) p + 1 = 0
fi
(ex + e– x) ± ÷(e x + e– x)2 – 4 p = ________________________ 2
x2 + 2xy = 8
( )
X = 0, we get
_____________
dy Y = y – x ___ dx It is given that,
_________
dy 2 y – x ___ = xy dx
(ex + e– x) ± ÷(e x – e– x)2 = ____________________ 2
dy ___ fi y – x ___ = ÷ x y dx
(ex + e– x) ± (ex – e– x) = ___________________ 2
= ex, e– x
( (
fi
) )
dy ___ x ___ = y – ÷x y dx
dy fi ___ = dx
y __ x –
dy dy fi ___ = ex, ___ = e– x dx dx Integrating, we get
__
÷
y __ x
(
)
__ dv __y fi v + x ___ = v – ÷ v , Let v = x dx __ dv fi x ___ = – ÷v dx dv__ dx fi ___ = – ___ x ÷v
Integrating, we get __
fi
2 ÷v = c – log |x|
fi
y 2 __ x = c – log |x|
__
÷
which is the required equation of the curve. 21. It is given that
÷ ( ) ( ( ) ) ( ( ) ) ________
______ dy 2 y 1 + ___ = ÷x 2 + y2 dx
fi
dy 2 y2 1 + ___ = (x2 + y2) dx
dy 2 (x2 + y2) fi 1 + ___ = _______ dx y2
x2 = __ 2 + 1 y
y = ex + c1, y + ex = c2
fi y – ex – c1 = 0, y + ex – c2 = 0 Hence, the solution is
(y – ex – c1) (y + ex – c2) = 0
23. The given differential equation is
( )
dy 2 dy ___ + 2x ___ = 3x2 dx dx
fi
p2 + 2px – 3x2 = 0
fi
(p + 3x) (p – x) = 0
fi
(p + 3x) = 0, (p – x) = 0
(
) (
)
dy dy fi ___ + 3x = 0, ___ – x = 0 dx dx fi dy = – 3x dx, dy = x dx Integrating, we get 3x2 x2 y + ___ – c1 = 0, y – __ – c2 = 0 2 2 Hence, the solution is
(
) (
)
3x2 x2 y + ___ – c1 y – __ – c2 = 0 2 2
Differential Equation
24. The given differential equation is
{ ( ) } 2
dy dy xy ___ – 1 = (x2 – y2) ___ dx dx
fi
xy (p2 – 1) = (x2 – y2) p
fi
xy p2 – xy = x2 p – y2 p
fi
xy p2 – x2 p – xy + y2 p = 0
fi
xp (yp – p) + y (yp – x) = 0
fi
(xp + y) (yp – x) = 0
fi
(xp + y) = 0, (yp – x) = 0
dy dy x ___ + y = 0, y ___ = x dx dx dy ___ dx fi ___ y + x = 0, y dy – x dx = 0 fi
which is passing through (– 2, 1), so, c = 1 Hence, the equation of the curve is (y + 3) = (x + 4)2. 27. It is given that dy ___ = x + x y dx dy ___ – x y = x dx which is a linear differential equation. fi
\
x2 – __
IF = e – Ú x dx = e 2
Hence, the solution is x2 – __
x2 – __
y ◊ e 2 = Ú x ◊ e 2 dx + c
dy ___ dx fi ___ y + x = 0, x dx – y dy = 0
fi
y ◊ e 2 = – e 2 + c
Integrating, we get
which is passing through (0, 1), so c = 2 Hence, the equation of the curve is
log |y| + log |x| = log c, x2 – y2 = a2
fi xy = c, x2 – y2 = a2 Hence, the solution is
(xy – c) (x2 – y2 – a2) = 0.
25. It is given that dy y ___ = x dx fi
x2 – y2 = c which is passing through (0, – 2), so c = – 4 Hence, the equation of the curve is
x2 – y2 = – 4
dy 26. Clearly, the slope of the tangent is ___ dx Also, the slope of the line segment joining the points (x, y) and (– 4, – 3) is y+3 _____ . x+4 It is given that dy y+3 ___ = 2 _____ x+4 dx
( )
x2 – __
x2 – __
x2 – __
y ◊ e 2 = – e 2 + 2
fi
y = 2e2 – 1
x2 __
28. The given differential equation is
x dx – y dy = 0
Integrating, we get
x2 – __
( )
d2y y3 ___ 2 – y = 1 dx
d2y 1 fi ___2 = y + __ 3 dx y
( )
d dy 1 fi ___ ___ = y + __ 3 dx dx y Integrating, we get
( )
y2 __ – 2
dy ___ = dx
As
y = 1, dy/dx = 0, so c = 0
( ) (
dx fi ___ = dy
y2 dy fi _____ = y4 – 1
Integrating, we get log |y + 3| = 2 log |x + 4| + log c
fi
)
1 1 __ y2 – __ 2 + c 2 y
)
dx ___ 2
2y2 dy fi _____ = d x y4 – 1
(y + 3) = (x + 4)2
1 ___ 2 = 2y
1 1 __ y2 – __ 2 2 y
dy 2 dx fi ______ = ______ (y + 3) (x + 4)
fi
(
2y2 dy _____________ 2 = d x (y – 1) (y2 + 1)
4.83
4.84 Integral Calculus, 3D Geometry & Vector Booster
(
)
1 1 fi _____ 2 + _____ dy = d x y – 1 y2 + 1
p x = 0, y = 1, c1 = __ 4 Hence, the curve is y–1 p 1 __ log _____ + tan– 1y = x + __ 2 y+1 4
Integrating, we get 1 __ log |2v + 1| + log|v – 1| = log c – 3 log x 2 y y c 1 fi __ log 2 __ x + 1 + log __ x – 1 = log __ 3 2 x ________ y__ y__ c fi x – 1 2 x + 1 = __ 3 x which is the required equation of the curve. 30. The equation of the normal at P (x, y) is dx Y – y = – ___ (X – x) dy
29. The equation of the normal at P (x, y) is
Putting
Y = 0, we get
dx – y = – ___ (X – x) dy
Integrating, we get
| ( ) | | ( ) |
| |
y–1 1 __ log _____ + tan– 1y = x + c1 2 y+1
( ( ) ) ( ÷ ( ) ) ( )
When
| |
Putting
dx Y– y = – ___ xy (X – x) Y = 0, we get
dx – y = – ___ xy (X – x)
fi
dy X = y ___ + x dx
fi
dy X = y ___ + x dx
Thus,
dy Q = y ___ + x, 0 dx
Thus,
dy B = y ___ + x, 0 dx
(
)
Also, it is given that,
)
PQ = k
( ( ( (
) )
dy 2 fi y ___ + y2 = k2 dx
Here, A = (x, 0) It is given that
dy x+y x + y ___ – x = _____ 2 dx
fi
dy x + y y ___ = _____ 2 dx
dy 2 fi y ___ = k2 – y2 dx
) )
dy 2 k2 – y2 fi ___ = ______ 2 dx y
÷
______
dy k2 – y2 fi ___ = ± ______ dx y2
dy x + y fi ___ = _____ 2y dx
(
)
fi
y dv 1 + v v + x ___ = _____ , Let v = __ x fi v dx
fi
dv 1 + v – 2v2 x ___ = __________ 2v dx
______
÷k 2 – y2 = c 31. It is given that dy ___ = dx dy fi ___ y = Integrating, we
v dv dx fi _____________ = – ___ x (2v + 1) (v – 1) (v – 1 + 1) dv dx fi _____________ = – ___ x (2v + 1) (v – 1)
(
)
dv 2dv dx 1 dv fi _______ + __ _____ – ______ = – ___ x (2v + 1) 3 v – 1 2v + 1 dv dv 3 dx _______ + _____ = – ____ x (2v + 1) v – 1
y dy ______ _______ = ± dx ÷k 2 – y2
Integrating, we get
v dv dx fi __________ = ___ x 1 + v – 2v2
fi
(
x
2y ___ x 2dx ____ x get
log |y| = log |x2| + log c
fi
y = c x2
which is passing through (1, 1), so, c = 1 Hence, the equation of the curve is y = x2
( )
Differential Equation
32. The equation of the normal at P (x, y) is Here,
dx Y – y = – ___ (X – x) dy dy dx A = x + y ___ , 0 , B = 0, y + x ___ dx dy
(
) (
)
P = (x, y)
It is given that, PB ___ = PA
(
) (
)
dy dx = x + y ___ ◊ y + x ___ dx dy
fi
dy dy 1 + ___ = x + y ___ dx dx
fi
dy (y – 1) ___ + (x – 1) = 0 dx
Integrating, we get 1 __ 2
(x – 1)2 + (y – 1)2 = c
÷ ( )
___________
dx 2 x2 + x ___ dy ____________ fi __________ = dv 2 y ___ + y2 dx
÷( )
which is passing through (5, 4), so c = 25 Hence, the equation of the curve is
1 __ 2
(x – 1)2 + (y – 1)2 = 25.
( ) ( ( ) )
fi
dv 2 dx 2 1 x2 + x ___ = __ y ___ + y2 4 dy dx
fi
x2 x2 + __ 2 = p
1 __ (p2y2 + y2) 4
y2 x2 fi __2 (p2 + 1) = __ (p2 + 1) 4 p 2 x2 y fi __2 = __ 4 p
(Tougher Problems for Jee-Advanced) 1. We have
x
y = C1 cos (2x + C2) – C83C 63 + C6 sin (x – C7)
fi y = C1 cos (2x + C2) – C93x + C6 sin (x – C7) Since the above equation has 5 arbitrary constants, so the order of the differential equation is 5. 2. The Equation of the family of parabolas is (y – k)2 = 4a (x – h),
fi
y2p2 = 4x2
fi
y p = ± 2x
where h and k are arbitrary constants
fi
y dv = ± 2x dx
Differentiating w.r.t. x, we get dy 2 (y – k) ___ = 4a dx
Integrating, we get y2 fi __ = c1 ± x2 2 fi
fi
y2 = 2 c1 ± 2x2
which is passing through (0, 4) so c1 = 8 Hence, the eqution of the curve is 2
y = 16 ± 2x
fi
y2 = 16 + 2x2
___
1 1 ____ + ___ = 1 OM ON
dy ___ = 0 dx
( )
d 2y dy 2 2a ___2 + ___ = 0 dx dx which is the required differential equation. 3. Given condition is
dy ___ = dx
2y ___ x
dy dx ___ fi ___ y = 2 ◊ x Integrating, we get
OM + ON = OM ◊ ON dy dx fi x + y ___ + y + x ___ dx dy
) (
d 2y (y – k) ___2 + dx
From Eqs (ii) and (iii), we get
Clearly, it is passing through (÷10 , – 6). 33. The equation of the normal at P (x, y) is dy (Y – y) = ___ (X – x) dx dy dx Thus, OM = x + y ___ and ON = y + x ___ xy dx Also, it is given that,
(
dy (y – k) ___ = 2a dx
)
...(i)
...(ii)
Again differentiaing w.r.t. x, we get
2
fi
4.85
log |y| = 2 log |x| + log |c|
fi
y = c x2
...(iii)
4.86 Integral Calculus, 3D Geometry & Vector Booster which is passing through the curve (1, 1), so c = 1 Hence, the equation of the curve is y = x2 4. The equation of the tangent is dy Y – y = ___ (X – x) ...(i) dx dx Equation (i) meets the x-axis at x – y ___ , 0 dy dy and the y-axis at 0, y – x ___ respectively. dx
(
)
(
)
dx x – y ___ = 2x dy dy y fi ___ = – __x dx dy ___ d x fi ___ y + x = 0
fi
x y = c
2
)
(
x2 + y2 Let _______ = v x
fi
fi
fi
dv v + x ___ = ev + v dx
fi
dv x ___ = ev dx dx e– v dv = ___ x
– e
fi
– e– v = log |c x|
fi fi
)
x2 + y2 = v x
fi
dz ___ dx dx dx y ___ = ___ dy dz
1 __ x = z
1 Let __ x = v
fi
dv – ___ + v = z dz
1 dx – __2 ___ = x dz
which is a linear differential equation. ...(i)
\
IF = e – Ú dz = e– z
Hence, the solution is
v ◊ e– z = Ú – z e– z dz + c
fi
v ◊ e– z = – (z – 1) e– z + c
fi
1 __ x = (ln y + 1) + c y
Hence, the required integral curve is 1 __ x = (ln y + 1) + e y fi
(
dy 1 ___ fi __ y dx =
( )
– e– v
c x =
1 which is passing through e, __ e , so, c = e
= log |x| + log c
c x = e
ln y = z
dy v = (z + 1) + c ez dv fi 2x + 2y ___ = v + x ___ dx dx 1 ln y fi __ x = (ln y + 1) + c e
Integrating, we get
Let
dv fi ___ – v = – z dz
x______ +y dy x2 + y2 2y ___ = e x + ______ – 2x x dx
– v
...(i)
fi
which is passing through (2, 4), so c = 8 Hence, the equation of the curve is xy = 8. 5. The given differential equation is
fi
dx 2 y ___ xy + x = x ln y
which is a linear differential equation.
log y + log x = log c
fi
1 dx fi __2 ___ + x dz
fi
2
dy x (1 – x ln y) ___ + y = 0 dx
dx fi ___ + x = x2z dz
Given condition is
(
)
2 x______ + y2 e– e– x
which is the required solution. 6. The given differential equation is
((ln y + 1) + e y) x = 1
7. The given differential equation can be written as
(
)
2 (x dx – y dv y dx – xdy ___________ = – ________ 2 2 (x – y ) (x2 + y2)
( ( ) )
d (x2 – y2) y fi ________ = – d tan– 1 __ x 2 (x2 – y )
dv ___ dz
Differential Equation
dt tan x ___ = (2t 2 + t + 2) dx dt dx fi ___________ = ___ x 2 (2t + t + 2)
Integrating, we get
fi
( ( ) )
y log |(x2 – y2)| + tan– 1 __ x = c
which is the required solution. 8. The given differential equation is d y + ___ (xy) = x (sin x + log x) dx
dt dx 1 fi __ ______________ = ___ x 2 2 15 1 t + __ + ___ 4 16 Integrating, we get 1 t + __ 2 4 ___ ____ tan– 1 _____ ___ = log |sin x| + C ÷15 15 ___ 16
{ ( )
fi
dy y + y + x ___ = x (sin x + log x) dx
fi
dy x ___ + 2y = x (sin x + log x) dx
(
which is the required solution. 11. The given differential equation is y y y cos __ x (xdy – ydx) + x sin __ x (xdy + ydx) = 0
dx ___
y ◊ x2 = Ú x2 (sin x + log x) dx + c
fi
y ◊ x2 = – x2 cos x + 2x sin x + 2 cos x
( )
( ) y y dy __y y dy __y fi __ x cos ( __ x ) ( ___ – x ) + sin ( __ x ) ( ___ + x ) = 0 dx dx y ( Let __x = u )
x3 x3 + __ log x – __ + c 3 9
(
2
(xdy + ydx) sin (xy) + (x ydx + xy dx) cos (xy) = 0 fi d (xy) sin (xy) + xy (xdy + ydx) cos (xy) = 0 fi d (xy) sin (xy) + xy d (xy) cos (xy) = 0 fi sin (xy) d (xy) + xy cos (xy) d (xy) = 0
)
Integrating, we get
On Integrating, we get
– cos (xy) + (xy) sin (xy) + cos (xy) = c
fi
(xy) sin (xy) = c
which is the required solution. 10. The given differential equation is 2
dy 2y cos x + y sin 2x + 2 cos x ◊ sin x ___ = ____________________________ dx sin2x
(
log |v sin v| = – 2log |x| + log c c fi v sin v = __ 2 x y__ y__ c fi x sin x = __ 2 x
( ) ( ) y c y sin ( __ x ) = __ x
fi
2
)
dy y2 y fi ___ = 2cosx _____ 2 + ____ + 1 dx sin x sin x
) ( dv fi (v cos v + sin v) ( x ___ ) + 2v sin v = 0 dx 2dx fi (____________ v cos v v sin v+ sin v ) dv + ____ x = 0
dv dv fi v cos v v + x ___ – v + sin v v + x ___ + v = 0 dx dx
9. The given differential equation is 2
)
4y + sin x 2 ___ fi ____ tan– 1 _________ ___ = log |sin x| + C ÷15 ÷15 sin x
\ IF = e2 Ú x = e2 log x = e log x2 = x2 Hence, the solution is
}
( ÷ )
dy 2y fi ___ + ___ x = (sin x + log x) dx which is a linear differential equation.
y Let ____ = t sin x
p p when x = 1, y = __ , then c = __ 2 2 Hence, the required solution is y p y sin __ x = ___ 2x 12. The given differential equation is
( )
÷ ÷
______________________________
dy x4y2 – x6 + 2x4y – x6y2 – 2x6y + x4 ___ ______________________________ = dy dt dx y2 – x2y2 + x3y2 – x5y2 ___ = t cos x + sin x ___ dx dx fi
4.87
dt t cos x + sin x ___ = 2 cos x (t2 + t + 1) dx
______________________________
x4 (y2 – x2 + 2y – x2y2 – 2x2y + 1) = _____________________________ y2 (1 – x2 + x3 – x5)
4.88 Integral Calculus, 3D Geometry & Vector Booster
÷
______________________________
x4 {(y2 + 2y + 1) – x2 (y2 + 2y + 1)} = ______________________________ y2 (1 – x2 + x3 – x5)
÷ ÷
___________________
(x + y)2 + 3 (x + y) + 1 = ____________________ (x + y)2 – 3 (x + y) + 2
dv v2 + 3v + 1 fi ___ – 1 = __________ , dx v2 – 3v + 1
+ 1)2 – x2 (y + 1)2} x2 {(y ___________________ = __ y (1 + x2) (1 + x3)
2 2 x2 {(y + 1) (1 – x )} = __ y _______________ (1 – x2) (1 + x3)
x2 (y – 1) = ________ _____ y ÷1 + x3
v2 – 3v + 1 fi __________ dv = 2 dx v2 + 1
_______________
)
x2 dx 1 fi 1 – _____ dy = _______ _____ y+1 ÷ 1 + x3 Integrating, we get _____
2 y – log |y + 1| = __ ÷ 1 + x3 + c 3
which is the required solution. 13. Given
x
log t f (x) = Ú ________ 2 dt, x ≥ 1 1 1 + t + t
Now,
fi
2v2 + 2 = __________ v2 – 3v + 1
2v 3 _____ fi 1 – __ dv = 2 dx 2 v2 + 1 Integrating, we get 3 v – __ log |v2 + 1| = 2x + C 2 3 fi y – __ log |(x + y)2 + 1| = x + C 2 which is the required solution. 15. Given f (0) = – 1 Put x = 0, y = 1, then f (1) = 1
Now,
f (x + h) + f (x) ◊ f (h) = f (xh + 1)
log t 1 f __ x = Ú ________ dt 1 1 + t + t2
fi
(f (x + h) – f (x)) + (f (x) ◊ f (h) – f (x))
1 1 Let t = __ y fi dt = – __2 dy y 1 y log __ y 1 1 __ __________ f x = Ú – __2 dy 1 1 1 1 + __ + __ y y y2
f (x + h) – f (x) ___________ f (x) f (h – 1) fi _____________ + h h f (xh + 1) – f (1) = ______________ ◊ x xh
1/x
( )
(
)
( )( ) ( )
( )
( Ú (
) )
y
dv v2 + 3v + 1 fi ___ = __________ + 1 dx v2 – 3v + 1
( ( ) )
y dv x2 dx fi _____ = _______ _____ y+1 ÷ 1 + x3
(
(Let v = x + y)
– log (y) = Ú ________ 2 (– 1) dy 1 y + y + 1 y
fi
= f (xh + 1) – f (1)
(
)
(
fi
f ¢ (x) + f (x) ◊ f ¢ (0) = x f ¢ (1)
fi
dy ___ + p y = q x dx
log (t) = Ú ________ 2 dt 1 t + t + 1
which is a linear differential equation.
= f (x)
Hence, the solution is
(
)
Hence, the result. 14. The given differential equation is dy ___ = dx
x2 + y2 + 3x + 3y + 2xy + 1 ________________________ x2 + y2 – 3x – 3y + 2xy + 2
\
)
)
f (xh + 1) – f (1) = lim ______________ x h Æ 0 xh
log (y) = _________ 2 dy 1 y + y + 1 x
(
(f (x + h) – f (x) f (h) – 1 lim _____________ + f (x) ________ h Æ 0 h h)
IF = e Úp dx = ep x
y ◊ epx = q Ú x epxdx + c
fi
y ◊ epx = q Ú x epxdx + c
fi
y ◊ ex = Ú x ex dx + c
4.89
Differential Equation
fi
y ◊ ex = (x + 1) ex + c
fi
y = (x – 1) + c e– x
fi
( ) ( )
when x = 1, y = 1, then c = 0 Thus, the equation of the given curve is
Integrating, we get
y = x – 1.
16. Put
x = 1, y = 1, then f (1) = 0.
Now,
f (x) (1 + h)) = x f (1 + h) + (1 + h) f (x)
fi
f (x + xh) – f (x) = xf (1 + h) + hf (x)
fi
f (x + xh) – f (x)
= x (f (1 + h) – f (1)) + hf (x)
(
)
f (xh + x) – f (x) fi ______________ ◊ x xh
( ) ( )
y y x x sin __ y d __ y + cos __ x d __ x dy + dx + ___ 2 = 0 y
( )
( )
y x 1 – cos __ y + sin __ x + x – __ y = c
which is the required differential equation. 18. The given differential equation is dy ___ = dx
(1 + y2) _________ x y (1 + x2)
y dy dx fi _____ 2 = ________ 1+y x (1 + x2)
(
)
y dy x 1 fi _______ = __ x – _____ 2 dx 2 (1 + y ) 1 + x x ( f (1 + h) – f (1)) hf (x) = ________________ + _____ h h Integrating, we get f (xh + x) – f (x) log |1 + y2| = 2 log |x| – log |1 + x2| + 2 log c fi lim ______________ x h Æ 0 xh fi (1 + y2) (1 + y2) = c2x2 f (1 + h) – f (1) = lim _____________ x + f (x) which is the required solution. h Æ 0 h 19. The given differential equation is
(
)
(
fi
x f ¢ (x) = x f ¢ (1) + f (x)
fi
x f ¢ (x) = x + f (x)
)
dy x ___ = x + y dx dy y fi ___ = 1 + __ x dx dy 1 ___ fi + – __ x y = 1 dx fi
Let
x y = v
fi
y dy __ 1 dv __x + ___ = x ___ dx dx
dv ___________ 1 ___ 1 fi __ 4 x dx = sin v + cos2 v
( )
which is a linear differential equation. \
y dy ________________ 1 __x = ___ = 4 dx sin (xy) + cos2 (xy)
dx ___ 1 IF = e – Ú x = e– log x = __ x
dx (sin4 v + cos2 v) dv = ___ x dx 1 1 fi __ (2sin2v)2 dv + __ (2cos2v) dv = ___ x 4 2 fi
Hence, the solution is y dx __x = Ú ___ x + c y __ fi x = log |x| + c
dx 1 1 fi __ (1 – cos2v)2 dv + __ (1 + cos2v) dv = ___ x 4 2 1 1 fi __ (1 – cos2v) dv + __ (1 + cos4v) dv 4 8
when x = 1, y = 0, then c = 0 Thus, the equation of the given curve is y = x log |x|
17. The given differential equation can be written as
(
)
( )
dx xdy x ___ y – ___ 2 sin __ y y
(
dy + ___ x –
)
( )
ydx y dy ___ 2 cos __ x + dx + ___ 2 = 0 x y
dx 1 + __ (1 + cos2v) dv = ___ x 2
Integrating, we get
(
)
(
)
sin4v 1 1 fi __ (v – sin2v) + __ v + _____ 4 8 4 fi
sin 2v 1 + __ v + _____ = log |x| + c 2 2
(
)
sin (4xy) 1 1 __ (xy – sin (2xy)) + __ xy + _______ 4 8 4
4.90 Integral Calculus, 3D Geometry & Vector Booster
(
)
which is a linear differential equation.
sin 2(xy) 1 + __ xy + _______ = log |x| + c 2 2
\
which is the required solution. 20. The given differential equation can be written as
(y2 + 1) dy + x2 dy + 2xy dx = 0
fi
(y2 + 1) dy + d (x2y) = 0
Hence, the solution is
y3 __ + y + x2y = c 3 which is the required solution. 21. The given differential equation is
Putting
(
fi
dy 1 y ___ + x = ___ 2 (x2 + y2)2 dx 2x
dy Y = y – x ___ dx It is given that,
( (
dy __ fi y – x ___ = ÷ xy dx dy __ fi x ___ + ÷ xy = y dx dy fi ___ + dx
2
d (x + y ) ___ dx fi _________ = 2 2 2 2 (x + y ) x
fi
Integrating, we get 1 1 – _______ 2 = c – __ x 2 (x + y )
fi
) (
y __ x
__
= – ÷v
Integrating, we get
(y + sin x cos2 (xy)) dx + x dy = 0
(
÷
y __ x =
dv__ ___ dx fi ___ + x = 0 v ÷
which is the required solution. 22. The given differential equation is
dy fi y + x ___ + sin x cos2 (xy) = 0 dx
__
__ dv __v v + x ___ + ÷v = v, v = x dx __ dv x ___ = ÷ v = 0 dx
1 _______ 1 fi __ = c x – (x2 + y2)
) )
dy 2 y – x ___ = xy dx
ydy + xdx ___ dx fi _________ = 2 2 2 2 (x y ) 2x 2
X = 0, we get
)
2 2 2 dy 1 x +y y ___ + x = __ ______ x 2 dx
y ◊ e p (x) x = Ú ( ep (x)x ◊ q (x) ) dx + c
24. The equation of tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx
Integrating, we get
IF = e Ú p (x) dx = e p (x) x
)
__
2 ÷v + log |x| = c
fi
y 2 __ x + log |x| = c
__
÷
which is the required solution. 25. We have
dy dv Let xy = v fi y + x ___ = ___ x x dx dx x Ú (1 – t) f (t) dt = Ú tf (t) dt, f (1) = 1 dv fi ___ + sin x cos2v = 0 0 0 dx Applying Newton and Leibnitz formula, we get dv x fi ___ = – sin x cos2v dx x (x – 1) f (x) + Ú (1 – t) f (t) dt = x f (x) fi sec2v dv = – sin x dx 0 Integrating, we get fi
x
fi
tan (xy) – cos x = c
which is the required solution. 23. The given differential equation is dy ___ + p (x) y = q (x) dx
(x – x – x) f (x) + Ú (1 – t) f (t) dt = 0 2
0
x
fi
(– x2) f (x) + Ú (1 – t) f (t) dt = 0 0
2
fi
(– x ) f ¢ (x) – 2x f (x) + (1 – x) f (x) = 0
fi
(x2) f ¢ (x) + (1 – 3x) f (x) = 0
Differential Equation
fi fi fi
dy (x2) ___ + (1 – 3x) y = 0 dx
( )
dy 1______ – 3y ___ dx = 0 y + x2 dy 1 3 ___ y + __ 2 – __ x dx = 0 x
(
)
Integrating, we get
j (y/x) dy y ______ fi ___ = __ + dx x j¢ (y/x) Thus, j (v) dv v + x ___ = v + _____ dx j¢ (v) j (v) dv fi x ___ = _____ dx j¢ (v)
1 log |y| – 3 log |x| – __ x = c
j¢ (v) dx fi _____ dv = ___ x j (v)
fi
y log __ 3 – x
Integrating, we get
when
x = 1, y = 1, then c = – 1
| |
1 __ x = c
Hence, the solution is
| | | |
y 1 log __ 3 – __ x + 1 = 0 x
fi
y 1–x 1 log __ 3 = __ x – 1 = _____ x x
y 1–x fi __3 = _____ x x
= e ( y=
fi
f (x) = x3e (
)
1_____ –x x
26. The given differential equation is
( )
dy 2 dy y ___ + (x – y) ___ – x = 0 dx dx
fi
dy yp2 + (x – y) p – x = 0, where ___ = p dx yp2 – yp + xp – x = 0
fi
yp (p – 1) + x (p – 1) = 0
fi
(yp – 1) (p – 1) = 0
fi
(yp + x) = 0, (p – 1) = 0
fi
x d x + y d y = 0, dy = dx
fi
Integrating, we get
x2 + y2 = a2, y = x + b
which is passing through (3, 4), so a2 = 25 and b = 1 Hence, the equations of the curves are
x2 + y2 = 25, y = x + 1
27. The given differential equation is
y y¢ = __ x +
fi
log |j (v)| = log |c x|
fi
|j (v) = c x
fi
y j __ x = c x
( )
which is the required solution. 28. Clearly, f (0) = 1
(
j (y/x) ______ j¢(y/x)
)
x+h f ______ – f (x) ◊ f (h) = 0 1 + xh
[ (
( )
1_____ –x x3e x
fi
log |j (v)| = log |x| + log c
)
Let y = vx
We have,
1–x _____ x
4.91
) )
] (
h – x2h f 1 + ______ – f (x) ◊ 1 + xh h – x2h fi ______________________ ______ 2 1 + xh h______ –xh ◊ h 1 + xh f (x) ◊ f (h) – f (x) = ______________ h
(
fi
f ¢ (x) ◊ (1 – x2) = f (x) ◊ f ¢ (0)
fi
f ¢ (x) ◊ (1 – x2) = f (x)
dy fi ___ (1 – x2) = y dx dy _______ dx fi ___ y = (1 – x2) Integrating, we get fi
| |
1–x log |y| = log _____ + log c 1+x
( )
1–x y = c _____ 1+x
when f (0) = 1, then we get, c = 1 1–x Thus, y = _____ 1+x which is the required solution.
( )
29. Let
x = 1 = y, then f (1) = 0
)
4.92 Integral Calculus, 3D Geometry & Vector Booster Now,
f (x (1 + h))
Integrating, we get = x f (1 + h) + (1 + h) f (x)
f (x + xh) – f (x) ______________ x {f (1 + h) – 0} _____ hf (x) fi x ◊ ______________ = + xh h h f (x + xh) – f (x) fi lim x ◊ ______________ x Æ 0 xh
(
) (
)
log |x| + log |y| = log |c|
fi
log |xy| log c
fi
x y = c
which is the required equation of the curve. 31. It is given that
x {f (1 + h) – 0} _____ hf (x) = lim ______________ + x Æ 0 h h
fi x f ¢ (x) = x f ¢ (1) + f (x)
÷ ( ) ( ( ) ) ( ( ) ) ( ( ) ) ( ) ( ) ( ) ( ) ________
______ dy 2 y 1 + ___ = ÷ x 2 + y2 dx
dy 2 y2 1 + ___ = (x2 + y2) dx
fi x f ¢ (x) = x + f (x) ( f ¢ (1) = 1) dy fi x ___ = x + y dx dy y fi ___ = 1 + __ x dx y dv fi v + x ___ = 1 + v, v = __ x dx dv fi x ___ = 1 dx dx fi dv = ___ x Integrating, we get
fi
fi v = log |x| + c
x2 ± y2 = a2 32. It is given that dy 1 ___ = __ dx y
dy 2 x 2 fi 1 + ___ = __ y + 1 dx dy 2 fi ___ = dx
dy x fi ___ = ± __ y dx fi
when x = 1, y = 0, then c = 0 y Thus, __ x = log |x|
fi ydy = dx Integrating, we get
fi y = x log |x| which is the required solution. 30. The equation of tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx
( (
) )
dx Y = 0, then A = x – y ___ , 0 dy
dy X = 0, then B = 0, y – x ___ dx It is given that, dx dy x – y ___ y – x ___ dy dx ________ = x and ________ = y 2 2 dy dx fi x – y ___ = 2x and y – x ___ = 2y dy dx dy dx fi y ___ = – x and x ___ = – y dy dx and
dx fi ___ x +
dy ___ y = 0
xdx ± y dy = 0
Integrating, we get
y fi __ = log |x| + c
Put
x 2 __ y
y2 __ = x + c 2 which is passing through (2, 2), so, c = 0 Hence, the eqution of the curve is
y2 = 2x
33. The tangent at P (x, y) is dy Y – y = ___ (X – x) dx If p be the length of perpendicular from the origin, then
| | | |
dy x ___ – y dx ________ p = __________ dy 2 ___ 1 + dx It is given that
÷ ( )
dy x ___ – y dx _________ __________ = x dy 2 ___ 1 + dx
÷ ( )
Differential Equation
) ( ( ) )
(
fi
2 dy dy 2 x ___ – y = x2 1 + ___ dx dx
fi
dy y – 2xy ___ – x2 = 0 dx 2
Hence, the result. dy y2 – x2 Also, ___ = ______ 2 xy dx y 2 __ x – 1 dy fi ___ = _______ x dx 2 __ y
( )
fi fi
( )
2
dv v – 1 v + x ___ = _____ , 2v dx dv v – 1 x ___ = ______ – v 2v dx
)
fi
fi
xdx + ydy _________ ______ = dx ÷ x 2 + y2
fi
2xdx + 2ydy ______ __________ = 2dx ÷ x 2 + y2
fi
d ÷x 2 + y2 = 2 dx
fi
fi x2 + y2 = c x which is the required equation of the curve. 34. The given curve is
y dy a = __ ___ 2 dx
______ dy x + y ___ = ÷ x 2 + y2 dx
fi ...(i)
...(ii)
fi
fi
(
)
dy y dy y2 = 2y ___ x + __ ___ 2 dx dx
( )
dy dy 2 y2 = 2xy ___ + y2 ___ dx dx
Replacing dy/dx by – dx/dy, we get
( ) ( )
dx dx 2 2 ___ y2 = 2xy – ___ xy + y xy
fi
dy 2 dy y ___ + 2xy ___ – y2 = 0 dx dx 2
( )
______
)
( )
______
)
| |
dy 2 xy ___ = y2 – x2 dx dy y2 – x2 ___ = ______ 2xy dx dy (y/x)2 – 1 ___ = _________ 2 (y/x) dx
(
)
fi
y dv v2 – 1 v + x ___ = _____ Let __ = v h 2v dx
fi
dv v2 – 1 x ___ = ______ – v 2v dx
fi
dv v2 + 1 x ___ = – _____ 2v dx
Eliminating a between Eqs (i) and (ii) we get
(
÷ ( )
(v2 + 1) x = c y 2 fi __ x + 1 x = c
dy 2y ___ = 4a dx
fi
______
dy 0 – y + x ___ dx ________ ___________ = x 2 dy 1 + ___ dx
fi
fi
(yp + x) = ÷ x 2 + y2
which is the required orthogonal trajectory. 35. The equation of the tangent to the curve at (x, y) is dy Y – y = ___ (X – x) dx It is given that
log |v2 + 1| + log |x| = log c
y2 = 4a (x + a)
fi
(
2v dx fi _____ dv + ___ x = 0 v2 + 1 Integrating, we get
(yp + x)2 = x2 + y2
÷x 2 + y2 = 2x + c
v2 + 1 = – ______ 2v
( ( )
fi
Integrating, we get
2v dx fi _____ dv = – ___ x 2 v +1
dy y2p2 + 2x y p – y2 = 0, p = ___ dx
(Let y = v x)
2
fi
4.93
2v dx fi _____ dv = – ___ x 2 v +1 Integrating, we get log |v2 + 1| = log c – log |x| c fi (v2 + 1) = __ x fi x2 + y2 = c x which is the required solution.
4.94 Integral Calculus, 3D Geometry & Vector Booster
Questions Asked in Past IIT-JEE Examination
3. The given differential equation is
dy ___ = sin (10x + 6y) dx
Let
fi
(
(
)
)
1 dv fi __ ___ – 10 = sin v 6 dx dy fi ___ = 6 sin v + 10 dx dv ___________ = dx Ú (6 sin v + 10) Ú
dv 1 fi __ Ú __________ = x + c 2 (3 sin v + 5)
(
)
2
sec (v/2) dv 1 fi __ Ú _______________________ = x + c 2 (5 + 5 tan2 (v/2) + 6 tan (v/2)) dt = x + c, Ú ____________ (5 + 5t2 + 6t)
(Let tan (v/2) = t))
dt 1 fi __ Ú ___________ = x + c 5 6 2 t + __ t + 1 5
(
)
dt 1 fi __ Ú ______________ = x + c 5 3 2 4 2 t + __ + __ 5 5
( ( ) ( ) ) 5t + 3 1 5 __ fi × __ tan ( _____ = x + c 5 4 4 ) 5 tan (5x + 3y) + 3 1 fi __ tan ( _______________ ) = x + c 4 4
3 3 4 fi (5x + 3y) = tan– 1 __ tan 4x + tan– 1 __ – __ 4 5 5
(
(
)
( ) )
( (
( ) ) )
( )
du equations ___ = p (x) u = f (x) and dx dv ___ + p (x) v = g (x), where p (x), f (x) and g (x) dx are continuous functions. If u (x1) > v(x1) for some x1 and f (x) > g (x) for all x > x1. Prove that any point (x, y) where x > x1 does not satisfy the equation y = u (x) and y = v (x). [IIT-1997]. 7. The equation of the tangent at any point P (x, y) is dy Y – y = ___ (X – x) dx dy It meets the axes at A x – y ___ , 0 and dx
(
(
)
) ( (
) (
)
(
dy B 0, y – x ___ respectively. dx
It is given that
( )
3 1 which is passing through origin so c = __ tan– 1 __ . 4 4 Thus, the equation of the curve is 5 tan (5x + 3y) + 3 3 1 1 __ tan– 1 _______________ = x + __ tan– 1 __ 4 4 4 4
6. Let u (x) and v (x) satisfy the differential
))
dy dx 1 1 \ Mid-point AB is __ x – y ___ , __ y – x ___ . 2 dy 2 dx
– 1
)
4. A curve y = f (x) passes through the point P (1, 1). The normal to the curve at P is a (y – 1) + (x – 1). If the slope of the tangent at any point on the curve is propor tional to the ordinate of the point, determine the equation of the curve. Also, otain the area bounded by the y-axis, the curve and the normal to the curve at P. [IIT-1996] 5. A spherical rain drop evaporates at a rate proportional to its surface area at any instant t. The differential equation giving the rate of change of the radius of the rain drop is....... [IIT-1997]
– 1
(
5 tan (5x + 3y) + 3 3 fi _______________ = tan 4x + tan– 1 __ 4 4
dv 1 fi __ Ú _________________ = x + c 2 _______________ 3 ◊ 2tan (v/2) 1 + tan2 (v/2) + 5
fi
dy dv 10 + 6 ___ = ___ dx dx dy 1 ___ dv fi ___ = __ – 10 dx 6 dx
fi
( )
)
5 tan (5x + 3y) + 3 3 fi tan– 1 _______________ = 4x + tan– 1 __ 4 4
...(i) 10 x + 6 y = v
(
(
)
dy dx 1 1 __ x – y ___ = x and __ y – x ___ = y 2 2 dy dx fi
dx x – y ___ = 2x dy
fi
dx – y ___ = x dy
Differential Equation
dy dx ___ fi ___ y = – x
( ( ) ) ( )
y 2 1 – __ x _________ = y 2 __ x
dy
fi
dx Ú ___y = – Ú ___ x
fi
log |y| = log c – log |x|
fi
x y = c
4.95
fi
dv 1 – v2 v + x ___ = – _____ , 2v dx
which is passing through (1, 1), so c = 1 Hence, the equation of the curve is
fi
dv 1 – v2 – 1 + v2 – 2v2 x ___ = – _____ – v = ___________ 2v 2v dx
x y = 1 8. The given curve is y = (c1 + c2) cos (x + c3) – c4ex + c5
fi
dv 1 + v2 x ___ = – _____ 2v dx
2v dx fi _____ 2 dv = – ___ x 1+v
fi
y = (c1 + c2) cos (x + c3) – c4ec 5◊ ex
fi
y = A cos (x + c3) – B ◊ ex
fi
Since the given curve has 3 arbitrary constants, so, the order of the differential equation represented by the given curve is 3.
2v dx dv = – Ú ___ Ú _____ x 1 + v2
fi
log |1 + v2| = log c – log |x| c (1 + v2) = __ x
9. The equation of the normal at any point P (x, y) is
dx Y – y = – ___ (X – x) dy
| |
...(i)
fi
fi x2 + y2 = c x which is passes through (1, 1), so c = 2 Hence, the equation of the curve is
It is given that
dx y + x ___ dy _________ __________ = |y| dx 2 ___ 1 + dy
10. The given differential equation is
÷ ( ) ( ) ( ( ) ) ( ) ( )
dx dx fi y + x ___ = y2 1 + ___ dy dy 2
2
dx 2 dx fi ___ (x2 – y2) + 2xy ___ = 0 dy dy
dx dx fi ___ = 0, (x2 – y2) ___ + 2xy = 0 dy dy
( )
2xy dx = 0, ___ = _______ dy (x2 – y2)
x2 + y2 = 2x
( ) ( ) ( ) ( )
dy 2 dy ___ – x ___ + y = 0 dx dx fi fi
dy y = x ___ + dx
dy 2 ___ dx
dy y = px + p2, where p = ___ dx
which is a Clairauts differential equation Thus, the solution is
y = c x + c2
fi
y = 2 x + 4
11. The given curve is
__
y2 = 2c (x + ÷c ) dy 2y ___ = 2c dx dy c = y ___ dx
...(i)
dx Now, ___ = 0 fi x = c dy
fi
which is passing through (1, 1), so c =1 Hence, the equation of the curve is x = 1
fi
2xy dx Also, ___ = _______ 2 2 dy (x – y )
From Eq. (i) and (ii), we get
2
2
(x – y ) = _______ 2xy
(Let y = vx)
...(ii)
( ÷ ) ____
dy dy y = 2y ___ x + y ___ dx dx 2
(
)
2 y dx fi __ ___ – x = 2 dy
dy y ___ dx
4.96 Integral Calculus, 3D Geometry & Vector Booster
(
y fi __ – 2
) ( )
dy 2 dy 3 x ___ = y ___ dx dx
Thus, the order of the differential equation is 3. 12. The given expression is x2 + y2 = 1 Differentiating w.r.t. x, we get fi fi
dy 2x + 2y ___ = 0 dx dy x + y ___ = 0 dx
Again differentiating, w.r.t. x, we get
( )
dy 2 d2y 1 + ___ + y ___2 = 0 dx dx
fi
1 + y¢2 + y ◊ y¢¢ = 0
fi
y ◊ y¢¢ + (y¢)2 + 1 = 0
13. The given differential equation is dy (1 + t) ___ – ty = 1 dt
dy fi _____ = y+1
– cos x _______ dx 2 + sin x
dy
fi
– cos x = _______ dx Ú _____ y + 1 Ú 2 + sin x
fi
log |y + 1| = log c – log |2 + sin x|
fi
(y + 1) (2 + sin x) = c
Put x = 0, y = 1, then c = 4 Hence, the equation of the curve is
(y + 1) (2 + sin x) = 4
p When x = __ , then 2 fi
4 y + 1 = __ 3 4 1 __ y = – 1 = __ 3 3
16. Given, dy (x + 1)2 + (y – 3) ___ = _______________ (x + 1) dx
( )
dy – t 1 fi ___ + ____ y = ____ t+1 t+1 dt which is a linear differential equation. t – Ú ____ dt t + 1 =
Thus, IF = e
(t + 1) – 1 – Ú ________ dt e t + 1
= e log (t + 1) – t = (t + 1) e– 1 Therefore, the solution is
( )
dy 2 + sin x ___ _______ = – cos x y + 1 dx
y ◊ (t + 1) e– 1 = Ú e– 1 dt + c – 1
– t
fi y ◊ (t + 1) e = – e + c Put t = 0, y = 1, then c = 2 Hence, the equation of the curve is y ◊ (t + 1) e– t = – e– t + 2 Put t = 1, then
y 3 = (x + 1) + ______ – ______ (x + 1) (x + 1)
dy y 3 fi ___ – ______ = (x + 1) – ______ (x + 1) dx (x + 1) which is a linear differential equation. dx – _____ 1 Thus, IF = e Ú x + 1 = e– log (x + 1) = ______ (x + 1) Therefore, the solution is y 3 _____ = 1 – _______ dx x+1 Ú (x + 1)2
(
)
y 3 fi _____ = x + _____ + c y+1 x+1 Put
x = 2 and y = 0, then c = – 3
2y ◊ e– 1 = – e– 1 + 2
Hence, the equation of the curve is y 3 _____ = x + _____ – 3 x+1 x+1
fi
2y = – 1 + 2e
fi
1 y = e – __ 2 14. A right circular cone with radius R and height H contains a liquid which evaporates at a rate propeortional to its surface in contact with air (proportonality constant k > 0). Find the after which the cone is empty fi
15. The given differential equation is
y = x2 + x + 3 – 3x – 3 = x2 – 2x.
Differential Equation ______
÷ 1 – y2 fi _______ dy = dx y
Hence, the required area 2
= Ú [0 – (x2 – 2x)] dx 0
(
______
)
3 2
x = x2 – __ 3 0
( )
Put fi
x cos y + y cos x = p x = 0, then y = p dy dy cos y – x sin y ___ – y sin x + cos x ___ = 0 dx dx dy Put x = 0, y = p, then ___ = – cos (p) = 1 dx dy dy dy 2 – sin y ___ – sin y ___ – x cos y ___ dx dx dx
( )
( ) ( )
d2y dy – x sin y ___ 2 – y cos x – sin ___ dx dx
dy d2y – sin x ___ + cos x ___ 2 = 0 dx dx
( )
( )
dy Put x = 0, y = p and ___ = 1, then dx d2y (– sin p – sin p – 1) + ___ 2 = 0 dx 2 d y fi ___2 = 1 dx 18. The equation of tangent at any point (x, y) to the curve y = f (x) is Y – y = f ¢ (x) (X – x)
(
)
dx It meets the x-axis at x – y ___ , 0 . dy It is given that
AP = 1
fi
AP 2 = 1
fi
dx 2 2 – y ___ xy + y = 1
( )
( ) ( ) ( ) ÷
dx y2 ___ = 1– y2 dy 2 dx 2 1 – y fi ___ = _____ dy y2 fi
2
fi
1 – y2 ÷_______ dy = Ú dx Ú y
fi
y ÷1 – y2 ________ =x+c Ú y2 dy
fi
t dt = x + c, Ú _____ t2 – 1
fi
1 2 dt = x + c Ú 1 + t_____ –1
fi
t–1 1 t + __ log ____ = x + c 2 t+1
_____
8 4 = 4 – __ = __ sq.u. 3 3 17. The given relation is
÷
______
y2 = ______ 1 – y2
2
(
(Let t2 = (1 – y2))
)
| |
|
_____
|
÷1 – y2 – 1 1 _____ = x + c fi ÷1 – y2 + __ log __________ 2 + 1 ÷ 1 – y2 _____
19. The given differential equation is
(x2 + y2) dy = xy dx
dy xy fi ___ = ______ dx x2 + y2
( ) ( )
y __ x dy fi ___ = _______ y 2 dx __ 1 + x
...(i)
which is a homogeneous differential equation.
(
)
dy dv Let y = v x fi ___ = v + x ___ dx dx fi
dv v v + x ___ = _____ dx 1 + v2
fi
dv v x ___ = _____ – v dx 1 + v2
fi
dv v – v – v3 v3 x ___ = _________ = – _____ 2 dx 1+v 1 + v2
( ) Ú ( )
1 + v2 dx fi ______ 3 dv = – ___ x v fi
_____
1 – y2 dx fi ___ = _____ dy y2
4.97
1 + v2 dx _____ dv = – Ú ___ x v3
fi
1 – ___ 2 + log |v| = c – log |x| 2v
fi
y x2 – ___2 + log __ x = c – log |x| 2y
| |
4.98 Integral Calculus, 3D Geometry & Vector Booster fi
x2 – ___2 + log |y| = c 2y
1 x = 1, y = 1, then c = – __ 2 Thus, the equation of the curve is When
x2 1 – ___2 + log |y| = – __ 2 2y
When
x = x0, y = e, then
x20 1 – ___2 + 1 = – __ 2 2e
fi
x20 3 1 ___ – 2 = – 1 – __ = – __ 2 2 2e
fi x20 = 3e2 fi
___
x0 = ÷3 e
20. The given differential equation is 2
y dx + y
= x dy
fi
ydx – xdy = – y2dy
ydx – xdy fi _________ = – dy y2 x fi d __ y = – dy fi
( ) Ú d ( __ xy ) = – Ú dy
x fi __y = c – y When
x = 1, y =1, then c = 2
Thus, the equation of the curve is x __y = 2 – y When
x = – 3, then
3 2 – y = – __ y
fi
3 y – 2 = __ y
fi
y2 – 2y – 3 = 0
fi
(y – 3) (y + 1) = 0
fi
y = 3, – 1
Since y > 0, so the value of y = 3. 21. The equation of any tangent at P (x, y) is
dy Y – y = ___ (X – x) dx
( )
)
dx It meets the x-axis at A x – y ___ , 0 dy dy and B 0, y – x ___ dx
(
Since P (x, y) divides the line AB in the ratio 3 : 1, we get dy 3 dx 1 fi __ y – x ___ = y, __ x – y, ___ = x 4 4 dx dy
(
)
(
)
dy x ___ = – 3y dx dy dx ___ fi ___ y = – 3 x fi
dy
fi
dx y = – 3 Ú ___ Ú ___ x
fi
log |y| = log c – 3 log |x|
fi
x3y = c
As f (1) = 1, we get c = 1 Thus, the equation of the curve is x3y = 1 1 fi y = __ 3 x which passes through (2, 1/8). dy 3 Also, ___ = – __4 dx x dx m = ___ = – 3 dy (1, 1)
( )
\ The equation of normal at (1,1) is 1 y – 1 = __ (x – 1) 3 fi 3y – 3 = x – 1 fi x – 3y + 2 = 0 22. The given differential equation is dy 2 ___ = _____ dx x + y Let x + y = v dy dv fi 1 + ___ = ___ dx dx dy dv fi ___ = ___ – 1 dx dx dv 2 fi ___ – 1 = __ v dx dv v + 2 fi ___ = _____ v dx
(
)
v fi _____ dv = dx v+2 fi
2 dv = Ú dx Ú ( 1 – _____ v + 2)
Differential Equation
fi
v – 2 log |v + 2| = x + C
fi
x + y – 2 log |x + y + 2| = x + C
fi
y – 2 log |x + y + 2| = C
When x = 1, y = 1, then fi
( )
e c = 1 – 2 log 4 = log ___ 16
Thus, the equation of the curve is fi
( ) ( )
e y – 2 log |x + y + 2| = log ___ 16 e y = log |x + y + 2|2 + log ___ 16
(
( ) )
e fi (x + y + 2)2 ___ = ey 16 16 fi |(x + y + 2)2 e– y)| = ___ e
( )
23. The given differential equation is ______
dy ÷1 – y2 ___ = _______ y dx y dv _____ fi ______ = dx ÷1 – y2 fi
y dv
_____ = Ú dx Ú ______ 2
÷1 – y
Integrating, we get _____
fi (x + c)2 + y2 = 1 which represents a circle with the centre (c, 0) and the radius 1. 24. Given t2 f (x) – x2f (t) lim ____________ = 1 t Æ x t–x
(
Hence, the solution is y dx __2 = – Ú ___4 + c x x y 1 fi __2 = ___ 3 + c x 3x As f (1) = 1, then c = 1 – 1/3 = 2/3 Therefore, the equation of the curve is y 3 1 __2 = ___ 3 + __ 2 x 3x fi
3x2 1 y = ___ + ___ 3x 2
25. The given differential equation is _____
_____
_____
_____
x ÷x 2 – 1 dy – y ÷y 2 – 1 dx = 0
fi
x ÷x 2 – 1 dy = y ÷y 2 – 1 dx
dy dx _____ = ________ _____ fi ________ 2 y ÷y – 1 x ÷x 2 – 1 dy
fi
dx _____ _____ = Ú _______ Ú ________ 2 2
fi
sec– 1 (y) = c + sec– 1 (x)
When
2 x = 2, y = ___ __ , then 3 ÷
y ÷y – 1
x÷x – 1
( )
2 c = – sec–1 (2) + sec–1 ___ __ 3 ÷ p p p fi c = – __ + __ = – __ 3 6 6 Thus, the equation of the curve is p sec– 1 (y) = sec– 1 (x) – __ 6 p – 1 fi y = sec sec (x) – __ 6 Thus, the statement (i) is trure Also, we can write Eq. (i) as fi
– ÷1 – y2 = x + c
)
(
)
2t f (x) – x2 f ¢ (t) fi lim ______________ = 1 t Æ x 1 fi
2x f (x) = x2 f ¢ (x) = 1
fi
dy 2x y – x2 ___ = 1 dx
(
( ) ( ) p 1 1 __ __ fi __ y = cos ( cos ( x ) – 6 )
p 1 1 cos– 1 __ y = cos– 1 __ x – __ 6 – 1
( ( ) ) ( ) p 1 + sin ( cos ( __ x ) ) sin ( __ ) 6
p 1 1 – 1 __ __ fi __ y = cos cos x cos 6
which is a linear differential equation.
3 __ ÷ 1 __ 1 ___ 1 1 __ fi __ y = x 2 + 2 1 – x2
\
IF =
dx ___ e– 2 Ú x =
1 e– 2 log x = __ 2 x
...(i)
)
dy x2 ___ – 2x y = – 1 dx dy 2 1 fi ___ – __ y = – __2 dx x x fi
4.99
– 1
__
÷
_____
Thus, the statement (ii) is false.
4.100 Integral Calculus, 3D Geometry & Vector Booster 26. Given differential equation is y¢ = y +1
Hence, the solution is y __x = – Ú x dx + c
dy fi ___ = y + 1 dx dy fi ___ y + 1 = dx
y x2 fi __x = c – __ 2 When
dy
x = 1, y = 1, then c = 3/2
Hence, the equation of the curve is
fi
= dx Ú y_____ +1 Ú
fi
log |y + 1| = x + c
y 3 __ x2 __x = __ – 2 2
when
x = 0, y = 1, then c = log 2
When x = – 3, then
Hence, the equation of the curve is
log |y + 1| = x + log 2
When
x = ln 2, then
y+1=4fiy=3
29. The given differential equation is
(x – 3)2 y¢ + y = 0
dy (x – 3)2 ___ + y = 0 dx dy y fi ___ = – _______ dx (x – 3)2 fi
Thus.
fi fi
1 C + _____ x – 3 =
1 _____
1 _____
e x – 3 ◊ ec = A ◊ ex – 3
Thus, the domain of the above function is R – {3} 28. The equation of tangent at any point (x, y) to the curve y = f (x) is Y – y = f ¢ (x) (X – x) Intercept on y-axis is y – x f ¢ (x) It is given that,
y – x f ¢ (x) = x3
dy y – x ___ = x3 dx dy y fi ___ – __ x = – x2 dx fi
which is a linear differential equation. Thus
IF =
dx ___ e – Ú x =
1 e– log x = __ x
y ◊ e g (x) = Ú (g (x) Ú g¢ (x)) e g (x) dx
fi
y ◊ e g (x) = e g (x) (g (x) – 1) + c
When x = 0, y = 0, then c = 1 Thus, the curve is
dy dx Ú ___y = – Ú (x_______ – 3)2 1 log |y| = C + _____ x–3 y = e
IF = e Ú g¢ (x) dx = e g (x)
Therefore, the solution is
dy dx _______ fi ___ 2 y = – (x – 3) fi
y ¢ (x) + y (x) g ¢ (x) = g (x) g ¢ (x)
dy fi ___ + g ¢ (x) y = g (x) g ¢ (x) dx which is a linear differential equation.
27. The given differential equation is
9 27 ___ 3x x3 18 y = ___ – __ = – __ + ___ = = 9 2 2 2 2 2
y ◊ e g (x) = eg (x) (g (x) – 1) + 1
When
x = 2, then y = (0 – 1) + 1 = 0.
30. Given equation is x
6 Ú f (t) dt = 3xf (x) – x3 – 5 0
fi
6f (x) = 3f (x) + 3x f ¢ (x) – 3x2
fi
3f (x) = 3x f ¢ (x) – 3x2
fi
f (x) = x f ¢ (x) – x2
dy y = x ___ – x2 dx dy fi x ___ – y = x2 dx dy y fi ___ – __x = x dx fi
which is a linear differential equation
dx ___ 1 IF = e – Ú x = e– log x = __ x
Hence, the solution is y __x = Ú dx + c
Differential Equation
y fi __x = x + c When x = 1, y = 2, then c = 1 Hence, the equation of the curve is y __x = x + 1 when x = 2, then y = 6. Thus, the value of f (2) is 6.
fi
dv v + x ___ = v + sec v dx
fi
dv x ___ = sec v dx
dv dx fi ____ = ___ x sec v fi
dv dx = Ú ___ Ú ____ x sec v
fi
dx Ú cos v dv = Ú ___ x
dy fi ___ – tan x ◊ y = 2x sec x dx
fi
sin (v) = log |x| + C
which is a linear differential equation.
fi
y sin __ x = log |x| + C
31. The given differential equation is
Thus,
y¢ – y tan x = 2x sec x
– Ú tan x dx
IF = e
log cos x
= e
= cos x
( )
( )
p which is passing through 1, __ , so, c = 1/2 6
Hence, the solution is
y ◊ cos x = Ú 2x ◊ sec x ◊ cos x dx + c
Hence, the equation of the curve is
fi
y ◊ cos x = Ú 2x ◊ dx + c
fi
y ◊ cos x = x2 + c
33. The given differential equation is
when x = 0, y = 0, then c = 0 Thus, the equation of the curve is
p p 2 1__ ___ When x = __ , then y ◊ ___ = 4 ÷2 16 p 2 y = ____ __ 8÷2
When
p 2p 2 x = __ , then y = ____ 3 9
dy xy x4_____ + 2x fi ___ – _____ 2 = ______ dx 1 – x ÷1 – x2 which is a linear differential equation.
( ) (
\
( )
__ 2p 2 dy fi ___ = ÷ 3 ◊ ____ + 9 dx
dy 2p 2 ___ 4p fi ___ = ____ __ + 3 dx 3÷3
( ) (
)
)
y ◊ ÷1 – x2 = Ú (x4 + 2x) dx + c
fi
x5 y ◊ ÷1 – x2 = __ + x2 + c 5
_____
_____
dy p 2 p fi ___ = ____ __ + ___ __ dx 8÷2 ÷ 2
x5 y ◊ ÷1 – x2 = __ + x2 5
fi
x5 x2 y = _______ _____ + ______ _____ 5÷1 – x2 ÷1 – x2
...(i)
which is a homogeneous differential equation. Let y = v x dy dv fi ___ = v + x ___ dx dx
__
3 ÷ ___ 2 x5 x2 Now, __ _______ _____ + ______ _____ dx 2 2 3 5 1 – x ÷ 1 – x ___ – 2
Ú
(
÷
÷
_____
= ÷ 1 – x2
which is passing through (0, 0), so c = 0 Hence, the equation of the curve is
dy p p 2 p __ Again, ___ – tan __ ◊ ____ __ = 2 ◊ __ ◊ ÷2 4 8÷2 4 dx
( )
1 __ log (1 – x2)
e 2
_____
4p ___ 3
dy y y 32. Given ___ = __ + sec __ x dx x
x – Ú _____ dx 1 – x2 =
IF = e
Hence, the solution is
dy p 2p 2 p Also, ___ – tan __ ◊ ____ = 2 ◊ __ ◊ 2 3 9 3 dx
( )
y 1 sin __ x = log |x| + __ 2
dy xy x4 + 2x ___ + _____ 2 = _______ _____ dx x – 1 ÷ 1 – x2
y ◊ cos x = x2
fi
4.101
)
4.102 Integral Calculus, 3D Geometry & Vector Booster __
(
Ú
)
÷
__
2–4 Clearly, g (0) = _____ < 0 22
3 ÷ ___ 2 x2 = __ ______ _____ dx 2 3 1 – x ÷ ___ – 2 3 ÷ ___ 2
(
)
2
x = 2 ◊ Ú ______ _____ dx 0 ÷ 1 – x2 =
p __ 3 2 ◊ sin2q dq, 0
Ú
p __ 3 (1 0
Ú
=
sin 2q 3 = q – _____ 2 0
(Let x = sin q)
– cos 2q) dq
( ) p 1 2p = ( __ – __ sin ___ 3 2 ( 3 ))
(
__
) )
g (0) ◊ g (– 1) < 0
(
)
Hence, g (x) has a root in between (– 1, 0). 35. Let the family of circles be
x2 + y2 – a x – a y + c = 0
...(i)
( )
d 2y dy 2 d 2y 2 + 2y ___2 + 2 ___ – a ___2 = 0 ...(ii) dx dx dx Eliminating a between Eqs (i) and (ii), we get
)
( )
dy 2x + 2y ___ d2y 2 2 d y dy dx ___ 2 + 2y ___2 + 2 ___ – __________ 2 = 0 dx dy dx dx 1 + ___ dx
3 p ÷ = __ – ___ 3 4
( )
dy (1 + ex) ___ + y ◊ ex = 1 dx
(
( (
Again differentiating w.r.t. x, we get
34. We have (1 + ex) y ¢ + y ex = 1 fi
)
Differentiating w.r.t. x, we get dy dy 2x + 2y ___ – a – a ___ = 0 dx dx
p __
(
and
3 1 2 1 + __ e – __ 1 – __ e e ___________ _______ g (– 1) = = > 0 1 2 1 2 1 + __ e 1 + __ e
)
(
(
)
(x + 2)2 4 \ _______ > __ > 1 3 3
)
dy ex 1 fi ___ + _______ ◊ y = _____ x dx (1 + ex) 1+e
2x + 2y y¢ fi 2 + 2y y¢¢ + 2 (y¢)2 – _________ y¢¢ = 0 1 + y¢
which is a linear differential equation.
x + y y¢ fi 1 + y y¢¢ + (y¢)2 – _______ y¢¢ = 0 1 + y¢
Thus,
IF =
ex x dx Ú _______ e (1 + e ) =
elog |1 + ex| = (1 + ex)
Multiplying both sides of Eq. (i) by IF and integrating, we get fi
y ◊ (1 + ex) = Ú dx + c
fi
y ◊ (1 + ex) = x + c
When
x = 0, y = 2, then c = 4
(
)
fi (y – x) y¢¢ + (1 + y¢ + y¢ 2) y¢ + 1 = 0 Thus,
P = (y – x) and Q = (1 + y¢ + y¢2)
and P + Q = (– x + 1 + y + y¢ + y¢2) 36. Given curve is dy (x2 + xy + 4x + 2y + 4) ___ – y2 = 0, x > 0 dx
Hence, the solution is
y ◊ (1 + ex) = x + 4
fi
dy {(x + 2)2 + y (x + 2)} ___ – y2 = 0 dx
fi
x+4 y = _______ (1 + ex)
fi
dy {(x + 2)2 + y (x + 2)} ___ = y2 dx
Thus,
y (– 4) = 0
fi
(1 + ex) – (x + 4) ex y¢ = _________________ (1 + ex)2 x
Let
dy ___________________ y d (x + 2) – (x + 2) dy fi ___ y = (x + 2)2 x
(1 + e ) – (x + 4) e g (x) = _________________ (1 + ex)2
(
)
dy y ______ fi ___ y = – d (x + 2) Integrating, we get
Differential Equation
y log |y| = c – ______ (x + 2)
When
x = 1, y = 3, then c = 1 + log3
Thus,
y y log __ + ______ = 1 3 (x + 2)
For
y = (x + 2), then
( )
( ) ( )
x + 2 log _____ = 0 3 x+2 fi _____ = 1 3
fi
x=1
For
y = (x + 2)2, then
(x + 2)2 log _______ + (x + 2) = 1 3
(
(
)
)
(x + 2)2 4 Now for x > 0, _______ > __ > 1 3 3
(
)
(x + 2)2 log _______ + (x + 2) > 2 3
So, it has no solution.
4.103
Chapter
5
Vectors
Concept Booster 1.1 Introduction A study of motion involves the introduction of a variety of quantities that are used to describe the physical world. For examples, distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories—vectors and scalars. A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is described by its magnitude only. The emphasis of this unit is to understand some fundamentals about vectors and to apply the fundamentals in order to understand motion and forces that occur in two dimensions.
For example, the rotation of a rigid body through finite angles have both magnitude and direction but do not satisfy the vector law of addition. Therefore, it is not a vector.
1.3 Representation
of
vectors
Directed line segment Any given protion of a given straight line where the two end-points are distinguished as the initial and the terminal points is called a directed line segment. The directed line segment with the initial point P and the ___›
terminal point Q is denoted by the symbol PQ or PQ .
1.2 Physical quantities The physical quantity is divided into two categories: (a) scalar quantities (b) vector quantities (a) Scalar quantities A quantity which has only magnitude and not related to any fixed direction in the space is called the scalar quantity. For example, mass, length, volume, density, time, temperature, etc., are all scalar quantities. If the unit of measurement is defined, a real number is sufficient to represent a scalar quantity. Thus, in this chapter, we shall represent scalars by real numbers. (b) Vector quantities A quantity which has magnitude as well as direction is called a vector. For exmaple, force, velocity, acceleration, displacement, momentum, etc. are all vector quantities. Notes: Quantities having magnitude and direction but not obeying the vector law of addition will not be treated as vectors.
The two end-points P and Q are not interchangeable. A directed line segment is called a vector if it has three following characteristics.
_ __›
(i) Length: The length of PQ is denoted by the symbol ___›
|PQ | . (ii) Support: The line of unlimited length of which a directed line segment is a part is called the support. (iii) Sense: The sense of the directed line segment is from its initial point to its terminal point. We genrally denote the vector by bold letter or by a single letter with an arrow or by a letter with a bar over its head, _ __› _› _ i.e. a, a , a denote the vector PQ .
5.2 Integral Calculus, 3D Geometry & Vector Booster
1.4 types
of
vectors
Like vectors
(i) Zero or Null vector A vector whose initial or terminal points are identical or coincident, is called a zero vector. It is also known as null vector. It is denoted by 0 and its direction is indeterminate. ___› _ ___› __› Thus AA , BB and CC are zero vectors. Geometrically, it represents a point.
Collinear vectors are called like vectors if their direction are the same.
(ii) Proper vector
Collinear vectors are called unlike vectors if their directions are opposite.
Any non-zero vector is called a proper vector. (iii) Unit vector If the magnitude of a vector is unity, it is called a unit vector. _› If |a | = 1, then a is called a unit vector. It is generally a denoted as and is defined as = ___ _› . |a | (iv) Co-initial vectors Two or more vectors are said to be co-initial vectors, if they have the same initial point.
_ __›
_ __›
Here, PQ , and QR vectors are like vectors. Unlike Vectors
___›
___›
Here, PQ and QR are unlike vectors. Difference between parallel and collinear vectors Every collinear vectors are parallel, whereas every parallel vectors need not be a collinear. Notes 1. Two non-zero vectors a and b are collinear (or parallel) if a = l b, " l ŒR. 2. If a = a1i + a 2 j + a3k and b = b1i + b2j + b3k are two collinear (or parallel) vectors, then a1 __ = b1
__› _ ___› ___› ___› _ __›
Here, OP , OQ , OR , OS , OT are co-initial vectors. (v) Co-terminal vectors Two or more vectors are said to be co-terminal vectors if they have the same terminal point.
a2 __ = b2
a3 __ . b3
(vii) Coplanar vectors Three or more vectors are said to be coplanar if they lie in the same plane, or they are parallel to the same plane.Three vectors a, b and c are coplanar if any one of their is a linear combination of the other two vectors. i.e
a = x b + y c
or
b = x a + y c
or
c = x a + y b
(viii) Free vector If the origin of a vector is not specified, it is called a free vector.
_ __› _ __› __› _ __› _ __› _
Here, PM , DM , CM , BM , AM are co-terminal vectors. (vi) Collinear or parallel vectors Two vectors are said to be collinear vectors if their supports are parallel or the same irrespective of their direction. Collinear vectors are also called parallel vectors.
___›
__› _ › _
_› ___›
_ ›
Here, OA , = a , OB = b , OC = c › _ › _
_ ›
Thus a , b , and c are free vectors.
Vectors
5.3
(ix) Localized vector For a vector of given magnitude and direction, if its initial point is fixed in space, the vector is called a localized vector.
It is also stated as, If two vectors are represented by the two sides of a triangle taken in order, their sum is represented by the third side of the triangle, taken in reverse order.
(x) Position vector
Parallelogram law of addition of vectors
Let O be the origin and P be a point in the space. The position vector of P is OP.
Let the parallelogram OACB, where _ __›
_›
_ __›
___›
___›
___›
___›
___›
_ ›
OA = a and OB = b
If a and b are the position vectors of the points A and B, then ___›
___›
___›
AB = OB – OA = b – a. (xi) Negative of a vector The vector, which has the same magnitude as the vector a but has the direction opposite to that of a, is called the negative of a and is written as – a. (xii) Equality of two vectors Two vectors are (a) the same (b) the same (c) the same
1.5 Algebra
of
= OA + OB
= a + b .
Properties of addition of vectors
vectors
(i) Addition of vectors is commutative, i.e. _›
1.5.1 Addition of vectors Let a and b be two given vectors. Take a point O in the space. ___›
_ ›
_›
This method of addition of two vectors is called the parallelogram law of addition vectors.
said to be equal if they have length or parallel supports sense.
It is possible to develop an algebra of vectors which is useful in the study of geometry, mechanics and other branches of applied mathematics.
___›
Thus, OC = OA + AB
_›
Let OA , = a, AB _ ›= b, so that the terminal point of a is the initial point of b .
_ ›
_ ›
_›
a + b = b + a (ii) Addition of vectors is associative, i.e. _ ›
_›
_ ›
_›
_ ›
_ ›
(a + b ) + c = a + (b + c )
(iii) Additive identity exists, i.e. _›
_ ›
_ ›
_›
_›
a + 0 = a = 0 + a (iv) Additive inverse also exists, i.e. _ ›
_ _›
_›
_ _›
_ ›
a + (– a ) = 0 (– a ) + a
Polygon law of vector addition Let OABCDE be a polygon where
___›
The vector OB is defined as the vector sum (or resultant) _› _ › of a and b . ___›
___›
___›
_ ›
_›
Thus, OB = OA + AB = a + b which is known as the triangle law of vector addition.
___›
__› _ › _
_› _ __›
__› _ › _
_›
___›
_ ›
OA = a , AB = b , BC = c , CD = d and DE = e .
5.4 Integral Calculus, 3D Geometry & Vector Booster Applying the triangle law, we have.
(i) If the rotation from OX to OY is in the anti-clockwise direction and OZ is directed upwards, the system is called the right handed system. (ii) If the rotation from OX to OY is in the clockwise direction and OZ is directed upwards, the system is called the left handed system.
_ _ __› _ __› _ › ___› › OB = OA + AB = a + b
___›
___›
___›
_ ›
_›
_ ›
___›
___›
___›
_›
_›
_›
___›
___›
___›
_›
_›
_›
OC = OB + BC = (a + b ) + c
_›
OD = OC + CD = (a + b + c ) + d _›
_›
OE = OD + DE = (a + b + c + d ) + e Thus, to find the sum of more than two vectors, a polygon is formed. Therefore, this method is known as the polygon law of vector addition. Note: If the initial point of the first vector and the final point of the last vector are the same, the sum of the vectors is a zero vector.
1.5.2 Subtraction of vectors If a and b be any two vectors, the subtraction of b from a is defined as the addition of – b to a and is written as a + (– b) = a – b, i.e. _›
_ ›
_ ›
_›
a + (– b ) = a – b
1.7 Position vector
of a
point
in a
space
Let O be a fixed point, known as the origin, and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis, respectively, in such a way that they form a right handed system. The plane XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane, respectively. Let P be a point in a space such that its distances from yz-, zx- and _ › › _ › _ xy-plane be a, b, c respectively and i , j , k are the vectors along x, y and z axes, respectively. Let
OA = a, OB = b and OC = c
Now, and
OP = OL + LP = OA + AL + LP = OA + OB + OC. = a i___________ + b j + c k. |OP| = ÷ a 2 + b2 + c2 .
1.8 Linear Combination 1.5.3 Multiplication of a vector by a scaler Let_ m be any scaler and a be any vector. Then its product › m a is called_ the multiplication of a vector by a scaler. › _› If a and be two vectors and m, n are scalers, then
_›
_›
_›
(i) m (a ) = m a = (a ) m _›
_›
(ii) m (n a ) = (m n) a (iii) (iv)
_› (m + n) a = _ › _› m (a + b ) =
1.6 Left
and
_› ma + _›
m a + m b
right handed orientation
__›
Z
Z
X
Y Z
Right Handed System
__›
__›
A system of vectors a1 , a2 , ..., an is said to be linearly independent if_there exist scalars x1, x2,...., xn (all zero) such _› _› › that x1a 1 + x2a 2 + ... + xna n = 0
X
O Y
X
x1a1 + x2a2 + ... + xnan = 0.
1.10 Linearly independent vectors
Y
O
1.9 Linearly dependent vectors A system of vectors a1, a2, ..., an is said to be linearly dependent if there exist scalars x1, x2, ..., xn (not all zero) such that
_› n a _ ›
Y
A vector r is said to be a linear combination of the vectors a, b, c, ..., if there exist scalars x, y, z, ... such that r = x a + y b + z c + ...
1.11 Section formulae Z X
Left Handed System
1.11.1 Internal section If a point R (r) divides the line segment joining the points P(r1) and Q (r2) internally in the ratio m : n, then
Vectors
m r2 + n r1 r = __________ . m+n
From the figure, we have by using triangle rule,
5.5
1.11.4 Centroid The point of intersection of the medians of a triangle is called its centroid. Let A (r1), B (r2) and C (r3) be the vertices of a triangle ABC and G (r) be its cen troid. Then the position vector r of the centroid is given by
r1 + r2 + r3 r = __________ . 3
m PR __ ____ = n RQ
OR – OP __ m fi _________ = n OQ – OR fi
r – r1 __ m ______ r 2 – r = n
fi
n (r – r1) = m (r2 – r)
fi
r (m + n) = m r2 + n r1
fi
m r2 + n r1 r = __________ m+n
1.11.2 External section If a point R(r) divides the line segment joining the points P(r1) and Q(r2) externally in the ratio m : n, then
1.11.5 Centroid of a tetrahedron Let _the position vectors of the points A, B, C and D are _› _› › _› a , b , c and d respectively. The position vector r of the centroid is
a + b + c + d r = ____________ 4
m r2 – n r1 r = _________ m – n . It can be proved by using triangle rule.
1.11.6 Incentre The point of intersection of the angle bisectors of a triangle is called its incentre.
1.11.3 Mid-Point formula If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio 1 : 1, then
r2 + r1 r = ______ . 2
Let the position vectors of the points A, B, C are r1, r2, r3, respectively, and r be the position vector of the incentre I. The position vector r of the incentre is given by
5.6 Integral Calculus, 3D Geometry & Vector Booster
a r1 + b r2 + c r3 r = ______________ a+b+c
1.11.7 Circumcentre The point of intersection of the perpendicular bisector of a triangle is called its circumcentre. Let the position vectors of the points A, B, C are r1, r2, r3, respectively, and r be the position vector of the circumcentre O. The position vector r of the incentre is given by (sin 2A) r1 + (sin 2B) r2 + (sin 2C) r3 r = ______________________________ sin 2A + sin 2B + sin 2C
1.13 straight line 1.13.1 Equation of a line passing thorugh a point and parallel to a vector The equation of a line passing through a point A with position vector r1 and parallel to a vector m is given by
r = r1 + l m
Now,
OP = OA + AP
fi
OP = OA + l M
fi
r = r1 + l m
1.11.8 Orthocentre The point of intersection of the perpendiculars of a triangle is called its orthocentre. Let the position vectors of the points A, B, C are r1, r2, r3, respectively and r be the position vector of the orthocentre H. The position vector of the orthocentre is
1.13.2 Equation of a line passing through two points A (r1) and B (r2) is
(tan A) r1 + (tan B) r2 + (tan C) r3 r = ___________________________ tan A + tan B + tan C
1.12 Bisector
of
angle
between
vectors A
and
r = r1 + l (r2 – r1)
Now,
OP = OA + AP
fi
OP = OA + l AB
fi
OA + l (OB – OA)
fi
r = r1 + l (r2 – r1).
b
(i) If a is not parallel to b, the vectors along the bisectors (internal) of angle between a and b is given by
___› OM =
(
_ ›
a l ___ _› + |a |
__›
)
b ___ _› |b |
(ii) If a is not parallel to b, the vectors along the bisectors (external) of the angle the between a and b is given by
(
_›
___› a ON = l ___ _› – |a |
_›
)
b ___ _› |b |
1.14 Plane 1.14.1 Vector Equation The vector equation of a plane passing through a point having position vector a and the normal to a vector n is given by r ◊ n = d
Vectors
N n
P
r
5.7
Let a and b be two non-zero vectors and q the angle between them. Its scalar product is denoted as a ◊ b and is defined as a ◊ b = ab cos (q ), 0 £ q £ p.
O
Let
OP = r and ON = n
It is given that ON ^ NP NP ◊ ON = 0 (OP – ON) ◊ ON = 0 (r – n) ◊ n = 0 r ◊ n – n ◊ n = 0 r ◊ n = n◊ n r ◊ n = d.
fi fi fi fi fi fi
2.2 Geometrical interpretation
of a ◊ b
1.14.2 Equation of a plane in normal form The vector equation of a plane normal to a unit vector and at a distance from the origin is r ◊ = d.
We have a ◊ b = a b cos (q ) = a [b cos (q )] = a OM a◊b Thus, OM = ____ a a◊ b Projection of b on a is ____ a .
Let O be the origin and ON be the perpendicular from O to the given plane such that ON = d . Let P be a point on the plane with position vector r so that OP = r Now, ON ^ NP
Thus, geometrically product of two vectors a and b is defined as the product of the magnitude of the first vector and the projection of the 2nd vector on the 1st vector.
2.3 Properties
of
dot product
NP ◊ ON = 0
fi
(OP – ON) ◊ ON = 0
= ba cosq
fi
(r – d ) ◊ d = 0
= ab cos q
fi
(r ◊d – d ◊ d ) = 0
fi fi
(r ◊ – d) = 0 r ◊ = d
(ii) If q acute, then a ◊ b > 0
Scalar product of two vectors 2.1 Definition If the product of two vectors is a scalar, it is known as scaler product or dot product of two vectors.
two vectors
(i) Dot product is commutative, i.e. a ◊ b = b◊ a
fi
which is the required equation of the plane.
of
Now
(iii) (iv) (v) (vi) (vii) (viii) (ix)
b ◊ a = ba cos (– q)
= a ◊b
If q obtuse, then a ◊ b < 0 If a and b are perpendiculars, then a ◊ b = 0 If a and b are like vectors, then a ◊ b = ab If a and b are unlike vectors, then a ◊ b = – a b Maximum value of a, b is ab Minimum value of a ◊ b is – ab a ◊ a = |a|2 = a2
5.8 Integral Calculus, 3D Geometry & Vector Booster a ◊b (x) Projection of a on b = ____ . |b|
Eliminating x, y and z from Eqs. (i), (ii) and (iii), we get
a. b (xi) Projection of b on a = ____ . |a| (xii) Orthonormal triad vectors If three vectors which are mutually perpendicular to each other, they are called orthonormal triad vectors. Let , , be three unit vectors which are mutually perpendicular to each other. Thus . = . = . and
a b c a⋅a a⋅b a⋅c = 0 b⋅a b⋅b b⋅c
2.4 Component vector A
of a
vector B
along and
perpendicular
to
. = 0 = ◊ = ◊
Therefore,
0 0
Here,
0
1 0
0
0 1
b = OM + MB Component of a vector b along a = OM
(xiii) If
a = a1 + a2 + a3
and
b = b1 + b2 + + b3 ,
then
a ◊ b = a1b2 + a2b2 + a3b3
|a| = ÷ a 21 + a22 + a 23
___________
|b| = ÷
Thus,
a1b1 + a2b2 + a3b3 ___________ ___________ cos (q) = ________________________ ÷a 21 + a22 + a23 ÷b 21 + b22 + b 23
(xiv) Any vector a can be written as a = (a ◊ ) + (a ◊ ) + (a◊ )
(xv) |a + b + c| 2 = |a|2 + |b|2 + |c| 2 + 2 S (a ◊ b) (xvi) If a, b, c are coplanar vectors, then
x a + y b + z c = 0
= (OB cos q) = (b cos q) b a◊ b = _____ a b a ◊ b = ____ a a◊ b __ a = ____ a a a ◊ b = ____ 2 a. |a| Component of a vector b perpendicular to a vector a = MB
( ) ( )
= OB – OM = b – OM
a◊ b = b – ____ 2 a |a|
( )
2.5 Physical significance two vectors
a b c a⋅a a⋅b a⋅c = 0 b⋅a b⋅b b⋅c
As we know that three coplanar vectors are always linearly dependent. Thus,
= (OM)
( ) ( )
___________ b21 + b22 + b23
and
OB = OM + MB
1
of the
dot prodcut
F
Taking dot product a and b with Eq. (i), we get x (a ◊ a) + y (a ◊b) + z (a ◊ c) = 0
...(ii)
x (b ◊ a) + y (b ◊ b) + z (b ◊ c) = 0
...(iii)
A force acting on a particle is said to be work done if the particle is displaced in a direction which is not perpendicular to the force applied.
...(i)
of
q A
B
Vectors
Let a force F acts on a particle P and the particle P gets displaced from the position A to B, where AB is not perpendicular to the applied force F. Let AB = d. \ Work done = F d (cos q) = F ◊ d = (Force) ◊ (Displacement). Note: The work done by the resultant of the number of forces F1, F2, ..., Fn in a displacement d of a particle is equal to the sum of the work done by the forces separately, i.e. Work done = F1 ◊ d + F2 ◊ d + ... + Fn ◊ d.
= (F1 + F2 + ... + Fn) ◊ d
= R ◊ d
5.9
Area of a parallelogram OACB = (base) (height) = (OA) (BM) = (a) (b sin q) = (ab sin q) = |a × b| Thus, |a × b| reprtesents the area of a parallelogram with adjacent sides represented by the vectors a and b.
3.3 Properties
of
vector product
of
two vectors
(i) Vector product is not commutative, i.e. a × b π b × a (ii) a × b = – b × a (iii) If a and b are parallel vectors, then a × b = 0 (iv) If a and b are collinear vectors, then a × b = 0
where R = F1 + F2 + ... + Fn
CROSS PRODUCT OF TWO VECTORS 3.1 Introduction If the product of two vectors is a vector, it is known as the vector product or cross product or outer product of two vectors.
(v) (m a) × b = a × (m b) = m (a × b), where m is any scalar. (vi) a × (b + c) = (a × b) + (a + c). (vii) Unit vector perpendicular to the plane of a and b is a×b = ± ______ . |a × b| (viii) A vector of the magnitude l and perpendicular to the plane of a and b is
(a × b) ± l _______ . |a × b|
(ix) If q be the angle between two vectors a and b, then |a × b sin (q) = ______ . |a||b| Let a and b be two_ vectors. The cross product of _two _› _ › › › vectors is denoted as a × b and is defined as a × b = (ab sin q) , _ › _› where q the angle between a _ and b and is unit vector › perpendicular to the plane of a and b. Here a, b, form a right handed system.
3.2 Geometrical Interpretation
of
×
Let OACB be a parallelogram with O origin. Let OA = a and OB = b and –AOB = q
(x) The area of DOAB is given by 1 __ |a × b|, 2 where OA = a, OB = b 1 (xi) The area of DABC = __ |a × b + b × c + c × a|, 2 position vectors of the points where a, b and c are the A, B and C, respectively. 1 (xii) The area of a parallelogram = __ |d1 × d2|, 2 where d 1 and d 2 are the diagonals of the parallelogram. (xiii) If A, B and C are three collinear, points then (a × b + b × c × c × a) = 0, where a, b and c are the position vectors of the points A, B and C, respectively. (xiv) A unit vector perpendicular to the plane of
5.10 Integral Calculus, 3D Geometry & Vector Booster –ABC is given by (a × b + b × c + c × a) = ± _____________________ . |a × b + b × c + c + a|
(xv) If a, b and a × b form a right handed system, any vector r is a linear combination of a, b and a × b, i.e. r = x a + y b + z (a × b), where x, y and z are scalars. (xvi) Orthonormal Triad Vectors If three vectors which are mutually perpendicular to each other, they are called orthonormal triad vectors. Let , , be three unit vectors which are mutually perpendicular to each other. Thus
×
and
×
=0= = ,
= abc sin q cos j
= (a b sin q) (c cos j)
= |a × b| (c cos j)
= (area of the IIgm OALB) × (height)
,
= Volume of a parallelopiped.
–
4.2 Properties
× ×
,
, 0
=
×
= ,
×
=
, – 0
– 0
(xvii) If a = a1 + a2 + a3 and b = b1 + b2 + b3 , then iˆ a × b = a1 b1
We know that [a, b, c] = a ◊ (b × c)
Therefore, ×
Here OA, OB and OC are the coterminous edges of a parallelopiped represented by the vectors a, b and c respectively.
kˆ a3 b3
ˆj a2 b2
( xviii) Lagranges identity For any two vectors a and b, a.a a.b (a × b)2 = (ab)2 – (a ◊ b)2 = . a.b b.b
Scalar Triple Product of Vectors
(i) Let
of
scalar triple product
vectors
a = a1i + a2 j + a3k,
b = b1i + b2 j + b3k
and
c = c1i + c2j + c3k
Then
[a b c] = a ◊ (b × c)
of
a1 = b1 c1
a2 b2 c2
a3 b3 c3
(ii) a ◊ (b × c) = b ◊ (c × a) = c ◊ (a × b) i.e.
[a b c] = [b c a] = [c a b]
(iii) a ◊ (b × c) = – b ◊ (a × c) i.e.
[a b c ] = – [b a c]
(iv) [ka b c] = k [a b c], where k ΠR
4.1 Introduction If the product of three vectors is a scalar, then it is known as scalar triple or box product or mixed product of three vectors. Let a, b and c are three vectors. Its scalar triple product is denoted as a ◊ (b × c) and is defined as a ◊ (b × c) = abc sin q cos j _ ›
_›
where q the angle between a and b and j the angle between c and a × b. Geometricel interpretation of [a, b, c]
(v) [la mb nc] = lmn [a b c] [a + b + c, d] = [a, c, d] + [b, c, d] If any two vectors are same, then [a, b, c] = 0 If any two vectors are parallel or collinear, then [a, b, c] = 0 If any one of them be a zero vector, then [a, b, c] =0 (x) If a, b, c are coplanar vectors, then [a, b, c] = 0
(vi) (vii) (viii) (ix)
Vectors
If a, b, c form a right handed system, then [a, b, c] > 0 If a, b, c form a left handed system, then [a, b, c] < 0 For any three vectors a, b and c, [a + b, b + c, c + a] = 2 [a, b, c] If a, b, c are coplanar vectors, then a + b, + b + c, c + a are also coplanar vectors If a and b are non-zero and non-collinear vectors, then [a, b, i] i + [a, b, j] j + [a, b, k] k = a × b (xvi) For any three vectors a, b, c, (xi) (xii) (xiii) (xiv) (xv)
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a
c⋅b
c⋅c
(xvii) If a, b, c, m, n are non-zero vectors, then a b c [a, b, c] ◊ [m × n] = a ⋅ m b ⋅ m c ⋅ m a ⋅n b⋅n c ⋅n
(xix) Any vector r can be expressed as a linear combination of three non-coplaner vectors a, b, c, then
r = ma + n b + p c
i.e.
[rbc] [rca] [rab] r = _____ a + _____ b + _____ c [abc] [abc] [abc]
(xx) If four points a, b, c and d are coplanar vectors, then [a b c] = [a b d] + [b c d] + [c a d]
VECTOR TRIPLE PRODUCT 5.1 Introduction If the product of three vectors is also a vector, then it is known as vector triple product. Let a, b and c be any three vectors, then its vector triple product is denoted as a × (b × c) and is defined as a × (b × c) = (a ◊ c) b – (a ◊ b) c
5.2 Geometrical r r r r
0 = l (a ◊ b) + m (a ◊ c)
l (a ◊ b) + m (a ◊ c) = 0
l (a ◊ b) = – m (a ◊ c)
(a ◊ b) (a ◊ c) _____ = – _____ = l (say) m l Therefore,
a × (b × c) = l [(a◊c) b – (a◊b) c]
Let a = i, b = i + j and c = i + j + k Now, i j k (b × c) = 1 1 0 = i – j 11 1
| |
2
Let fi fi fi
significance of
a × (b × c)
= a × (b × c) ^ a and r ^ (b × c) ^ a and r lies in the plane of b and c is orthogonal to a and coplanar with b and c
5.3 Explanation
of
Vector Triple Product
Since r is coplanar with b and c, then we can write r = l b + m c Thus,
a × (b × c) = l b + m c
a ◊ (a × (b × c)) = l (a ◊ b) + m (a ◊ c)
5.11
and a × (b × c) = i × (i – j) = – i × j = – k Also, (a ◊ c) b – (a ◊ b) c = i + j – i – j – k = – k Thus,
– k = l × – k
fi Hence,
l=1 a × (b × c) = (a◊c) b – (a◊b) c
5.4 Properties
of
vector triple product
(i) A unit vector perpendicular to a and coplanar with b and c is given by
a × (b × c) ± ___________ . |a × (b × c)|
If a, b and c be three non-zero vectors, then a × (b × c) + b × (c × a) + c × (a × b) = 0 If a, b and c be three non-zero vectors, then a × (b × c) + b × (c × a) + c × (a × b) are coplaner. The vector product of three vectors is not associative, i.e. a × (b × c) π (a × b) × c (v) If a, b and c be three non-zero vectors, then [(a × b), (b × c), (c × a)] = [a, b, c]2 (ii) (iii) (iv)
Proof: We have
(b × c) × (c × a)
= d × (c × a) (Let d = (b × c)]
= (d ◊ a) c – (d ◊ c) a
= (b × c ◊ a) c – (b × c ◊ c) a
= [a, b, c] c – 0
= [a, b, c,] c
Thus, [(a × b), (b × c), (c × a)]
= (a × b) ◊ ((b × c) × (c × a))
5.12 Integral Calculus, 3D Geometry & Vector Booster
= (a × b) ◊ [a, b, c] c
= ((a × b) ◊ c) [a, b, c]
= [a, b, c] [a, b, c]
= [a, b, c]2
(vi) If a, b and c be coplanar vectors, then (a × b), (b × c), (c × a) are also coplanar. Proof: As we know that, if a, b, c are coplanar vec-
tors, then [a, b, c] = 0 We have [(a × b), (b × c), (c × a)]
= [a, b, c]2 = 0
Thus, (a × b), (b × c), (c × a) are coplanar. (vii) For any vector a, i × (a × i) + j × (a × j) + k × (a × k) = 2a Proof: We have i × (a × i) = (i ◊ i) a – (i ◊ a) i = a – (i◊a) i = a – a1i, where a = a1i + a 2 j + a3k Similarly, j × (a × j) = a – a2 j and
k × (a × k) = a – a3k
Thus,
i × (a × i) + j × (a × j) + k × (a × k)
= a – a1i + a – a2 j + a – a3k
= 3 a – (a1i + a2 j + a3 k) = 3 a – a = 2 a (viii) If a and b be two non-zero vectors, then i × ((a × b) × i) + j × ((a × b) × j)
+ k × [(a × b) × k] = 2 (a × b).
Proof: Replace a by a × b in property (vii), we get the required result. (ix) If a × (b × c) = (a × b) × c, then (c × a) × b = 0 Proof: We have
a × (b × c) = (a × b) × c
fi
a × (b × c) = – c × (a × b
fi
(a ◊ c) b – (a ◊ b) c = – (c ◊ b) a + (c ◊ a) b
fi
– (a ◊ b) c = – (c ◊ d) a
fi
(c ◊ b) a – (a ◊ b) c = 0
fi
(c × a) × b = 0
(x) If a, b, c and d are four coplanar vectors then, [a, b, c] = [a, b, d] + [b, c, d] + [c, a, d]
Proof: Let the position vectors of the points A, B, C and D are a, b, c and d, respectively. Thus, AB = b – a, AC = c – a, AD = d – a Given A, B, C and D are coplanar, so, AB ◊ AC, AD are coplanar Therefore, AB ◊ (AC × AD) = 0 fi (b – a) ◊ ((c – a) × (d – a)) = 0 fi (b – a) ◊ (c × d – c × a – a × d) = 0 fi (b ◊ c × d) – (b ◊ c × a) – b ◊ (a × d) – (a ◊ c × d) = 0 fi [b, c, d] – [b, c, a] – [b, a, d] = [a, c, d] fi [b, c, d] – [b, a, d] – [a, c, d] = [b, c, a] fi [b, c, a] + [a, b, d] + [c, a, d] = [b, c, a] fi [b, c, a] = [b, c, d] + [a, b, d] + [c, a, d] fi [a, b, c] = [a, b, d] + [b, c, d] + [c, a, d] Hence, the result.
5.5 Scaler product
of
four vectors
If a, b, c and d are four non-zero vectors, then the scalar product of (a × b) and (c × d) is called scalar product of four vectors and is denoted as (a × b) ◊ (c × d) and is defined as (a ⋅ c) (a ⋅ d) (b ⋅ c) (b ⋅ c)
(a × b)◊(c × d) =
We have
(a × b) ◊ (c × d) = (a × b) ◊ p, [Let p = (c × d)]
= a ◊ (b × p) = a ◊ (b × (c × d))
= a ◊ ((b ◊ d) c – (b ◊ c) d)
= (a ◊ c) (b ◊ d) – (a ◊ d) (b ◊ c)
=
(a ⋅ c) (a ⋅ d) (b ⋅ c) (b ⋅ c)
5.5.1 If a and b lie in a plane normal to the plane containing c and d, then (a × b) ◊ (c × d) Proof: it is given that (c × d) is perpendicular to the plane containing c and d. But a and b are normal to the plane containing c and d Thus, c and d lie in the plane containing a and b. Also (a × b) is perpendicular to the plane containing a and b. Therefore, (a × b) is perpendicular to (c × d) also. Hence,
(a × b) ◊ (c × d) = 0
Vectors
5.5.2 If a, b, c and d are four non-zero vectors, then (a × b) ◊ (c × d) + (b × c) ◊ (a × d) + (c × a)◊(b × d) =0 Proof: We have, (a × b) ◊ (c × d) + (b × c) ◊ (a × d) + (c × a) ◊ (b × d) =
(a ⋅ c) (a ⋅ d) (b ⋅ a) (b ⋅ d) (c ⋅ b) (c ⋅ d) + + (b ⋅ c) (b ⋅ d) (c ⋅ a) (c ⋅ d) (a ⋅ b) (a ⋅ d)
= (a ◊ c) (b ◊ d) – (a ◊ d) (b ◊ c)
+ (b◊a) (c◊d) – (c◊a) (b◊d)
+ (c◊b) (a◊d) – (a◊b) (c◊d)
= 0
the plane containing a and b and the other one parallel to the plane containing c and d. 5.7.1 If a, b, c and d are four non-zero vectors, then a × [b × (c × d)] = (b ◊ d) (a × c) – (b◊c) (a × d) Proof: We have,
a × [b × (c × d)]
= a × [(b ◊ d) c – (b ◊ c) d]
= (b ◊ d) (a × c) – (b ◊ c) (a × d) Hence, the result. 5.7.2 If a, b, c and d are coplanar vectors then (a × b) × (c × d) = 0 Proof: As we know that (a × b) is perpendicular
5.6 Vector product
of
four vectors
If a, b, c and d are four non-zero vectors, the vector product of (a × b) and (c × d) is called the vector product of four vectors and is denoted as (a × b) × (c × d) and is defined as (a × b) × (c × d) = [a, b, d] c – [a, b, c] d
5.13
= [a, c, d] b – [b, c, d] a
to a and b vectors and (c × d) is also perpendicular to c and d vectors. Since a, b, c and d are coplanar vectors, so (a × b) and (c × d) are perpendicular to the same plane. Thus, (a × b) and (c × d) are parallel. Therefore, (a × b) × (c × d) = 0 5.7.3 If b, c and d are three non-coplanar vectors, then (a × b) × (c × d) + (c × d) + (a × d) × (b × c) is parallel to a.
Proof: We have,
Proof: We have,
(a × b) × (c × d)
(a × b) × (c × d) = [a, b, c] c – [a, b, c] d Again, (a × c) × (d × b) = – (d × b) × (a × c) = – [d, b, c] a + [d, b, a] c = – [b, c, d] a – [a, b, d] c Also, (a × d) × (b × c) = – (b × c) × (a × d)
= p × (c × d), [Let a × b = p]
= (p ◊ d) c – (p ◊ c) d
= ((a × b) ◊ d) c – ((a × b) ◊ c) d
= [a, b, d] c – [a, b, c] d
Also,
(a × b) × (c × d)
= (a × b) × q, [Let q = (c × d)]
= (a◊q) b – (b◊q) a
= [a◊(c × d)] b – [b◊(c × d)] a
= [a, c, d] b – [b, c, d] a Hence, the result.
5.7 Geometrical interpretation
of
(a × b) × (c × d )
(a × b) × (c × d) is a vector perpendicular to (a × b) and (c × d). But (a × b) is perpendicular to the plane containing a and b. Thus, (a × b) × (c × d) represents a vector coplanar with a and b. Similarly (a × b) × (c × d) represents a vector coplanar with c and d. Hence, (a × b) × (c × d) represents a vector parallel to the line of intersections of two planes, one being parallel to
= – [b, c, d] a + [b, c, a] d
= – [b, c, d] a + [a, b, c] d
...(i)
...(ii)
...(iii)
Adding Eqs. (i), (ii) and (iii), we get (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) = – 2 [b, c, d] a Therefore, (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) is parallel to a.
5.8 Reciprocal system
of
vectors
Let a, b and c be three non-zero vectors such that [a, b, c] π 0. The reciprocal system of vectors are denoted as a¢, b¢, c¢ and are defined as b×c c × a a×b a¢ = _______ , b¢ = _______ and c¢ = _______ . [a, b, c] [a, b, c] [a, b, c]
5.14 Integral Calculus, 3D Geometry & Vector Booster
Properties
of
reciprocal system
of
v ectors
If a, b, c and a¢, b¢, c¢ be reciprocal system of vectors, then (i) a ◊ a¢ = 1 = b ◊ b¢ = c ◊c¢ Proof: We have,
(
)
(b × c◊a) c – (b × c ◊ c) a = ____________________ [a, b, c]2
a ◊ (b × c) = _________ [a, b, c]
[a, b, c] c – 0 = ____________ [a, b, c]2 c = _______ [a, b, c] a Similarly, b¢ × c¢ = _______ [a, b, c]
[a, b, c] = ________ = 1 [a, b, c] Similarly, b ◊ b¢ = 1 and c ◊ c¢ = 1 (ii) a ◊ a¢ + b ◊ b¢ + c ◊ c¢ = 0 (iii) a ◊ b¢ = 0 = a ◊ c¢, b ◊ c¢ = 0 = b ◊ a¢ and c ◊ a¢ = 0 = c ◊ b¢
(
)
c×a a ◊ b¢ = a ◊ _______ [a, b, c]
(
)
(
(
)
)
[a¢, b¢, c¢] = a¢◊(b¢ × c¢)
)
c×a b×c a×b = _______ ◊ ________ × _______ [a, b, c] [a, b, c,] [a, b, c] 1 = ________ [a × b, b × c, c × a] [a, b, c]3 [a, b, c,]2 = _________ [a, b, c]3 1 = _______ [a, b, c] Thus, [a¢, b¢, c¢] [a, b, c] = 1 (v) a × a¢ + b × b¢ + c × c¢ = 0 a+b+c (vi) a¢ × b¢ + b¢ × c¢ + c¢ + a¢ = _________ [a, b, c] Proof: We have,
b×c c×a a¢ × b¢ = _______ × ________ [a, b, c] [a, b, c,]
Hence,
a + b + c a¢, × b¢ + b¢ × c¢ + c¢ × a¢ = _________ [a, b, c]
(x) If a, b and c be three non-coplanar vectors and a¢, b¢ and c¢ be its reciprocal system of vectors, then any vector r can be expressed as r = (r ◊ a¢) a + (r ◊ b¢) b + (r ◊ c¢) c
Proof: We have,
(
b c¢ × a¢ = _______ [a, b, c]
(ix) If a¢, b¢ and c¢ be the reciprocal system of a, b and c, then a, b and c is the reciprocal system of a¢, b¢ and c¢,
a×b a◊c¢ = a ◊ _______ [a, b, c]
a ◊ (a × b) = _________ = 0 [a, b, c] (iv) [a, b, c] [a¢, b¢, c¢] = 1
and
(vii) (a + b + c) ◊ (a¢+ b¢ + c¢) = 3. (viii) The system of unit vectors i, j, k be its own reciprocal.
a ◊ (c × a) = _________ = 0 [a, b, c] Also,
d × (c × a) = __________ [a, b, c]2 (d ◊ a) c – (d ◊ c) a = ______________ [a, b, c]2
b×c a ◊ a¢ = a ◊ _______ [a, b, c]
Proof: We have
(b × c) × (c × a) = _______________ [a, b, c]2
or r = (r ◊ a) a¢ + (r ◊ b) b¢ + (r ◊ c) c¢ Proof: Since a, b, c be three non-coplanar vectors, any vector r can be expressed as a linear combination of a, b and c. Thus, r = x a + y b + z c, where x, y, z Œ R Taking dot product of (b × c), we have r ◊ (b × c) = x a ◊ (b × c) + y b ◊ (b × c) + z c ◊ (b × c) fi r ◊ (b × c) = x a ◊ (b × c) + 0 + 0 = x a ◊ (b × c) fi r ◊ (b × c) = x a ◊ (b × c) r ◊ (b × c) r ◊ (b × c) fi x = ________ = ________ a ◊ (b × c) [a, b, c] fi x = r ◊ a¢ Similarly, y = r◊b¢, = z = r ◊ c¢ (xi) If a, b and c be three non-coplanar vectors, any vector r can be expressed as a linear combination of a, b and a × b, i.e. r = x a + y b + z (a × b).
Vectors
5.15
Exercises (Problems based on Fundamentals)
DOT PRODUCT OF VECTORS 1. If a = 3 i – 2 j + k and b = 2 i + 3 j, find the angle between a and b. 2. If a = i + 2 j – 3 k and b = 3 i + j + 2 k, prove that (a + b) is perpendicular to (a – b). 3. If |a| = 3, |b| = 4, |c| = 5 such that (a + b + c) = 0, find the value of (a ◊ b + b ◊ c + c ◊ a). 4. If a, b and c are mutually perpendicular unit vectors, find |a + b – c|. 5. Find the projection of the vector a = 3 i + j + 2 k on the vector b = j – 2 j = i – j. 6. Find the angle between the vectors a and b such that 2a + b = i + j and a + 2 b = i – j. 7. Find the unit vector perpendicular to each of the vectors i – 2 j + k and 2 i + j – 3 k. 8. Find a unit vector in xy-plane that makes an angle of 45° with the vector i + j and an angle of 60° with the vector 3 i – 4 j. 9. If a, b and c be mutually perpendicular vectors of equal magnitude, show that (a + b + c) is equally inclined to a, b and c. ___ 10. Find a vector of magnitude ÷51 which makes equal angles with the vectors 1 1 a = __ (i – 2 j + 2 k), b = __ (– 4 i – 3 k) and c = j. 3 5 11. Find a vector of magnitude 4, which is equally inclined to vectors a = i + j, b = j + k and c = k + i. 12. Find a unit vector in the plane of (i + 2 j + k) and (i + j + 2 k) and perpendicular to (2 i + j + k). 13. If and are unit vectors and q the angle between q 1 them, show that sin __ = __ | – |. 2 2 14. If , and are unit vectors perpendicular to each other, find the value of ( – ) ◊ ( – ). 15. Let and be two unit vectors and q the angle between them. If ( + ) is a unit vector, find sin (q). 16. If and be two unit vectors inclined to x-axis at angles 30° and 120° respectively, find the value of | + |. 17. If a and b are two vectors and their lengths are a a b 2 a–b2 and b, show that __ 2 – __ 2 = _____ ab a b 18. Let u = i + j, v = i – j, and w = i + 2 j + 3 k. If is a unit vector such that u ◊ = 0 and v ◊ = 0, then find |w ◊ |.
( )
(
) ( )
19. If a, b and c be unit vectors such that a is perpendicular to the plane of b and c and the angle between p b and c is __ , find |a + b + c|. 3 20. If |a| = 3, |b| = 4, |c| = 5 and a ^ (b + c) b ^ (c + a) and c ^ (a + b), find |a + b + c|. 21. Let and are two unit vectors and the angle between them is 60°, find the value of | + / – |. ___
17 ÷ 22. If |a| = 1, |b| = 4, |c| = ____ such that (a + b + 3 c) 3 = 0, find the value of (a ◊ b + b ◊ c + c ◊ a). 23. Constant forces F1 = i – j + k, F2 = – i + 2 j – k and F3 = j – k act on a particle at a point A. Find the work done when the particle is displaced from the point A to B where A = 4 i – 3 j – 2 k and B = 6 i + j – 3 k. 24. Prove that, by vector method, b2 + c2 – a2 cos (A) = ___________ . 2bc 25. 26. 27.
Prove that, by vector method, a = b cos (C) + c cos (B). Prove that, by vector method, cos (A – B) = cos A ◊ cos B + sin A ◊ sin B Prove that, by vector method,
cos (A + B) = cos (A) cos (B) – sin (A) sin (b)
Cross product of two vectors 28. If a = 2 i – j + k, and b = i – 3 j – 5 k, find a × b. 29. Find the area of a triangle whose adjacent sides are determined by the vectors a = 3 i + 4 j and b = 3 i – 3 j + k. 30. Find the area of a parallelogram whose adjacent sides are determined by the vectors a = i + 2 j + 3 k and b = 3 i – 2 j + k. 31. Find the area of a parallelogram whose diagonals are given by the vectors a = i + j + k and b = 3 i – 4 j + k. 32. Find the area of a triangle whose vertices are (3, – 1, 2), (1, – 1, – 1) and (4, – 3, 1). 33. Find a unit vector perpendicular to each of the vectors 2 i – j + k and 3 i – 4 j – k. 34. Prove that a × (b + c) + b × (c + a) + c × (a + b) = 0. 35. If a × b = c × d and a × c = b × d, show that (a – d) is parallel to (b – c). 36. If (a + b + c) = 0, prove that a × b = b × c = c × a.
5.16 Integral Calculus, 3D Geometry & Vector Booster 37. Prove that |a × |2 + |a × |2 + |a × |2 = 2a2. 38. Let , and be three unit vectors such that p a ◊ b = a ◊ c = 0 and the angle between b and c is __ , 6 prove that a = ± 2 (b × c). 39. If a × b = a × c, a π 0 and b π c, prove that b = c + a, " Œ R. 40. Let A = 2 i + k, B = i + j + k and C = 4 i – 3 j + 7 k. Find a vector R satisfying R × b = R × C and R ◊ A = 1. 41. Find the number of vectors of unit length perpendicular to the vectors a = i + j and b = j + k. 42. Let a = x i + y j + z k and b = j, find a vector c for which a, b and c form a right handed system. 43. Let a = i + j and b = 2 j – k. Find the point of intersection of the lines r × a = b × a and r × b = a × b. 44. If a and b are two non-zero vectors such that r ◊ a = 0 and r × a = b, find r. 45. Let a = i + j + k, c = j – k such that a ◊ b = 3 and a × b = c, find b. 46. Find the angle between the vectors [a – (a ◊ b) b] and (a × b). 47. Find the area of a square whose one diagonal is (3 i + 4 j). 48. A constant force F = 3 i + 2 j – 4 is applied at the point (1, – 2, 2). Find the vector moment of F about the point (2, – 1, 3). 49. Forces (5 i + k) and (– 5 i – k) act at the points P (9, – 1, 2) and Q (3, – 2, 1), respectively. Find the moment of the couple. 50. Find the torque about the point 2 i + j + k of a force 4 i + k acting through the point (i – j + 2 k). 51. In a D ABC, a = BC, b = CA and c = AB, by using vector method, prove that a b c ____ = ____ = _____ . sin A sin B sin C
Find [a, b + c, a + b + c]. If a, b, and c be three non-coplanar vectors, find [a + b + c, a + b, a + c]. Find the value of [a – b, b – c, c – a]. If a, b and c are three non-zero and non-coplanar vectors such that a ◊ (b × c) = |a| |b| |c|, find the angle between a and b. 60. For non-zero vectors a, b and c, if a ◊ (b × c) = 4, 1 find the value of __ (b × c) ◊ (a + b + c). 2 56. 57. 58. 59.
61. If the vectors 2 i – j + l k, i – j + 2 k and 3 i – 2 j + k are coplanar vectors, find the value of l. 62. Find the volume of a parallelopiped whose coterminous edges are a = 3 i – 2 j + k, b = i + j + k and c = i + j – 2 k. 63. Find the volume of a parallelopiped whose coterminous edges are a + b = i + j – k, b + c = 2 i + 3 j – 4 k and c + a = 3 i + 4 j – 7 k. 64. The volume of a parallelopiped determined by the vectors a, b and c is 5. Find the volume of the parallelopiped determined by the vectors 3 (a + b), (b + c) and 2 (c + a). 65. If the volume of the parallelopiped formed by the vectors i + a j + k, j + a k and a i + k, a > 0, is minimum, find the value of a. 66. Let V1 be the volume of the parallelopiped formed by three non-zero vectors a, b and c and V2 be the volume of the parallelopiped formed by the vectors p = a – 2 b + 2 c, q = 3 a – b + c and r = a – b + V1 2 c. Find ___ . V2
( )
67. If the vectors a i + j + k, i + b j + k and i + j + c k, (a, b, c π 1) are coplanar, then find the value 1 1 1 of _____ + _____ + _____ . 1–a 1–b 1–c 68. For any three vectors a, b and c, prove that
52. Prove, by the vector method,
sin (A – B) = sin A cos B – cos A sin B.
53. Prove, by vector method,
sin (A + B) = sin A cos B + cos A sin B
SCALER TRIPLE PRODUCT OF VECTORS 54. For any three vectors a, b and c, prove that [a + b, b + c, c + a] = 2 [a, b, c]. 55. If ,
and
be unit coplanar vectors, find
[2a – b, 2b – c, 2c – a].
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a c⋅b c⋅c 2
69. The edges of a parallelopiped are of unit length and parallel to non-coplaner unit vectors a, b and c such 1 that a ◊ b = b ◊ c = c ◊ a = __ . Find the volume of the 2 parallelopiped. 70. If a and b be non-zero and non-collinear vectors, prove that [a, b, i] i + [a, b, j] j + [a, b, k] k = a × b. 71. If a, b, c, m and n are non-zero vectors, prove that
Vectors
a b c [a, b, c] (m × n) = a ◊ m b ◊ m c ◊ m a ◊n b◊n c ◊n 72. If a vector r can be expressed as a linear combination of three non-coplanar vectors a, b and c, prove that [r b c] r = ______ a + [a b c]
[r c a] ______ b+ [a b c]
[r a b] ______ c [a b c]
73. If a and b be unit vectors such that 1 [a, b, c] = __ , find the angle between a and b. 4 74. If a = x (a + b) + y (b × c) + z (c × a) and [a, b, c] 1 = __ , find (x + y + z). 8 75. Find the volume of the tetrahedron with edges a = i + j + k, b = i – j + k and c = i + 2 j – k. 76. If a, b and c are unit vectors satisfying |a – b|2 + |b – c|2 + |c – a|2 = 9, find |2a + 5b + 5c|. 77. The three vectors i + j, j + k, k + i taken two at a time form three planes. The three unit vectors drawn perpendicular to three planes form a parallelopiped. 4 Prove that its volume is ___ __ c.u. 3 ÷ 78. If a, b and c are three non-coplanar vectors such that a + b + c = a d and b + c + d = b a, find a + b + c + d + 4.
VECTOR TRIPLE PRODUCT 79. Find a unit vector which is orthogonal to 3 i + 2 j + 6 k and is coplanar with the vectors 2 i + j + k and i – j + k. 80. If a = i + j + k, a ◊ b = 1, a × b = j – k, find b. 81. If a, b and c are three non-parallel unit vectors such 1 that (a × (b × c)) = __ b, find the angles which a 2 makes with b and c. 82. If a, b and c are three non-coplanar unit vectors such that 1 (a × (b × c)) = ___ __ (b + c), 2 ÷ where b and c are non parallel, find the angle between a and b. 83. If a × (a × c) + b = 0 such that |a| = 1 |b| and |c| = 2, find the angle between a and c.
If q be the angle between b and c, find the value of sin q. 86. If a and b be mutually perpendicular unit vectors satisfying r ◊ a = 0, r ◊ b = 1 and [r, a, b] = 1, find r. 87. If |a| = 7, |b| = 1 = |c| and a × (b × c) + b × (c × a) 1 = __ a, where a and b are non-collinear vectors, find 2 |a × c|. 88. Simplify: (d + a) ◊ a × b × (c × d)]. 89. Solve: p x + (x × a) = b. 90. If [x a b] = 0, x ◊ a = 0, x ◊ b = 1, a ◊ b = 0, find x. 91. Find the value of [a × (b + c), b × (c – 2a), c × (a + 3b)] if [a, b, c] = 2. 92. Find the value of [(a – b), (a – b – c), (a + 2b – c)] if [a, b, c]. 93. Find the value of [(2a × 3b), (3b × 4c), (4c × 2a)] if [a, b, c].
Vector equations 94. Solve: x × b = a × b, where a and b are non-zero and non-collinear vectors. 95. Solve: x × a = b and x ◊ a = 0. 96. Solve: x × b = a × b and x ◊ c = 0, where c is not perpendicular to b. 97. Solve: x a + y b + z c = d. 98. Solve: x + y = a, x × y = b, x ◊ a = 1
84. If a × (b × c) is perpendicular to (a × b) × c, prove that (a ◊ c) = 0. 85. Let a, b and c be three non-zero vectors such that
1 (a × b) × c = __ |b| |c| |a. 3
5.17
_›
Mixed Problems (More than one options are correct) _ ›
› _› _
1. If a and b are two vectors such that a , b < 0 and › _› _
_›
_ ›
_›
_ ›
|a ◊ b | < |a × b | , the angle between a and b is p (a) p (b) __ 3 p 3p __ ___ (c) (d) 4 _ 4 _› › _› 2. Let a , b and c are three vectors of equal magnitudes. p The angle between each pair of vectors is __ such that 3 _ __ › _› _› |a + b + c| = ÷ 6 , the value of (|a | + 2) is (a) 2 (b) 3 (c) 1 (d) 0 3. If a, b and c are three unit vectors inclined to each other at angle q, the maximum value of q is p p (a) __ (b) __ 3 2 2p 5p ___ ___ (c) (d) 3 6 4. The number of vectors of unit length perpendicular _ › _› to vectors a = (1, 1, 0) and b = (0, 1, 1) is
5.18 Integral Calculus, 3D Geometry & Vector Booster
(a) 1 (b) 2 (c) 3 (d) infinite 5. Let a, b and g be distinct real numbers. The points with position vectors a , b , g , b , g , a and g , a , b (a) are collinear (b) form an equilateral triangle (c) form a scalene triangle (d) form a right-angle triangle. _›
_›
6. If P and Q have position vectors a and___ b relative to › the origin O such that X and Y divide PQ internally and externally respectively in the ratio 2 :1, the ___› vector XY is _ 3 _ › _› 4 _› › (b) __ (a – b ) (a) __ (b – a ) 3 2 _ 5 _› _› 4 › _› __ __ (c) (b – a ) (d) (b – a ) 3 6 7. If _the_› three points with position vectors _› _ _ › › _› › (1, a , b )_, (a , _›2, b ) and (a , b , 3) are collinear, the › value of a + b is (a) 1 (b) 4 (c) 5 (d) none. 8. The acite angle between the medians drawn from the acute angle of an isosceles right-angled triangle is 3 2 (a) cos– 1 __ (b) cos– 1 __ 3 4 4 2 – 1 __ – 1 __ (c) cos (d) cos 3 5
( ) ( )
_›
( ) ( )
_›
_›
_›
__›
12. Let u = + , v = – and w = – 2 + 3 _ __› › If is a unit vector such that u ◊ , the value of |w ◊ | is (a) 1 (b) 2 (c) 3 (d) 0 _›
_ ›
_ ›
_ ›
_›
_ ›
13. If a + b _+ c = _0, |a | = 3, |b | = 5, |c | = 7, the angle › › between a and b is p 2p (a) __ (b) ___ 3 6 5p p (c) ___ (d) __ 3 3 _› _›
_›
14. Let a , b and c be three vectors of the lengths 3, 4 _› _› _› _› and 5 respectively. Let a _be perpendicular to b _+ c , b › › _› _› _› _› _› _› to a + c and c to a + b , the value of |a + b + c | is __ __ (a) 2 ÷5 (b) 2 ÷2 __
__
(c) 10 ÷5
(d) 5÷2
_›
15. If a vector a of length 50 is collinear with the _› vector b = 6 – 8 – and makes an acute angle with the positive z-axis, then _›
_›
_›
_›
(a) a = 4b
(c) b = 4 a
_›
(b) a = – 4 b (d) none.
_ › 16. Let and b be two non-parallel unit vectors in a _ › _› plane. If (a a + b ) bisects the internal angle between _› _› a and b , then a is _› a
9. If e 1 and e 2 are two unit vectors and q the angle q between them, then cos __ is 2 _› _› 1 _› 1 _› __ (a) |e 1 + e 2| (b) __ |e 1 – e 2| 2 2 _› 1 _› 1 _› _› __ __ (c) |e 1 × e 2| (d) |e 1 ◊ e 2| 2 2
(a) 1/2
(b) 1
(c) 2
(d) – 1
10. If the vectors (3 p + q ) , (5 p – 3 q ) and (2 p + q ) , _› _› (4 p – 2 q ) are pairwise mutually perpendicular vectors _› _› such that q the angle between p and q , the value of sin (q ) is
( )
_›
___
÷55 (a) ____ 4 3 (c) ___ 16
_›
_›
_›
_›
_›
___
÷55 (b) ____ 8 ____ 247 ÷ (d) _____ 16
11. The set of values of c for which the angle between the vectors c x – 6 + 3 , x – 2 + 2c x is acute for every x in R is 4 4 (a) 0, __ (b) 0, __ 3 3 11 4 4 (c) ___ , __ (d) 0, __ 9 3 3
( ) ( )
[ ] [ )
17. Let , and are three unit vectors such that ( + + ) is also a unit vector. If pairwise angle between , , are q1, q2 and q3, spectively, the value of cos (q1) + cos (q 2) + cos (q3) is (a) 3 (c) 1
_› _›
(b) – 3 (d) – 1 _›
18. Given three vectors a , b and_ c each two of which are › _› _› non-collinear. Further (a + b ) is collinear with c and _› _› _› _› _› (c +__ b ) is collinear with_ a and each length of a , b _ › _› _› › _› _› is ÷ 2 . The value of (a ◊ b + b ◊ c + c ◊ a ) is
(a) 3
(b) – 3
(c) 0
(d) not defined.
19 ___ (a) – _____ 5÷43
(b) 0
19.
_ _ _ › _ › _ › _› _› _› › _› _› › If p =_ 3a – 5b , q = 2a + b , r = a + 4b and s = _ _ _ › › › › – a + b are four vectors such that sine of ( p and q ) _ _ › › is 1 and sine of (r and s ) is also 1, the value of _ › _ › cosine of (a and b ) is
Vectors
19 ___ (d) _____ 5÷43
(c) 1
_› _ › 20. If a and b are non-zero and non-collinear and the _› _› _ _ › › linear combination (2x – y) a + 4b = 5 a + (x – 2y) b
holds good for all real x and y, then the value of x + y is (a) – 3 (b) 1 (c) 17 (d) 3. _ _ › › 21. If p the position vector of the orthocentre and q the position vector of the centroid of _the triangle ABC _ › › and origin O is the circumcentre. If p = k q , the value of k is (a) 3 (b) 2
(c) 1/3
(d) 2/3
(c) 275 (d) 300 28. The set of values of m for which the vectors + m , + + (m + 1) and coplanar, is
–
5.19 +
+ m are non-
(a) R
(b) R – {1}
(b) R – {– 2}
(d) j
_ › 29. Given a = x + y + z , _ › _ _ › › _ › b = – + , c = + 2 , (a ◊ c ) = _› _ › p between a and b is __ , then 2 _› _› _› 2 _› _› _› (a) [a , b , c ] = |a | (b) [a , b , _ _ › _› _ › 2 › _› (c) [a , b , c ] = 0 (d) [a , b ,
4 and the angle _ _› 2 › c ] = |a | _ _ › › c ] = |a | 2
22. If the vectors – 2x – 3y and +3x + 2y are orthogonal to each other, the locus of the point is (a) a/an circle (b) ellipse (c) a/an parabola (d) straight line.
30. Let a, b, c be three distinct non-negative numbers. If the vectors a + a + a , + and c + c + b lie in a plane, then c is
23. A tangent is drawn to the curve x2 y = 8 at a point A (x1, y1) where x1 = 2. The tangent cuts the x-axis
(a) AM of a and b
(b) GM of a and b
(c) HM of a and b
(d) equal to zero.
___› ___›
at a point B. The value of (AB ◊ OB ) is
(a) 3 (c) 6
(b) – 3 (d) – 6.
_ _ › › 24. If e 1 and e 2 be two unit vectors and q the angle
( )
q between them, then sin __ is 2
_› 1 _› (a) __ |e 1 + e 2 | 2 _ › 1 _ › (c) __ |e 1 × e 2| 2
_› 1 _› (b) __ |e 1 – e 2 | 2 1 _ › _ › (d) __ |e 1 ◊ e 2| 2
_ › _ _› _› _ _ _ _ › › › › › 25. If u = a – b , v = a + b and |a | = 2 = |b | _› _› _› _› = (a ◊ b ) , the value of |u × v | is
(a) 0 (c) 4
(b) 2 (d) 16.
(a) (5, 2, 2)
(
)
2 5 __ 2 (c) __ , __ , 3 3 3
( ) 5 2 2 __ (d) ( __ , __ , . 3 3 3)
5 2 __ 2 (b) __ , __ , 3 3 3
_›
_›
__›
_›
_›
__›
_›
__›
_›
_›
_›
__›
(a) u ◊ (v × w )
(b) (v × w ) ◊ u
(c) v ◊ (u × w )
(d) (u × v ) ◊ w
32. The vector perpendicular to the vectors _› _› a = (2, – 3, 1), b = (1, – 2, 3) and satisfies the _›
_›
condition c ◊ ( + 2 + 7 ), the vector c is
(a) (7, 5, 1)
(b) (– 7, – 5, 1)
(c) (1, 1, – 1)
(d) (1, 2, 3)
_ › _› _› 33. If a = ( + + ) and b = ( + 2 + ), the vector c _ › _ _› _› › _› such that a ◊ c = 2 and a × c = b is
1 1 (a) __ ( 3 + 2 + 5 ) (b) __ ( – + 2 + 5 ) 3 3 1 1 (c) __ ( + 2 + 5 ) (d) __ ( 3 + 2 + ) 3 3 34. The altitude of a parallelopiped whose coterminous
_›
(a) 225
__›
and C = ( +
_ ›
_ ›
+ 3 ) with A and B as the sides of
the base of the parallelopiped, is
(b) 250
_›
edges are the vectors A = ( + + ) B = (2 + 4 – )
_ _ › › _› _› 27. If |a | = 1, |b | = 2 and and the angle between a and b _› _› _ _ › › 2p is ___ , the value of {(a + 3 b ) × (3 a – b ) }2 is
3
31. For three vectors which of the following expressions is not equal to any of the following three?
_› _ › 26. If a = (1, 1, 1) are given vectors, then a vector b _› _ _ _ › › › _› satisfying the conditions a × b = c and a ◊ b = 3
is
_› _ › __› u , v , w ,
2 ___ (a) ____ ÷19
4 ___ (b) ____ ÷19
3 ___ (c) ____ ÷19
2÷2 ___ (d) ____ . ÷19
__
5.20 Integral Calculus, 3D Geometry & Vector Booster 2. Let a and b are two vectors satisfying the conditions a × b = 0. Does it imply that one of the vectors a + a + b are coplanar and a, b and c are distinct, and b must be a null vector. Give one example in then support of your answer. (a) a3 + b3 + c3 = 1 (b) a + b + c = 1 [Roorkee-JEE, 1984] 3. Constant forces P = 2 i – 5 j + 6 k and Q = – i + 1 __ 1 __ 1 (c) __ (d) a + b + c = 0. 2 j – k act on a particle. Determine the work done a + b + c = 1 when the particle is displaced from a point A with _› _› __› _› _› __› If u , v , w are vectors such that u + v + w = 0, then position vector 4 i – 3 j – 2 k to a point B with a _› _› _› __› __› _› position vector 6 i + j – 3 k. Îu + v , v + w , w + u ˚ is [Roorkee-JEE, 1984] (a) 1 (b) 0 4. Prove that for any two vectors a and b (c) –1 (d) None. _ (a × b)2 = a2b2 – (a ◊ b)2 _› › _› If a , b and c are three mutually perpendicular [Roorkee-JEE, 1985] _ _ _ › _ › _ › › _ › _› › _ › _ › vectors, then (r ◊ a ) a + (r ◊ a ) b + (r ◊ a ) c is 5. If (a × b) = (b × c) π 0, show that a + c = k b, where _› _› k is a scalar. (a) r (b) 2 r [Roorkee-JEE, 1985] _› _› (c) – r (d) – 2 r 6. Find the value of the constant S such that the scalar _› _› product of the vector (i + j + k) with the unit vector Let a = + and b = 2 – . The point of intersec_ _ _ parallel to the sum of the vectors (2i + 4 j – 5k) and › › › _› _› _› _› _› tion of the lines r × a = b × a and r × b = a × b (S i + 2 j + 3k) is equal to 1. [Roorkee-JEE, 1985] is 7. Find a such that the vectors (2 i – j + k), (i + 2 j – 3k) (a) (3, – 1, 1) (b) (3, 1, – 1) and (3i + a j + 5k) are coplanar. (c) (– 3, 1, 1) (d) (– 3, – 1, – 1) [Roorkee-JEE, 1986] _› _› _› _› _› _› _› 8. If a and b are non-null vectors and |a + b| = |a – b|, If |a | = 1 = |b | and |c | = 2 and a × (a × c ) + b _ show that a and b are perpendicular to each other. _› › _› = 0 , the acute angle between a and b is [Roorkee-JEE, 1986] ___ p p 9. Find a vector of magnitude 51 which makes equal ÷ (a) __ (b) __ 4 6 1 angles with the three vectors a = __ (i – 2 j + 2k), b p __ 3 (c) (d) p 2 1 __ = (– 4i – 3k) and c = j. _ › _› 5 If b and _c are two mutually perpendicular unit vec[Roorkee-JEE, 1987] › tors and a is any vector, then _ › 10. Three vectors a = (12, 4, 3), b = (8, – 12, – 9) and _› _ _ ( › _› _› › _ › _ › a ◊ b × c ) ( _› _ › ) c = (33, – 4, – 24) define a parallelopiped. Evaluate b × c is (a ◊ b ) b + (a ◊ c ) c + _________ _› _ ( b × c › 2 ) the length of its edges, area of the faces and its volume. _› (a) 0 (b) a [Roorkee-JEE, 1988] _ › _› __ a (c) 2 a (d) b × c c × a 2 11. It is given that x = _______ , y = _______ and [a, b, c] [a, b, c] a × b z = _______ , where a, b and c are non-coplanar [a, b, c]
35. If the vectors a + b + c , b + c + a and c
36. 37. 38.
39.
40.
1. Three force vectors P, Q and R of 15 KN each acts along AB, BC and CA, respectively. The position vectors OA, OB and OC are given in metres as OA = 2 i – 4 j + 3 k, OB = 5 i + 3 j – 2k and OC = – 2 i + 2 j + 3 k. Find the resultant force vector S of the vectors P, Q and R. [Roorkee-JEE, 1983]
vectors. Show that x, y and z also form a non-coplanar system. Find the value of x ◊ (a + b) + y ◊ (b + c) + z ◊ (a + b).
[Roorkee-JEE, 1989]
12. It is given that r × b = c × b, r ◊ a = 0, a ◊ b π 0 What is the geometrical meaning of these equations
Vectors
separately? If the above three statements hold good simultaneously, determine the vector r in terms of a, b and c. [Roorkee-JEE, 1989] 13. Let a, b, c are three unit vectors, a + b + c = 0. Then a ◊ b + b ◊ c + c ◊ a is (a) – 3/2 (b) 0 (c) – 1 (d) 1 [Roorkee-JEE, 1990] 14. Let a = x i + y j + z k, b = j. For what c, a, b, c form a right handed system? (a) 2 i – x k (b) 0 (c) – z i × x k (d) y j [Roorkee-JEE, 1990] 15. Let a, b, c are in the same plane. Which of the following is correct? (a) a ◊ b × c = 0 (b) a ◊ b × c = 1 (c) a ◊ b × c = 2 (d) a ◊ b × c = 3
[Roorkee-JEE, 1990]
16. Given that the vectors A, B, C form a triangle such that A = B + C. Find__a, b, c and d such that the area of the triangle is 5 ÷6 , where A = a i + b j + c k, B = d i + 3 j + 4 k, C = 3 i + j – 2 k. [Roorkee-JEE, 1990] 17. If a and b be two unit vectors, the vector (a + b) × (a × b) is (a) perpendicular to (a – b) (b) parallel to (a – b) (c) equal to 2 (a – b) (d) equal to 2 (a – b) [Roorkee-JEE, 1991] 18. If the position vectors of points A and B with respect to origin O are a = 2 i – j + 2 k and b = i + 2 j – 2k, respectively, the projection of the vector a + b + a × b on a line perpendicular to the plane OAB is (a) 15 (b) 10
___ (c) ÷75
___ (d) ÷ 65
[Roorkee-JEE, 1991] 19. If two vectors a and b are such that a ◊ b = 0, the solution of x + a = b for all x (a) is unique (b) is possible only, when a = b (c) does not exist (d) will be infinitely many. [Roorkee-JEE, 1991]
5.21
20. Let be a unit vector and b be a non-zero vector not parallel to . Find the angles of the triangle, two sides of each are represented by the vectors __
÷ 3 ( × b) and b – ( ◊ b) [Roorkee-JEE, 1991] 21. Unit vector in the xy-plane that makes an angle of 45° with the vector i + j and an angle of 60° with the vector 3 i – 4 j is
(a) i
i – j __ (c) ____ 2 ÷
i+j __ (b) _____ 2 ÷ (d) none
[Roorkee-JEE, 1992] 22. Find the value of l that the vectors 2 i – j + k, i + 2 j – 3k and 3i + l j + 5k are coplaner (a) – 4 (b) 0 (c) 2 (d) 4 [Roorkee-JEE, 1992] 23. In parallelogram ABCD, the interrior bisectors of the consecutive angles B and C intersect at P. Find –BPC. [Roorkee-JEE, 1992] 24. If the sides of an angle are given by the vectors a = i – 2 j + 2k and b = 2i + j + 2k, the internal bisector is 1 (a) 3i + j + 4k (b) __ (3i – j + 4k) 3 1 (c) __ (– i – 3j) (d) 3i – j – 4k 3 [Roorkee-JEE, 1993] 25. Let a, b and c are three vectors such that a × b = c, b × c = a, then (a) a = 1, b = c (b) c = 1, a = b (c) b = 2, c = 2a (d) b = 1, c = a
[Roorkee-JEE, 1993]
26. Let a, b and c are in the same plane. Which of the following is correct? (a) a ◊ b × c = 0 (b) a ◊ b × c = 1 (c) a ◊ b × c = 2 (d) a ◊ b × c = 1 [Roorkee-JEE, 1994] 27. If two vectors a and b are such that a ◊ b = 0, the solution of x × a = b for all x. (a) is unique (b) is possible only, when a = b (c) does not exist (d) will be infinitely many.
[Roorkee-JEE, 1994]
5.22 Integral Calculus, 3D Geometry & Vector Booster 28. Solve the following system of simultaneous equations for vectors x and y x + y = a, x × y = b, x ◊ a = 1
(a) p – 4q
7 1 (b) __ p + __ r 5 5
(c) 2p – 3q – r
(d) – 4p – 2r
[Roorkee-JEE, 1998] 37. The vector c directed along the bisectors of the angle between the vectors a__= 7i – 4 j – 4k and b = – 2i – j + 2k, where |c| = 3 ÷6 is given by
[Roorkee-JEE, 1994]
29. The vector – i + j – k bisects the angle between vector c and 3 i + 4 j. Determine the unit vector along a.
1 (a) – ___ (3i – 8jz + 12k) 15 1 (b) – ___ ( 11i + 10j + 2k ) 15 1 (c) – ___ ( 3i + 8j + 12k ) 15 (d) none.
[Roorkee-JEE, 1995]
30. Find the scalers a and b if a × (b × c) + (a ◊ b) b = (4 – 2b – sin a) b + (b 2 – 1) c and (c ◊ c) a = c, where b and c are non-collinear. [Roorkee-JEE, 1995] 31. |(a × b)◊ c| = |a| |b| |c|, where a, b, c are non-zero vectors holds, if (a) a ◊ b = 0 (b) b ◊ c = 0
(c) c ◊ a = 0
(d) a ◊ b = b ◊ c = c ◊ a [Roorkee-JEE, 1996]
32. Let x, y and z be unit vectors such that x + y + z 3 a, x × (y × z) = b (x × y) × z = c, a ◊ x = __ , a ◊ y = 2
= 7 __ 4
and |a| = 2. Find x, y and z in terms of a, b, and c. [Roorkee-JEE, 1996] __
33. Vectors x, y and z each of magnitude ÷ 2 makes an angle of 60° with each other. If x × (y × z) = a, y × (z × x) = b and (x × y) = c, find x, y, z in terms of a, b and c. [Roorkee-JEE, 1997] 34. Which of the following is a true statement? (a) (a × b) × c is coplaner with c (b) (a × b) × c is perpendicular to a (c) (a × b) × c is perpendicular to b (d) (a × b) × c is perpendicular to c. 35.
[Roorkee-JEE, 1998] If A, B, C and D are any four points, then AB ◊ CD + BC ◊ AD + CA◊ BD is equal to (a) 1 (b) 0 (c) – 1 (d) None.
[Roorkee-JEE, 1998] 36. If p = 2a – 3b, q = a – 2b + c, r = – 3a + b + 2c, where a, b, c are non-zero are non-coplanar vectors, then the vector – 2a + 3b – c is equal to
(a) i – 7 j + 2k
(b) i + 7 j – 2k
(c) – i + 7 j – 2k
(d) i – 7 j – 2k
[Roorkee-JEE, 1998] 38. a, b, c three non-coplanar vectors such that r1 = a – b + c, r2 = b + c – a, r3 = c + a – b, and r = 2a – 2b + 4c, if r = x1r1 + x2r2 + x3r3, then
7 (a) x1 = __ 3 (c) x1 + x2 + x3 = 4
(b) x1 + x3 = 3 (d) x2 + x3 = 2
[Roorkee-JEE, 1998] 39. If x × y = a, y × z = b, x ◊ b = c, x ◊ y = 1 and y ◊ z = 1, find x, y and z in terms of a, b and c. [Roorkee-JEE, 1998] 40. If a, b, c are non-coplanar vectors and d is a unit vector, find the value of |(a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)| independent of d. [Roorkee-JEE, 1999] 41. If a = i + j – k, b = – i + 2 j + 2k and c = – i + 2 j – k, find a unit vector normal to the vectors a + b and b – c. [Roorkee-JEE, 2000] 42. Given that the vectors a and b are perpendicular to each other, find vector v in terms of a and b satisfying the equations v ◊ a = 0, v ◊ b = 1 and [v, a, b] = 1 [Roorkee-JEE, 2000] 43. a, b, c are three unit vectors such that a × (b × c) 1 = __ (b + c). Find the angle between vectors a and b 2 given that vectors b and c are nonparallel. [Roorkee-JEE, 2000] 44. The diagonals of a parallelogram are given by the vectors 2i + 3 j – 6k and 3i – 4 j – k. Determine its sides and the area also. [Roorkee-JEE, 2001] 45. Find the value of l that a, b, c are all non-zero and (– 4i + 5 j) a + (3i – 3 j + k) b (i + j + k) c = l (a i + b j + c k)
[Roorkee-JEE, 2001]
Vectors
46. Find the vector r which is perpendicular to a = i – 2 j + 5k, b = 2i + 3 j – k and r ◊ (2i + j + k) + 8 = 0.
_›
_›
5.23
_›
9. Let a and b are two unit vectors such that |a | = 1, _›
_› _›
_›
_›
_›
_›
[Roorkee-JEE, 2001]
|b | = 4, a ◊ b =_›2. If _c = (2a × b ) – 3b , find the › angle between b and c .
47. Two vertices of a triangle are – i + 3 j and 2i + 5 j and its orthocentre is at i + 2 j. Find the position vector of the third vertex.
10. Given that a , b , p and q are four vectors such
where m is a scaler, prove that
[Roorkee-JEE, 2001]
_›
_›
_›
(Tougher Problems for JEE-Advanced)
_›
_›
_› _› _ ›
_›
that a + b = m p , b ◊ q = 0 and (b ) 2 = 1, _› _ › _ ›
_›
_› _ › _ ›
_› _ ›
|(a ◊ q ) p – (p ◊ q ) a | = |p ◊ q | _›
11. Find a vector v , which is coplanar with with the
_
› _› vectors + – 2 and + – 2 and is orthogonal 1. If a and b be two unit vectors, find the range of _› to the vector 2 + + . _› 3 _› _› __ _ › |a + b | + 2 |a – b | . It is given that the projection of v along the vector 2 __ p 2. Find the unit vector which makes an angle __ – + is equal to 6÷3 . 4 2 1 _› 1 with axis of z and is such that ( + + ) is a unit _› _› 12. Let a = , b = 1 and c = 1 . 0 vector. 0 –1 – 3 3. If r and s are non-zero vectors and the sacalar b is Find the numbers a, b and g such that _› _› chosen in such a way that |r + b s | is minimum, find –2 _› _ _ › › _› _› a a = b b + g c = –5 the value of |r + b s |2 + |b s |2. _ 6 › _› u = – 2 + 3 , 13. Let a 3-dimentional vector V satisfies the condition _› __› _ _ › › 4. Given that v = 2 – + 4 and w = + 3 + 3 , 2V + V × ( + 2 ) = 2 + 2 . and _ › __ _ _ _ _ _ _ _ › › › › › › If 3 |V | = ÷ m , m Œ N, find m. (u ◊ R – 10) + (v ◊ R 20) + (w ◊ R – 20) = 0 _› __› _› _› 14. Let A = – 2 + 3 , B = 2 + – and C = + . Find the vector R .
( )
[ ] [ ]
[ ]
[ ]
5. Suppose the_ vectors a,_ b, c on a_ › plane satisfy the _ › › › conditions |a | = |b | = |c | = |a + b | = 1; c ◊ a = 0 _› _ ›
_›
_›
and _b ◊ c > 0. Find the angle between (2a + b ) › and b .
6. Find the minimum area of the triangle whose vertices are A (– 1, 1, 2), B (1, 2, 3) and C (t, 1, 1), where t is a real number. 7. Suppose the vectors a, b, c on a plane satisfy the _›
_›
_›
_›
and b ◊ c > 0. If the vector c is a linear combination _›
_›
l a + m + b , find the ordered pair (l ◊ m).
_›
_ ›
8. Given that a and b are two unit vectors such that the
( )
_› _ _ › › 1 angle between a and b is cos– 1 __ . If c be a vector 4 _ _ › › _› _› _› in the plane of a and b such that |c | = 4 c × b = _› _› _ _ _ › › › 2a × b and c = l a × m b , find the
__›
_ ›
_ _›
__›
15. Find the angle between any edge and face, which is not containing the edge of a regular tetrahedron and also find the angle between the two faces of a regular tetrahedron.
INTEGER TYPE QUESTIONS
_›
_›
conditions |a | = |b | = |c | = |a + b | = 1; c ◊ a = 0 _› _ ›
_ ›
If the vector B × C = x A + y B + z C , where x, y and z are scalars, find the value of (100x + 10y + 8z).
1. If [a, b, c] = 1, find the value of
a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) _________ + + (c × a)◊ b (a × b)◊ c (b × c)◊ a
2. If a, b and c are non-coplanar vectors and p, q, r are defined as
b×c c×a a×b p = ______ , q = ______ , r = ______ , [b c a] [c a b] [a b c]
(i) values of l
find the value of (a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r
(ii) sum of the values of m
(iii) product of all values of m.
and c = l +
_›
3. Let a = _›
_›
– 2 – 3 , b = 2 + 3 – _›
+ (2 l – 1) . If c
5.24 Integral Calculus, 3D Geometry & Vector Booster _›
_›
is parallel to the plane of the vectors a and b , find the value of (l2 + 1).
_› _› _› _› 4. Let r be _a vector perpendicular to ( a + b + c ) _ › _› › _› _› _› _› _› where [a , b , c ] = 2. If r = P (b × c ) + Q (c × a ) + _› _ › R (a × b ) , find the value of (P + Q + R + S). _› _› _› _› _› _› 5. Let |a | = 1, |b | = 1, |c | = 2 and a ^ (b + c ), _ _ _ › › › _› _› _› _› _› _› b ^ (c + a ) , c ^ (a + × b ) such that m = |a + b + c | ,
Linked Comprehension Type (For Jee-Advanced Examination Only) Passage I _ _ _ › _ › › › Let the vectors x , y and z are coplanar such that x = a + _ ›
– , y =
2
find the value of (m + 1).
6. If a, b and c are the p th, q th and r th terms of an _› HP and u = (q – r) + (r – b) + (p – q) and _› v =
__ a + __ + b value of (m +
_› _› __ c such that m = |(u ◊ v ) + 2|, find the 4). _›
_›
7. Let a = – 2 + 3 , b = 2 + 3 – + (2 l – 1) . _›
_›
If c is parallel to the plane of a and
_›
and c = l + _› b ,
find the value
of (l + l + 2). › _› _
_›
8. Let a , b and c be three vectors of magnitudes 3, 4 _›
_›
› _› _
_›
_›
and 5 respectively and a ^ b + c , b ^ c + a and _ › _ _› › c ^ a + b such
(
that m =
_ › _ _› › |a + b + c | ,
find the value
)
m of ______ __ . 2 + 3 ÷
_ ›
_ ›
+ , v =
__›
–
and w =
+ 2 – 3 . If
is a unit vector such that ◊ = 0 and v ◊ = 0, find the value of_ | ◊ |. › _› 11. If a and b are non-zero and non-collinear and _› _› _› the linear combination (2x – y) a + 4 b = 5 a + _› (x – 2y) b holds for real x and y, find the value of (x + y + 2). _ ›
_ ›
12. Given the vectors u = 2 – – , v = – – 2 __› and w = – 3 . If the volume of the parallelopiped _ __› › _ › having – c u , v and c w are concurrent edges is 8, find the positive integral value of c. _› p =
_›
13. Consider three vectors + + , q = 2 + 4 _› – and r = + + 3 and s_ be a unit vector. If _› _› _› _› _› › (p × q ) × r = u p + v q + w r , find the value of (u + v + w + 4). _›
_›
14. Consider_ three vectors p = _ + + , q = 2 + 4 › › – and r = + + 3 and s be a unit vector. Find the value of _› _›
_›
_›
_› _›
_›
_›
(
_› _›
_›
_›
|(p ◊ s ) (q × r ) + (q ◊ s ) (r × p ) + (r ◊ s ) (p × q )|
+
–c .
)
1 1 1 (i) The value of _____ + _____ + _____ is 1–a 1–b 1–c
(a) 1
(b) 2
(c) 0
(d) – 1
(
)
a b c (ii) The value of _____ + _____ + _____ is 1–a 1–b 1–c
(a) – 1 (c) – 3
(b) – 2 (d) 0 _› _ ›
_› _ ›
_›
(iii) The value of Îx + y , y + z , z + z ˚ is
(a) 1 (c) 0
(b) 2 (d) –1
Passage II
_ ›
_ ›
Consider the three vectors_ p = + + , q = 2 + 4 – _ › › and r = + + 3 also s be a unit vector.
9. If the three points with position vectors (1, a, b), (a, 2, b) and (a, b, 3) are collinear in the space, find the value of (a + b).
10. Let u =
_ ›
and z =
_›
2
+b –
_› _› _›
(i) p , q , r are (a) (b) (c) (d)
linearly dependent vectors can form the sides of a possible triangle _› _› _› (q – r ) is orthogonal to p any one can be expressed as a linear combination of the other two.
_›
_›
_›
_›
_›
_›
(ii) If (p × q ) × r = u p + v q + w r , the value of (u + v + w + 2) is (a) 6 (b) 4 (c) 0 (d) 2 (iii) The value of _› _› _› _› _› _› _› _› _› _› _› _› (p ◊ s ) (q × r ) + (q ◊ s ) (r × p ) + (r ◊ s ) (p × q ) is
|
|
(a) 4 (c) 18
(b) 8 (d) 2.
Passage III _› _› _› _› Consider the three vectors p , q and r such that p = _›
_ _› › (p × r )
+ , q = – + and 2, then _› _› _› (i) the value of Î p , q , r ˚ is
=
_› q +
4 (a) – __ 3
8 (b) – __ 3
(c) 2
(d) 1
_› c p and
_› _› (p ◊ q )
+ =
Vectors _›
_› _ ›
_› _ ›
_›
(ii) the value of Î p + q , q + r , r + p ˚ is
4 (a) – __ 3
8 (b) – __ 3
16 (c) – ___ 3
(d) 2
(iii) the value of Î p × q, q × r, r × p˚ is
16 (b) ___ 9
4 (a) __ 3 64 (c) ___ 9
(d) 1
Passage IV The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors , , such that 1 ◊ = ◊ = ◊ = __ . 2
(i) The value of Î × , ×
× ,
1 (a) __ 2
1 (b) __ 4
1 (c) __ 3
1 (d) __ 9
× ˚ is
(ii) The volume of the parallelopiped with coterminous edges are given by , , is
1 (a) __ 3
1__ (b) ___ 3 ÷
1__ (c) ___ 2 ÷
1 (d) __ . 2
(iii) The volume of the tetrahedron formed by the vectors , , is
1__ (a) ____ 6÷2
1__ (b) ____ 3÷2
1__ (c) ____ 4÷2
1 __ (d) _____ . 12÷2
Passage V _›
If a + b + g = 2 and a = a + b + g and also _ _ › › × ( × a ) = a . (i) The value of (a 2 + 1) is (a) 2 (b) 1 (c) –1 (d) 0. 2 (ii) The value of (b + b + 1) is (a) 1 (b) 2 (c) –1 (d) 0 (iii) The value of (g 2 – g) is (a) – 2 (b) 0 (c) 4 (d) 2
Passage Vi _›
_›
__›
Let u , v , w be three _› _› _› 3 _› _› = a , a ◊ u = __ , a ◊ v = 2 (i) The value of (a) 3/4 (c) 1/2 (ii) The value of (a) – 3/4 (c) – 1/4 (iii) The value of (a) 1/2 (c) – 1/2 Passage VII _›
_›
_›
_›
5.25 __›
unit vectors such that u + v + w _› 7 __ and |a | = 2. 4
_› __› a ◊ w is
_ › __› u ◊ w is
_› _› u ◊ v is
(b) 1/4 (d) –1/2 (b) 1/2 (d) 1/6 (b) 3/4 (d) – 3/4.
_›
Let a , b , and c be three vectors with magnitude 4 such _› _ _› _› _› _› › 1 that a ◊ b = b ◊ c = c ◊ a = __ . 2 (i) The height of the parallelopiped whose adjacent › _› _ _› edges by the vectors a , b and c is
÷
÷ 2 (d) 4 × __ ÷ 3
__
__
2 (a) __ 3
2 (c) 3 × __ 3
2 (b) 2 × __ 3
÷
__
__
(ii) The volume of_ the prism whose adjacent edges by _› › _› the vectors a , b and c is __
__
(a) 2 ÷2
(b) 3 ÷2
(c) 4 ÷2
(d) 16 ÷2
__
__
(iii) The volume of the_tetrahedron whose adjacent edges _› › _› by the vectors a , b and c is __
__
4÷2 (a) ____ 3
8÷2 (b) ____ 3
16÷2 (c) _____ 3
32÷2 (d) _____ 3
__
Passage VIII _›
_›
__
_›
Let a , b and c are three non-parallel unit vectors such __ _› _› _› 1 › _› that a × ( × ) = __ (b + c ) . 2 _ _ › › (i) The angle between a and c is (a) 90° (b) 60° (c) 30° (d) none. _ ›
_›
(ii) The angle between a and b is (a) 90° (b) 60° (c) 120° (d) 0°
5.26 Integral Calculus, 3D Geometry & Vector Booster _ ›
_›
_ ›
_ ›
(a – b) × c – (a + b) (d) y = __________________ 2 _› (iii) The vector z is
(iii) The value of |a × b | 2 + |a × c | 2 is
(a) 1 (c) 0
(b) 2 (d) 4.
› _› _
_›
_›
_›
_›
Let a , b and c be three vectors such that |a | = 1, |c | , |b | = _›
_›
___
4 and |b × c | = ÷ 15 and also
(i) The angle between the
( ) ( )
_›
_› _› b – 2 c = l a . _ › _› vectors b and c is
( ) ( )
1 1 (a) sin–1 __ (b) cos–1 __ 4 4 1 1 (c) tan–1 __ (d) cos–1 __ 4 4 (ii) The value of l is (a) ± 4 (b) ± 3 (c) ± 2 (d) ± 6 (iii) The angle between the vectors a and b is
( ) ( ) ( ) ( )
7 (a) p – cos– 1 __ 8 7 (b) p + cos– 1 __ 8 7 (c) – p + cos– 1 __ 8 1 (d) p – cos– 1 __ . 8
Passage X
If x × (y × z) = a, y × (z × x) = b and (x × y) = c.
_›
(a + b) × c – (a + b) (a) x = ___________________ 2 (a + b) × c + (a + b) (b) x = ___________________ 2 (a – b) × c + (a + b) (c) x = __________________ 3 (a + b) × c – (a – b) (d) x = __________________ 3 _ ›
(ii) The vector y is
b + a + (a – b) × c (d) z = _________________ 2.
Passage XI Let x, y and z be three unit vectors such that _›
_›
_›
_›
_›
8 4 (c) x = a + __ b – __ c 3 3 8 4 (d) x = – a + __ b – __ c 3 3 _ ›
(ii) The vector y is (a) y = a – 4c
(b) y = – 4c
(c) y = a + 4c + 3b
(d) y = 2b – 4c _›
(iii) The vector z is 2 (a) z = __ (c – b) 3 1 (b) z = __ (c – b) 3 4 (c) z = __ (c – b) 3 5 (d) z = __ (c – b). 3
(a – b) × c + (a + b) (a) y = __________________ 2
Passage XII
(a + b) × c + (a + b) (b) y = __________________ 2
(a – b) × c + (a – b) (c) y = __________________ 2
_›
_› _› _› _› _› _› _› 3 _› _› 7 (x × y ) × z , c , a ◊ x = __ , a ◊ y = __ and |a | = 2. 2 4 _ › (i) The vector x is 8 4 (a) x = a + __ b + __ c 3 3 8 4 (b) x = a – __ b + __ c 3 3
_› _›
x + y + z = a , x × (y × z ) = b
__
(i) The vector x is
Vectors x, y and z each of magnitude ÷ 2 makes an angle of 60° with each other.
b – a + (a – b) × c (a) z = _________________ 2 b + a – (a + b) × c (b) z = _________________ 2 b – a + (a + b) × c (c) z = __________________ 2
Passage IX
_ ›
_ ›
_ › _ ›
_ ›
_› _ › _ ›
If x + y = a , x × y = b , x ◊ a = 1.
_ ›
(i) The vector x is _› 1 _› (a) x = __ 2 ( a + a _› 1 _› (b) x = __ 2 ( a – a
_ › _› a × b
)
_›
_ ›
a × b )
Vectors
_ _› 1 _› _› › (c) x = – __ 2 ( a + a × b ) a
_ _ › › 1 ( _ › (a) y = a + __ 2 a + a _› _› 1 _› (b) y = a – __ 2 ( a + a
_› _ › a × b ) _› a ×
_› b
)
_ _› _› 1 ( _› _› › ) (c) y = – a + __ 2 a + a × b a
_ _› _› 1 ( _› _› › ) (d) y = a + __ a – a × b . _›
a2
_›
+
+
_› (b) – a _› (d) 3a .
_›
and b =
Column I (A) The projection of
30°
_›
_ › _› a on b is _›
_
_ ›
_ ›
__
_
180°
120°
4. Match the following Columns: Column I
Column II
(A) If a be any vector, the value of (P) _›
_›
_›
2 |a |
_›
|(|a ◊ |2 + |a ◊ |2 + |a ◊ |2)1/2 | is
+ 2 + 2
› (B) The projection of b on a is
Column II (P) (Q)
÷
___
26 ___ 2 ___
÷26 ____ 4
__ _ ›
(B) If a be any vector, the value of | (Q) ÷2 |a | _›
× (a × )|2 × )|
_›
_›
– (a × )|2 – (a __ _›
formed (R) (C) The area of a _triangle _› › by the vectors a and b is
1 __ 3
|a | (C) If a be any vector, the value (R) ÷3 _ _ _ › › › 2 2 of (|(|a × | + |a × | + |a × |2)1/2 |
(D) The area of a parallelogram (S)
1 ___ __ 3 ÷
(D) If a be any vector, the value of (S)
having diagonals
(B) If a, b and c be unit vectors (Q) such that a is _perpendicular to _› _› › b and c and |a + b + c | = 1, the angle between b and c is
› _› › › (D) If |_a ◊ b | = ÷3 | a × b | , the angle (S) between a and b is
1. Match the following Columns:
Let a =
60°
_ ›
Matrix Match (For Jee-Advanced Examination Only)
(A) If a, b and c be three vec- (P) _› unit _ _ › › tors such that a + _b – c _=› 0, › the angle between a and b is
_
_› (a) a _› (c) 2 a
Column II
+ b + c = 0 and |a | = (R) (C) If a _› _› 1, |b | = _5 |c | =_› 7, the angle › between a and b is
_›
(iii) The vector (x + y ) is
3. Match the following Columns: Column I
_ _› 1 _› › _› (d) x = __ 2 ( a + b × a ) a _› (ii) The vector y is
5.27
_ › _› a and b is
_›
_›
_›
|a |
_›
|(a ◊ ) + (a ◊ ) + (a ◊ ) | is
2. Match the following Columns: Column I
_›
_›
_›
_›
Column II
(A) If |a + b | = |a – b |, the angle (P) _ › _› between a and b is _
_
› › _› › (B) If |_a + 2b | = |a – 2b | , the angle (Q) _ › _› between a and b is
_ › _› |a + b |
(C) If the angle
_
_ _› › _› = 2b | and |a | = |b | , _› _ › between a and b is _› |a –
_
(R)
› › _› _› › (D) If |_a + 2b | = |a – 3b | , and |a | = 1, (S) _ › _› the angle between a and b is
60°
5. Match_ the following Columns: _› › _› If a , b and c are three non-coplaner vectors such › _ _› _ ›
that [a , b , c ] = 2. 45° 0°
90°
Column I
Column II
(A) The value of _ _ [ _a › + b › , b › + _c › , _c › + _a › ] is
(P)
12
(B) the value of
(Q)
16
[ _a › +
_› 2 b ,
_› 2 b +
_›
_›
_›
3 c , 3 c + a ] is
5.28 Integral Calculus, 3D Geometry & Vector Booster (C) The value of _ ›
_›
_›
_ › _ ›
_ ›
[ a + b , b + c , c + a ] is
(D) The value of
_ › _› _ _ _ › › _ › › a × 2 b , 2 b × c , c × a is
[
(R)
9
(S)
4
__›
+ 4 and C = 3 + triangle is 5 ÷6 sq.u. Column I
]
6. Match the following Columns: If a, b, c and a¢, b¢, c¢ be the reciprocal systems of vectors. Column I
Column II
(A) The value of (a + b + c) is
(P)
15
(B) The value of (a + c + d) is
(Q)
11
(C) The value of (b + c + d) is
(R)
19
(D) The value of (a + b + c + d) is
(S)
14
10. Match_ the following Columns: _ _ › › › If a , b and c be unit vectors satisfying
(B) The value of (Q) a¢ (a + b) + b¢ ◊ (b + c) + c¢ ◊ (c + a) is
0
|a – b | 2 + |b – c | 2 + |c – a | 2 = 9.
(C) The value of (a + b + c) ◊ (a¢ + b¢ + c¢) is
(R)
1
(A) The value of |2 a + 5 b + 4 c (P) | is
1
(D) The value of (a × a¢ + b × b¢ + c × c¢) is
(S)
3
› › (B) The value of |2 a + 3 b + 3c (Q) | is
5
› › (C) The value of |2 a + b + c | is
(R)
4
(D) The value of |2 a + 7 b + 7 c (S) | is
3
(P)
_›
Column I
Column II (P)
1
(B) The value of (b 2 + 1) is
(Q)
2
(C) The value of (l + 1) is
(R)
0
(D) The value of (b + g ) is
(S)
3
_› c =
Let + , + , and a a + b b. If the _› vectors – 2 + , 3 + 2 – and the vector c are coplanar. Column I
Column II
(A) The value of ( a + 3b + 2 ) is
(P)
1
a (B) The value of __ + 4 is b
(Q)
4
b 2 The value of 3 __ + __ is a 3 (D) a b The value of 3 __ + __ + 11 is b a
(R)
6
(
(C)
)
( )
( )
_›
(S)
_›
__›
_›
_›
_›
_›
_
_›
_
_ ›
_›
Column II
_
_›
_›
11. Match the following Columns: Column II
(A) The volume of the parallelopiped (P) › _ _ › _ › determined by the vectors a , b , c is 2. The volume of the parallelopiped › _ › _ determined by the vectors 2 (a × b ) , _› _› _› _› 3 (b × c ) and (c × a ) is
100
(B) The volume of the parallelopiped (Q) _› determined by the vectors a , _› _ › b , c is 5. The volume of the parallelopiped determined by the
30
_›
_›
_›
_›
vectors 3 (a + b ), (b + c ) and _› _› 2 (c + a ) is
_›
_›
24
_›
(2 a + 3 b ) and (a – b ) is
2
_›
__›
_›
(C) The area of a triangle with (R) adjacent sides determined by _› _› the vectors a and b is 20. The area of the triangle with adjacent sides determined by the vectors
Given that A , B , C form a triangle such that A = B + C . If A = a +
_›
_›
_›
9. Match the following Columns: _›
_›
Column I
8. Match the following Columns: _› b =
_ ›
_
(A) The value of (a + 1) is
_› a =
_ ›
Column I
7. Match the following Columns: _ › If _a + b + g = 2 and a = a + b + g , + ( × a) › = 0 .
Column II
2
(A) The value of a ◊ a¢ + b ◊ b¢ + c ◊ c¢ is
– 2 and the area of the
__
+ b + c and B = d + 3
(D) The area of a parallelogram with (S) adjacent_sides determined by the _› › vectors a and b is 30. The area of the parallelogram with adjacent _› sides by the vectors (a _› determined _› + b ) and a is
60
Vectors
Questions asked in Previous Years’ JEE-Advanced examinations 1. The value of A {(B + C) × (A + B + C)} is (a) 0 (b) [ABC] + [BCA] (c) [ABC] (d) none [IIT-JEE, 1981]
2. Let A, B, C be there unit vectors. Suppose that A B = 0 = A C and the angle between B and C is p /6. Then A = ± 2 (B × C). Is it true or false? [IIT-JEE, 1981] 3. For non-zero-vectors a, b, c |(a × b) c| = |a||b||c| holds if and only if (a) a b = 0, b c = 0 (b) c a = 0, b c = 0 (c) c a = 0, a b = 0 (d) a b = b c = c a = 0 [IIT-JEE, 1982] 4. A1, A2, ..., An are the vertices of a regular polygon with n sides and O its centre. Show that n – 1
S (OAi × OAi + 1) = (1 – n) (OA2 × OA1 + 1). i = 1
[IIT-JEE, 1982]
5. Find all values of l such that (x, y, z) π (0, 0, 0) and x (i, j, 3k) + y (3i – 3j + k) + z (– 4i + 5j) = l (x i + y j + z k), where i, j, k are unit vectors along the co-ordinate axes. [IIT-JEE, 1982] 6. The points with position vectors 60i + 3j, 40i – 8j, a i – 52j are collinear if (a) a = – 40 (b) a = 40 (c) a = 20 (d) None. [IIT-JEE, 1983] 7. If X A = 0 = X B = X C for some non-zero vector X, then [A B C] = 0. Is it true or false? [IIT-JEE, 1983] 8. The volume of the parallelopiped whose sides are given by OA = 2i – 3j, OB = i + j – k and OC = 3i – k is (a) 4/13 (b) 4 (c) 2/7 (d) None. [IIT-JEE, 1983] 9. If c be given non-zero scalar and A and B be given non zero vectors such that A ^ B. Find the vector X which satisfies the equations A X = c, A × X = B [IIT-JEE, 1983] 10. A vector A has components A1, A2, A3 in a right handed rectangular cartesian co-ordinate system
5.29
OXYZ. The co-ordinate system is rotated about the p x-axis through an angle __ . Find the components of 2 A in the new co-ordinate system in terms of A1, A2, A3 [IIT-JEE, 1983] 11. The points with position vectors a + b, a – b and a + k b are collinear for all real values of k. Is it true or false? [IIT-JEE, 1984] 12. Let a = a1i + a2 j + a3k, b = b1i + b2 j + b3k and c = c1i + c2 j + c3k be three non-zero vector such that c is a unit vector perpendicular to both the vectors a and b. If the angle between a and b be p /6, then
| |
a 1 b 1 c1
a 2 b 2 c2
a3 b 3 is equal to c3
(a) 0 (b) 1 1 (c) __ a21 + a22 + a 23 b21 + b 22 + b 23 4 3 (d) __ a21 + a22 + a23 b21 + b22 + b 23 c21 + c22 + c 23 4 [IIT-JEE, 1986] 13. A vector a has components 2p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If, with respect to the new system, a has component p + 1 and 1, then (a) p π 0 (b) p = 1 or p = – 1/3 (c) p = – 1 or p = 1/3 (d) p = 1 or p = – 1 [IIT-JEE, 1986] 14. The position vectors of the points A, B, C and D are 3i – 2j – k, 2i + 3j – 4k, – i + j + 2k and 4i + 5j + l k respectively. If the points A, B, C and D lie on a plane, find the value l. [IIT-JEE, 1986] 15. The number of vectors of unit length perpendicular to vectors a = (1, 1, 0) and b = (0, 1, 1) is (a) 1 (b) 2 (c) 3 (d) infinite [IIT-JEE, 1987]
(
) (
)
(
) (
) (
)
16. If the vectors a i + j + k, i + b j + k and i + j + c k, (a, b, c π 1) are coplanar, the value of 1 1 1 _____ + _____ + _____ is ... 1–a 1–b 1–c [IIT-JEE, 1987]
5.30 Integral Calculus, 3D Geometry & Vector Booster 17. Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively, are given by ... [IIT-JEE, 1987] 18. If A, B, C, D are four points in the space, prove that |AB × CD × BC × AD × CA × BD| = 4 (area of D ABC) [IIT-JEE, 1987] 19. Let a, b, c be three non-coplanar vectors and p, q r are vectors defined by the relations b×c c×a a×b p = _______ , q = _______ and r = _______ , [a, b, c] [a, b, c] [a, b, c] the value of the expression (a + b) p + (b + c) q + (c + a) r is equal to
(a) 0 (c) 2
(b) 1 (d) 3
[IIT-JEE, 1988] 20. The components of a vector a along and perpendicular to a non-zero vector b are ... and ... respectively. [IIT-JEE, 1988] 21. Let OACB be a parallelogram with O at the origin and OC a diagonal. Let D be the mid-point of OA. Using vectors method, prove that BD and CO intersect in the same ratio. Determine the same ratio. [IIT-JEE, 1988] 22. For any three vectors a, b, c, (a – b) {(b – c) × (c – a)} = 2a (b × c). Is it true or false? [IIT-JEE, 1989] 23. If vectors a, b and c are coplanar, show that
|
|
b c a a a a c = 0 a b b a b b b c
[IIT-JEE, 1989]
24. In a triangle OAB, E is the mid-point of BO and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : OD, using vector method. [IIT-JEE, 1989] 25. Let A = 2i + k, B = i + j + k and C = 4i – 3j + 7k. Determine a vector R satifying R × B = C × B and R A = 0. [IIT-JEE, 1990] 26. Determine the value of c so that for all real x, the vectors c x i – 6j – 3k and x i + 2j + 2c x k make an obtuse angle with each other. [IIT-JEE, 1991]
27. Given that a = (1, 1, 1), c = (0, 1 –1,) a b = 3 and a × b = c, then b = ... [IIT-JEE, 1991] 28. A unit vector coplanar with i + j + 2k and i + 2 j + k and perpendicular to i + j + k is ... [IIT-JEE, 1992] 29. Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie on a plane, then c is (a) the AM of a and b (b) the GM of a and b (c) The HM of a and b (d) equal to zero. [IIT-JEE, 1993] 30. Let a = 2i – j + k, b = i + 2j – k, and c = i + j – 2k be three vectors. A vector in the plane of b and c, __ 2 whose projection on a is __ , is 3 (a) 2i + 3j – 3k (b) 2i + 3j + 3k
÷
(c) –2i – j + 5k
(d) 2i + j + 5k
[IIT-JEE, 1993] 31. In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find BP the ratio ___ , using vector method. PE [IIT-JEE, 1993] 32. Let p and q be the position vectors of P and Q, respectively, with respect to O and |p| = p, |q| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular, then
(a) 9p2 = 4q2
(b) 4p2 = 9q2
(c) 9p = 4q
(d) 4p = 9q
[IIT-JEE, 1994] 33. Let a, b and g be distinct real numbers. The points with position vectors a i + b j + g k, b i + g j + a k and g i + a j + b k (a) are collinear (b) form an equilateral triangle (c) form a scalene triangle (d) form a right angled triangle. [IIT-JEE, 1994] 1 34. The vector d = __ (2i – 2j + k) is 3
(a) a unit vector.
Vectors
p (b) makes an angle __ with the vector. 3 (2i – 4j + 3k) 1 (c) parallel to the vector – i + j – __ k 2
(
)
(d) perpendicular to the vector (3i + 2j – 2k) [IIT-JEE, 1994] 35. A unit vector perpendicular to the plane determined by the points P (1, –1, 2), Q (2, 0, –1) and R (0, 2, 1) is ... [IIT-JEE, 1994] 36. If the vectors a, b, c, d are not coplanar, prove that the vector (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) is parallel to a. [IIT-JEE, 1994] 37. Let a = i – j, b = j – k, c = k – i. If d be a unit vector such that a d = 0 = [b, c, d], then d equals
( (
)
(
)
i + j – 2k __ (a) ± _________ ÷6
i + j – k __ (b) ± ________ 3 ÷
i + j + k __ (c) ± ________ 3 ÷
(d) ± k
)
[IIT-JEE, 1995] 38. If , b and c be non-coplanar unit vectors such that 1 a × (b × c) = ___ __ (b + c), the angle between a and 2 ÷ b is 3p p (a) ___ (b) __ 4 4 p (c) __ (d) p 2 [IIT-JEE, 1995] 39. Let u, v and w be the vectors such that u + v + w = 0. If |u| = 3, |v| = 4, |w| = 5, the value of u v + v w + w u is (a) 47 (b) – 25 (c) 0 (d) 25 [IIT-JEE, 1995] 40. If a, b and c are three non-coplaner vectors,
(a + b + c) ◊ {(a + b) × (a + c)} is
(a) 0
(b) [a, b, c]
(c) 2 [a, b, c]
(d) – [a, b, c]
[IIT-JEE, 1995] 41. A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the line determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is ... [IIT-JEE, 1996]
5.31
42. If b and c any two non-collinear unit vectors and a be any vector, then a (b × c) (a b) b + (a c) c + _________ (b × c) = ... |(b × c)| [IIT-JEE, 1996] 43. The position vectors of the vertices A, B, C of a tetrahedron ABCD are i + j + k, i and 3i, respectively. The altitude from vertex D to the opposite face ABC, meets the median line through A of the D ABC at E. If the length of the side AD is 4 and the volume of __ 2÷2 ____ the tetrahedron is . Find the position vector of 3 E for all its possible positions. [IIT-JEE, 1996] 44. Let p, q and r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation p × [(x – q) × p] + q × [(x – r) × q] + r × [(x – p) × r] = 0, then x is given by 1 (a) __ (p + q – 2r) 2 1 (c) __ (p + q + r) 3
45.
46.
47. 48. 49.
50.
1 (b) __ (p + q + r) 2 1 (d) __ (2p + q – r) 3 [IIT-JEE, 1997] Let OA = a, OB = 10a + 2b and OC = b where O, A and C are non-collinear points. Let p denotes the area of the quadrilateral OABC and q denotes the area of the parallelogram with OA and OC as adjacent sides. If p = kq then k = ... [IIT-JEE, 1997] Let a, b, and c be three vectors having magnitudes 1, 1 and 2, respectively. If a × (a × c) + b = 0, the acute angle between a and c is ... [IIT-JEE, 1997] If A, B and C are vectors such that |B| = |C|. Prove that |(A + B) × (A + C)| × (B × C) (B + C) = 0 [IIT-JEE, 1997] Let a, b and c be non-coplaner unit vectors, equally inclined to one another at an angle q. If a × b + b × c = p a + q b + r c, find scalars p, q and r in terms of q. [IIT-JEE, 1997] If a = i + j + k, b = 4i + 3j + 4k and c = i + a j __ + b k are linearly independent vectors and |c| = ÷ 3 , then (a) a = 1, b = 1 (b) a = 1, b = ± 1 (c) a = –1, b = ± 1 (d) a = ± 1, b = 1 [IIT-JEE, 1998] For the vectors u, v and w, which of the following expressions is not equal to any of the remaining three?
5.32 Integral Calculus, 3D Geometry & Vector Booster (b) (v × w) ◊ u (d) (u × w) ◊ w [IIT-JEE, 1998] 51. Which of the following expressions are meaningful?
(a) u ◊ (v × w) (c) v ◊ (u × w)
(a) u ◊ (v × w)
(b) (u ◊ w) ◊ w
(d) u × (v ◊ w) [IIT-JEE, 1998] 52. For any two vectors u and v, prove that (i) (u ◊ v)2 + |u × v|2 = (u2 v2)
(c) (u ◊ v) × w
2
2
(ii) (1 + |u| ) (1 + |v| )
53.
= |u + v + (u × v)|2 + (1 – (u ◊ v))2 [IIT-JEE, 1998] Let a and b be two non-collinear unit vectors. If u = a – (a b) b and v = a × b, then |v| is (a) |u| (b) |u| + |u a| (c) |u| + |u b| (d) |u| + u (a + b) [IIT-JEE, 1999]
planes determined by the pair of vectors a, b and c, d respectively, then the angle between P1 and P2 is
(a) 0
p (c) __ 3
[IIT-JEE, 2000] 59. If a, b and c are unit coplanar vectors, then the value of [2a – b 2b – c 2c – a] is
(a) 0
(c) – ÷3
that a ◊ c = |c|, |c – a| = 2 ÷2 and the angle between (a × b) and c is 30°, the value of |(a × b) × c| is (a) 2/3 (b) 3/2 (c) 2 (d) 3. [IIT-JEE, 1999] 55. Let a = 2 i + j + k and b = i + 2 j – k and a unit vector c be coplanar. If c is perpendicular to a, then c is
1__ (a) ___ (– j + k) 2 ÷
1__ (b) ___ (– i – j – k) 2 ÷
1__ (c) ___ (– i – 2j) ÷5
1__ (d) ___ (i – j – k) 3 ÷
[IIT-JEE, 1999] 56. Let u and v be unit vectors. If w be a vector such that 1 w × (w × u) = v, prove that |(u × v) w| £ __ and that 2 the equality holds if and only if u is perpendicular to v. [IIT-JEE, 1999] 57. If the vectors a, b and c from the sides BC, CA and AB, respectively, of a triangle ABC, then (a) a ◊ b + b ◊ c + c ◊ a = 0
(b) a × b + b × c = c × a (c) a ◊ b = b ◊ c = c ◊ a
(d) a × b + b × c + c × a = 0
[IIT-JEE, 2000] 58. Let the vertices a, b, c and d be such that (a × b) × (c × d) = 0. Let P1 and P2 be the two
__
(b) 1
__
(d) ÷ 3
[IIT-JEE, 2000]
60. If a, b and c are unit vectors, then |a – b|2 + |b – c|2 + |c – a|2 does not exceed
(a) 4 (c) 8
(b) 9 (d) 6
[IIT-JEE, 2000] 61. Let a = i – k, b = x i + j + (1 – x) k and c =
54. Let a = 2i + j – 2k and b = i + j. If c be a vector such __
p (b) __ 4 p (d) __ 2
y i + x j + (1 + x – y) k, then [a b c] depends on (a) only x (b) only y (c) neither x nor y (d) both x and y. [IIT-JEE, 2001]
62. Find 3-dimensional vectors v1, v2, v3 satisfying v1 ◊ v1 = 4, v1 ◊ v2 = – 2 and v1◊v3 = 6, v2 v2 = 2, v2 ◊ v3 = 5 and v3◊v3 = 29. [IIT-JEE, 2001] 63. Let A (t) = f1 (t) i + f2 (t) j and B (t) = g1 (t) i + g2 (t) j, t Œ [0, 1], where f1, f2, g1, g2 are continuous functions. If A (t) and B (t) are non-zero vectors for all t and A (0) = 2i + 3j, A (1) = 6i + 2j, B (0) = 3i + 2j, B (1) = 2i + 6j, show that A(t), B(t) are parallel for some t. [IIT-JEE, 2001] 64. Let V be the volume of the parallelopiped formed by the vectors a = a1i + a2 j + a3k, b = b1i + b2 j + b3k and c = c1i + c2 j + c3k. If ar, br, cr where r = 1, 2, 3, are non-negative real numbers and 3
S (ar + br + cr) = 3L, r = 1
Show that V £ L3.
[IIT-JEE, 2002] 65. If a and b are two unit vectors such that a + 2b and 5a – 4b are perpendicular to each other, the angle between a and b is p p (a) __ (b) __ 4 3
Vectors
( )
( )
1 (c) cos–1 __ 3
2 (d) cos–1 __ 7 [IIT-JEE, 2002] 66. Let v = 2 i + j – k, w = i + 3 k and u is a unit vector, the maximum value of [u v w] is
(a) –1
(c) ÷59
__ ___ (b) ÷ 10 + ÷6 ___ (d) ÷ 60
___
[IIT-JEE, 2002] 67. The value of a be such so that the volume of a parallelopiped formed by the vectors i + a j + k, j + a k, a i + k becomes minimum, is (a) –3 (b) 3 __ __ (c) 1/÷3 (d) ÷ 3 [IIT-JEE, 2003] 68. If u, v, w are three non-coplaner unit vectors and a, b, g are the angle between u and v, v and w and w and u respectively, and x, y, z are unit vectors along the bisectors of the angle a, b, g respectively. Prove that [x × y, y × z, z × x]
( )
( ) ( )
g b a 1 = ___ [u v w]2 sec2 __ sec2 __ sec2 __ 2 2 2 16 69. 70. 71.
[IIT-JEE, 2003] If a, b, c and d are distinct vectors satisfying relation a × b = c × d and a × c = b × d. Prove that a ◊ b + c ◊ d + a ◊ c + b ◊ d [IIT-JEE, 2004] If a = i + j + k, a. b = 1 and a × b = j – k, then b is (a) i – j + k (b) 2 j – k (c) i (d) 2 i – k [IIT-JEE, 2004] The unit vector which is orthogonal to the vector 5 i + 2 j + 6 k and is coplanar with the vectors 2 i + j + k and i – j + k is 1 ___ (a) ____ (2i 41 ÷ 1 ___ (b) ____ (2i ÷29 1 ___ (c) ____ (3j 10 ÷ 1 ___ (d) ____ (2i ÷69
– 6j + k)
73. Let a, b, c be three non-coplanar vectors and a ◊ b a ◊ b b1 = b – ____ a, b2 = b + ____ 2 a, 2 |a| |a| c ◊ a c ◊ b c1 = c – ____ 2 a + ____ 2 b1, |a| |c| b1 ◊ c c ◊ a c2 = c – ____2 a – ____ 2 b1, |a| |b1| c ◊ a b ◊ c c3 = c – ____ 2 a + ____ 2 b1, |c| |c| c ◊ a b ◊ c and c4 = c – ____ 2 a – ____ 2 b1, |c| |b1| the triplet of pairwise orthogonal vectors is (a) (a, b1, c1) (b) (a, b1, c2) (c) (a, b2, c2) (d) (a, b1, c3) [IIT-JEE, 2005]
74. Let a = i + 2 j + k, b = i – j + k and c = i + j – k. A vector in__ the plane of a and b, whose projection on c is 1/÷3 , is
(a) 4 i – j + k
(b) 3 i + j + k
(c) 2 i + j + 2k
(d) 4 i + j – 4 k
[IIT-JEE,2006]
75. The number of distinct real values of g which the vectors – g 2 i + j + k, i – g 2 j + k and i + j – g 2 k are coplanar is (a) 0 (b) 1 (c) 2 (d) 3 [IIT-JEE, 2007] 76. Let a, b and c be unit vectors such that a + b + c = 0, Which one of the following is correct? (a) a × b = b × c = c × a = 0 (b) a × b = b × c = c × a π 0 (c) a × b = b × c = a × c π 0 (d) a × b, b × c, c × a are mutually perpendicular. [IIT-JEE, 2007]
– 5j) – k) – 8j + k)
[IIT-JEE, 2004] 72. Let v be a unit vector along the incident ray, w be a vector along the reflected ray and a be a unit vector along the outward normal to the plane mirror at the point of incidence P. Express w in terms of a and v. [IIT-JEE, 2005]
5.33
77. Let the vectors PQ, QR, RS, ST, TU and UP represent the sides of a regular hexagon. Statement I: PQ × (RS + ST) π 0 Statement II: PQ × RS = 0 and PQ × ST π 0
[IIT-JEE, 2007]
78. Let two non-collinear unit vectors a and b form an acute angle. A point P moves so that at any time t, the position vector OP (where O is the origin) is given by a cos t + b sin t, where P is the farthest from origin O, let M be the length of OP and u be the unit vector along OP. Then
5.34 Integral Calculus, 3D Geometry & Vector Booster
a+b (a) u = ______ and M = (1 + a ◊ b)1/2 |a + b| a–b (b) u = ______ and M = (1 + a ◊ b)1/2 |a – b| a+b (c) u = ______ and M = (1 + 2 a ◊ b)1/2 |a + b|
a–b (d) u = ______ and M = (1 + 2 a ◊ b)1/2 |a – b| [IIT-JEE, 2008] 79. The edges of a parallelopiped are of length and parallel to non-coplanar unit vectors a, b, c such 1 that a ◊ b = b ◊ c = c ◊ a = __ . The volume of the 2 parallelopiped is 1__ 1__ (a) ___ (b) ____ 2 2÷2 ÷
___
(a) 8/9
(b) ÷ 17 / 9
(c) 1/9
(d) 4÷5 / 9
__
[IIT-JEE, 2010] 85. Let non-zero vectors a, b and c such that a ◊ b = 0, (b – a) ◊ (b + c) = 0 and 2 |b + c| = |b – a|. If a = m b + 4 c, the possible values of m are ...
__
3 ÷ (c) ___ 2
1__ (d) ___ 3 ÷
[IIT-JEE, 2008] 80. If a, b, c and d are unit vectors such that 1 (a × b) ◊ (c × d) = 1 and (a × c) = __ , 2 then (a) a, b, c are non-coplanar (b) b, c, d are non-coplanar (c) b, d are non-parallel (d) a, d are parallel and b, c are parallel
[IIT-JEE, 2009]
81. The volume of the parallelopiped with its edges represented by the vectors i + j, i + 2j and i + j + p k is... [IIT-JEE, 2009] 82. Angle between vectors a and b where a, b and c are unit vectors satisfying a + b + R3c = 0 is ... [IIT-JEE, 2009] 83. Let P, Q, R, S be the points on the plane with position vectors – 2 i – j, 4 i, 3 i + 3j and – 3 i + 2j respectively. The quadrilateral PQRS must be a
(a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c) rectangle but not a square (d) rhombus but not a square.
[IIT-JEE, 2010]
84. Two adjacent sides of a parallelogram ABCD are given by AB = 2 i + 10 j + 11 k and AD = – i + 2 j + 2 k. The side AD is rotated by an acute angle a in the plane of the parallelogram so that AD becomes AD ¢. If AD ¢ makes a right angle with the side AB, the cosine of the angle a is given by
[IIT-JEE, 2010] 86. If a and b be vectors in the space given by i – 2 j 2i + j + 3k ___ a = ______ __ and __________ , then the value of 14 ÷ ÷5 (2a + b) ◊ {(a × b) × (a – 2 b)} is [IIT-JEE, 2010] 87. Let a = i + j + k, b = i – j + k and c = i – j – k be three vectors. A vector v in the plane of a and b, 1 whose projection on c is ___ __ , is given by 3 ÷ (a) i – 3 j – 3 k (b) – 3 i – 3 j + k (c) 3i – j + 3k (d) i + 3 j – 3 k [IIT-JEE, 2011] 88. The unit vector(s) which is/are coplanar with vectors i + j + 2 k, i + 2 j + k and perpendicular to the vector i + j + k is given by (a) j – k (b) – i + j (c) i – j (d) – j + k [IIT-JEE, 2011] 89. Let a = i – k, b = – i + j and c = i + 2 j + 3 k be three given vectors. If r be a vector such that r × b = c × b and r ◊ a = 0, the value of r ◊ b is ... [IIT-JEE, 2011] __
__
90. If a = j + ÷3 k, b = – j + R3 k and c = 2 ÷3 k form a triangle, the internal angle of the triangle between a and b is. [IIT-JEE, 2011] 91. If a, b and c are unit vectors satisfying |a – b|2 + |b – c|2 + |c – a|2 = 9, then |2a + 5b + 5c| is. [IIT-JEE, 2012]
___
92. If a and b are vectors such that |a + b| = ÷ 29 and a × (2i + 3j + 4k) = (2i + 3j + 4k) × b, a possible value of (a + b) ◊ (– 7i + 2j + 3k) is
(a) 0 (c) 4
(b) 3 (d) 8.
[IIT-JEE, 2012] 93. Let PR = 3i + j – 2k and SQ = i – 3j – 4k determine diagonals of a parallelogram PQRS and PT = i + 2j + 3k be another vector. The volume of the parallelopiped determined by the vectors PT, PQ and PS is
Vectors
(a) 5 (c) 10
(b) 20 (d) 30
[IIT-JEE, 2013] 94. The volume of the parallelopiped determined by the vectors a, b and c is 2. The volume of the parallelopiped determined the vectors (a × b), 3 (b × c) and (c × a) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 95. The volume of the parallelopiped determined by the vectors a, b and c is 5. The volume of the parallelopiped determined the vectors 3 (a + b), (b + c) and 2 (c + a) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 96. The area of a triangle with adjacent sides determined by the vectors a and b is 20. The area of the triangle with adjacent sides determined by the vectors (2a + 3b) and (a – b) is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013] 97. The area of a parallelogram with adjacent sides determined by the vectors a and b is 30. The area of the parallelogram with adjacent sides determined by the vectors (a + b) and a is (a) 100 (b) 30 (c) 24 (d) 60 [IIT-JEE, 2013]
__
98. Let x, y and z be three vectors each of magnitude ÷ 2 p and the angle between each pair of them is __ . If a 3 be a non-zero vector perpendicular to x and (y × z) and b is another non-zero vector perpendicular to y and (z × x), then (a) b = (b ◊ z) (z – x) (b) a = (a ◊ y) (y – z) (c) (a ◊ b) = (a ◊ y) (b ◊ z) (d) a = (a ◊ y) (z – y) [IIT-JEE, 2014]
99. If a, b and c are three non-coplaner unit vectors such p that the angle between every pair of them is __ . If 3 a × b + b × c = pa + qb + rc, where p, q and r are scalears, find the value of
(
)
p2 + 2q2 + r2 ____________ is... q2
[IIT-JEE, 2014] _›
› ___› _
___› _ ›
1. (d) 6. (d) 11. (d) 16. (b)
2. (b) 7. (b) 12. (c) 17. (d)
3. (c) 8. (c) 13. (d) 18. (b)
4. (b) 9. (a) 14. (d) 19. (d)
5. (b) 10. (b) 15. (b) 20. (b)
___›
100. Let PQR be a triangle. a = QR , b = RP , c = PQ _›
_›
_› _ ›
__
If |a | = 12, |b | = 4÷3 , and b ◊ c = 24, which of the following is (are) true? _
| c › | 2 – |_a › | = 12 (a) ____ 2
_›
| c | 2 + |_a › | = 30 (b) ____ 2 _ __ › _› _› _› (c) |a × b + c × a | = 48÷3
› _ › _
(d) a ◊ b = – 72
[IIT-JEE, 2015] 2
101. In R , if the magnitude of the projection vector of __
__
the 3 + is ÷ 3 and a = 2 + __ vector a + b on ÷ ÷ 3 b, find the possible values of |a |. [IIT-JEE, 2015] _› _› _› 102. Suppose that p , q and r are three non-coplanar vectors in R3. _› _› _› Let the components of a vector s , along p , q _› and r be 4,_3 and 5, respectively. If _the components _› _› _› _› _› › › of _a vector s along ( – p + q + r ), ( p – q + r ) and _ _ ( – p › – q › + r › ) are x, y and z, respectively, find the value of 2x + y + z. [IIT-JEE, 2015] 103. Let
__›
= u1 + u2 + u3 be a unit vector in R3 and w
_ › 1 = ___ __ ( + + ). Given that there exists a vector v ÷6 _› _› _› _› in R3 such that |u × v | = 1 and . (u × v ) = 1.
Which of the following statement (s) is (are) correct? _ › (a) There is exactly one choice for such v . _› (b) There are infinitely many choices for such v . _› (c) If u lies in the xy-plane, then |u1| = |u2|. _›
(d) If u lies in the xz-plane, then 2 |u1| = |u3|. [IIT-JEE, 2016]
Answers LEVEL II
5.35
21. (a) 26. (b) 31. (c) 36. (b)
22. (a) 27. (c) 32. (a) 37. (a)
23. (**) 28. (a) 33. (b) 38. (b)
24. (b) 29. (d) 34. (d) 39. (a)
25. (a) 30. (c) 35. (d) 40. (b)
5.36 Integral Calculus, 3D Geometry & Vector Booster
LEVEL-IV
COMPREHENSIVE LINK PASSAGES
1. [– 5, 5]
2. – __ – __ + ___ __ 2 2 ÷ 2
3. |r | 2
4. – + 2 + 5 p 5. __ 2
_›
__
3 ÷ 6. ___ . 2
1 2__ 7. ___ __ , ___ 3 ÷ 3 ÷
8. (i) 2
5p 9. ___ 6
(
)
(ii) –1 (iii) 12
(ii) (b) (ii) (b) (ii) (c) (ii) (c) (ii) (a) (ii) (c) (ii) (d) (ii) (c) (ii) (a) (ii) (b) (ii) (b)
(iii) (c) (iii) (a) (iii) (c) (iii) (a) (iii) (d) (iii) (b) (iii) (c) (iii) (a) (iii) (a) (iii) (c) (iii) (c)
P-XII :
(i) (a)
(ii) (b)
(iii) (a)
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
12. a = – 1, b = – 2, g = 3. 13. m = 6 14. 101 1 1 15. cos–1 ___ __ , cos–1 __ 3 3 ÷
( )
( )
INTEGER TYPE QUESTIONS 1. 3 6. 6 11. 3
2. 3 7. 2 12. 2
3. 3 8. 8 13. 2
4. 4 9. 4 14. 4
5. 7 10. 3
Hints
1. Let q the angle between a and b.
Now,
a ◊ b = a b cos (q)
fi
a ◊ b cos (q) = ____ a b
fi
6 ___ – 6 +___0 cos (q) = _________ =0 14 ÷ 13 ÷
fi
p q = __ 2
p Hence, the angle between a and b is __ . 2 2. We have, a + b = (i + 2j – 3k) + (3 i + j + 2k)
(i) (a) (i) (c) (i) (b) (i) (a) (i) (b) (i) (a) (i) (d) (i) (b) (i) (b) (i) (a) (i) (a)
MATRIX MATCH
11. 9 (– + )
P-I : P-II : P-III : P-IV : P-V : P-VI : P-VII : P-VIII : P-IX : P-X : P-XI :
= (4 i + 3j – k)
and
(A)Æ(R), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(S), (A)Æ(P), (A)Æ(S), (A)Æ(S), (A)Æ(S),
(B)Æ(S), (B)Æ(S), (B)Æ(R), (B)Æ(P), (B)Æ(P), (B)Æ(S), (B)Æ(P), (B)Æ(P), (B)Æ(P), (B)Æ(P),
(C)Æ(P), (C)Æ(P), (C)Æ(P), (C)Æ(Q), (C)Æ(S), (C)Æ(S), (C)Æ(S), (C)Æ(P), (C)Æ(Q), (C)Æ(P),
(D)Æ(P) (D)Æ(P) (D)Æ(Q) (D)Æ(S) (D)Æ(Q) (D)Æ(Q) (D)Æ(Q) (D)Æ(P) (D)Æ(R) (D)Æ(Q)
11. (A)Æ(R), (B)Æ(S),
(C)Æ(P),
(D)Æ(Q)
solutions Also, a – b = (i + 2j – 3k) – (3i + j + 2k) Now,
= (– 2 i + j – 5k) (a + b) ◊ (a – b) = – 8 + 3 + 5 = 0
Thus, (a + b) is perpendicular to (a – b). 3. Given,
(a + b + c) = 0
fi | (a + b + c) |2 = 0 fi |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 9 + 16 + 25 + 2(a ◊ b + b ◊ c + c ◊ a) = 0 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 50 fi (a ◊ b + b ◊ c + c ◊ a) = – 25
Vectors
|(a + b – c)|
Now,
= |a|2 + |b|2 + |c|2 + 2 (a ◊ b – a ◊ c + b ◊ c)
fi
=1+1+1+0
fi
=3 __ |a + b – c| = ÷ 3
4. We have, 2
fi and
5. As we know that the projection of a on b
a ◊ b = ____ . |b| 3_________ –2+2 3 = __________ = ___ __ = 1 + 4 + 1 ÷6 ÷
÷
__
3 __ 2
6. Given 2 a + b = i + j and a + 2b = i – j Solving, we get
1 1 a = __ (i + 3j) and b = __ (i – 3j) 3 3
Since,
a ◊ b = a b cos (q )
fi
a ◊ b cos (q ) = ____ a b 1 __ (1 – 9) 9 8 4 ________ cos (q ) = = – ___ = – __ 10 10 5 ___ 9
fi
fi
( )
4 (q ) = cos–1 – __ 5
7. Let r = a i + b j + c k. r We have to find = __ . |r| Here, r is perpendicular to the given vectors, so a – 2 b + c = 0 and 2a + b – 3c = 0 Solving, we get a b c _____ = _____ = _____ 6–1 2+3 1+4
r ◊ a = r a cos (45°) _______
__ 1 __ x+y=÷ x 2 + y2 ◊ ÷2 ◊ ___ 2 ÷ ______
x+y=÷ x 2 + y2
...(i)
r ◊ b = r b cos (60°) ______
1 3 x – 4y = ÷x 2 + y2 ◊ 5 ◊ __ 2 ______ 2 fi __ (3 x – 4 y) = ÷ x 2 + y2 5 From Eqs. (i) and (ii), we get fi
x i + y j r _______ = __ = ________ |r| ÷x 2 + y2
13 y i + y j _________ = __________ ÷169y 2 + y2
9. Given Let
...(ii)
x = 13 y
Thus,
13 i + j ____ = ______ . 170 ÷ a ◊ b = 0 = a ◊ c = b ◊ c |a| = |b| = |c| = l
Now, (a + b + c) ◊ a = |a + b + c||a| cos (q1) fi |a + b + c||a| cos (q1) = (a ◊ a + a ◊ b + a ◊ c) fi |a + b + c||a| cos (q1) = |a|2 l |a| fi cos (q1) = _________ = _________ |a + b + c| |a + b + c| l |b| Similarly, cos (q2) = _________ = _________ |a + b + c| |a + b + c| and
l |c| cos (q3) = _________ = _________ |a + b + c| |a + b + c|
Thus,
cos (q1) = cos (q2) = cos (q3)
fi
(q1) = (q2) = (q3).
___
a b __ c fi __ = __ = 5 5 5
10. Let r = x i + y j + z k such that |r| = ÷ 51 . It is given that
fi
Thus,
a = b = c = l (say) l (i + j + k) (i + j + k) r __ __ = __ = __________ = _________ . |r| l ÷3 ÷3
8. Let the vector be r = x i + y j. r We have to find = __ . |r|
Let
a = i + j and b = 3i – 4j
5.37
cos (q1) = cos (q2) = cos (q3)
a ◊ r b ◊ r c ◊ r fi ____ = ____ = ____ |a||r| |b||r| |c||r| a ◊ r b ◊ r ___ c◊r fi ____ = ____ = |a| |b| |c| fi
a ◊ r = b ◊ r = c ◊ r ( |a| = 1 = |b| = |c|)
x – 2y + 2z ________ y – 4x – 3z __ fi __________ = = 3 1 5
5.38 Integral Calculus, 3D Geometry & Vector Booster x – 2y + 2z ________ y – 4x – 3z __ fi __________ = = = l (say) 3 1 5 Solving, we get Also,
___
x + y + z = 51
fi
25l2 + l2 + 25l2 = 51
fi
51 l2 = 51
fi Also,
fi
l =1
fi
l = ± 1
fi
r = ± (– 5i + j + 5k)
fi
cos (q1) = cos (q2) = cos (q3)
c ◊ r ___ |c|
a ◊ r = b ◊ r = c ◊ r
fi
2x + y + z = 0 z __ 5 z __ = l (say) 1
= ± ( j + k)
13. We have, 1 |a| = ___ __ = |b| = |c| |( – )|2 2 ÷ = | | + | |2 –2 ( ◊ ) fi x+y=y+z=x+z fi
(
x + y = y + z = x + z = l (say)
Also, fi
= [1 + 1 – 2 ( ◊ )]
= 2 – 2 ( ◊ )
l x = – __ = y = z 2
= 2 – 2 cos q
= 2 (1 – cosq)
|r| = 4
2
2
2
x +y +z =4
l2 l2 __ l2 fi __ + __ + = 4 4 4 4 3l2 fi ___ = 4 4 16 fi l2 = ___ 3 fi
4__ l = ± ___ 3 ÷
4__ Therefore, r = ± ___ (i + j + k) 3 ÷ 12. Let
)
Solving, we get
a = (i + 2j + k),
( ) q fi | – | = 2 sin ( __ ) 2 q 1 fi sin ( __ ) = __ | – | 2 2 14. Given
...(i)
r ◊ c = 0
y x __ = ___ = 0 – 5 y x fi __ = ___ = 0 – 1 r Therefore, = __ |r|
a ◊ r b ◊ r ____ c ◊ r fi ____ = ____ = |a| |r| |b|r| |c| |r|
fi
3x – y – z = 0
Solving Eqs. (i) and (ii), we get
11. Let r = x i + y j + z k such that |r| = 4. It is given that
b ◊ r ____ = |b|
| |
r ◊ (a × b) = 0
x y z fi 1 2 1 = 0 112
2
a ◊ r fi ____ = |a|
c = (2i + j + k)
2
fi
and
such that r = (x i + y j + z k). It is given that r, a and b lie in the same plane, so
|r| = ÷ 51 2
b = (i + j + 2k)
and r be a vector
x = – 5l, y = l, z = 5l
2
q = 2 × 2 sin2 __ 2
◊ = 0 = ◊ = ◊
and | | = 1 = | | = | | We have,
( – ) ◊ ( – )
= ( ◊ – ◊ – ◊ + ◊ )
= (0 – 0 – ◊ + 0)
...(ii)
Vectors
= – | |2
= – 1.
15. Given
| |=1=| |=| + |
Also, fi fi
| + |=1
fi
| + |2 = 1 2
Also,
| | + | + 2 ( ◊ ) = 1
fi
1 + 1 + 2 ( ◊ ) = 1
fi
2 cos (q ) = 1 – 2 = – 1
fi
1 cos (q ) = – __ 2
Thus,
sin (q ) = ÷ 1 – cos2 (q )
fi
3 ÷ 1 ___ = sin (q ) = 1 – __ 4 2
fi ÷x 2 + y2 + z2 = 1
__________
÷
__
______
__
3 ÷ 16. It is given that, a ◊ i = |a| |i| cos (30°) = ___ 2 1 and b ◊ i = |b| |i| cos (120∞) = – __ 2 Clearly, the angle between a and b is 90°.
=1+1+0
=2
__
17. Given,
)
a2 __ 4 + a
b2 __ 4 – b
2 (a ◊ b) ______ 2 2 a b
1 = __ 2 + a
1 __ 2 – a
2 (a ◊ b) ______ 2 2 a b
)
Now,
|(w ◊ )| = |– 3| = 3.
19. Given
a ◊ b = 0 = a ◊ c
Also,
p 1 b ◊ c = |b||c| cos __ = __ 3 2
u ◊ = 0 x+y=0 x = – y
( )
|a + b + c|2 = |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) 1 = 1 + 1 + 1 + 2 0 + __ + 1 2 =4 |a + b + c| = 2.
(
...(i)
)
2 (a ◊ b + b ◊ c + c ◊ a) = 0 |a + b + c|2
= |a|2 + |b|2 +|c|2 + 2 (a ◊ b + b ◊ c + c ◊ a)
= 9 + 16 + 25 + 0
Thus,
= 50 2 |a + b + c| = 5.
| |
2
| |2 + | |2 + 2( ◊ ) = _________________ | |2 + | |2 – 2 ( ◊ )
1 + 1 + 2 cos (60°) = ________________ 1 + 1 – 2 cos (60°)
1 + 1 + 1 = _________ 1+1–1
= 3.
= x i + y j + z k such that | | = 1.
Now, fi fi
r = ± k
| + | + 2 _______ _____ = – | – |2
a–b 2 = _____ a b 18. Let
Thus,
21. We have,
a2 + b2 – 2 (a ◊ b) = _______________ a2 b2
(
0 + 0 + z2 = 1 z = ± 1
20. Given Now,
fi | + | = ÷ 2
b 2 __ 2 = b
fi fi
Thus,
| + |2 = | |2 + | |2 + 2( ◊ )
(
x2 + y2 + z2 = 1
We have,
a __ 2 – a
fi
Now,
( ◊ ) = 0
| |=1 __________
2
fi
Thus,
...(ii)
From Eqs. (i) and (ii), we get x=0=y
We have,
v ◊ = 0 x–y=0 x = y
5.39
5.40 Integral Calculus, 3D Geometry & Vector Booster
| |
__ + Thus, _____ = ÷ 3 –
fi b2 + c2 + 2 b c cos (p – A) = a2
22. Given,
fi 2 b c cos (A) = b2 + c2 – a2 (a + b + 3c) = 0
fi
(a + b + c) = (– 2c)
fi b2 + c2 + 2 b c cos (A)= a2
We have,
|a + b + c|2 = |(– 2c)|2
b2 + c2 – a2 fi cos (A) = ___________ 2 b c Hence, the result. _ __›
__› _› _
___›
___›
_ ›
_ __›
_›
25. Let BC = a , CA = b and AB = c
fi |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 4|c|2 68 17 fi 1 + 16 + ___ + 2 (a ◊ b + b ◊ c + c ◊ a) = ___ 9 9 68 – 17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = _______ – 17 9 51 fi 2 (a ◊ b + b ◊ c + c ◊ a) = ___ – 17 9 17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = ___ – 17 3 34 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – ___ 3
We have,
17 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – ___ 3 23. Let the resultant force be F such that
fi b + c = – a
F = F1 + F2 + F3
i.e.
F = 2j – k
Now,
AB = 2 i + 4 j – k
= (2 j – k) ◊ (2i + 4j – k)
= 8 + 1 = 9.
24.
_›
_›
_ ›
_›
_ ›
_ ›
_›
fi a + b + c = 0 › _ › _
_ ›
_ › _ ›
_ › _ ›
fi a ◊ b + a ◊ c = – a ◊ a fi
a b cos (p – C ) + a c cos (p – B ) = – a2
fi
– a b cos (C) – a c cos (B) = – a2
fi
\ Work done = (Force) ◊ (Displacement)
___›
BC + CA + AB = 0
a = b cos (C) + c cos (B)
26.
_ › ___› ___› _ _ › ___› › Let BC = a , CA = b and AC = c
_ __›
Here, OP = a + b ___›
and OQ = c + d
___› ___› ___› We have BC + CA + AB = _ › _ _ › › fi a + b + c = 0 _ › _› _› fi b + c = – a _›
_ ›
_ ›
fi ( b + c ) = (– a ) 2 2
fi b2 + c2 + 2 b ◊ c = a2
___› ___›
0
___
___
| › | | › |
Now, OP ◊ OQ = OP OQ cos (– POQ) Now,
(a + b ) ◊ (c + d )
______
______
= ÷a 2 + b2 ÷c 2 + d2 cos (A – B) fi
cos (A – B)
Vectors
( a + b ) ◊ ( c + d ) ______ ______ = __________________ ÷ c 2 + d2 ÷ a 2 + b2 a c + b d ______ ______ = _______________ 2 ÷ c 2 + d2 ÷ a + b2 d a c b ______ ______ = _______ ______ ◊ _______ + ________ _______ ◊ _______ 2 2 2 2 2 2 2 ÷ c + d ÷a + b ÷c + d2 ÷ a + b = cos A ◊ cos B + sin A ◊ sin B 27.
30. Now,
5.41
| |
i a × b = 1 3
j 2 – 2
k 3 1
= 8 i + 8 j – 8 k
= 8 (i + j – k)
Area of the parallelogram = |a × b| __
= 8÷3 sq.u.
31. Now,
| |
i a × b = 1 3
j 1 – 4
k 1 1
= 5 i + 2 j – 7 k
Area of the parallelogram Here, and
___› OP =
___› OQ =
1 = __ |a × b| 2
___________ 1 = __ × ÷ 25 + 4 + 49 2
78 ÷ = ____ sq.u. 2
32. Let and Now, and
a = 3 i – j + 2 k, b=i–j–k c = 4 i – 3 j + k. a b = – 2 i – 3 k a c = i – 2 j – k
\
i j a b × a c = – 2 0 1 – 2
a + b c + d
___
Now, ___› ___›
___
___
| › | | › |
OP ◊ OQ = OP OQ cos (– POQ) fi
a c – b d = (OP) (OQ) cos (A + B)
fi
a c – b d cos (A + B) = _________ (OP) (OQ)
a c b d = ___ ◊ ____ – ___ ◊ ____ OP OQ OP OQ
= cos (A) cos (B) – sin (A) sin (B)
28. We have,
| |
i a × b = 2 1
j – 1 – 3
k 1 –5
= 8 i + 11 j – 5 k
1 29. Area of a triangle = __ |a × b| 2 Now,
| |
i a × b = 3 5
j 4 7
k 0 0
= (21 – 20) k =k
1 Thus, the area of the triangle = __ sq.u 2
| |
k – 3 – 1
= – 6 i – 5 j + 4 k
1 Area of the triangle = __ |a b × a c| 2 ___________ 1 = __ ÷36 + 25 + 16 = 2 33. Let a = 2 i – j + k and b = 3 i – 4 j – k. A unit vector perpendicular to a and b (a × b) = ± _______ |a × b| Now,
| |
i (a × b) = 2 3
j – 1 – 4
k 1 – 1
= 5i + 5j – 5k
= 5 (i + j – k)
___
77 ÷ ____ sq.u 2
5.42 Integral Calculus, 3D Geometry & Vector Booster (a × b) Thus, the unit vector = ± ______ |a × b|
and Therefore,
1__ = ± ___ (i + j – k). 3 ÷
|(a × k)|2 = a12 + a22
|
|2
| a × |2 + a × + | a × | 2
34. We have,
= (a22 + a32 + a12 + a32 + a12 + a22)
a × (b + c) + b × (c + a) + c × (a + b)
= 2 (a12 + a22 + a32)
= 2a2
= a × b + a × c + b × c + b × a + c × a + c × b = a × b – c × a + b × c – a × b + c × a – b × c = 0 35. Given
a × b = c × d
...(i)
and
a × c = b × d
...(ii)
(b × c) = ± ______ |b × c|
Subtracting Eq. (ii) from Eq. (i), we get
38. Given a ◊ b = 0 = a ◊ c Thus, a is perpendicular to b and c. A unit vector perpnedicular to b and c
a×b–a×c=c×d–b×d
fi
a × (b – c) = (c – b) × d
fi
a × (b – c) – (c – b) × d = 0
fi
a × (b – c) + (c – b) × d = 0
fi
a × (b – c) – d × (b – c) = 0
fi
(a – d) × (b – c) = 0
(b × c) = ± __________ p |b| |c| sin __ 6 (b × c) = ± ______ 1/2
( )
= ± 2 (b × c).
(a + b + c) = 0
fi
b × (a + b + c) = b × 0 = 0
fi
b×a+b×b+b×c=0
fi
b×a+b×c=0
39. Given, fi \ fi fi 40. Given, fi \ fi fi Also,
fi
b × c = – b × a
– 6 l – 6l = 1
fi
fi
b × c = a × b
fi
1 l = – ___ 12
Thus,
1 R = – __ (–3 i + 4 j – 6k). 2
Thus, (a – d) is parallel to (b – c). 36. Given, (a + b + c) fi
a × (a + b + c) = a × 0 = 0
fi
a×a+a×b+a×c=0
fi
a×b+a×c=0
fi
a × b = – a × c
fi
a × b = c × a
...(i)
Also,
...(ii)
From Eqs. (i) and (ii), we get
a×b=b×c=c×a
37. Let
a = a1i + a2j + a3k
Now,
(a × i)
= (a1i × i + a2j × i + a3 k × i)
= (– a2k – a3j)
Thus,
|(a × i)|2 = a22 + a32
41. Now,
2
2
Similarly, |(a × j)| = a1 + a3
2
a×b=a×c a × (b – c) = 0 (b – c) is parallel to a (b – c) = l a b – c + l a, l ŒR R× R× R is R= R=
B=R×C (B – C) = 0 parallel to (B – C) l (B – C) l (– 3 i + 4 j – 6 k)
R ◊ A = 1
| |
i a × b = 1 0
j 1 1
k 0 1
=i–j+k
Unit vector perpendicular to a and b
...(i)
Vectors
a×b = ± ______ |a × b|
= ± (i – j + k) Thus, the number of vectors of unit length = 2. 42. As we know that, if a, b and c form a right handed system, then
| |
c=a×b
fi
i c = x 0
43. Given,
j y 1
k z = – z i + x k 0
5.43
Comparing, we get
(z – y) = 0, (x – z) = 1, (y – x) = – 1
Solving, we get fi
5 2 x = __ , y = __ = z 3 3
Thus,
1 b = __ (5 i + 2 j + 3 k) 3
46. The vector [a – (a ◊ b) b] lies in the plane of a and b and the vector (a × b) is perpendicular to each of a and b Thus, the angle between (a – (a ◊ b) b) and (a × b) = 90°.
r × a = b × a and r × b = a × b
fi
r × a = – a × b and r × b = a × b
fi
r × a = – r × b
fi
r×a+r×b=0
fi
r × (a + b) = 0
\
r is parallel to (a + b)
fi
r = l (a + b)
fi
r = l (a + b)
fi
r × b = l (a × b + b × b)
fi
r × b = l (a × b)
Here,
fi
a × b = l (a × b)
Given a + b = 3 i + 4 j
fi
l=1
Now,
|a + b|2 = |a|2 + |b|2 + 2 (a ◊ b)
Therefore, r = (a × b)
fi
|a + b|2 = |a|2 + |b|2 = 2 |a|2
fi
2 |a|2 = |a + b|2 = |a + b|2 = (32 + 42) = 25
fi
25 |a|2 = ___ 2
= 3 i + j – k
44. Given that
r × a = b and r ◊ a = 0
fi
r×a=b
fi
a × (r × a) = a × b
fi
(a ◊ a) r – (a ◊ r) a = a × b 2
47.
|a| = |b| and (a ◊ b) = 0
25 \ Area of a square = |a|2 = ___ . 2
48. Let
A = (1, – 2, 2) and B = (2, – 1, 3)
Thus,
r = BA = OA – OB = – i – j – k
fi
|a| r – 0 = a × b
Therefore, Momentum about the point B
fi
a×b r = _____ |a|2
=r×F
45. Given
a × b = c and a ◊ b = 3
Let
b = x i + y j + z k
i j k = – 1 – 1 – 1 3 2 – 4
Thus,
x+y+z=3
= 6 i – 7 j + k
49. Here,
PQ = – 6 i – j – k
| |
i and 1 x fi
j 1 y
k 1 z
(z – y) i + (x – z) j + (y – x) k = j – k
| |
Thus, moment of the couple
=M
= PQ × F
5.44 Integral Calculus, 3D Geometry & Vector Booster
| |
___›
i j k = – 6 – 1 – 1 5 0 1
Here, OP = a + b
= – i + j + 5 k
Now,
OP × OQ = (a d – b c)
50. Here,
r = (i – j + 2 k) – (2 i + j + k)
fi
[(OP) (OQ) sin (–POQ)] = (a d – b c)
fi
r = i – 2 j + k
fi
(a d – b c) – sin (A – B) = _________ (OP) (OQ)
a d ___ b c = ___ ◊ ____ ◊ ◊ ____ OP OQ OP OQ
= cos A ◊ sin B – sin A ◊ cos B
fi
sin (A – B) = sin A ◊ cos B – cos A ◊ sin B
___›
and OQ = c + d
Thus, torque about the point
| |
=r×F
i = 1 4
= – 2 i – 3 j + 8 k.
51. Given
D ABC, a = BC, b = CA, c = AB
j k 1 – 2 0 1
Hence, the result. 54. We have, [a + b, + b + c, c + a] = (a + b) ◊ (b + c) × (c + a) = (a + b) ◊ (b × c + b × a + c × c + c × a) = (a + b) ◊ (b × c + b × a + c × a) = (a ◊ b × c + a ◊ b × a + a ◊ c × a)
_ ›
_›
+ (b ◊ b × c + b ◊ b × a + b ◊ c × a)
= [a, b, c] + [a, b, c]
_ ›
Clearly, a + b + c = 0
= 2 [a, b, c]
55. We have,
_ _ › › _ _ _ _› › › › fi a × b = b × c = c × a
_ _ › › _ _ _ _› › › › fi a × b = b × c = c × a
|
| |
|
|
|
fi | a b sin (p – C) | = | b c sin (p – A) | = | c a sin (p – B) |
fi
a b sin C = b c sin A = c a sin B
a b sin C _______ b c sin A _______ c a sin B fi _______ = = a b c a b c a b c sin C ____ sin A ____ sin B fi _____ = a = c b sin C sin A ____ sin B _____ fi ____ = = c a b Hence, the result. 52.
[2 a – b, 2b – c, 2c – a] = (2 a – b) ◊ (2b – c) × (2c – a) = (2 a – b) ◊ (4 b × c – 2 b × a + c × a) = (2 a – b) ◊ (4 b × c + 2 a × b + c × a) = (8 a ◊ b × c – b ◊ c × a) = 7 [a, b, c] = 0, a, b and c are coplaner vectors 56. We have, [a, b + c, a + b + c] 57.
= a ◊ (b + c) × (a + b + c) = a ◊ (b × a + b × c + c × a + c × b) = (a ◊ b × c + a ◊ c × b) = (a ◊ b × c – a ◊ c × b) = 0. We have, [a + b + c, a + b, a + c] = (a + b + c) ◊ (a + b) × (a + c) = (a + b + c) ◊ (a + c + b × a + b + c) = (a ◊ b × c + b ◊ a × c + c ◊ b × a)
Vectors
58.
= (a ◊ b × c – a ◊ b × c + c ◊ b × a) = [a, b, c] – [a, b, c] + [a, b, c] = [a, b, c] We have, [a – b, b – c, c – a] = (a – b) ◊ (b – c) × (c – a) = (a – b) ◊ (b × c – b × a + c × a) = (a – b) ◊ (b × c + a × b + c × a) = (a ◊ b × c – b ◊ c × a) = [a, b, c] – [a, b, c] = 0.
59. We have,
a ◊ (b × c) = |a| |b| |c|
fi
(a × b) ◊ c = |a| |b| |c|
fi
|a| |b| |c| sin (q ) cos (q ) = |a| |b| |c|
fi
sin (q ) cos (j) = 1
fi
sin (q ) = 1, cos (j) = 1
fi
p q = __ , j = 0 2
p Hence, the angle between a and b is q = __ 2 60. We have, 1 __ (b × c) ◊ (a + b + c) 2 1 = __ ((b × c) ◊ a + (b × c) ◊ b + (b × c) ◊ c) 2 1 = __ ((b × c) ◊ a + 0 + 0) 2 1 = __ (b × c) ◊ a 2 1 = __ [a, b, c] 2 1 = __ × 4 = 2 2 61. Since the given vectors are coplanar, so
| |
2 1 3 fi fi
–1 – 1 – 2
l 2 = 0 1
6 – 5 + l (– 2 + 3) = 0 1+l=0
fi l = – 1 62. The volume of the parallelopiped
| |
= [a, b, c]
3 = 1 1
– 2 1 1 1 1 – 2
= 3 (– 3) + 2 (– 3) + 0
= |– 3| = 3 c.u.
5.45
63. We have,
2 (a + b + c) = 6 i + 8 j – 12 k
fi
(a + b + c) = 3 i + 4 j – 6 k
Thus, a = i + j – k, b = k, c = 2 i + 3 j – 5 k Hence, the volume of the parallelopiped
| |
1 = 0 2
1 0 3
64. Given
[a, b, c] = 5
Now,
[3 (a + b), (b + c), 2 (c + a)]
– 1 1 = 1 c.u. – 5
= 3.2 [(a + b), (b + c), 2 (c + a)]
= 6 [(a + b), (b + c), 2 (c + a)]
= 6 × 2 [a, b, c]
=6×2×5
= 60. 65. The volume of the parallelopiped,
| |
1 V = 0 a
a 1 0
1 a 1
= 1 + a (a2 – 1)
= a3 – a + 1
dV fi ___ = 3a2 – 1 da For maximum or minimum, dV ___ = 0 da gives fi
3a2 – 1 = 0
fi
1__ a = ± ___ 3 ÷ –
+
+
1__ 1__ ___ – ___ 3 ÷3 ÷
1 By sign scheme, the point of minima is a = ___ __ . 3 ÷ 67. Since the given vectors are coplanar, so
| | |
1 1 b 1 = 0 1 c – a 1 –a a 1 C Æ C 2 – C1 fi 1 b = 0, 2 0 – 1 C3 Æ C3 – C1 1 c–1 0
a 1 1
|
(
)
5.46 Integral Calculus, 3D Geometry & Vector Booster __
fi a (b – 1) (c – 1) (a – 1) (c – 1)
3 ÷ __ = ____ 2÷2
fi (1 – b) (1 – c) + (1 – a) (1 – c)
70. Let
a × b = a1i + a2j + a3k
+ (1 – a) (1 – b) = 0 Dividing both the sides by (1 – a) (1 – b) (1 – c), we get a 1 1 _____ + _____ + _____ = 0 1–a 1–b 1–c a 1 1 fi 1 + _____ + _____ + _____ = 1 1–a 1–b 1–c 1 1 1 fi _____ + _____ + _____ = 1 1–a 1–b 1–c
We have,
[a, b, i] i + [a, b, j] j + [a, b, k] k
68. Let a = a1i + a2j + a3k,
= a1i + a2j + a3k
b = b1i + b2j + b3k and c = c1i + c2 j + c3k We have, [a, b, c]2
= (a × b) Hence, the result. 71. Let a = a1i + a2 j + a3k,
+ (1 – b) (1 – b) = 0.
(
)
| | | |
= [a, b, c] [a, b, c]
a 1 b 1 = c1
a 2 b 2 c2
a3 b3 × c3
a 1 b 1 c1
a 2 b 2 c2
a 3 b 3 c3
a12 + a2 2 + a32 a1b1 + a2 b2 + a3 b3 a1c1 + a2 c2 + a3 c3 b1a1 + b2 a 2 + b3 a3 b12 + b2 2 + b32 b1c1 + b2 c2 + b3 c3 c1a1 + c2 a2 + c3 a3 c1b1 + c2 b2 + c3 b3 c12 + c2 2 + c32
|
|
a ◊ a a ◊ b a ◊ c = b ◊ a b ◊ b b ◊ c c ◊ a c ◊ b c ◊ c
|
69. As we know that,
|
=
1 2 1 2
1 2
1 2
1
1
1 2
= a1i
Similarly, [a, b, j] j = a2j and
[a, b, k] k = a3k
Thus,
b = b1i + b2 j + b3k and c = a = c1i + c2 j + c3k Also, let m = m1i + m2 j + m3k and
n = n1i + n2 j + n3k
Now,
[a, b, c] (m × n)
| | | |
b a c = a ◊ m b ◊ m c ◊ m a ◊ n b ◊ n c ◊ n
|
a 2 b 2 c2
a3 b 3 × c3
a 1 = b 1 c1
i m 1 n1
|
j m 2 n2
k m 3 n3
r ◊ (b × c) = x a ◊ (b × c) + y b ◊ (b × c) + z c ◊ (b × c)
1
( ) (
72. Given r = x a + y b + z c Taking scalar product of (b × c)
a ◊ a a ◊ b a ◊ c [a, b, c]2 = b ◊ a b ◊ b b ◊ c c ◊ a c ◊ b c ◊ c 1
[a, b, i] i = ((a × b) ◊ i) i
)
1 1 1 __ 1 = 1 – __ + __ __ – 2 2 4 2
fi
r ◊ (b × c) = x a ◊ (b × c)
fi
[r, b, c] = x [a, b, c]
fi
[r, b, c] x = _______ [a, b, c]
[r, c, a] [r, a, b] Similarly, y = _______ & z = _______ [a, b, c] [a, b, c]
3 1 1 __ = __ – __ = 2 8 8
Hence,
3 ÷ Thus, [a, b, c] = ____ __ 2÷2
73. Given,
__
Therefore, the volume of the parallelopiped
[r b c] r = ______ a + [a b c]
[r c a] ______ b + [a b c]
1 [a b, a × b] = __ 4
fi
1 (a × b) ◊ (a × b) = __ 4
[r a b] ______ c [a b c]
Vectors
fi fi fi fi fi 74. Given fi fi
1 |(a × b)|2 = __ 4 1 __ |(a × b)| = 2 1 a b sin (q ) = __ 2 1 sin (q ) = __ 2 p q = __ . 6 = x (a × b) + y (b × c) + z (c × a) ◊ c = x (a × b) ◊ c + y (b × c) ◊ c + z (c × a) ◊ c
Now, |2a + 5b + 5c| = |2a + 5 (b + c)| = |2a + 5 (– a) = |– 3 (a)| = |– 3| |a| = 3. 77. Let Now,
= x (a × b) ◊ c = x [a b c] ◊ c x = _______ [a, b, c]
a = i + j, b = j + k and c = k + i
| |
i (a × b) = 1 0
j 1 1
k 0 = (i – j + k) 1
(a × b) 1 = _______ = ___ __ (i – j + k) |(a × b)| ÷3
◊ c ◊ b Similarly, y = _______ , z = _______ [a, b, c] [a, b, c]
similarly,
(b × c) 1 = _______ = ___ __ (i + j – k) |(b × c)| ÷3
Thus,
x+y+z
and
(c × a) 1 = _______ = ___ __ (i + j – k) |(c × a)| ÷ 3
◊ c ◊ a ◊ b = _______ + _______ + _______ [a, b, c] [a, b, c] [a, b, c]
Volume of a parallelopiped
◊ (a + b + c) = ____________ [a, b, c]
1 1 1 1__ ___ = ___ __ ◊ ___ ◊ __ 1 ÷3 – 1 3 ÷3 ÷
◊ (a + b + c) = ____________ 1/8
= 8 a ◊ (a + b + c).
1 1__ = ____ 1 3÷3 – 1
2 = ____ __ (1 + 1) 3÷3
75. Volume of the tetrahedron
|
| | 0 2 0
1 = __ [a, b, c] 6
1 1 = __ × 1 6 1
4__ = ____ c.u 3÷3
78. Given
a + b + c = a d
and
b + c + d = b a
1 1 1 – 1 2 – 1
|
– 1 1 1 – 1 1 1
1 (C2 Æ C2 + C1) – 1 1
| |
1 = __ (– 1 + 2 + 3) 6
fi
d = – b – c – b a
fi
a d = – a b – a c – a b a
2 = __ c.u. 3
fi
a + b + c = a b – a c – a b a
76. Given |a – b|2 + |b – c|2 + |c – a|2 = 9 fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3 fi (a2 + b2 + c2) + 2 (a ◊ b + b ◊ c + c ◊ a) = 3 – 3 fi |a + b + c|2 = 0 fi (a + b + c) = 0
5.47
Comparing, the co-efficients of the vectors, we get
a = – 1, a b = – 1
fi
a = – 1, b = 1
Now. a + b + c + d + 4 = a d + d + 4 = – d + d + 4 = 4. 79. Let a = 3 i + 2 j + 6 k, b = 2 i + j + k
5.48 Integral Calculus, 3D Geometry & Vector Booster c = i – j + k.
and
a × (b × c) We have to find ± ___________ . |a × (b × c)|
| | | |
Now,
i (b × c) = 2 1
j 1 – 1
i a × (b × c) = 3 2
j 2 – 1
(21 i – 7 k) ________ = ± _________ ÷441 + 49
80. We have, a × (a × b) = (a ◊ b) a – (a ◊ a) b = a – |a|2 b = a – 3 b
| | j 1 1
k 1 – 1
= – 2 i + j + k Thus, from Eq. (i), we get
3 b = (i + j + k) – (– 2 i + j + k)
fi
3 b = 3 i
fi
b=i
81. We have, fi
1 (a × (b × c)) = __ b 2 1 (a ◊ c) – (a ◊ b) c = __ b 2 1 = __ , (a ◊ b) = 0 2
1 1 (a ◊ c) = ___ __ , (a ◊ b) = ___ __ 2 2 ÷ ÷
fi
1 1 a c cos(q1) = ___ __ and a b cos (q2) = ___ __ 2 2 ÷ ÷
fi
1 1 cos (q1) = ___ __ and cos (q2) = ___ __ 2 2 ÷ ÷
fi
(3 i – 7 k) ___ = ± ________ 10 ÷
i a × (a × b) = 1 0
fi
fi
(21 i – 7 k) ___ = ± _________ 7÷10
Now,
1 cos (q1) = __ and cos (q2) = 0 2 p p (q1) = __ , (q2) = __ 3 2
82. We have, 1 (a × (b × c)) = ___ __ (b + c) 2 ÷ 1 fi (a ◊ c) b – (a ◊ b) c = ___ __ (b + c) 2 ÷
k 6 – 3
3 b = a – a × (a × b)
fi
Hence, the result.
= 21 i – 7 k Thus, a × (b × c) ± ___________ |a × (b × c)|
fi
1 a c cos (q1) = __ and a b cos (q2) = 0 2
fi
k 1 1
= 2 i – j – 3 k Also,
fi
...(i)
p p q1 = __ and q2 = __ 4 4 p q2 = __ 4
p Hence, the angle between a and b is __ 4 83. Given, a × (a × c) + b = 0 fi a × (a × c) = – b fi |a × (a × c)| = |– b| p fi |a| |(a × c)| sin __ = 1 2 fi |a| |(a × c)| = 1
( )
fi
|(a × c)| = 1
fi
a c sin q = 1 1 sin q = __ 2 p q = __ 6
fi fi
p Hence, the angle between a and c is __ . 6 84. We have, a × (b × c) ◊ (a × b) × c = 0 fi {(a ◊ c) b – (a ◊ b) c}◊{(c ◊ a) b – (c ◊ b) a} = 0 fi ((a ◊ c) b)2 – (a ◊ c) (c ◊ b) (b ◊ a) – (a ◊ b) (c ◊ a) (c ◊ b) + (a ◊ b)(c ◊ b) (c ◊ a) = 0 fi ((a ◊ c) b)2 – (a ◊ c) (c ◊ b) (b ◊ a) = 0 fi (a ◊ c) {(a ◊ c) b2 – (a ◊ b) (b ◊ c) = 0 fi (a ◊ c) = 0, {(a ◊ c) b2 – (a ◊ b) (b ◊ c)} = 0 Hence, (a ◊ c) = 0
Vectors
85. Given, 1 (a × b) × c = __ |b| |c| a 3 1 fi c × (a × b) = – __ |b| |c| a 3 1 fi (c ◊ b)a – (a ◊ a) b = – __ |b| |c| a 3 1 __ fi (c ◊ b) = – |b| |c| a 3 1 fi |c| |b| cos (q) = – __ |b| |c| 3 1 fi cos q = – __ 3 __ _____ 1 2÷2 fi sin q = 1 – __ = ____ 9 3 Hence, the result. 86. Given,
÷
a ◊ b = 0
r = m a + n b + p (a × b) ...(i) r ◊ a = m (a ◊ a) + n (a ◊ b) + p (a ◊ (a × b)) m=0 r ◊ b = m (a ◊ b) + n (b ◊ b) + p (b ◊ (a × b)) 1 = 0 + n (b ◊ b) + 0 n=1 r ◊ a × b = m (a ◊ a × b) + n (b ◊ a × b)
Let Now, fi Also, fi fi Again,
+ p [(a × b) ◊ (a × b)]
fi
1 = 0 + 0 + p |(a × b)|2
fi
p=1
Hence, from Eq. (i),
r = b + (a × b)
87. We have,
1 a × (b × c) + b × (c × a) = __ a 2
1 fi (a ◊ c) b – (a ◊ b) c + (b ◊ a) c – (b ◊ c) a = __ a 2 1 fi (a ◊ c) = 0, (a ◊ b) = 0, (b ◊ a) = 0, (b ◊ c) = – __ 2
Thus,
|(a × c)|
( )
p = a c sin __ 2 = 7. 1. 1
= 7.
88. We have, (d + a) ◊ [a × b × (c × d)]
= (d + a) ◊ (a × [(b ◊ d) c – (b × c) d]
= (d + a) ◊ [(a × c) (b ◊ d) – (a × d) (b ◊ c)]
= [d ◊ (a × c) (b ◊ d) + a ◊ (a × c) (b ◊ d)
– d ◊ (a × d) (b ◊ c) – a ◊ (a × d) (b ◊ c)]
= [a, c, d] (b ◊ d)
89. Given, fi
px + (x × a) = b p (a ◊ x) + a ◊ (x × a) = a ◊ b
fi
p (a ◊ x) + 0 = a ◊ b
fi
p (a ◊ x) = a ◊ b (a ◊ b) p = _____ (a ◊ x)
fi
px + (x × a) = b p (a × x) + a × (x × a) = a × b
fi
p (a × x) + (a ◊ a) x – (a ◊ x) a = a × b
fi
(a ◊ a) x = (a ◊ x) a – a × b – p (a × x)
fi
(a ◊ b) (a ◊ a) x = _____ p a – a × b – p (a × x)
fi
(a ◊ b) (a ◊ a) x = _____ p a – a × b – p (p x – b)
fi
(a ◊ b) (a2 + p2) x = _____ p a – a × b – p b
fi
(a ◊ b) _____ p a – a × b – p b x = ___________________ (a2 + p2)
90. Given fi fi
[x a b] = 0 x, a, b are coplanar vectors x = m a + n b
(a ◊ c) = 0
fi
a c cos (q1) = 0
fi
x ◊ a = m (a ◊ a) + n (b ◊ a)
fi
cos (q1) = 0
fi
x ◊ a = m (a ◊ a) + n (b ◊ a)
fi
0 = m |a|2 + 0
fi
m=0
Also,
x = m a + n b
fi
x ◊ b = m (a ◊ b) + n (b ◊ b)
fi
( )
p = cos __ 2
( )
p q1 = __ 2
...(i)
Also, fi
Now,
5.49
5.50 Integral Calculus, 3D Geometry & Vector Booster fi
1 = 0 + |b|2
fi
(x × a) × a = b × a
fi
1 n = ___ 2 |b|
fi
a × (x × a) = – b × a = a × b
fi
(a ◊ a) x – (a ◊ x) a = a × b
Thus,
b x = m a + n b = ___ 2 = |b|
fi
|a|2 x = a × b
fi
a × b x = _____ 2 . |a|
b ___ |b|
91. We have, [a × (b + c), b × (c – 2a), c × (a + 3b)] = [a × b + a × c, b × c – 2b × a, c × a + 3c × b] = [a × b – c × a, b × c + 2a × b, c × a – 3b × c] = [x – z, y + 2x, z – 3y] where x = a × b, y = b × c, z = c × a
= (x – z) ◊ (y + 2x) × (z – 3y)
96. Given, fi
r×b=a×b (r – a) × b = 0
fi fi
(r Рa) = l b, l ΠR r = a + l b
fi r ◊ c = (a ◊ c) + l (b ◊ c)
= (x – z) ◊ (y × z – 6 x × y + 2x × z) = x ◊ (y + z) + 6 z ◊ (x + y)
fi
(a ◊ c) l = – _____ (b ◊ c)
Thus,
(a ◊ c) r = a – _____ b. (b ◊ c)
= 7 x ◊ (y × z) = 7 [x y z] 2
= 7 [a b c] 92.
97. Given,
= 28 We have, [(a – b), (a – b – c), (a + 2 b – c)] = (a – b) ◊ (a – b – c) × (a + 2 b – c)
x a + y b + z c = d Taking scalar product of (b × c), we get x a ◊ (b × c) + y b◊ (b × c) + z c◊ (b × c) = d ◊ (b × c)
= (a – b) ◊ [2 a × b – a × c – b × a + b × c – c × a – 2c × b] = a ◊ (b × c) – 2a ◊ (c × b) + b ◊ (a × c) + b ◊ (c × a) = a ◊ (b × = 3 [a, b, = 15 93. We have,
c) + 2a ◊ (b × c) – b ◊ (c × a) + b ◊ (c × a) c]
94. Given,
= 576 × 4 = 2304
[(2a × 3b), (3b × 4c), (4c × 2a)] = [x × y, y × z, z × x] where 2a = x, 3b = y, 4c = z = [x, y, z]2 = [2 a, 3 b, 4 c]2 = 4. 9. 16 [a, b, c]2
x×b=a×b fi (x – a) × b = 0 fi (x – a) = l b, fi x = a + l b. 95. We have, x×a=b
( )
fi
x a ◊ (b × c) + 0 + 0 = d ◊ (b × c)
fi
x a ◊ (b × c) = d ◊ (b × c)
fi
d ◊ (b × c) [d b c] x = ________ = ______ a ◊ (b × c) [a b c]
Similarly, we can easily find that [d c a] [d a b] y = ______ and z = ______ [a b c] [a b c] 98. We have, x+y=a fi
(x + y) ◊ a = a ◊ a
fi
x ◊ a + y ◊ a = a ◊ a
fi
1 + y ◊ a = |a|2
...(i)
Also, x × y = b
where l ΠR
fi
a × (x × y) = a × b
fi
(a ◊ y) x – (a ◊ x) y = a × b
fi
(|a|2 – 1) x – y = a × b [from Eq. (i)] ...(ii)
Again x + y = a Adding Eqs. (ii) and (iii), we get
(|a|2) x = (a × b + a)
...(iii)
Vectors
(a × b + a) x = __________ |a|2
fi
Thus,
From Eq. (iii), we get (a × b + a) y = a – x = a – __________ |a|2
Hence,
| | j 3 3
k 0 4
= 12 i – 8 j + 6 k π 0
(a × b + a) (a × b + a) x = __________ , y = a – __________ 2 |a| |a|2
a = 2 i + 3 j and b = 3 j + 4 k i a × b = 2 0
5.51
3. Resultant of P and Q
AB = 3 i + 7 j – 5 k
= (2 i – 5 j + 6 k) + (– i + 2 j – k) = (i – 3 j + 5 k). Displacement, = AB = (6 i + j – 3 k) – (4 i + 3 j – 2 k) = (2 i + 4 j – k)
BC = – 7 i – j + 5 k
Workdone = Force ◊ Displacement
CA = 4 i – 6 j
= (i – 3 j + 5 k) ◊ (2 i + 4 j – k) = |2 – 12 – 5| = |– 15| = 15units. 4. We have,
1. We have,
and
Unit vectors along A B, B C, C A are 1 ___ ____ (3i + 7j – 5k) 83 ÷ 1 ___ ____ (– 7i + j + 5k) ÷75
(a × b)2 = (a b sin q )2
= a2 b2 sin2 q
1 ___ and ____ (4 i – 6 j). ÷52
= a2 b2 (1 – cos2q )
= a2 b2– a2 b2 cos 2q
Since the forces P, Q, R are of 15 KN each, so
= a2 b2 – (a ◊ b)2
15 P = ____ ___ (3 i + 7 j – 5 k) 83 ÷
15 Q = ____ ___ (– 7 i – j + 5 k) ÷75
= (b × c)
= – (c × b)
15 and R = ____ ___ (4 i – 6 k) ÷52 Thus, the resultant S is given by
fi
(a × b) + (c × b) = 0
fi
(a × c) × b = 0
15 15 = ____ ___ (3 i + 7 j – 5 k) + ____ ___ (– 7 i – j + 5 k) 83 ÷ ÷75 15 ___ (4 i – 6 j) + ____ ÷52
) (
)
3 6 7 7 4 1 = 15 ____ ___ – ____ ___ + ____ ___ i + ____ ___ – ____ ___ + ____ ___ j 83 ÷75 83 ÷75 ÷52 ÷52 ÷ ÷
(
5 ___ + – ____ + 83 ÷
(a × b) = (b × c) π 0
fi a × c = k b, where k is a scalar. 6. Sum of the vectors
= P + Q + R
[ (
5. Given,
)]
= (2 i + 4 j – 5 k) + (S i + 2 j + 3 k)
= (2 + S) i + 6 j – 2 k Unit vector parallel to the sum of the vectors
(2 + S) i + 6 j – 2 k _______________ = _________________ ÷(2 + S)2 + 36 + 4
(2 + S) i + 6 j – 2 k = _________________ ___________ 44 ÷ S 2+ 4S +
5 ____ ___ k ÷75 Also, it is given that
2. As a × b = 0 does not imply that the vectors a and b are null vectors, let
(2 + S) i + 6 j – 2 k ___________ (i + j + k) ◊ ________________ = 1 44 ÷ S 2 + 4S +
5.52 Integral Calculus, 3D Geometry & Vector Booster ___________
fi
(2 + S) + 6 – 2 = ÷ S 2 + 36 + 44
Thus, the vector r is
fi
(S + 6)2 = (S2 + 4S + 44)
fi
S2 + 12 S + 36 = (S2 + 4S + 44)
10. We have |a| = ÷ 144 + 16 + 9 = 13
fi
8S = 8
___________
fi S=1 7. Since the vectors (2i – j + k),
____________
____
|b| = ÷ 64 + 144 + 81 = ÷ 289 = 17
and
|c| = ÷ 1089 + 16 + 576 = ÷ 1681 = 41
_______________
_____
Thus, the length of the edges are 13, 17 and 41 respectively. \ Area of the faces = |a × b|, |b × c|, |c × a|
(i + 2j – 3k) and (3i + aj + 5k)
| |
are coplanar, so,
2 –1 1 1 2 –3 = 0 3 a 5 fi
= ± (– 5i + j + 5k).
3 (3 – 2) – a (– 6 – 1) + 5(4 + 1) = 0
= 220, 435, 455
Volume of the parallelopiped = [a, b, c]
|
fi
3 + 7a + 25 = 0
fi
7a = – 28
fi
a = – 4.
8. Given,
|a + b| = |a – b|
fi
|a + b|2 = |a – b|2
fi
a2 + b2 + 2a ◊ b = a2 + b2 – 2a ◊ b
= 3696 11. Since a, b and c are non-coplanar vectors, so [a, b, c] π 0.
fi
4 a ◊ b = 0
fi
a ◊ b = 0
|
12 4 3 = 8 – 9 –12 33 – 4 –24
Now,
[x, y, z] = x ◊ (y × z)
c×a b×c a×b = _______ ◊ _______ × _______ [a, b, c] [a, b, c] [a, b, c]
...(i)
Let the vector r makes an angle q with the vectors a, b and c.
1 = ________ [b × c, c × a, a × b] [a, b, c]3
1 = ________ 3 [a, b, c]2 [a, b, c]
1 = _______ π 0. [a, b, c]
Thus, a and b are perpendicular to each other. 9. Let r = x i + yj + z k fi
x2 + y2 + z2 = 51
r ◊ a r ◊ b r ◊ c cosq = ____ = _____ = ____ |r| |a| |r| |b| |r| |c| r ◊ a r ◊ c r ◊ b r ◊ c fi ____ = ____ and _____ = ____ |r| |a| |r| |c| |r| |b| |r| |c| Thus,
1 1 __ (– 4x – 3z) __ (x – 2y + 2z) y y 3 5 ___ ___ ___ fi _____________ = ____ ___ ; ___________ = ____ ÷51 ÷51 ÷51 ÷51 fi
x – 5y + 2z = 0; 4x + 5y + 3z = 0
Solving, we get x ___ = – 5
y __ = 1
z __ = l (say) 5
From Eqs. (i) and (ii), we get
25l2 + l2+ 25l2 = 51
fi
51l2 = 51
fi fi
l2 = 1 l = ± 1
...(ii)
(
)
Thus, x, y and z are non-coplanar vectors. Now, x ◊ (a + b) + y ◊ (b + c) + z ◊ (a + b) (b × c) ◊ (a + b) + (c × a) ◊ (b + c) + (a × b) × (c + a) = ______________________________________________ [a, b, c] (b × c) ◊ a + (c × a) ◊ b + (a × b) ◊ c = _______________________________ [a, b, c] [a, b, c] + [a, b, c] + [a, b, c] = __________________________ [a, b, c] 3[a, b, c] = _________ [a, b, c] = 3
Vectors
12. Given,
r×b=c×b
fi
(r – c) × b = 0
fi (r – c) is parallel to b. or r is coplanar with the vectors b and c. Also, r ◊ a = 0 fi r is perpendicular to a. Again a ◊ b π 0 fi neither a nor b is a null vector or a is not perpendicular to b. 13. Given, a+b+c=0 fi (a + b + c)2 = 0 fi a2 + b2 + c2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 1 + 1 + 1 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3 – 3 fi (a ◊ b + b ◊ c + c ◊ a) = ___ 2 14. As we know that, if a, b, c form a right handed system, so a×b=c fi c=a×b fi
| |
i j k c = x y z 01 0 = x k – z i = – z i + x k
15. Since a, b and c are lie in the same plane, so they are coplanar. Thus, [a, b, c] = 0 fi
a ◊ (b × c) = 0
16. Given,
A=B+C
= 10 i – (2d – 12) j + (9 – d) k __
Given the area of the triangle is 5÷6 . __ 1 fi __ |A × B| = 5 ÷6 2 __
fi |A × B| = 10 ÷6
________________________
fi 100 + (2d + 12)2 + (9 – d)2 = 600 fi (2d + 12)2 + (9 – d)2 = 500 fi 4d 2 + 48d + 144 + 81 – 18d + d 2 = 500 fi 5 d 2 + 30 d – 275 = 0 fi d 2 + 6 d – 55 = 0 fi (d + 11) (d – 5) = 0 fi d = –11, 5 Hence,
a = 8, b = 4, c = 2, d = 5
or
a = – 8, b = 4, c = 2, d = –11
17. We have, (a + b) × (a × b) = a × (a × b) + b × (a × b)
= (a ◊ b)a – (a ◊ a) b + (b ◊ b)a – (b ◊ a)b
=a–b+a–b
= 2 (a – b) 18. Ans. (c) 19. Ans. (d) 20. Also, Let , be unit vectors on form_ a right handed system › Let b = b1 + b2 + b3 C
fi ai + bj + ck = di + 3j + 4 k + 3i + j – 2k
A
R
B
fi a = d + 3, b = 4, c = 2.
We have,
Then P = ÷ 3 ( × b )
| | | |
i j k = a 4 2 d3 4
= 10 i – (4a – 2d ) j + (3a – 4d) k
= 10 i – (4d + 12 – 2d) j + (3d + 9 – 4d)k
__
2 fi ÷100 + (2d + 12) + (9 – d)2 = 10÷6
Q
i j k (A × B) = a b c d3 4
5.53
P
_›
_›
__ __
= ÷ 3 (b2 – b3 ) __›
_›
_›
and Q = b – ( ◊ b ) = (b2 + b3 ) _› __›
P Q Then cos A = _____ _ › _ _› =0 |P ||Q |
A = 90°
_›
_›
__›
Also, R = P – Q
so that , ,
5.54 Integral Calculus, 3D Geometry & Vector Booster __
__
23. Let BP as x-axis and BY as y-axis.
= (÷3 b 2 – b3) – (÷3 b 2 – b2) Now, cos B =
_› _› P R _____ _ › _ › |P | |R |
(
__
) ( ) __ 3 ( b22 + b23 ) – 2 ÷3 b2 b3 – 1 __________________ B = cos __ 2 2 ÷3 ( b2 + b 23 )
Let i and j be unit vectors along x-axis and y-axis, respectively.
C = 180° – 90° – B = 90° – B
Here,
3 b22 + b 23 – 2 ÷3 b2 b3 = __________________ __ 2 2 ÷3 b2 + b23 Thus, Also,
(
)
Hence, the result,
p a + b = __ 2 Let q the angle between BP and CP fi
Question asked in Past IIT-JEE Examinations 21. Let x be a unit vector such that x = ai + bj Thus, |x| = 1
Now,
fi
|a i + b j| = 1
fi
÷a 2 + b2 = 1
fi
a2 + b2 = 1
_______
...(i)
Also (ai + bj) ◊ (i + j) = |(ai + bj)| ◊ |(i + j)| cos (45°) fi fi Again,
__ 1 a + b = 1 ◊ ÷2 ◊ ___ __ 2 ÷
a + b = 1
...(ii)
(ai + bj) ◊ (3i – 4j) = |(ai + bj)| ◊ |(3i – 4j)| cos (60°) 1 fi 3a – 4b = 1.5 ◊ __ 2 5 fi 3a – 4b = __ 2 On solving (ii) and (iii), we get 13 1 a = ___ and b = ___ 14 14 1 Thus, x = ___ (13i – j) 14
| |
22. Since the given vectors are coplanar, so 2 –1 1 1 2 – 3 = 0 3 l 5 fi 3(3 – 2) – l (– 6 – 1) + 5 (4 + 1) = 0 fi 3(3 – 2) – l (– 6 – 1) + 5 (4 + 1) = 0 fi 3 + 7l + 25 = 0 fi 7l + 28 = 0 fi l = – 4
2a + 2b = p
...(ii)
CP = i cos (a + b) + j sin (a + b)
( )
( )
p p = i cos __ + j sin __ = j 2 2 p Thus, BP ◊ CP = i ◊ j = __ . 2 Therefore, the angle between the angle bisectors of angles B and C is 90°. 24. As we know that, a vector in the direction of the bisector of the angle between two vectors a and b is a b 1 1 __ + ___ = __ (i – 2j + 2k) + __ (2i + j + 2k) |a| |b| 3 3
1 = __ (3i – j + 4k) 3
25. We
a=b×c = b × (a × b)
= (b ◊ b) a – (b ◊ a) b
= |b|2a – (b ◊ a) b
Comparing, we get
|b|2 = 1, (b ◊ a) = 0
fi Also,
|b| = 1, a ^ b c=a×b
= (b × c) × b
= (b ◊ b) c – (b ◊ c) b
= |b|2 c – (b ◊ c) b Comparing, we get |b|2 = 1,(b ◊ c) = 0 fi
|b| = 1, b ^ c
Thus,
|b| = 1 and a = c
26. Ans. (a) 27. Ans. (d)
Vectors
28. Given,
x×y=b
fi
(x × y) × a = b × a
fi
a × (x × y) = – b × a
31. Given
p b = and a = (4n + 1) __ , n ŒI 2 |(a × b) ◊ c| = |a| |b| |c|
fi
|a| |b| |c| sinq cosj = |a| |b| |c|
fi
sinq cosj = 1
fi
sinq = 1 and cosj = 1 p q = __ and j = 0 2
fi
=a×b
fi
(a ◊ y) x – (a ◊ x) y = a × b
fi
(a ◊ y) x – y = a × b
...(i)
fi
Also,
x + y = a
...(ii)
Similarly, |(a × b) ◊ c| = |a| |b| |c|
fi
x ◊ a + y ◊ a = a ◊ a
fi
1 + y ◊ a = a2
fi
y ◊ a = a2 – 1
...(iii)
From Eqs. (i) and (iii), we get
2
(a – 1) x – y = a × b
...(iv)
Adding and subtracting Eq. (ii) from Eq. (iv), we get
1 x = __ 2 (a + a × b) a 1 y = a – __ 2 (a + a × b). a
fi
|(b × c) ◊ a| = |a| |b| |c|
fi
b^c
Again
|(a × b) ◊ c| = |a| |b| |c|
fi
|(c × a) ◊ b| = |a| |b| |c|
fi
a ^ c
Thus,
a ^ b, a ^ c and b ^ c
32. Given,
x+y+z=a
fi
(x + y + z) ◊ a = a ◊ a
fi
(x ◊ a + y ◊ a + z ◊ a) = |a|2
30. We have,
fi
3 __ + 2
a × (b × c) + (a ◊ b) b = (4 – 2b – sina) b + (b 2 – 1)c
fi
fi (a ◊ c) b – (a ◊ b) c + (a ◊ b) b
Also,
and 29. Ans. (b)
2
= (4 – 2b – sina) b + (b – 1) c
Comparing, we get
(a ◊ c + a ◊ b) = 4 – 2b – sina
...(i)
2
and
(b – 1) = – (a ◊ b)
fi
b 2 = 1(a ◊ b)
Also
(c ◊ c) a = c
fi
(c ◊ c) (a ◊ c) = (c ◊ c)
...(ii)
fi (a ◊ c) = 1 From Eqs. (i) and (iii), we get
...(iii)
...(iv)
1 + (a ◊ b) = 4 – 2b – sina
Adding Eqs. (ii) and (iv), we get
(
)
7 __ + z ◊ a = 4 4 13 3 (a ◊ z) = 4 – ___ = __ 4 4
...(i)
x+y+z=a
fi
x ◊ x + y ◊ x + z ◊ x + a ◊ x
fi
3 1 + y ◊ x + z ◊ x = __ 2
fi
1 y ◊ x + z ◊ x = __ 2
...(ii)
3 Similarly, x ◊ y + z ◊ y = __ 4
...(iii)
and
1 x ◊ z + y ◊ z = – __ 4
...(iv)
Adding Eqs. (ii), (iii) and (iv), we get
1 x ◊ y + y ◊ z + z ◊ y = __ 2
4 – 2b – sina + b 2 = 2
Hence,
fi
b2 – 2b + 2 = sina
3 1 x ◊ y = __ , y ◊ z = 0, z ◊ x = – __ 4 4
Now,
x × (y × z) = b
fi (b – 1)2 + 1 = sina It is possible only when
fi
(x ◊ z) y – (x ◊ y) z = b
fi
3 1 – __ y – __ z = b 4 4
(b – 1) = 0 and sina = 1
5.55
...(v)
5.56 Integral Calculus, 3D Geometry & Vector Booster Also,
(x × y) × z = c
fi
fi
z × (x × y) = – c
Solving Eqs. (iii) and (vi), we get
fi
(z ◊ y) x – (z ◊ x) y = – c
fi
1 0x + __ y = – c 4
fi
y = – 4c
...(vi)
From Eqs. (v) and (vi), we get fi
3 c – __ z = b 4 4 z = __ (c – b) 3
Finally, x + y + z = a fi
x=a–y–z
fi
8 4 x = a + __ b + __ c 3 3 __ __ 1 x ◊ y = ÷2 ◊ ÷2 cos (60°) = 2 ◊ __ 2
Thus,
x ◊ y = 1 = y ◊ z = z ◊ x
fi
x ◊ x = 2 = y ◊ y = z ◊ z
Now,
x × (y × z) = a
fi
(x ◊ z) y – (x ◊ y) z = a
fi
y – z = a
Also,
y × (z × x) = b
fi
(y ◊ z) z – (y ◊ z) x = b
fi
z–x=b y – x = a + b
Again,
(x × y) = c
fi
x × (x × y) = x × c
fi
(x ◊ y) x – (x ◊ x) y = x × c
fi
x – 2y = x × c y × (x × y) = y × c
fi
(y ◊ y) x – (y ◊ x) y = y × c
fi
2x – y = y × c
Subtracting Eq. (iv) from Eq. (v), we get
x + y = (y – x) × c
(a + b) × c – (a + b) x = __________________ 2
(a + b) × c + (a + b) y = __________________ 2
and
b – a + (a + b) × c z = _________________ 2
34. Let
r = (a × b) × c
fi
r ^ (a × b) and r ^ c
fi
[(a × b) × c] ^ c
–2a + 3b – c = lp + mq + nr
fi ...(i)
where l, m, n ŒR – 2a + 3b – c = l (2a – 3b) + m (a – 2b + c) + n (– 3a + b + 2c) = (2l + m – 3n) a
+ (– 3l – 2m + n) b + (m + 2n) c
Comparing, we get ...(ii) ...(iii)
(2l + m – 3n) = – 2,
(– 3l – 2m + n) = 3
and
(m + 2n) = –1
Solving, we get ...(iv)
Finally, x × y = c fi
and A = i, B = i + j, C = i + j + k Now, AB ◊ CD + BC ◊ AD + CA ◊ BD = –1 + 0 + 1 = 0. 36. Since a, b and c are non-zero and non-coplanar vectors, so any vector can be expressed as a linear combination of these three vectors, i.e.
Adding Eq. (i) and (ii), we get
...(vi)
35. Let D be the origin.
33. We have,
x + y = (a + b) × c
7 1 Therefore, –2a + 3b – c = – __ q + __ r. 5 5 37. According to the question,
...(v)
7 1 l = 0, m = – __ , n = __ 5 5
fi fi
(
a c = l __ + |a|
( (
)
b ___ , |b|
where l ŒR
)
7i – 4j – 4k – 2i – j + 2k c = l ___________ + ___________ 9 3 i – 7j + 2k c = l __________ 9
)
Vectors
It is given that, __ |c| = 3÷6 fi
|c|2 = 54
fi
1 + 49 + 4 54 = l2 _________ 81
(
l = 81
l = ± 9
fi
y + y × a x = _________ |y|2
(
)
i – 7j + 2k Therefore, c = ± 9 __________ 9
|y|2 x = y + y × a
)
2
fi
fi
= ± (i – 7j + 2k)
38. Given r = x1r1 + x2r2 + x3r3
fi (2a – 3b + 4c)
( ) ( ) ( ) a×b a×b ( _____ ) – ( _____ ) × a c c = ____________________ a_____ ×b ( c ) b× a b× a – _____ c – ____ × a c ____________________ = b×a2 _____ c
2
c ((a × b) + (a × b) × a) = ______________________ (a × b)2
= (a – b + c) x1 + (b + c – a) x2 + (c + a – b) x3
Finally, y × z = b
= (x1 – x2 + x3) a + (– x1 + x2 – x3) b
fi
y × (y × z) = y × b
+ (x1 + x2 + x3) c
fi
(y ◊ z)y – (y ◊ y)z = y × b
Comparing, we get
fi
y – |y|2 z = y × b
fi
|y|2z = y – y × b
fi
y – y × b z = _________ |y|2
x1 – x2 + x3 = 2
fi
– x1 + x2 – x3 = – 2
fi x1 + x2 + x3 = 4 On solving, we get,
7 1 x1 = __ , x2 = 1, x3 = – __ 2 2
39. Given,
x×y=a
fi
b × (x × y) = b × a
fi
(b ◊ y)x – (b ◊ x) y = b × a
Also,
y×z=b
fi
y ◊ (y × z) = y ◊ b
fi
0 = y ◊ b
fi
y ◊ b = 0 – (b ◊ x) y = b × a
fi
b×a b×a y = – _____ = – _____ c (b ◊ x)
Again
x×y=a
fi
y × (x × y) = y × a
fi
(y ◊ y) x – (y ◊ x) y = y × a
fi fi
...(ii)
...(iii)
c ((a × b) – (a × b) × b) = ______________________ . (a × b)2
r = a (b × c) + b (c × a) + g (a × b) where a = (a ◊ d), b = (b ◊ d), g = (c ◊ d)
Now,
r ◊ a = a [a, b, c]
r ◊ b = b [a, b, c]
r ◊ c = g [a, b, c]
Thus, r ◊ (a + b + c) = (a + b + g ) [a, b, c] fi r ◊ (a + b + c) = (a ◊ d + b ◊ d + c ◊ d) [a, b, c] fi r ◊ (a + b + c) = (a + b + c) ◊ d [a, b, c] fi (r – d [a, b, c]) (a + b + c) = 0
|y| x – (y ◊ x)y = y × a |y| x – y = y × a
fi
2 2
) ( ) ( )
40. Let r = (a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)
From Eqs. (i) and (ii), we get
(
[\ (y ◊ z) = 1]
a×b a × b _____ c – _____ × b c ____________________ = a×b 2 _____ c
...(i)
5.57
[ (y ◊ x) = 1]
fi (r – d [a, b, c]) = 0 [ coplanar]
a, b, c are non-
5.58 Integral Calculus, 3D Geometry & Vector Booster fi r – d [a, b, c] = 0 fi r = d [a, b, c] fi |r| = |d [a, b, c]|
(a × b) b v = ___ + ________ 2 |b | |(a × b)|2
Therefore, 43. Given,
r ◊ (a + b) = 0 and r ◊ (b – c) = 0
1 a × (b × c) = __ (b + c) 2 1 fi (a ◊ c) b – (a ◊ b) c = __ (b + c) 2 Comparing, we get 1 1 a ◊ c = __ , a ◊ b = – __ 2 2 1 1 fi cos q1 = __ and cos q2 = – __ 2 2 p 2p __ ___ fi q1 = and q2 = 3 3 2p Thus, the angle between a and b is ___ . 3 44. Area of a given parallelogram 1 = __ |d1 × d2| 2 i j k 1 = __ 2 3 – 6 2 3 – 4 –1
fi
3y + z = 0 and 3z = 0
fi
y = 0 and z = 0
fi |r| = |[a, b, c]| ( d is a unit vector) which is independent of d. Thus, |(a ◊ d) (b × c) + (b ◊ d) (c × a) + (c ◊ d) (a × b)|
= [a, b, c]
41. Let
r = xi + yj + zk
Now,
a + b = (i + j – k) + (– i + 2j + 2k) = (3j + k)
and b – c = (– i + 2j + 2k) – (– i + 2j – k) = 3k It is given that,
Since r is a unit vector, so
x2 + y2 + z2 = 1
2
| | | |
|
1 = __ (– 27i – 16i – 17k) 2 1 _______________ = __ ÷729 + 256 + 289 2 _____ 1 = __ × ÷ 1274 2 39.5 = ____ = 19.75 2
fi
x =1
fi
x = ± 1
Thus,
r = ± i
45. We have,
fi
rˆ = ± i
(– 4i + 5j) a + (3i – 3j + k) b (i + j + 3k)c
Hence, the result. 42. Here, v, b and a × b are coplanar.
= l (ai + bj + ck)
Thus,
v = a b + b (a × b)
(– 4a + 3b + c)i + (5a – 3b + c)j + (b + c)k = l (ai + bj + ck) Comparing, we get
fi
v ◊ b = a b ◊ b + b (a × b) ◊ b v ◊ b = a |b|2 + 0
(– 4a + 3b + c) = l a,
fi fi
Now,
1 = a |b|2 + 0 1 a = ___ 2 |b| [v, a, b] = 1
fi
v ◊ (a × b) = 1
fi
l3 + 4l2 – 25l = 0
fi
(a b + b (a × b)) ◊ (a × b) = 1
fi
l (l2 + 4l – 25) = 0
fi
(a b ◊ (a × b) + b (a × b) ◊ (a × b)) = 1
fi
l = 0 and (l2 + 4l – 25) = 0
fi
(a 0 + b (a × b)2) = 1 1 b = ________ |(a × b)|2
fi
fi
– 4 ± ÷116 l = 0 and l = _________ 2
fi
l = 0 and l = – 2 ± ÷29
fi
(5a – 3b + c) = l b and (b + 3c) = l c.
|
|
3 – 4 – l 1 Thus, –3 – l 5 1 0 1 3–l
____
___
Vectors
46. Let Given,
r = xi + yj + zk
x – 2y + 5z = 0
r ◊ a = 0, r ◊ b = 0
fi
2x + 3y – z = 0 y z x Thus, ______ = ______ = ____ 2 – 15 10 + 1 3+ 4 y z x ____ fi = ___ = __ = l (say) – 13 11 7 Also,
r ◊ (2i + j + k) + 8 = 0
fi
2x + y + z + 8 = 0
...(i)
...(ii)
– 26l + 11l + 7l + 8 = 0
fi
– 8l + 8 = 0
fi
l=1
(Tougher Problems for Jee Advanced) 1. We have,
› _› _
|a + b | 2 = a2 + b2 + 2 (a ◊ b )
= 1 + 1 + 2cosq
= 2 (1 + cosq)
q = 2.2 cos2 __ 2
2
( ) q = 4 cos ( __ ) 2
( ) 3 q __ | a + b | = 3 cos ( __ ) 2 2 _ › _› q |a + b | = 2 cos __ 2 _›
_›
fi Also,
Hence, r = – 13i + 11j + 7k 47. Let the position vector of the third vertex be x i + y j.
_ ›
_›
fi
From Eqs (i) and (ii), we get
_›
_ ›
› _› _
|a – b | 2 = a2 + b2 – 2 (a ◊ b )
= 1 + 1 – 2 cosq
= 2 (1 – cosq)
( ) q |a – b | = 2 sin ( __ ) 2 q 2 |a – b | = 4 sin ( __ ) 2 q = 4 sin2 __ 2
_›
fi
_›
fi
_›
_›
Thus, Here,
AC = (x + 1)i + (y – 3)j
fi
BO = i + 3j
Since
AC ^ BO
fi
AC ◊ BO = 0
fi
(x + 1) + 3(y – 3) = 0
fi
x + 3y = 8
Also,
BC = (x – 2)i + (y – 5) j
and
OA = 2i – j
Again,
BC ◊ OA
fi
2 (x – 2) – (y – 5) = 0
fi
2x – y = 1
|
|
( )
|
= [– 5, 5]. _ ›
...(i)
...(ii)
5 17 x = __ , y = ___ 7 7 5 17 = __ i + ___ j. 7 7
|
_ › _› 3 _› _ › Hence, the range of __ a – b + 2 |a – b | 2
Hence, the position vector of the thrid vertex is
( )
_› _› q q 3 _› _› __ a + b + 2|a – b | = 3cos __ + 4sin __ 2 2 2
On solving Eqs (i) and (ii), we get
5.59
2. Let a = x + y + z , where It is given that, _› _› 1 a ◊ k = ___ __ 2 ÷ 1 fi z = ___ __ 2 ÷ _›
x2 + y2 + z2 = 1
Also,
|a +
+ |=1
fi
|(x + 1) + (y + 1) + z | = 1
fi
(x + 1) + (y + 1)2 + z2 = 1
fi
1 (x + 1)2 + (y + 1)2 + __ = 1 2 1 2 2 (x + 1) + (y + 1) = __ 2
fi
2
5.60 Integral Calculus, 3D Geometry & Vector Booster _ _ › › _› p Thus, the angle between (2a + b ) and b is __ . 2 ___› 6. Now, AB = 2 + +
1 x2 + y2 + 2 (x + y) + 2 = __ 2 1 1 __ __ + 2 (x + y) + 2 = 2 2
fi fi
_ __›
fi
x + y = –1
...(i)
Also,
1 x2 + y2 = __ 2
...(ii)
and AC = (t + 1) – Thus,
1 ___› ___› ) ar (D ABC) = __ ( AB × AC 2
|
|
| | |
From Eq. (i) and (ii), we get 1 1 x = – __ = y, z = ___ __ 2 ÷2
1 = __ 1 1 2 2 (t + 1) 0 –1
_› kˆ 1 Thus, a = – __ ( + ) + ___ __ 2 2 ÷
1 = __ ( – + (t + 3) – (t + 1) ) 2
3.
_ _› _› It is given that |r + bs | is minimum __› __› _ › Hence, the value of |r + bs | 2 + |bs | 2
when b = 0
_›
4. Let R = a + b + c .
› _› _ (u ◊ R –
10) +
› _› _ (v ◊ R –
1 = __ ÷ 2
1 = __ ÷2t 2 + 8t + 11 2
Let 20) +
___________________ 2 1 + (t + 3) + (t + 1)2 ____________
_ ›
= |r | 2
It is given that
|
__› _› (w ◊ R )
=0
z = 2t2 + 8t + 11
dz fi __ = 4t + 8 dt
fi (a – 2b + 3c) + (2a + b + 4c)
fi
fi
d 2z ___ 2 = 4 > 0 dt z is minimum
fi
Area is minimum.
+ (a + 3b + 3c) = 10 – 20 – 20
Comparing the co-efficients, we get a – 2b + 3c = 10, 2a + b + 4c = – 20 and a + 3b + 3c = – 20 Solving, we get a = –1, b = 2, c = 5 _›
Hence, R = – + 2 + 5 .
5. It is given that,
fi
_ › _› |a + b | = 1 _ › _ › |a + b | 2 = 1
fi
a2 + b2 + 2 (a ◊ b ) = 1
fi
1 + 1+ 2 (a ◊ b ) = 1
_› _›
_› _›
fi
_› _› (a ◊ b )
Now,
(2a + b ) ◊ b = (2a ◊ b + b ◊ b )
_›
For maximum or minimum, dz __ = 0 gives 4t + 8 = 0 dt t = – 2 Thus, the minimum area is
1 = – __ 2
_›
_ ›
_›
_ ›
|a + b | = 1
fi
|a + b | 2 = 1
fi
|a | 2 + |b | 2 + 2 (a ◊ b ) = 1
fi
1 + 1 + 2 (a ◊ b ) = 1
› _› _
› _› _ 2(a ◊ b )
( )
_ › _ ›
+
_› |b | 2
1 = 2 – __ + 1 2
= –1 + 1
= 0.
_›
_ ›
› _ › _
_› _›
_
_› › 1 (a ◊ b ) = – __ . 2 It is also given that,
fi
_ › _ ›
=
7. Given,
3 1 ___________ ___ = __ ÷8 – 16 + 11 = __ 2 2 ÷
_ ›
_ ›
_›
c = l a + m b _›
› _› _
fi
|c | 2 = l2 + m2 + 2l m (a ◊ b )
fi
1 1 = l2 + m2 + 2l m – __ 2
( )
= l2 + (2l)2 – l (2l)
|
Vectors
fi fi fi Again
fi
2 m = 2l = ___ __ 3 ÷
10. Given, _› _ _ › › a + b = m p
fi
(
)
1 2__ (l, m) = ___ __ , ___ 3 ÷ 3 ÷ 8. It is given that,
Thus,
( )
1 cosq = cos (cos–1) __ = 2
Now.,
1 __ 4
___›
__›
_ ›
_›
_ ›
› _ › _
( )
_›
_ › __› (2a × b )
_› _ › Now, b ◊ c =
_› _› a ◊ c
_› – 3 |b | 2
_›
=
= – 6
2 2
2
= 4a b sin q + 144
= 64 sin2q + 144
3 = 64 × __ + 144 = 48 + 144 = 192 4
_› _›
11. Let a =
+
–2 +
_› b = _› and c =
–2
| | | | | |
– 2 +
+
_›
____
__
|c | = ÷ 192 = 8÷3
_› _ › Again, b ◊ c =
– 48
= – 3 – 3 – 3
Also,
|
_ › _› _› a × (b × c ) = 1 1 – 2 –3 – 3 – 3
| |
= 3 1 1 – 2 –1 –1 –1
= 3 1 1 – 2 –1 –1 –1
= – 48
_› _› _ _ › › |c | 2 = |(2a × b ) |2 + 9 |b | 2 – 0
_›
= |(p ◊ q ) |
– 3b
› _› _ – 3(a ◊ b )
_ ›
9. Given, c =
fi
_› _›
_ ›
Sum of the roots = –1 and the product of the roots = 12 Hence, sum of the values of m = –1 and the product of all values of m = 12.
Also,
_ › _ ›
l = 2, m = m
1 16 = 4 + m2 + 4m __ 4 m2 + m – 12 = 0
_ › _ › _ ›
= |(p ◊ q ) | |b |
Now,
fi
_ › _ › _ ›
_ ›
|(a ◊ q ) p – (p ◊ q ) a | = |(p ◊ q ) | |m p – a |
|c | 2 = 4 |a | 2 + m2|b | 2 + 4m (a ◊ b )
_ › _ ›
_› _ ›
_›
fi
_› _ ›
_ › _ ›
fi a ◊ q + b ◊ q = m (p ◊ q )
_› _› b × c = 1 – 2 1 – 2 1 1
\ c = 2a + mb
fi
__
3 ÷ – 48 __ 3__ cosj = ________ = – ____ = – ___ 2 4 × 8÷3 2÷3 5p j = ___ 6
_› _ ›
_ _ _› _ › › _› × b Also, c × b = 2a _ _ _› _ › › _› fi (c – 2a ) × b = 0 _› __› _› \ (c – 2a ) is parallel to b _› __› _› fi (c – 2a ) = m b _› __› _› fi c = 2a + m b _› _ › = l a + m b
Thus,
|b | |c | cosj = – 48
fi a ◊ q = m (p ◊ q )
_› _›
a ◊ b = ab cos (q) fi
_› _ ›
3l2 = 1 1 l2 = __ 3 1 l = ___ __ 3 ÷
fi
5.61
|
= 3 1 1 – 2 0 0 – 3
12. It is given that,
= – 9 ( – )
[ ]
– 2 _ › _› _› a a + b b + g c = – 5 6
fi
– 2 1 2 1 a 0 + b 1 + g 1 = – 5 0 – 3 –1 6
( ) ( ) ( ) ( )
5.62 Integral Calculus, 3D Geometry & Vector Booster fi
a + 2b + g = – 2
fi
b + g = – 5
fi
– 3a – g = 6
and 3x – y + z = 2 Solving, we get 1 1 x = 1, y = __ , z = – __ 2 2 Hence, the value of
Solving, we get a = 1, b = – 2, g = 3.
13.
_› Let V =
a + b + c
_› X =
+2
Now,
fi
| |
_›
= 2 (a – c) + (2b + c) + (2a – b + 2c)
2 (a – c) + (2b + c) + (2a – b + 2c)
= (2 + )
Comparing the coefficients of , , and , we get a – c = 1, 2b + c = 0, 2a – b + 2c = 1. Solving, we get Now,
7 1 2 a = __ , b = __ , c = – __ 9 9 9 _›
_ ›
9 |V | 2 = m
fi
49 + 1 + 4 m = 9 × __________ 81
= 2 ( –
= 3
2. We have,
(a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r (a + b) ◊ (b × c) (b + c) ◊ (c × a) = _____________ + _____________ [a, b, c] [a, b, c] (c + a) ◊ (a × b) + _____________ [a, b, c]
_›
3. It is given that, c is parallel to the plane of the _ › _› vectors a and b , i.e.
c ^ (a × b) fi (a × b) ◊ c = 0
+ ) x ( – 2 + 3 ) + y (2 +
= 2 ( –
= 1 + 1 + 1
__› _ __› › z C = B × C
[a, b, c] ________ [a, b, c] [a, b, c] = ________ + + _______ [a, b, c] [a, b, c] [a, b, c]
= 3
14. It is given that, _› y B +
1. We have,
[a, b, c] _______ [a, b, c] ________ [a, b, c] = _______ + + [a, b, c] [a, b, c] [a, b, c]
( ) 54 = 9 × ( ___ ) = 6 81
= 101
a ◊ (b × c) + 0 b ◊ (c × a) + 0 c ◊ (a × b) + 0 = _____________ + ____________ + ____________ [a, b, c] [a, b, c] [a, b, c]
__
_› x A +
3 |V | = ÷ m
fi
= 100 + 5 – 4
a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) _________ + + (c × a) ◊ b (a × b) ◊ c (b × c) ◊ a
2V + V × X
fi
_›
= –2c + c + (2a – b) _›
(100 x + 10 y + 8 z)
Integer type Questions
_› _› V × X = a b c 120
15. Do yourself
_ ›
and Y = 2 +
– 2x + y + z = – 2
– ) + z ( + )
+ )
|
|
3 1 –2 fi 2 3 = 0 – 1 l 1 (2l – 1) fi (6l – 3 + 1) + 2 (4l – 2 + l) + 3(2 – 3l) = 0
Comparing the co-efficients of the unit vectors, we get
fi (6l – 2) + 2 (5l – 2) + 3 (2 – 3l) = 0
fi
x + 2y = 2
fi (6l – 2) + (10l – 4) + 3(6 – 9l) = 0 7l = 0
Vectors
fi l=0 Hence, the value of (l2 + 2) is 2. 4. We have, r ◊ (a + b + c) = 0 fi r ◊ a + r ◊ b + r ◊ c = 0 Now, r ◊ a = P [b, c, a]
Thus, m = 0 Hence, the value of (m + 4) = 4.
b×c c×a a×b 7. Clearly, p = _______ , q = _______ , r = _______ [a, b, c] [a, b, c] [a, b, c] _›
_›
_›
_›
_›
Hence, the value of (a + b + c) ◊ (p + q + r )
r ◊ b = Q [c, a, b]
and r ◊ c = R [a, b, c] Adding, we get
a ◊ b × c _______ b ◊ c × a _______ c ◊ a × b = _______ + + [a, b, c] [a, b, c] [a, b, c]
[a, b, c] [a, b, c] [a, b, c] = _______ + _______ + _______ [a, b, c] [a, b, c] [a, b, c]
=3
P [b, c, a] + Q [c, a, b] + R [a, b, c] = 0
fi
(P + Q + R) = 0
Hence, the value of (P + Q + R + 5) = 5. 5. It is given that _ ›
_›
_›
_ ›
_›
_› _ _›
_›
8. We have, _›
_ ›
a ^ (b + c ) , b ^ (c + a ) , c ^ (a + × b ) _› _› _› fi a ◊ (b + c ) = 0, _ › _ _ › › b ◊ (c + a ) = 0 _ › _› _› and c ◊ (a + b ) = 0 _ › › _ _› _ _› _› › Thus, a ◊ b + b ◊ c + c ◊ a = _ › _› _› Now, |a + b + c | 2
_›
_›
_›
_›
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi – 2 (a ◊ b + b ◊ c + c ◊ a) = 3 fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3
0
Now,
|a + b + c|2
= a2 + b2 + c2 + 2 (a ◊ b + b ◊ a + c ◊ a)
=3–3=0
= a2 + b2 + c2
Thus,
|a + b + c| = 0
= 1 +1 + 4
fi
a + c = – a
2
2
Hence, the value of
=6 Hence, the value of (m2 + 1) = 7.
6. In an AP,
_›
1 tp = A + (p – 1) d = __ a
( ) 1 c–b q – r = __ ( _____ d cb )
(
Now,
= 0
= |3a|
=3
+a +b ,
9.
_› Let p =
a + 2 + b
_ › and r = a + b + 3 __› Now, pq = (a – 1) + (2 – a) __› and qr = + (b – 2) + (3 – b)
)
(
(a – 1) + (2 – a) = l{(b – 2) + (3 – b) }
)
= {(q – r) + (r – p) + (p – q) } ◊ __ a + __ + __ b c 1 1 1 __ 1 1 __ 1 1 = _____ __ – __ + – __ + – __ abcd b c c a a b
(
Since the vectors are collinear, so
1 a–c r – p = __ _____ ac d u ◊ v
_›
_› q =
1 b–a p – q = __ _____ d ab
Thus,
_ ›
|2a + 5b + 5c | = |2a + 5 (b + c)| = |2a – 5a|
1 tq = A + (q – 1) d = __ b 1 tr = A + (r – 1) d = __ c
_›
|a – b |2 + |b – c |2 + |c – a | 2 = 9
› _› _ _ _ › _ › › _ › = a + b + c + 2 (a ◊ b + b ◊ c + c ◊ a ) 2
5.63
)
fi
(a –1) = 0, (3 – b) = 0
fi a = 1, b = 3 Hence, the value of (a + b) = 4. _›
10. Let n = a + b + c .
5.64 Integral Calculus, 3D Geometry & Vector Booster _ ›
Thus, _›
n = ___ _› |n |
Comparing the co-efficients, we get
Now, u ◊ = 0 fi a + b = 0 fi
_› v ◊
=0fia–b=0
Hence, the value of
(u + 4v + w) = 17
_ ›
_› | ◊
(u + 2v + w) = 7
and (u – v + 3w) = – 8 Solving, we get
Solving, we get a = 0, b = 0 and c = 1 Thus, n =
u = – 3, v = 5 and w = 0
Hence, the value of (u + v + w + 4) = – 3 + 5 + 0 + 4 = 6.
| = 3.
11. Clearly,
2x – y = 5
14. Let s = x (p + q ) + y (q + r ) + z (r + p )
and
x – 2y = 4
Now, p ◊ s = y (p ◊ (q × r ) ) = y [p , q , r ]
_›
3x = 6
fi
x=2
and so
y = – 1
_›
_ › › _
Solving, we get
_›
_›
_ › _ ›
_›
_ ›
_›
_›
_ › _ › _ ›
› _› _
p ◊ s y = ________ _› _› _› [p , q , r ]
fi
_ › › _
_ _
Hence, the value of (x + y + 2) = 3.
› › q ◊ s r ◊ s Similarly, x = ________ _ › _ › and z = ________ _ › _ › _ _ › › [p , q , r ] [p , q , r ]
12. It is given that,
Thus,
|– cu , v , cw | = 8
fi
– 2 1 1 c2 1 2 = 8 –1 1 0 –1
fi
– 2 1 –1 c2 1 –1 3 = 8 1 0 0
| | | |
fi
c2(3 – 1) = 8
fi
c2 = 4
fi
c=2
(C3 Æ C1 + C3)
_ ›
_› _›
_›
_›
_› _›
_›
_›
› _› _
_›
_›
(r ◊ s ) (p × q ) + ____________ _› _› _ › [p , q , r ] › _› _
_›
_›
_› _›
_›
_›
_›
+ (q ◊ s ) (r × p ) + (r ◊ s ) (p × q )
_›
_› _›
_›
_›
+ (r ◊ s ) (p × q ) |
_ ›
_ ›
_› _ › _ › _ ›
= |s [p , q , r ] |
= |s | |[p , q , r ] |
=1×4
=4
_›
_› _› _›
Questions asked in Past Iit-Jee Examinations
_ ›
1. We have,
A ◊ {(B + C) × (A + B + C)} = A ◊ {B × A + B × B + B × C + C × A + C × B
_ ›
(p × q ) × r = up + vq + wr
7 + 17 – 8 = ( +
_›
fi s [ p , q , r ] = (p ◊ s ) (q × r )
_› _›
= 7 + 17 – 8
_ ›
_›
fi |(p ◊ s ) (q × r ) + (q ◊ s ) (r × p )
(p × q) × r = – 5 3 2 1 13
It is given that,
_› _›
_›
_› _ › _ › _ ›
(p × q) = 1 1 1 = – 5 + 3 + 2 2 4 – 1
Thus,
_›
| | | |
13. Now,
_› _›
_ (p ◊ s ) (q × r ) (____________ q ◊ s ) (r × p ) › fi s = ____________ + _› _› _ _› _› _ › › [p , q , r ] [p , q , r ]
__› _ › _ _›
+ C × C} = A ◊ {B × A + B × C + C × A + C × B}
+ ) u
+ (2 + 4 + ) v + ( +
+ 3 ) w
= (u + 2v + w) + (u + 4v + w) + (u – v + 3w)
= A ◊ (B × A) + A ◊ (B × C) + A ◊ (C × A)
+ A ◊ (C × B)
Vectors
= 0 + A ◊ (B × C) + 0 + A ◊ (C × B)
fi
x + 3y – 4z = l x
= [A B C] – [A B C]
x – 3y + 5z = l y
and
3x + y = l z
fi
(1 – l)x + 3y – 4z = 0
x – (3 + l) y + 5z = 0
and
3x + y – l z = 0
= 0
2. We have,
( )
p 1 |(B × C)| = BC sin __ = __ . 2 6
Now,
(B × C) A = ± ________ = ± 2 (B × C). |(B × C)|
3. We have |a| |b| |c| = |(a × b) ◊ c| = Volume of a parallelopiped having three adjacent sides as a, b, c. = Volume of a rectangular parallelopiped having a, b, c as adjacenat sides.
|
Eliminating x, y, z, we get
|
3 1 – l – 4 – (l + 3) 5 = 0 1 3 – 5 1 fi
l3 + 2l2 + l = 0
fi
l (l2 + 2l + 1) = 0
fi
l (l + 1)2 = 0
fi a ◊ b = b ◊ c = c ◊ a = 0
fi
l = –1, 0
OP = p = 60i + 3j,
4. Let |OAi| = R, for every i
5.65
6. Let
OQ = q = 40i – 8j
and
OR = r = a i – 52 j
Since the points are collinear, so
Let p be a unit vector perpendicular to the plane of the regular polygon.
( ) ] [ 2p = [ R sin ( ___ n ) ] p
2p OAi × OAi + 1 = (OAi) (OAi + 1) sin ___ n p
fi
-20 - 11 =0 a - 60 - 55
fi
-20 1 =0 a - 60 5
fi
– 100 – a + 60
fi
a = – 40.
X ◊ A = 0 = X ◊ B = X ◊ C
2
Thus,
( ( ) ) 2p × OA ) = – ( R sin ( ___ n ) )p
n – 1 2p S (OAi × OAi + 1) = (n – 1) R2sin ___ n p
i = 1
Also,
(OA2
1
2
(
fi
60 3 1 R Æ R 2 – R1 -20 -11 0 = 0 2 R3 Æ R3 – R1 a - 60 -55 0
We have, for 1 £ i £ n –1
60 3 1 40 -8 1 = 0 a -52 1
7. Since,
fi
A, B, C are coplanar
n – 1
Thus,
[A, B, C] = 0
i = 1
Thus, S (OAi × OAi + 1) = (1 – n) (OA2 × OA1) 5. Given x (i + j + 3k) + y (3i – 3j + k) + z (– 4i + 5j)
= l (x i + y j + z k)
8. V = [OA OB OC]
2 -3 0 = 1 1 -1 3 0 -1
)
5.66 Integral Calculus, 3D Geometry & Vector Booster 9. Given,
= 2 (–1 – 0) + 3 (–1 + 3) = – 2 + 6 = 4.
Now,
A ◊ B = 0, A ◊ X = c, A × X = B
A × (A × B) = A × B
fi
(A ◊ X) A – (A ◊ A) X = A × B 2
fi
cA – A X = A × B
fi
A2X = cA – A × B
cA – A × B X = ___________ A2 10. Let i, j, k be the unit vectors along the positive directions of x, y and z axes respectively. fi
= |c| |a × b| cos (0) = |a × b| 1 = __ (ab) 2
Thus, a1 b 1 c1
a2 b2 c2
A1 = r sin q cos j,
and
A2 = r sin q sin j A3 = r cos q.
(
(
and
)
)
B3 = r cos q = A3.
11. The given statement is true. Let the position vectors of the points A, B and C are a + b, a – b and a + kb respectively. Then AB = – 2b Similarly, BC = (k + 1)b Clearly, AB is parallel to BC for all k in R Thus, A, B and C are collinear p 1 12. Given (a × b) = (a b) sin __ = __ (ab) 2 6 Now, [c a b] = c ◊ (a × b)
( )
= [a b c]2
1 = __ a2 b2 4 1 = __ (a 21 + a22 + a 23) (b21 + b 22 + b23) 4 13. As the length of the vectors remain the same.
(2p)2 + 1 = (p + 1)2 + 1
fi
4p2 = (p + 1)2
fi
p + 1 = ± 2p 1 fi p = 1, – __ . 3 14. Given A, B, C and D are coplanar, so AB, AC and AD are also coplanar. AB = OB – OA
= (2i + 3j – 4k) – (3i – 2j – k)
= – i + 5j – 3k.
Also,
p When the x-axis is rotated through an angle of __ , the 2 new components of A are B1, B2, B3 respectively. p Thus, B1 = r sin q cos j + __ 2 = – r sin q cos = – A1 p B2 = r sin q sin j + __ = r sin q cos j 2 = A1
2
Now,
Here,
a3 b3 c3
AC = OC – OA
= (– i + j – 2k) – (3i – 2j – k)
= 2i + 3j + 3k.
Finally,
AD = OD – OA
= (4i + 5j + lk) – (3i – 2j – k)
Given,
= i + 7j + (l + 1) k
fi
[AB AC AB] = 0 -1 5 -3 2 3 3 =0 1 7 (l + 1)
fi 1 (15 + 9) – 7 (– 3 + 6) + (1 + l) (– 3 – 10) = 0 fi 24 – 21 – 13 (l + 1) = 0 fi 13 (l + 1) = 3 3 fi (l + 1) = ___ 13 3 10 fi l = ___ – 1 = – ___ 13 13 (a × b) 15. Number of unit vectors = ± ________ |(a × b)|
Vectors
= 2 |a × b + b × c + c × a| Also, area of a triangle D ABC 1 = __ |a × b + b × c + c × a| 2 From, (i) and (ii), we get,
16. Given
a 1 1 1 b 1 =0 1 1 c
fi
a 1- a 1- a C Æ C 2 – C1 1 b -1 0 = 0 2 |AB × CD + BC × AD + CA × BD| C3 Æ C3 – C1 1 0 c -1 = 4 ar (D ABC)
(
)
fi a (b – 1) (c – 1) – (1 – a) (c – 1)
19. (a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r
+ (1 – a) (1 – b) = 0
(a + b) × (b × c) (b + c) ◊ (c × a) = ______________ + _____________ [a b c] [a b c]
fi a (1 – b) (1 – c) + (1 – a) (1 – c) + (1 – a) (1 – b) = 0 Divide both the sides by (1 – a) (1 – b) (1 – c), we get a 1 1 _____ + _____ + _____ =0 1– 1–b 1–c a 1 1 fi 1 + _____ + _____ + _____ = 1 1– 1–b 1–c 1 1 1 fi _____ + _____ + _____ =1 1– 1–b 1–c _›
Also,
4a + 3b = a __ = 3 _›
0 0 b ___ = l – 4
(c + a) ◊ (a × b) + _____________ [a b c] a ◊ (b × c) _________ b ◊ (c × a) _________ c ◊ (a × b) = _________ + + [a b c] [a b c] [a b c] [a b c] _______ [a b c] _______ [a b c] = _______ + + [a b c] [a b c] [a b c]
= (Projection of a on b)
a ◊ b = ____ |b|
Let a = x + y be the required vectors a ◊ b Projection of a along b = ____ = 1 b 4x + 3y ______ _______ = 1 ÷16 + 9 4x + 3y = 5 Also, projection of a along c is 2 a ◊ c ____ = 2 c a x + b y _______ ________ = 2 ÷a 2 + b 2
...(i)
= 3 20. Component of a along b
17. Let c = a + b _› _ › b ◊ c =
5.67
( ) a ◊ b b = ( ____ ) ___ |b| |b|
( )
a ◊ b = ____ 2 b |b|
Component of a perpendicular to b a ◊b = a – ____ 2 b |b|
|b|2 a – (a ◊ b) b = _____________ |b|2
b × (a × b) = __________ . |b|2
...(i)
3x – 4y = 10 ...(ii) Solving (i) and (ii), we get, x = 2, y = – 1 Hence, the required vector is 2 – 18. Let the position vectors of the points A, B, C and D be a, b, c and d respectively w.r.t the origin O Now, AB = b – a, AD = d – a BC = c – b, BD = d – b CD = d – c, CA = a – c Now, |AB × CD + BC × AD + CA × BD|
( )
21. Do Yourself By the help of internal section formula 22. (a – b) ◊ {(b – c) × (c – a)} = (a – b) ◊ (b × c – b × a – c × c + c × a) = (a – b) ◊ (b × c – b × a + c × a) = (a ◊ b × c – a ◊ b × a + a ◊ c × a) – (b ◊ b × c – b ◊ b × a + b ◊ c × a) = (a ◊ b × c) – (b ◊ c × a) = [a b c] – [a b c] = 0
5.68 Integral Calculus, 3D Geometry & Vector Booster 23. Given
[a b c] = 0.
Let
a = a1i + a2 j + a3 k
b = b1i + b2 j + b3k
and
c = c1i + c2 j + c3k
=0
a3 i b3 ¥ a1 c3 b1
j a2 b2
25. Given,
R × B = C × B and R ◊ A = 0.
Now,
(R – B) × C = 0
fi
R = B + l C
Also,
R ◊ A = 0
fi
(B + l C) × A = 0
fi
B ◊ A + l (C ◊ A) = 0
fi
(2 + 1) + l (8 + 7) = 0
k a3 = 0 b3
fi
15l + 3 = 0 1 fi l = – __ 5 Hence, the vector
4 c < 0 and – __ < c < 0 3
27. Given, Now,
( a, b, c are coplanar vectors.) 24. Do yourself By the help of internal section formula and the vector equation of a line.
fi
( )
a b c a a ◊a b◊a a ◊ a a ◊ b a ◊ c = b a ◊ b b ◊ b b◊a b◊b b◊c c a ◊c b◊c a2 b2 c2
c < 0 and c (3c + 4) < 0
4 Hence, the value of c = – __ , 0 3
Then
a1 = b1 c1
fi
1 R = (i + j + k) – __ (4i – 3j + 7k) 5 1 __ = (i + 8j – 2k) 5
a ◊ b = 3 and a × b = c a × (a × b) = a × c
fi
(a ◊ b) a – (a ◊ a) b = a × c
fi
3a – 3b = a × c
fi
3b = 3a – a × c
= (i + j + k) – (– 2i + j + k)
= 3i
fi b=i 28. We have,
i j k (b × c) = 1 1 1 1 1 2 =i–j
Therefore,
i j k a × (b × c) = 1 2 1 1 -1 0 = i + j – 3k
Thus,
29. Given
a × (b × c) ____ 1 = ___________ = ___ (i + j – 3k). |a × (b × c)| ÷ 11 a a c 1 0 1 =0 c c b
fi
a (0 – c) – a(b – c) + c(c) = 0
fi
c2 = ab
26. Let
a = c x i – 6j – 3k
and
b = x i + 2j + 2c x k
Given,
a ◊ b < 0
fi
c x2 – 12 – 6c x < 0
fi
c x2 – 6c x – 12 < 0
= (i + 2j – k) + l(i + j – 2k)
fi
c < 0 and D < 0
= (1 + l)i + (2 + l) j – (2l + 1) k
fi
c < 0 and 36c2 + 48c < 0
fi
c < 0 and 3c2 + 4c < 0
2 Now, the projection of v on a = __ . 3
Thus, c is the GM of a and b. 30. Let v be a vector in the plane of b and c. Then
v = b + lc
÷
__
Vectors
÷
__
(
v ◊ a 2 fi ____ = __ |a| 3 __
– l – 1 = 2
fi
l = – 3
)
1 So, d is parallel to – i + j – __ k . 2
2(1 + l) – (2 + l) – (2l + 1) ___ ÷2 __ fi __________________________ = __ ÷3 ÷6 fi
Now,
d ◊ (3i + 2j – 2k) 1 = __ (2i – 2j + k) ◊ (3i + 2j – 2k) 3 1 = __ (6 – 4 – 2) = 0 3
Therefore, v = – 2i – j + 5k 31. Do yourself By the help of internal section formula and basic Geometry. 32. Given OP = p and OQ = q
Thus, d is parallel to (3i + 2j – 2k). QP × QR 35. Unit vector = ± _________ |QP × QR| 36. We have,
1 = ± ___ __ (2 + ÷6
+ )
3p + 2q 3p – 2q OR = ________ and OS = _______ 1 5
Now,
= p × (c × d), p = (a × b)
OR ^ OS = 0
= (p ◊ d)c – (p ◊ c) d
fi
OR ◊ OS
= ((a × b) ◊ d)c – ((a × b) ◊ c) d
Thus,
(
) (
)
3p + 2q _______ 2p – 2q fi ________ ◊ = 0 1 5 fi
9p ◊ p = 4q ◊ q 2
5.69
(a × b) × (c × d)
= [a b d]c – [a b c] d. Similarly,
(a × c) × (b × d) = [a c d]b – [a b c] d
and (a × d) × (b × c) = [a d c]b – [a d b]c
2
fi
9p = 4q .
Therefore, x = (a × b) × (c × d) + (a × c) × (d × b)
33. Let
OP = a i + b j + g k,
OQ = b i + g j + a k
= [a b d] c – [a b c] d + [a c d] b – [a b c] d
and
OR = g i + a j + b k
+ [a d c] b – [a d b]c = – 2[a b c] d
Now, PQ = OQ – OP
= (b – a) i + (j – b) j + (a – g ) k
= (g – a) i + (a – b) j + (b – g ) k
Again,
QR = OR – OQ
Thus,
= (g – b) i + (a – g )j + (b – a) k |PQ| = |QR| = |PR| _________________________ (b – a)2 + (g – b)2 + (a – g )2
=÷
Therefore, D PQR is an equilateral triangle. 34. Given, 1 d = __ (2i – 2j + k) 3 1 Clearly |d| = __ ◊ 3 = 1 3 Thus, d is a unit vector. 2 1 Also, d = – __ – i + j – __ k 3 2
(
= – 2[b c d] a Thus, x is parallel a.
Also, PR = OR – OP
+ (a × d) × (b × c)
)
37. Given fi Thus,
[b c d] = 0 b, c, d are coplanar vectors d = b + l c
= (j – k) + l (– i + k)
Also,
= (– l) i + j + (l – 1) k a ◊ d = 0
fi
– l – 1 = 0
fi
l = –1
Therefore, d = i + j – 2k Hence,
(i + j – 2k) d __ = ± ___ = ± _________ . |d| ÷6
38. Given
1 a × (b × c) = ___ __ (b + c) 2 ÷
5.70 Integral Calculus, 3D Geometry & Vector Booster fi
1 (a ◊ c) b – (a ◊ b) c = ___ __ (b + c) 2 ÷
Comparing the components of b and c, we get, fi fi fi
1__ 1 (a ◊ b) = – ___ and (a ◊ c) = ___ __ 2 2 ÷ ÷ 1__ a b cos q = – ___ 2 ÷ 1__ cos q = – ___ 2 ÷ 3p q = ___ 4
Volumne of the tetrahedron __
39. Given, u+v+w=0 fi u2 + v2 + w2 + 2(u ◊ v + v ◊ w + w ◊ u) = – 50 fi 2 (u ◊ v + v ◊ w + w ◊ u) = – 50 fi (u ◊ v + v ◊ w + w ◊ u) = – 25 40. Given, (a + b + c) ◊ {(a + b) × (a + c)} = (a + b + c) ◊ {(a × a) + (a × c) + (b × a) + (b × c)} = (a + b + c) ◊ {(a × c) + (b × a) + (b × c)} = a ◊ (b × c) + b ◊ (a × c) + c ◊ (b × a) = [a b c] – [a b c] – [a b c] 41. Let N1 = normal to the plane parallel to i and i + j = i × (i + i) = k
and N2 = normal to the plane parallel to i – j, i + k
= (i – j) × (i + k)
= – i – j + k.
Note that a is parallel to N1 × N2 So, we can consider a = N1 × N2 = k × (– i – j + k)
= (i – j) Let q the angle between a and i – 2j + 2k. Thus, fi
2 cos q = 1 + _____ __ = 2 . 3 ÷ p __ q = 4
1 ___ __ 2 ÷
42. If b is perpendicular to c, the given expression is equal to a. Otherwise can not be calculated. 43. Let DE = p.
__
fi
|BC × BA| p = 4÷2
fi
2÷2 p = 4÷2
fi
p=2
__
__
Here, D ADE is a right-angled triangle. Thus,
AE 2 = AD 2 – DE 2 = 16 – 4 = 12
fi
AE = 2÷3
__
OM = OB + OC Now, _______________ = 2i 2 Thus, fi
= – [a b c]
__
2÷2 1 fi __ |BC × BA| p = ____ 3 6
fi (u + v + w)2 = 0
2÷2 ABCD = ____ 3
AM = OM – OA = 2i – i – j – k = (i – j – k) __
AM = ÷3
If E lies on AM produced, then M is the mid-point of AE. Then 1 __ (e + (i + j + k)) = 2i 2 fi e = 3i – j – k If E lies on MA produced, then A divides the join of E and M in the ratio 2:1. Thus, 2(2i) + e ________ =i+j+k 2 + 1 fi
e = – i + 3j + 3k
44. Let
|p| = |q| = |r| = l
Also,
p ◊ q = 0 = q ◊ r = r ◊ p
Let Now,
x = a p + b q + g r, (where a, b, g, Œ R) p × ((x – q) × p) = (p ◊ p)(x – q) – [p ◊ (x – q)] p = (p ◊ p)(x – q) – (p ◊ x – p ◊ q) p
Vectors
= (p ◊ p)(x – q) – (p ◊ x) p 2
48. Given,
2
= p (x – q) – (a p ) p
= l2 (x – q) – l2 a p
= l2 [(x – q) – a p] Similarly,
...(i)
a ◊a a ◊b a ◊c [a b c] = b ◊ a b ◊ b b ◊ c c◊a c◊b c◊c
q × [(x – r) × q] = l2 [(x – r) – b q]
...(ii)
r × [(x – p) × r] = l2 [(x – p) – g r]
...(iii)
1 = l l
= (1 – l2) – l (l – l2) + l (l2 – l)
= 1 – l2 – l2 + l3 + l3 – l2
= 2l3 – 3l2 + 1.
fi
[a b c]2 = 2l3 – 3l2 + 1
fi
[a b c] = ÷ 2l 3 – 3l2 + 1
2
Adding Eqs. (i), (ii) and (iii), we get fi 45. Given,
0 = l2 (3x – (p + q + r) – x) 1 x = __ (p + q + r). 2 p = ar (quad OABC)
= ar (DOAB) + ar (DOCB) 1 1 = __ |a × (10a + 2b)| + __ |b × (10 a + 2b)| 2 2 1 1 = __ |2a × b| + __ |10b × a| 2 2 = |a × b| + 5|b × a|
= 6 |a × b|
Also,
q = ar (parallelogram OABC) = |a × b|
Here,
__________
Taking dot product of a × b + b × c = pa + pb + rc with a, b and c we get p + q cos q + r cos q = m ...(i)
p cos q + q + r cos q = 0
...(ii)
p cos q + q cos q + r = m
...(iii)
k = 6.
Adding, we get
a × (a × c) = – b
fi fi
|a × (a × c)| = |– b| p |a| |(a × c)| sin __ = 1 2 |a| |(a × c)| = 1
fi
|(a × c)| = 1
fi
a c sin q = 1 1 sin q = __ 2 p __ q = 6
fi
fi fi
( )
47. We have, [(A × C + B × A + B × C) (B × C) (B + C)] = (A × C + B × A + B × C) ◊ [(B × C) × (B + C)] = (A × C + B × A + B × C) ◊ [(B × C) + (C × B)] = (A × C + B × A + B × C) ◊ [(B × C) – (B × C)] = (A × C + B × A + B × C) ◊ 0 = 0.
(where l = cos q)
m = (1 – cos q ) ÷1 + 2 cos q
fi
fi
l l , 1
= m, (say)
p = 6q
a × (a × c) + b = 0
l 1 l
____________
Thus,
46. Given,
5.71
(p + q + r) (1 + 2 cos q )
fi
2m (p + q + r) = ___________ (1 + 2 cos q)
...(iv)
Multiplying Eq. (iv) by cos q and subtracting from Eq. (i), we get,
2m cos q m – __________ = p (1 – cos q) 1 + 2 cos q
m 1 fi p = ___________________ = ___________ __________ (1 + 2 cos q)(1 – cos q) ÷1 + 2 cos q Similarly, – 2 cos q 1 r = ___________ __________ and q = ___________ __________ ÷1 + 2 cos q ÷1 + 2 cos q 49. Given that _› _› a = ( + + ), b = (4 + 3 + 4 ) _› and c = ( + a +_ b ) are linearly independent › _› _› vectors c = l a + m b for some scalers l and m not all zeroes. + a + b = ( + 4m) + (l + 3m) + (l + 4m) l + 4m = 1 l + 3m = a Thus, l + 4m = b
5.72 Integral Calculus, 3D Geometry & Vector Booster On solving, we get, b =__1 Also, given that, |c| = ÷ 3 2 2 1 + a + b = 3 a = ± 1 50. Ans. (c) 51. Ans. (a) 52. We have,
55. Given c is coplanar with a and b. Thus,
(i) (u ◊ v)2 + |(u × v)|2
= u2 v2 cos2q + u2v2 sin2q
= u2v2 (cos2q + sin2q)
= u2v2
1 = __ |a × b| 2 3 = __ , where (a × b) = 2i – 2 j + k 2
(ii) |u + v + (u × v)|2 + (1 – (u ◊ v))2
c = a + l b
= (2i + j + k) + l (i + 2j – k)
= (2 + l) i + (1 + 2l)j + (1 – l) k
Also, fi
c is perpendicular to a c ◊ a = 0
fi
2(2 + l) + (1 + 2l) + (1 – l) = 0
fi
4 + 2l + 1 + 2l + 1 – l = 0
fi
3l + 6 = 0
= (u + v)2 + |(u × v)|2 + 2(u + v) ◊ (u × v)
fi
l+2=0
+ (1 – (u ◊ v))2
fi
l = – 2
So,
c = – 3j + 3k
2
2
2
= (u + v) + |(u ◊ v) | + (1 – (u ◊ v)) 2
2
2 2
2
= u + v + 2u ◊ v + u v sin q 2 2
+ 1 – 2u ◊ v + u v cos q
2
2
2 2
= u + v + u v + 1 = (1 + u2) (1 + v2) = (1 + |u|2) (1 + |v|2). 53. We have |v| = |a × b| = a b sin q = sin q Also,
|u|2 = a2 – 2(a ◊ b) (a ◊ b) + (a ◊b)2 b2
= 1 – 2 (a ◊ b)2 + (a ◊ b)2
= 1 – (a ◊ b)2 = 1 – cos2q = sin2q
fi
|u| = sin q = |v|
Again,
u ◊ b = (a ◊ b) – (a ◊ b)(b ◊ b)
= (a ◊ b) – (a ◊ b) b2
= (a ◊ b) – (a ◊ b) = 0
Thus, 54. Given,
2
|v| = |u| + |u ◊ b| __
|c – a| = 2÷2
__
fi |c – a|2 = ( 2÷2 )2 = 8 2 2 fi c + a – 2c ◊ a = 8 fi c2 + 9 – 2 c = 8 fi c2 – 2c + 1 = 8 fi (c – 1)2 = 0 fi c=1 We have, |(a × b) × c| = |a × b| |c| sin (30°)
1 = ___ __ (– j + k) 2 ÷ 56. Let u and v are not parallel. Let w be a linear combination of u, v and u × v Then w = a u + b v + g (u × v).
Thus,
w × u = b (v × u) + g ((u × v) × u) = b (v × u) + g ((u ◊ u)v – (v ◊ u)u)
Now,
w+w×u=v
fi
a u + b v + g (u × v) + g v – g (v ◊ u)u + b (v × u) = v
fi (a – g (v ◊ u)) u + (b + g ) v + (g – b) (u × v) = v fi (a – g (v ◊ u)) = 0, (b + g ) = 1, (g – b ) = 0 1 1 fi b = g = __ , a = __ (v . u) 2 2 Also, fi
(u × v) ◊ w = a (0) + b (0) + g (u + v) = g (u × v) |(u × v) ◊ w|2 = g 2 |(u × v)|2
= g 2 (u2v2 – (u . v)2)
1 = __ (1 – (u ◊ v)2) 4
fi
1 £ __ 4 1 |(u × v) ◊ w| £ __ 2
and the equality holds iff u ◊ v = 0 ¤ u and v are perpendicular to each other.
Vectors
57. Given
a+b+c=0
fi
a + b = – c
fi
a × (a + b) = – a × c
fi
(a × b) = c × a
Also,
a + c = – b
fi
b × (a + c) = – b × b
fi
b × c = – b × a = a × b
...(i)
...(ii)
From Eqs. (i) and (ii), we get
a×b=b×c=c×a
5.73
=1+x–x =1 = neither depends on x nor y.
62. Thre vectors v1, v2, v3 involve 9 unknown quantities. Since, we are given six equations involving 9 unknown quantities, it is not possible to have uniquely 3 vectors-satisfying these conditions. As v1 ◊ v2 = 4, let us consider v1 = 2k. Also, let v2 = ai + bj + ck and v3 = xi + yj + zk.
58. Let N1 = normal vector to plane P1 = a × b
Now,
v1 ◊ v2 = – 2 fi 2c = – 2 fi c = – 1
and N2 = normal vector to plane p2 = c × d Now,
Now,
v1 . v3 = 6 fi 2z = – 6 fi z = – 3
Now,
v2 ◊ v2 = 2, v2 ◊ v3 = 5, v3 ◊ v3 = 29
\
a2 + b2 + c2 = 2 fi a2 + b2 = 2 – 1 = 1
x2 + y2 + z2 = 29 fi x2 + y2 = 20
and
ax + by + cz = 5
fi
ax + by = 5 – 3 = 2.
Put
b = 0, we get,
a2 = 1 fi a = ±1
and
ax = 2 fi x = ± 2
Now,
x2 + y2 = 20 fi y2 = 20 – 4 = 16
fi
y = ± 4.
N1 × N2 = (a × b) × (c × d) = 0
Thus, N1 and N2 are parallel. Therefore, the angle between P1 and P2 is 0. 59. Given [a b c] = 0 Now, [2a – b 2b – c 2c – a]
= (2a – b) ◊ {(2b – c) × (2c – a)}
= (2a – b) ◊ {4(b × c) – 2(b × a) + (c × a)}
= 8 (a ◊ (b × c)) – (b ◊ (c × a))
= 8 [a b c] – [a b c]
= 7 [a b c]
Thus, one possible set of vectors is v1 = 2k, v2 = i + 2k, v3 = – 2i + 4j + 3k. 63. A (t) is parallel to B (t) for some t in [0, 1]
= 0. 60. We have, |a – b|2 + |b – c|2 + |c – a|2 = 2(a2 + b2 + c2) – 2(a ◊ b + b ◊ c + c ◊ a) = 6 – 2(a ◊ b + b ◊ c + c ◊ a) Also,
...(i)
(a2 + b2 + c2) + 2 (a ◊ b + b ◊ c + c ◊ a)2 = (a + b + c)2 ≥ 0 fi
3 + 2 (a ◊ b + b ◊ c + c ◊ a) ≥ 0
fi
2 (a ◊ b + b ◊ c + c ◊ a) ≥ – 3
fi
– 2 (a ◊ b + b ◊ c + c ◊ a) ≥ 3
...(ii)
|a – b|2 + |b – c|2 + |c – a|2 £ 6 + 3 = 9 1 0 -1 61. [a b c] = x 1 (1 - x ) y x 1+ x - y
Let h(t) = f1(t) g2(t) – f2(t) g1(t)
h(1) = f1 (1) g2(1) – f2(1) g1(1)
= 6.6 – 2.2 = 32 > 0 since h is a continuous function and h(0) h(1) < 0 Then there is some t in [0, 1] for which h(1) = 0 Thus, A(t) is parallel to B (t) for this t. 64. We have, V = [a b c]
(C3 Æ C3 + C1)
h(0) = f1(0) g2(0) – f2(0) g1(0)
= 2.2 – 3.3 = – 5 < 0
From Eqs (i) and (ii), we get
1 0 0 = x 1 1 y x 1+ x
f1 (t) f2 (t) _____ = _____ for some t in [0, 1] g1 (t) g2 (t)
a1 = b1 c1
a2 b2 c2
a3 b3 c3
= a1(b2c3 – b3c2) + a2(b3c1 – b1c3) + a3 (b1c2 – b2c1)
5.74 Integral Calculus, 3D Geometry & Vector Booster = (a1b2c3 + a2b3c1 + a3b1c2) – (a1b3c2 + a2b1c3 + a3b2c1)
We assume that (a1b2c3 + a2b3c1 + a3b1c2) ≥ (a1b3c2 + a2b1c3 + a3b2c1)
i j k = 2 1 -1 = 3i – 7j – k 1 0 3 _________
As we know that, AM ≥ GM
1 fi __ (a1 + b2 + c3) ≥ (a1 b2 c3)1/3 3
Thus, the maximum value of ___
1 fi (a1 b2 c3) ≥ ___ (a1 + b2 + c3)3 27 1 Similarly, (a2 b3 c1) ≥ ___ (a2 + b3 + c1)3 27 1 and (a3 b1 c2) ≥ ___ (a3 + b1 + c2)3 27 Thus,
[u v w] = ÷ 59 .
67. Given
1 a 1 V= 0 1 a a 0 1
= 1 + a (a2 – 1)
(a1 b2 c3) + (a2 b3 c1) + (a3 b1 c2)
= a3 – a + 1
1 £ ___ [(a1 + b2 + c3)3 + (a2 + b3 + c1)3 + (a3 + b1 + c2)3]
dV fi ___ = 3a2 – 1 da
27
1 £ ___ [(a1 + b2 + c3) + (a2 + b3 + c1) + (a3 + b1 + c2)]3 27 Since,
x3 + y3 + z3 £ (x + y + z)3 for x, y, z ≥ 0
3 1 3 = ___ S (ar + br + cr) 27 r =1
1 = ___ (3L) 3 27
= L3
Thus, 65. Given,
V £ L3.
fi
5a2 – 4 (a ◊ b) + 10 (a ◊ b) – 8b2 = 0
fi
5 + 6(a ◊ b) – 8 = 0
fi
6(a ◊ b) = 3
fi
1 (a ◊ b) = __ 2
fi fi fi
(
)
___
r = |r| = ÷ 9 + 49 + 1 = ÷ 59
dV For maximum or minimum, ___ = 0 da fi
3a2 – 1 = 0
fi
1 a2 = __ 3 1__ a = ± ___ 3 ÷
fi
1__ 1__ ___ ___ 3 ÷3 ÷
(a + 2b) ◊ (5a – 4b) = 0
1 Hence, the value of a = ___ __ . 3 ÷ 68. Given x = u + v, y = v + w, z = w + u u + v ˆ _______ v + w ˆ _______ w+u Thus, xˆ = ______ y = , z = . |u + v| |v + w| |w + u| We have,
1 a b cos q = __ 2
( )
p 1 cos q = __ = cos __ 2 3 p q = __ . 2
66. We have, [u v w] = u ◊ (v × w) = u ◊ r
(u + v)2 = u2 + v2 + 2u ◊ v
= 1 + 1 + 2 cos a
= 1 + 1 + 2 cos a
= 2(1 + cos a)
fi
( ) ( )
a = 2 ◊ 2 cos2 __ 2 a |(u + v)| = 2 cos __ 2
= u r cos q
a Similarly, |(v + w)| = 2 cos __ 2
= r cos q, where r = u × v
and
|(w + u)| = 2 cos __
Vectors
Thus,
x = 1/2 (u + v) sec __,
y = 1/2(v + w) sec __
and
z = __(w + u) sec g/2
Now,
[x × y y × z z × x] = [x y z]2
Thus,
a × (b × c) 1 = __________ = ____ ___ (3j – k) |a × (b × c)| ÷ 10
72. Note that a is directed along the internal bisector of the triangle formed by the vectors - v and w Thus,
k a = – v + w
fi
k2a2 = v2 + w2 – 2 (v ◊ w)
fi fi
k2 = 1 + 1 – 2vw cos (p – 2q) k2 = 2 + 2 cos 2q
fi
k2 = 2(1 + cos 2 q) = 2 . 2 cos2q
g 2 b a 1 = __ [u v w] sec __ sec __ sec __ 4 2 2 2
fi
k = 2 cos q
fi
k = 2(a ◊ (– v))
g b a 1 = ___ × sec2 __ sec2 __ sec2 __ [a b c]2 2 2 2 16
Thus, w = v + ka = v – 2(a ◊ (– v)). 73. Ans. (b)
[
( )
( ) ( ) ]
g 2 b a 1 = __ [u + v v + w w + u] sec __ sec __ sec __ 8 2 2 2
[ [
( ) ( ) ] ( ) ( ) ( ) ] [ ( ) ( ) ( ) ] ( )
g b a 1 = __ ◊ 2[u v w] sec __ sec __ sec __ 8 2 2 2
69. Given, fi fi fi fi \ fi fi
2
a × b = c × d and a × c = b × d a×b–a×c=c×d–b×d a × (b – c) = (c – b) × d = d × (b – c) a × (b – c) – d × (b – c) = 0 (a – d) × (b – c) = 0 (a – d) is parallel to (b – c) (a – d) ◊ (b – c) π 0 a ◊ d + c ◊ d π a ◊ c + b ◊ d.
Hence, the result. 70. We have, a × (a × b) = (a ◊ b)a – (a ◊ a)b a × (a × b) = a – a2b = a – 3b 3b = a – a × (a × b) 3b = (i + j + k) – (– 2i + j + k) 3b = 3i b=i a × (b × c) 71. To find = ___________ |a × (b × c)|
fi fi fi fi fi
We have, Now,
i j k b × c = 2 1 1 = 2i – j – 3k 1 −1 1 i j k a × (b × c) = 5 2 6 2 −1 −3
= 27j – 9k
= 9 (3j – k)
5.75
a ◊ b We have a ◊ b1 = a ◊ b – ____ 2 (a ◊ a) = 0. |a|
b1 ◊ c a ◊ c a ◊ c2 = a ◊ c – ____ 2 (a ◊ a) – ____ 2 (b1 ◊ a) |a| |b1| = a ◊ c – a ◊ c = 0
and
b1 ◊ c c ◊ a b1 ◊ c2 = b1 ◊ c – ____ 2 (b1 ◊ a) – ____ 2 (b1 ◊ b1) |a| |b|
= b1 ◊ c2 – 0 – b1 ◊ c = 0
74.
Thus, the required triplet is (a b1 c2). Ans.(c) Let v be a vector. A vector v in the plane of a and b So, v, a, b are coplaner vectors. Thus, v = a + l b fi v = (1 + l) i + (2 – l) j + (1 + l) k fi v = (1 + l) i + (2 – l) j + (1 + l) k
1 the projection of v on c is ___ __ 3 ÷ v ◊ c 1 fi ____ = ___ __ |c| 3 ÷ Also,
(1 + l) + (2 – l) – (1 + l) ___ 1 __ fi ________________________ = __ 3 ÷3 ÷ fi
(2 – l) = 1
fi l=1 Therefore, v = 2i + j + 2k 75. Ans. (c)
Given
−l 2 1 1 1 −l 2 1 =0 1 1 −l 2
5.76 Integral Calculus, 3D Geometry & Vector Booster fi
– l2 (l4 – 1) – 1(l2 – 1) + 1(l2 + 1) = 0
= cos2t + sin2t + (a ◊ b) sin 2 t
fi
– l2 (l4 – 1) + (l2 + 1) + (l2 + 1) = 0
= 1 + (a ◊ b) sin 2 t
fi
6
2
– l + 3l + 2 = 0 6
2
Thus,
________ p |OM| = |OP|max = ÷ 1 + (a ◊ b) , t = __ 4
fi
1 1 1 OM = ___ __ a + ___ __ b = ___ __ (a + b) 2 2 2 ÷ ÷ ÷
fi
l – 3l – 2 = 0
fi
a3 – 3a – 2 = 0, a = l2
fi
a2 (a – 2) + 2a(a – 2) + 1(a – 2) = 0
fi
(a – 2) (a2 + 2a + 1) = 0
Therefore,
fi
(a – 2) (a + 1)2 = 0
79. We have,
fi
(a – 2) = 0, (a + 1)2 = 0
fi
a = 2, a = – 1
fi
2
l = ± ÷2
__
Thus, the number of real values of l is 2. Ans. (b) Given a+b+c=0 fi a + b = – c fi a × (a + b) = – a × c fi (a × b) = c × a Also, a + c = – b fi b × (a + c) = – b × b fi b × c = – b × a = a × b From (i) and (ii), we get, a×b=b×c=c×a
77. We have RS + ST = RT
As PQ and TR are not parallel to each other.
PQ × (RS × ST) π 0
Thus, statement-I is true. Next, since PQ and RS are not parallel, so,
2
2
fi
a⋅a a⋅b a⋅c 1 [a b c] = b ⋅ a b ⋅ b b ⋅ c = 2 c ⋅a b⋅c c ⋅c 1 2
l = 2, l = – 1 l =2
76.
1
2
fi
PQ × RS π 0
Thus, statement-II is false. 78. Ans. (a) Given OP = a cos t + b sin t |OP|2 = |a cos t|2 = |b sin t|2 + 2 (a ◊ b) sin t cos t
OM a+b = _____ = ______ |OM| |a + b|
1 2 1 1 2
1 2 1 2 1
( ) ( ) ( 3 1 __ 1 2 1 = ( __ – __ – = __ = __ 4 8 8) 4 2
)
1 1 1 __ 1 1 1 __ 1 = 1 1 – __ – __ __ – + __ __ – 4 2 2 4 2 4 2
1 [a, b, c] = ___ __ 2 ÷
...(i)
fi
...(ii)
80. Let q1 and q2 be the angles between a and b, and c and d and n1 and n2 are the unit vectors perpendicular to the plane of a and b, and c and d. Now,
a × b = (a b sinq1) n1
and
c × d = (c d sinq2) n2
Let j the angle between n1 and n2. Given
(a × b) ◊ (c × d) = 1
fi
(sin q1) (sin q2) (n1 ◊ n2) = 1
fi
(sin q1) (sin q2) (cos j) = 1
p It is possible only when q1 = __ = q2, j = 0 2 Since j = 0, so n1 and n2 are parallel. Thus, a, b, c and d are coplaner vectors. Also, fi fi
1 a ◊ c = __ 2
( )
p 1 cos q3 = __ = cos __ 2 3 p q3 = __ . 3
( )
p Since the angle between a and c is __ and b and d 3 p is also __ . So b and d are non-parallel. 3
Vectors
81. We have,
1 1 0 V= 1 2 0 1 1 p
= p (2 – 1)
= p.
82. Given,
2
– 2 + 20 + _________ 22 ____________ = _______________________ 4 + 100 + 121 ÷ 1 + 4 + 4 ÷
40 = _____ = 15 . 3
(
8 __ 9
p Note that a = __ – q 2 p Thus, cos a = cos __ – q 2
__
a + b = – ÷3 c
__
(a + b) = (– ÷3 c )
fi
a2 + b2 + 2a ◊ b = 3c2
) (
)
= sin q
÷ ( ) 81 – 64 71 17 = _______ = ___ ÷ 81 ÷ 81 = ÷ ___ 9 .
8 2 = ÷1 – cos q = 1 – __ 9 2
_______
fi
85. Given,
a = mb + 4c
fi
a ◊ b = mb ◊ b + 4b ◊ c
fi
mb2 + 4b ◊ c = 0
83.
________
_________
2
fi
1 + 1 + 2a ◊ b = 3 1 fi a ◊ b = __ 2 1 fi a b cos q = __ 2 p 1 fi cos q = __ = cos __ 2 3 p __ fi q = 3 Let M be the mid-point of PR. Then the position vector of M is
( )
___
___
...(i)
Also,
a ◊ a = m(a ◊ b) + 4(a ◊ c)
fi
|a|2 = 4(a ◊ c)
...(ii)
1 1 __ (– 2i – j + 3i + 3j) = __ i + j 2 2
Given,
(b – a) ◊ (b + c) = 0
Let N be the mid-point of QS. Then the position vector of N is
fi
(b ◊ b) + (b ◊ c) – (a ◊ b) – (a ◊ c) = 0
fi
|b|2 + (b ◊ c) – (a ◊ c) = 0
1 1 = __ (4i – 3i + 2j) = __ i + j 2 2 Thus, PQRS is a parallelogram. ______
___
= ÷36 + 1 = ÷37
And QR = |PR| = |3i + 3j – 4i| = |– i + 3j|
_____
...(iii)
Again,
Also, PQ = |PQ| = |4i – (– 2i – j)| = |6i + j|
5.77
___
= ÷1 + 9 = ÷10
2|b + c| = |b – a|
fi
4|b + c|2 = |b – a|2
fi
4(b2 + c2 + 2b ◊ c) = (b2 + a2 – 2(a ◊ b))
fi
4(b2 + c2 + 2b ◊ c) = (b2 + a2)
fi
3b2 + c2 + 8(b ◊ c) = a2
...(iv)
2
Therefore, PQRS is not a rhombus.
Eliminating (b ◊ c) and a from Eqs (i), (ii), (iii) and (iv), we get
Also, PR = |PR| = |3i + 3j – (– 2i – j)|
(2m2 – 10m) |b|2 = 0
fi
(2m2 – 10m) = 0
fi
m(m – 5) = 0
fi
m = 0, 5.
___
= |5i + 4j| = ÷ 41
And QS = |QS| = |– 3i + 2j – (– 4i)|
_____
__
= |i + 2j| = ÷ 1 + 4 = ÷ 5
So, PQRS is not a rectangle. Therefore, PQRS is a parallelogram, which is neither a rhombus nor a rectangle. 84. Let q be the angle between AB and AD. AB ◊ AD Thus, cos q = ________ |AB| |AD|
86. Given |a| = 1 = |b| and a . b = 0 Now, (2a + b) ◊ {(a × b) × (a – 2b)}
= (2a + b) ◊ {(2b – a) × (a × b)}
= (2a + b) ◊ {(2b × (a × b) – a × (a × b))}
= (2a + b) ◊ {(2(b ◊ b)a – 2(b ◊ a)b – (a ◊ b)a + (a ◊ a)b)}
5.78 Integral Calculus, 3D Geometry & Vector Booster
= (2a + b) ◊ (2b2a + a2b)
90. We have,
= (2a + b) ◊ (2a + b)
|a| = 2 = |b| and a ◊ b = –1 + 3 = 2
= (4a2 + b2 + 4a ◊ b)
Let q be the angle between them.
=4+1+0
=5 87. Given v, a, and b are coplaner, so, v = a + lb = (i + j + k) + l (i – j + k)
= (1 + l) i + (1 – l) j + (1 + l) k
1 Also, the projection of v on c is ___ __ . 3 ÷ v ◊ c 1 fi ____ = ___ __ |c| 3 ÷
|
a2 + b2 – c2 cosq = ___________ 2ab
Then
Thus, 91. Given
4 + 4 – 12 1 = __________ = – __ 8 2 2p q = ___ . 3 |a – b|2 + |b – c|2 + |c – a|2 = 9
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 9 fi 6 – 2 (a ◊ b + b ◊ c + c ◊ a) = 9
|
(1 + l) – (1 – l) – (1 + l) 1 __ fi _______________________ = ___ __ 3 ÷3 ÷
fi 2 (a ◊ b + b ◊ c + c ◊ a) = – 3
fi
fi |a + b + c|2 = 0
(1 – l) = 1
fi 2 (a2 + b2 + c2) – 2 (a ◊ b + b ◊ c + c ◊ a) = 3 – 3 fi (a + b + c) = 0
fi Thus,
l=2 v = 3i – i + 3k
88. Let
a = i + j + k,
Now,
b = i + j + 2k
= |2a + 5 (– a)|
and
c = i + 2 j + k
= |– 3 (a)|
= |– 3| |a|
Now,
i j k (b × c) = 1 1 2 1 2 1
=3
Also,
= – 3i + j + k i j k a × (b × c) = 1 1 1 –3 1 1
|2a + 5b + 5c| = |2a + 5 (b + c)|
92. Given, a × (2i + 3j + 4k) = (2i + 3j + 4k) × b fi a × (2i + 3j + 4k) = – b × (2i + 3j + 4k) fi (a + b) × (2i + 3j + 4k) = 0 fi (a + b) = l (2i + 3j + 4k)
= – 4 j + 4k
fi |(a + b)| = |l (2i + 3j + 4k)|
= 4(– j + k).
fi ÷29 l2 = ÷29
a × (b × c) Thus, ___________ = (– j + k). |a × (b × c)| 89. We have, r×b=c×b
___
___
fi |l| = 1 fi l = ±1 Thus,
(a + b) = ± (2i + 3j + 4k)
Now,
(a + b) ◊ (– 7i + 2j + 3k)
= ± (– 14 + 6 + 12)
fi
a × (r × b) = a × (c × b)
fi
(a ◊ b) r – (a ◊ r) b = a × (c × b)
fi
(a ◊ b) r = a × (c × b)
fi
– r = a × (c × b) = 3i – 6j – 3k
fi
r = – 3i + 6j + 3k
= ± 4 93. When diagonal are given, then area of a parallelogram 1 PQRS = __ |PQ × SQ| 2
Thus,
r . b = 3 + 6 + 0 = 9.
Volume of the parallelopiped
5.79
Vectors
= ((PQ × PS)) ◊ PT)
1 = 2 × __ × (PR × SQ) ◊ PT 2
= [(PR × SQ)] ◊ PT)
3 1 −2 = 1 −3 − 4 1 2 3
[
]
= 30.
98. It is given that a is in the direction of x × (y × z) fi
a = l1(x × (y × z)) = l2((x ◊ z)y –(x ◊ y)z)
( ( ) ( ) )
1 1 fi a = l1 2 × __ y – 2 × __ z = l1 (y – z) 2 2 fi a ◊ y = l1(y ◊ y – y ◊ z) a = a ◊ y(y – z)
Thus,
= |3 (– 9 + 8) – (3 + 4) – 2 (2 + 3)|
Similarly, b = (b ◊ z)(z – x)
= |– 3 – 7 – 10|
Now,
= 20
= (a ◊ y)(b ◊ z)(1 – 1 – 2 + 1)
94. Given
[a b c] = 2
= – (a ◊ y)(b ◊ z).
99. Given a × b + b × c = pa + qb + rc
Now,
[2(a × b) 3(b × c) (c × a)]
fi
= 6[(a × b) (b × c) (c × a)]
fi
= 6[a b c]2
= 24
95. Given
[a b c] = 5
Now,
[3 (a + b) (b + c) 2(c + a)]
= 6 [(a + b) (b + c) (c + a)]
= 6 × 2 [a b c]
= 60
1 96. Given __ |a × b| 2 fi
a ◊ b = (a ◊ y)(b ◊ z){(y – z) ◊ (z – x)}
|a × b| = 0
a ◊ [(a × b) + (b × c)] = p + q (a ◊ b) + r (a ◊ c)
a ◊ (b × c) = p + q (a ◊ b) + r(a ◊ c) q r fi a ◊ (b × c) = p + __ + __ 2 2 q r fi p + __ + __ = a ◊ (b × c) = [a, b, c] 2 2 p r fi __ + q + __ = 0 2 2 p q __ + __ + r = [a, b, c]2 2 2 Also,
...(i) ...(ii) ...(iii)
a ⋅a a ⋅b a ⋅c [a, b, c] = b ⋅ a b ⋅ b b ⋅ c c⋅a c⋅b c⋅c 2
1 Now, __ |(2a + 3b) × (a – b)| 2
1
1 = __ |– 2a × b + 3b × a| 2
1 = __ |– 2a × b – 3a × b| 2
1 = __ |– 5a × b| 2
5 = __ |a × b| 2
= 100
fi
97. Given
|a × b| = 30
Now,
|a × (a + b)|
= |a × a + a + b|
= |0 + a × b|
= |a × b|
=
1 2 1 2
1 2 1 1 2
1 2 1 2 1
( ) ( ) ( 3 1 __ 1 2 1 = ( __ – __ – = __ = __ 4 8 8) 4 2
)
1 1 1 __ 1 1 1 __ 1 = 1 1 – __ – __ __ – + __ __ – 4 2 2 4 2 4 2
1 [a, b, c] = ___ __ 2 ÷
From Eqs (i), (ii) and (iii), we get
q = r = – q
(
)
2 p2 + 2q2 + q2 q____________ + 2q2 + q2 Now, ____________ = =4 q2 q2
5.80 Integral Calculus, 3D Geometry & Vector Booster 100. We have,
fi
_› _ _ › › |b + c | = |a | _ › _› _› |b + c | 2 = |a | 2 _ › _› _ _ _ › › › |b | 2 + |c | 2 + 2b ◊ c = |a | 2
fi
48 + |c | 2 + 48 = 144
fi
|c | 2 = 144 – 96 = 48
fi
4÷3
=
_›
_› |c |
_›
_›
_› _›
› _› _ (a ◊ b )
= – 72
Again, _› _› _› _› a + b + c = 0
_›
_›
_ ›
_›
_›
_ ›
|a × b + c × a | = |a × b + a × b | _›
_›
_›
_›
_›
9 7 x = 4, y = __ , z = – __ 2 2
Hence, the value of 2x + y + z = 9 103. We have, _› ◊ ( × v ) _ › fi | | | × v | cos a = 1 fi cos a = 1 _› fi ^ and ^ v
_ › _› _› _› fi a × b = c × a
_›
= 2|a × b |
_ › › 2 = 2 a2b2 – (a ◊ b )
= 2÷144 × 48 – (– 72)
Hence, infinitely many such vectors v exist.
= 2÷144 × 48 – 72 × 72
If
=
fi
=
=
=
=
____________ _
÷________________
2
_________________
Hence, the result. 101. It is given that
|
__
______________ 2÷72(2 × 48 – 72) ___________ 2÷72(96 – 72) _______ 2÷72 × 24 __ 2 × 24 ÷3 __ 48 ÷3
|
__ a ÷3 + b ________ = ÷ 3 2
_›
_›
Solving, we get
_›
_›
+ (x + y + z)r
144 + 48 + 2(a ◊ b ) = 48
_› _›
_›
+ z (– p – q + r ) _ _ › › = (– x + y – z)p + (x – y – z)q
(x – y – z) = 3 (x + y + z) = 5
_›
= |c | 2
_›
= x (– p + q + r ) + y (p – q + r )
fi fi
_›
_›
_ ›
_›
|a | 2 + |b | 2 + 2(a ◊ b ) = |c | 2
fi
_›
_›
(– x + y – z) = 4
fi
Thus,
...(ii)
fi
fi
_ › _› |a + b | 2
fi
a=2+÷ 3 b
_ › 48 Now, – |a | = ___ – 12 = 24 – 12 = 12 2 2 Also,
Given
__
s = 4p + 3q + 5r
__
_ › _› |a + b |
...(i)
102. We have,
_ ›
fi
__
a ÷3 + b = ± 2÷3
From Eqs (i) and (ii), we get a = – 1 or 2 Thus, |a| = 1 or 2
_›
_› |c | = _› |c | 2 ___
__
_›
As it is given that there exists a vector v . fi
must be perpendicular to
_›
= u1 + u2 __›
◊ w = 0
fi fi fi
(u1 + u2) = 0 u1 = – u2 |u1| = |– u2| = |u2|
If
u = u1 + u3 _ › __›
fi u ◊ w = 0 fi
u1 + 2u3 = 0
fi
|u1| = 2|u3|
Chapter
6
3D-Co-ordinate Geometry
Concept Booster 1.1 Introduction In earlier classes, we learnt about points, lines, cicles and conic sections in two-dimensional geometry. In 2D-geometry, a point represented by an ordered pair (x, y) for which both x and y are real numbers. In the space, each body has length, breadth and height, i.e. each body exits in three-dimensional space. Thus, three independent quantities are required to represent any point in a space and so three axes are required to represent these quantities.
1.2 Rectangular
1.3 Position
vector of a point in a space
Let O be a fixed point, known as the origin, and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis, respectively, in such a way that they form a right handed system.
co-ordinate system
The cartesian system of three lines which are mutually perpendicular to each other is called rectangular co-ordinate system. When three mutually perpendicular planes intersect at a point, the mutually perpendicular lines are obtained and these lines also pass through that point. If we assume the point of intersection as the origin, these three planes are known as co-ordinate planes and the three lines are known as co-ordinate axes.
The plane XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane, respectively. Let P be a point in a space such that its distances from yz-, zx- and xy-planes be a, b and c, respectively and i, j and k are the vectors along x, y and z axes, respectively, i.e.
= OA + AL + LP
Octants
= OA + OB + OC.
Every plane bisects the space. Hence three-co-ordinate plane divide the space into eight parts. Each part is called an octant.
= ai + bj + ck.
OA = a, OB = b and OC = c
Now,
OP = OL + LP
__________
and
|OP| = ÷ a 2 + b2 + c2 .
Notes: 1. Equation 2. Equation 3. Equation 4. Equation 5. Equation 6. Equation
of of of of of of
x-axis: y = y-axis: x = z-axis: x = xy-plane: z yz-plane: x zx-plane: y
0, z = 0 0, z = 0 0, y = 0 =0 =0 =0
6.2 Integral Calculus, 3D Geometry & Vector Booster
1.4 Distance
r – r1 m fi _____ = __ n r2 – r
between two points in space
From the figure, it is clear that
PQ = OQ – OP
= r2 – r1
P
Q r2
r1 O
fi
n (r – r1) = m (r2 – r)
fi
r (m + n) = mr2 + nr1
fi
mr2 + nr1 r = ________ m+n
|PQ| = |r2 – r1|
Let
P = (x1, y1, z1) and Q = (x2, y2, z2)
fi
(xi + yj + zk)
Then
PQ = OQ – OP
m (x2i + y2 j + z2k) + n (x1i + y1j + z1k) = _________________________________ m+n
= (x2 – x1) i + (y2 – y1) j + (z2 – z1) k
Thus, the distance between two points, ____________________________
PQ = ÷ (x 2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
1.4.1 Distance from Origin Let O be the origin and P (x, y, z) be any point, then
__________ x2 + y2 + z2 .
OP = ÷
1.4.2 Distance of a Point from Co-ordinate Axes _______
mx2 + nx1 my2 + ny1 mz2 + nz1 fi x = ________ , y = ________ , z = ________ m+n m+n m+n
1.5.2 External Section If a point R(r) divides the line segment joining the points P(r1) and Q(r2) externally in the ratio m : n, then mr2 – nr1 r = ________ . m – n mx2 – nx1 x = ________ , m – n
The distance from a point to x-axis = ÷ b 2 + c2
my2 – ny1 y = ________ , m – n
The distance from a point to y-axis = ÷ a 2 + c2
mz2 – mz1 z = _________ m – n
_______
_______
and the distance from a point to z-axis = ÷ a 2 + b2
1.5.3 Mid-Point Formula If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio 1 : 1, then r2 + r1 r = ______ . 2 Thus, the co-ordinates of R are
1.5 Section
(
)
x1 + x2 y______ 1 + y2 z______ 1 + z2 _______ , , 2 2 2
formulae
1.5.1 Internal Section
1.5.4 Centroid
If a point R(r) divides the line segment joining the points P(r1) and Q(r2) internally in the ratio m : n, then
mr2 + nr1 r = ________ . m+n
PR fi ____ = RQ
m __ n
OR – OP m fi _________ = __ n OQ – OR
The point of intersection of the medians of a triangle is called the centroid of the triangle. Let A (r1), B (r2) and C (r3) be m the vertices of a triangle ABC PR where ___ = __ n and G (r) be its centroid. Then RQ the position vector of the centroid is
r1 + r2 + r3 r = __________ . 3
3D-Co-ordinate Geometry
Let the position vectors of OA, OB and OC are r1, r2 and r3, respectively. Let the position vectors of G be r. Now, the position vector of D is r2 + r3 OD = ______ 2
= OG = r r2 + r3 2 ◊ ______ + 1 r1 2 = ______________ . 1+2
1.6.3 Refult-1 Let OP = r = ai + bj + ck and i, j, k be the unit vectors along the x-axis, y-axis and z-axis, respectively Now,
r i = a
fi
|r| |i| cos a = a a cos a = __ |r| a l = __ |r|
fi
Thus, the co-ordinates of the centroid G are
(
A set of three numbers a, b, c, which are proportional to the direction cosines l, m, n respectively of a line, are called the direction ratios.
fi
r1 + r2 + r3 = __________ 3
1.6.2 Direction Ratios
l __ m __ n Thus, __ a = b = c .
As we know that, the centroid divides the median in the ratio 2 : 1. Thus, the position vector of G,
)
x1 + x2 + x3 y1__________ + y2 + y3 z1_________ + z2 + z3 __________ , , 3 3 3
Similarly,
b c m = __ and n = __ |r| |r|
i.e.
a l = ____________ __________ , 2 c2 ÷ a + b2 +
and
c n = ____________ ___________ . 2 c2 ÷ a + b2 +
Centroid of a Tetrahedron Let the co-ordinates of A, B, C and D are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) and (x4, y4, z4), respectively. Then the co-ordinates of its vertices are
(
6.3
a = ____________ ___________ 2 c2 ÷ a + b2 +
1.6.4 Refult-2
If l, m, n be the direction cosines of a line, then x1 + x2 + x3 + x4 y1_____________ + y2 + y3 + y4 z1_____________ + z2 + z3 + z4 ______________ , , l 2 + m2 + n2 = 1 4 4 4
)
1.6.5 Refult-3 Any vector r can be expressed as Let
r = |r| (li + mj + nk) r = ai + bj + ck.
r fi __ = |r|
1.6 Direction line
cosines and direction ratios of a vector or a
1.6.1 Direction cosines If a line makes an angle a, b, g with the positive direction of the co-ordinate axes. Then cos a, cos b, cos g are known as the direction cosines of the given line and are, generally, denoted as l, m and n, respectively. Thus, l = cos a, m = cos b, n = cos g.
a b c __ i + __ j + __ k |r| |r| |r|
= li + mj + nk
fi r = |r| (li + mj + nk)
1.6.6 Refult-4 If a vector r having direction cosines l, m, n, the projection of r on the co-ordinate axes are given by
l|r|, m|r|, n|r|.
1.6.7 Refult-5 The projection of the segment joining the points P (x1, y1, z1) and Q (x2, y2, z2) on a line having direction cosines l, m, n is given by l (x2 – x1) + m (y2 – y1) + n (z2 – z1).
6.4 Integral Calculus, 3D Geometry & Vector Booster
1.6.8 Refult-6
= (l21 + m 21 + n 21) (l22 + m 22 + n 22)
If the co-ordinates of P and Q are (x1, y1, z1) and (x2, y2, z2), the direction ratios of the line PQ are a = x2 – x1, b = y2 – y1, c = z2 – z1 and the direction cosines of the line PQ are
fi
(l1m2 – l2m1)2 + (m1n2 – n2m1)2
x2 – x1 y2 – y1 z2 – z1 l = ______ , m = ______ , n = ______ |PQ| |PQ| |PQ|
fi
1.6.9 Direction Cosines of the Axes
fi
Since the positive x-axis makes angles 0°, 90°, 90° with the axes of x, y and z, respectively, then the direction cosines of x-axis are (1, 0, 0). Similarly, the direction cosines of y-axis and z-axis are (0, 1, 0) and (0, 0, 1) respectively.
l1 fi __ = l2
m = a1i + b1j + c1k
and
n = a2i +
Then
m ◊ n = m n cos q
2j
+ c2k
a1a2 + b1b2 + c1c2 __________ __________ = ________________________ 2 ÷a 1 + b 21 + c21 ÷a 22 + b 22 + c22 a1a2 + b1b2 + c1c2 = _______________ l1l2
___________
l1 = ÷ a 21 + b 21 + c21 , l2 = ÷ a 22 + b 22 + c22
a1 a2 fi cos q = __ __ + l1 l2
b1 2 __ __ + l1 l2
m1 ___ m = 2
n1 __ n . 2
Straight Line A straight line in a space can be determined uniquely if (i) it passes through a fixed point and is parallel to a fixed line. (ii) it passes through two fixed points. (iii) it is the intersection of two given non-parallel planes.
___________
where
l1m2 = l2m1; m1n2 = n2m1; n1l2 = l2n1
2.1 Definition
m ◊ n _____ m ◊ n fi cos q = ____ mn = |m| |n|
(l1m2 – l2m1) = 0, (m1n2 – n2m1) = 0, (n1l2 – l2n1) = 0
1.6.10 Angle between Two Vectors Let
+ (n1l2 – l2n1)2 = 0
c1 c2 __ __ l1 l2
= l1l2 + m1m2 + n1n2
(i) Condition of perpendicularity
2.2 Equation
of a line passing thorugh a point and
parallel to a vector
The equation of a line passing through a point A with position vector r1 and parallel to a vector m is r = r1 + lm Now, fi
OP = OA + AP = OA + lm
Let r = x i + y j + z k
r1 = x1 i + y1 j + z1 k
and
m = a i + b j + c k r – r1 = l m
Here,
q = 90°
Then
fi
cos q = 0
fi (x – x1) i + (y – y1) j + (z – z1) k
fi
l1l2 + m1m2 + n1n2 = 0
(ii) Condition of parallelism Here,
q = 0°
fi
cos q = 1
fi
lll2 + m1m2 + n1n2 = 1
fi
(l1l2 + m1m2 + n1n2)2 = 1
fi
(l1l2 + m1m2 + n1n2)2
= l (a i + b j + c k)
x – x1 y_____ – y1 z_____ – z1 Thus, ______ = = c . a b Notes: 1. In case of a line, the direction ratios and the direction cosines are the same. 2. Any point on the line can be considered as
(al + x1, bl + y1 + cl + z1)
3D-Co-ordinate Geometry
2.3 Equation of a Line Passing Through Two Points A (r1) and B (r2) is r = r1 + (r2 – r1) OP = OA + AP
Now,
fi OP + OA + l AB fi r = r1 + l (r2 – r1). r = x i + y j + z k
\
r1 = x1 i + y1 j + z1 k
r2 = x2 i + y2 j + z2 k
Then
(r – r1) = l(r2 – r1)
2.5 Skew lines Two lines are said to be skew lines if they are neither parallel nor intersecting. Clearly, the skew lines can never be coplnar.
2.5.1 Shortest Distance Between Two Skew Lines
fi OA + l (OB – OA)
Let
6.5
Let
L1 : r = r1 + l u
and
L1 : r = r2 + m v
Let
OA = r1 and OB = r2
Here,
L1 || u and L2 || v
fi (x – x1) i + (y – y1) j + (z – z1) k
= (x2 – x1) i + (y2 – y1) j + (z2 + z1) k
x – x1 fi ______ = x2 – x 1
y – y1 ______ = y2 – y 1
z – z1 ______ z2 – z 1
fi PQ ^ L1 and PQ ^ L2 fi PQ ^ u and PQ ^ v fi PQ || u × v Thus,
2.4 Angle
between
two straight lines
Let
L1 : r = r1 + l m
and
L2 : r = r2 + m n
Then
m ◊ n = m n cos q
The shortest distance = PQ
= Projection of AB on PQ
AB ◊ PQ = _______ |PQ|
(r2 – r1) ◊ (u × v) = ______________ . |(u × v)|
Notes 1: If two straight lines intersect, the shortest distance between them is zero.
m ◊ n fi cos q = ____ mn
The Plane
Let
m = a1 i + b1 j + c1 k
and
n = a2 i + b2 j + c2 k
3.1 Definition
Thus,
a1a2 + b1b2 + c1c2 __________ __________ cos q = ________________________ 2 2 + c21 ÷a 22 + b22 + c22 ÷ a 1 + b 1
A plane is a surface such that if any two points on it are taken, the line joining them lies completely on it.
1. Condition of perpendicularity When
\
p q = __ 2 a1a2 + b1b2 + c1c2 = 0
2. Condition of parallelism When
q=0
a1 \ __ a2 =
b1 __ + b2
c1 __ c 2
3.2 General Form A first degree equation represents a plane. The general equation of a plane is given by a x + b y + c z + d = 0.
Notes: 1. Equation 2. Equation 3. Equation 4. Equation given by
of xy-plane is z = 0. of yz-plane is x = 0 of zx-plane is y = 0. of a plane passing through the origin is a x + b y + c z = 0.
6.6 Integral Calculus, 3D Geometry & Vector Booster
3.3. Equation
of a plane passing through a point (x, y, z,)
The equation of a plane passing through a point (x1, y1, z1) is a (x – x1) + b (y – y1) + c (z – z1) = 0 Let the equation of the plane be a x + b y + c z + d = 0 ...(i) which is passing through (x1, y1, z1). Thus, a x1 + v y1 + c z1 + d = 0 ...(ii) Subtracting Eqs. (ii) from Eq. we get a (x – x1) + b (y – y1) + c (z – z1) = 0
3.4 Equation of a plane passing non-collinear points
through three
The equation of any plane passing through (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by x – x1 y – y1 z – z1 x – x y z = 0 2 – z1 2 1 2 – y1 x3 – x1 y3 – y1 z3 – z1
|
|
The equation of any plane passing through (x1, y1, z1) is given by a (x – x1) + b (y – y1) + c (z – z1) = 0 ...(i) When it is passing through (x2, y2 , z2) and (x3, y3, z3), then a (x2 – x1) + b (y2 – y1) + c (z2 – z1) = 0 ...(ii) a (x3 – x1) + b (y3 – y1) + c (z3 – z1) = 0 ...(iii)
|
Eliminating a, b, c from Eqs (i), (ii) and (iii), we get x – x1 y – y1 z – z1 x y z = 0 2 – z1 2 – x1 2 – y1 x3 – x1 y3 – y1 z3 – z1 which is the required equation of the plane.
3.5 Coplanarity
|
and
B ◊ b + D = 0
C ◊ c + D = 0
D fi B = – __ b D fi C = – __ c .
Putting the values of A, B and C in Eq. (i), we get D D D __ __ – __ a x – b y – c z + D = 0 y __z x __ fi __ a + b + c – 1 = 0 y __z x __ fi __ a + b + c = 1
which is the required equation of the plane.
3.7 Normal
to a plane
A line perpendicular to a plane is called the normal to the plane. Clearly every line in a plane is perpendicular to the normal to the plane.
3.8 Vector
form
The vector equation of a plane passing through a point having position vector a and normal to a vector n is given by r n = d
of four points
The points P (x1, y1, z1), Q (x2, y2, z2), R (x3, y3, z3) and S (x4, y4, z4) are coplanar, then x – z1 2 – x1 y 2 – y1 z2 x y z = 0 3 – z1 3 – x1 3 – y1 x4 – x1 y4 – y1 z4 – z1
3.6 Intercept
Let the equation of the plane be passing through A (a, 0, 0), B (0, b, 0), and C (0, 0, c). A x + B y + C z + D = 0 ...(i) D So, A ◊ a + D = 0 fi A = – __ a
|
form of a plane
|
The equation of a plane in intercept form is given by y __z x __ __ a , b , c = 1, where a, b, c are the lengths of x, y and z axes, respectively.
Let OP = r and ON = n Now, ON ^ NP fi NP ON = 0 fi (OP – ON) ON = 0 fi (r – n) n = 0 fi r n – n n = 0 fi r n = n n fi r ◊ n = d.
3.9 Equation
of a plane in normal form
The vector equation of a plane normal to a unit vector and at a distance from the origin is given by r ◊ = d Let O be the origin and ON be the perpendicular from O to the given plane such that ON = d .
3D-Co-ordinate Geometry
Let P be a point on the plane with position vector r so that OP = r. ON ^ NP
Now,
fi NP ON = 0 fi (OP – ON) ON = 0 fi (r ◊ d ) ◊ d = 0
(ii) Castesian form: a1b1 + a2b2 + a3b3 __________ __________ cos q = ________________________ ÷a 21 + a 22 + a23 ÷b 21 + b22 + b 23 where and
n1 = a1 i + a2 j + a3 k n2 = b1 i + b2 j + b3 k.
Condition of Perpendicularity
fi (r ◊d – d ◊ d ) = 0 fi (r ◊ – d) = 0
fi r ◊ = d
Condition of Parallelism
which is the required equation of the plane.
a3 a1 a2 __ __ = __ = . b1 b2 b3
3.10 Normal
form of a
Plane
If l, m, n be the direction cosines of the normal to a given plane and p be the length of the perpendicular from the origin to the plane, the equation of the plane is given by l x + m y + n z = p. As we know that the normal form of a plane is given by r = d Let r = x i + y j + z k and = l i + m j + n k The equation of the plane becomes l x + m y + n z = p.
3.11 Theorem Equation of a plane passes through a point A with position vector a and is parallel to the given vectors b and c. b P c r
a O
Let r be the position vector of any point P in the plane, then AP = OP – OA =r–a Since, the vectors r – a, b, c are coplanar So,
[r – a b c] = 0
fi [r b c] – [a b c] = 0 fi [r b c] = [a b c] which is the required equation of the plane.
3.12 Angle
3.13 Angle
a1b1 + a2b2 + a3b3 = 0
between a line and a plane
The angle between a line and a plane is the angle between the line and the normal to the plane. Let the equation of the plane be
a1x + b1y + c1z + d = 0
between two planes
The angle between two planes is defined as the angle between their normals. Let r n1 = d1 and r n2 = d2 and q be the angle between them. Then n1 n2 (i) Vector form: cos q = ______ . |n1| |n2|
...(i)
and the equation of the line be – y1 z – z1 x – x1 y_____ _____ = = _____ a2 c2 b2 Thus, cos (90° – q)
A
6.7
fi
a1a2 + b1b2 + c1c2 __________ __________ = ________________________ 2 ÷a 1 + b21 + c21 ÷a 22 + b22 + c 22
a1a2 + b1b2 + c1c2 __________ __________ sin q = _______________________ ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Condition of Perpendicularity a1 __ b1 __ c1 __ a2 = b2 = c2 Condition of parallelism a1a2 + b1b2 + c1c2 = 0
3.14 Equation
of a plane parallel to a given plane
The equation of a plane parallel to another plane a x + b y + c z + d = 0 is a x + b y + c z + k = 0.
3.15 Equation
of a plane parallel to the axes
3.15.1 Equation of a plane parallel to x -axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of x-axis is x y __z __ = __ = 1 0 0 Since the plane (i) is parallel to the line (ii), so, a ◊1 + b . 0 + c . 0 = 0 fi a=0
...(i) ...(ii)
6.8 Integral Calculus, 3D Geometry & Vector Booster
3.16 Equation of a (x2, y2, z2) and ratios a, b, c
Putting the value of a in Eq. (i), we get b y + c z + d = 0. which is the required equation of the plane.
3.15.2 Equation of a plane parallel to y -axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of y-axis is x y __z __ = __ = 0 1 0 Since the plane (i) is parallel to the line (ii), so, a ◊ 0 + b ◊ 1 + c ◊ 0 = 0 fi b=0 Putting the value of b in Eq. (i), we get a x + c z + d = 0. which is the required equation of the plane.
...(ii)
fi c=0 Putting the value of a in (i), we get a x + b y + d = 0. which is the required equation of the plane.
3.15.4 Equation of a plane parallel to xy -plane The equation of xy-plane is given by z=0 So the equation of any plane parallel to xy-plane is z+k=0
A x + B y + C z + D = 0
\
A (x – x1) + B (y – y1) + C (z – z1) = 0
and
A (x – x2) + B (y – y2) + C (z – z2) = 0
Also,
A ◊ a + B ◊ b + C ◊ c = 0
...(i) ...(ii)
Hence, the result.
3.17 Equation of the plane passing through a point (x1, y1, z1) and parallel to two lines having direction ratios (a1, b1, c1) and (a2, b2, c2) is x – x1 a1 a2
y – y1 b1 b2
z – z1 c1 = 0 . c2
The equation of any plane passing through (x1, y1, z1) is A (x – x1) + B (y – y1) + C (z – z1) = 0 Since the plane (i) is parallel to two lines, so A a1 + B b1 + C c1 = 0 A a2 + B b2 + Cc2 = 0
x
x1
y
y1
z
...(ii) ...(iii)
z1
a1
b1
c1
a2
b2
c2
=0
which is the required equation of the plane.
3.18 Equation
3.15.6 Equation of a plane parallel to zx -plane
...(i)
Eliminating A, B, C from Eqs (i), (ii) and (iii), we get
So the equation of any plane parallel to yz-plane is x+k=0
...(i)
which is passing through (x1, y1, z1) and (x2, y2, z2)
The equation of yz-plane is given by x=0
z – z1 z2 – z1 = 0 . c
Let the equation of the plane be
3.15.5 Equation of a plane parallel to yz -plane
y – y1 y2 – y1 b
Eliminating, A, B and C from above three equations, we get x – x1 y – y1 z – z1 x2 – x1 y2 – y1 z2 – z1 = 0 a b c
3.15.3 Equation of a plane parallel to z-axis Let the equation of any plane be a x + b y + c z + d = 0 and the equation of z-axis is x y z __ = __ = __ 0 0 1 Since the plane (i) is parallel to the line (ii), so, a ◊ 0 + b ◊ 0 + c ◊ 1 = 0
x – x1 x2 – x1 a
...(i)
plane passing thorugh (x1, y1, z1), parallel to the line having direction
of a plane passing through the line of
intersection of planes
The equation of zx-plane is given by
The equation of a plane passing through the line of intersection of planes
y=0 So the equation of any plane parallel to zx-plane is
a1x + b1y + c1z + d1 = 0
is a1x + b1y + c1z + d1 + l (a2x + b2y + c2z + d2) = 0.
y + k = 0.
and a2x + b2y + c2z + d2 = 0
3D-Co-ordinate Geometry
3.19 Distance
3.24 Condition
of a point from a plane
The length of the perpendicular from a point P (x1, y1, z1) to the plane a x + b y + c z + d = 0 is ax1 + b y1 + c z1 + d __________ _________________ + c2 ÷ a 2 + b2
3.20 Distance
|
|
between two parallel planes
of coplanarity of two lines
Let
x – x1 y_____ – y1 z_____ – z1 L1: ______ a = = c b1 1 1
and
x – x1 y_____ – y2 z_____ – z2 L2: ______ a = = c b2 2 2
If two lines L1 and L2 are coplanar, then
The distance between two parallel planes a x + b y + c z + d1 = 0 and a x + b y + c z + d2 = 0 is given by d – d2 ____________ ___________ . 2 c2 ÷ a + b2 +
3.21 Sides
and the equation of the plane containing them is
|
|
of a plane
We know that a plane divides the three-dimensional space in two equal parts. Two points A (x1, y1, z1) and B (x2, y2, z2) are on the same side or opposite side of the plane a x + b y + c z + d = 0 if a x1 + b y1 + c z1 + d __________________ > 0 or < 0. a x2 + b y2 + c z2 + d
3.22 Intersection
of a line and a plane
x – x1 y_____ – y1 z_____ – z1 Let L: _____ = = c a b and P: A x + B y + C z + D = 0 Any point on the line (i) can be considered as (x1 + al, y1 + bl, z1 + cl)
x2
...(iii)
A (x1 + al) + B (y1 + bl) + C (z1 + cl) + D = 0 fi (A x1 + B y1 + C z1 + D) + l (Aa + Bb + Cc) = 0
y2
y1
z2
z1
a1
b1
c1
a2
b2
c2
x1
y
y1
z
=0.
z1
a1
b1
c1
a2
b2
c2
=0
or x
...(i) ....(ii)
x1
x
If it lies on the plane (ii), then
3.25 Equation planes
x2
y
y2
z
z2
a1
b1
c1
a2
b2
c2
=0.
of the planes bisecting the angle between the
The equation of the planes bisecting the angle between the planes a1x + b1y + c2z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by
|
| |
|
(A x1 + B y1 + C z1 + D) fi l = – ____________________ (Aa + Bb + C c)
a1x + b1y + c1y + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Substituting the values of l in Eqs (iii), we get the required co-ordinates of the point of intersection.
3.26 Bisector planes
3.23 Condition
Let the equation of the two planes be
If the line
for a line to lie in a plane
x – x1 ______ y – y1 z_____ – z1 L: ______ = m = n l
lies in the plane P: a x + b y + c z + d = 0, then
(i) a x1 + b y1 + c z1 + d = 0
(ii) a l + b m + c n = 0 Since the line L = 0 lies in a plane P = 0, therefore the point (x1, y1, z1) on L = 0 will also lie in the plane P = 0. So,
a x1 + b y1 + c z1 + d = 0
Also, since the line L = 0 lies in the plane P = 0, therefore, the normal to the plane is also normal to the line. Thus,
a l + b m + c n = 0.
6.9
of the acute and obtuse angles between two
a1 x + b1y + c1z + d1 = 0
...(i)
and
a2 x + b2y + c2z + d2 = 0
...(ii)
First we make d1 > 0, d2 > 0 (i) If a1a2 + b1b2 + c1c2 > 0, the origin lies in the obtuse angle between two planes and the equation of the obtuse angle bisector is a1x + b1y + c1z + d1 a2x + b1y + c2z + d2 __________ __________ __________________ = __________________ 2 2 2 + c 1 ÷a 22 + b 22 + c22 ÷ a 1 + b1 and the acute angle bisector is a1x + b1y + c1z + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ 2 2 2 ÷a 1 + b 1 + c1 ÷a 22 + b 22 + c22
6.10 Integral Calculus, 3D Geometry & Vector Booster (ii) If a1a2 + b1b2 + c1c2 < 0, the origin lies in the acute angle between the two planes and the equation of the bisector of the acute angle is
is
x – x1 l1 l2
y – y1 m1 m2
z – z1 n1 = 0 . n2
a1x + b1y + c1z + d1 __________________ a2x + b2y + c2z + d2 __________ __________ __________________ = 2 2 2 ÷a 1 + b1 + c 1 ÷a 22 + b22 + c22
The equation of the plane containing L1 = 0 is a (x – x1) + b (y – y1) + c (z – z1) = 0
...(i)
and the obtuse angle is
where
...(ii)
a1x + b1y + c1z + d1 a2x + b2y + c2z + d2 __________ __________ __________________ = __________________ . ÷a 21 + b21 + c 21 ÷a 22 + b22 + c 22
Since L2 = 0 is parallel to the plane (i), therefore the normal to the plane (i) is also normal to the line L2 = 0.
3.27 Foot
Eliminating a, b, c from Eqs (i), (ii) and (iii), we get
of perpendicular of a point w.r.t. a plane
The foot of the perpendicular M of a point P (x1, y1, z1) w.r.t. a plane a x + b y + c z + d = 0 is given by x2 – x1 y______ z2 – z1 (ax1 + by1 + cz1 + d) 2 – y1 ______ = = ______ = – __________________ a c b (a2 + b2 + c2)
3.28 Image
of a point w.r.t. a palne
Thus,
3.30 Equation
a l1 + b m2 + c n2 = 0 x – x1 l1 l2
y – y1 m1 m2
...(iii)
z – z1 n1 = 0 . n2
of the plane containing two given lines
The equation of the plane containing two lines
The image Q of a point P (x1, y1, z1) w.r.t. a plane a x + b y + c z + d = 0 is x2 – x1 y______ z2 – z1 2 (ax1 + by1 + cz1 + d) 2 – y1 ______ = = ______ = – ____________________ a c b (a2 + b2 + c2) where Q = (x2, y2, z2).
x – x1 y_____ – y1 z_____ – z1 L1: _____ = m = n l1 1 1
and
x – x2 y_____ – y2 z_____ – z2 L2: _____ = m = n l2 2 2
3.29 Equation of the plane containing a given line and parallel to a given line is The equation of the plane containing a x – x1 ______ y – y1 L1: ______ = m = l1 1 and also parallel to another line x – x2 y_____ – y2 L2: _____ = m = l2 2
a l2 + b m1 + c n1 = 0
given line z – z1 _____ n 1
x – x1 l1 l2
y – y1 m1 m2
z – z1 n1 = 0 n2
x – x2 l1 l2
y – y2 m1 m2
z – z2 n1 = 0 . n2
or
z – z2 _____ n2
Exercises
(Problems based on Fundamentals)
ABC of 3D-Geometry 1. Find the distance between the points P (3, 4, 5) and Q (– 1, 2, – 3). 2. Find the shortest distance of the point P (a, b, c) from x-axis. 3. If the line passing through the points (5, 1, a) and 17 13 (3, b, 1) crosses the yz-plane at 0, ___ , – ___ , find 2 2 the value of a + b + 2. 4. Show that the points P (0, 7, 10), Q (– 1, 6, 6) and C (– 4, 9, 6) are the vertices of an isosceles rightangled triangle.
(
)
5. Find the locus of a point, the sum of distances from (1, 0, 0) and (– 1, 0, 0) is 10. 6. Find the ratio in which the plane x – 2y + 3z = 17 divides the line joining the points (– 2, 4, 7) and (3, – 4, 8). 7. Find the co-ordinates of points which trisects the line joining the points P (– 3, 2, 4) and Q (0, 4, 7). 8. If the centre of a tetrahedron OABC are (1, 2, –1), where A = (a, 2, 3), B = (1, b, 2), C = (2, 1, c) respectively, find the distance of P (a, b, c) from the origin. 9. Find the ratio in which the line joining the points P (– 2, 3, 7) and Q (3, – 5, 8) is divided by the yz-plane.
3D-Co-ordinate Geometry
10. The vertices of a triangle are A (5, 4, 6), B (1, – 1, 3) and C (4, 3, 2). The internal bisector of –BAC meets BC in D. Find AD. Direction cosines and direction ratios 11. If a line is equally inclined with the axes, find its direction cosines. 12. Find the angle at which the vector (4 i + 4 j + k) is inclined to the axes. 13. If a line makes an angle a, b, g with the co-ordinate axes, find the value of (i) S sin2 a (ii) S cos (2a). 14. A line OP makes with the x-axis and y-axis at an angle of 120°, 60°, respectively. Find the angle made by the line with the z-axis. 15. A vector r is inclined to x-axis and to y-axis at 60°. If |r| = 8, find the vector r. 16. Find the projection of the line joining (1, 2, 3) and (– 1, 4, 2) on the line having direction ratios (2, 3, – 6). 17. What is the angle between the lines whose direction
(
__
) (
__
)
÷3 3 ÷ 3 1 3 1 ___ cosines are – __ , __ , – ___ and – __ , __ , ? 4 4 2 4 4 2 18. Find the angle between any two diagonals of a cube. 19. Find the angle between one diagonal of a cube and a diagonal of one face. 20. A line makes angles a, b, g with the four diagonal of a cube, find the value of
S cos2 a (ii) S sin2 a (iii) S cos(2a) (i)
21. Find the angle between the lines whose direction cosines are satisfy the equations l + m + n = 0 and l 2 + m2 – n2 = 0. 22. Find the direction cosines of the two lines which are connected by the relation l – 5m + 3n = 0 and 7l 2 + 5m2 – 3n2 = 0. 23. Find the angle between the lines which are connected by the relation l + m + n = 0 and 2l m + 2l n + m n = 0. 24. Show that the straight lines whose direction cosines are given by the equations al + b m + cn = 0 and ul2 + v m2 + w n2 = 0 are perpendicular if a2 (u + v) + b2 (u + w) + c2 (u + v) =0 a2 b2 __ c2 and parallel if __ u + __ v + w = 0. 25. If a variable line in two adjacent positions has direction cosines (l, m, n) and (l + d l, m + d m, n + d n), show that the small angle dq between the two positions is given by (dq)2 = (d l)2 + (d m)2 + (d n)2.
6.11
Striaght Line 26. Find the equation of a line passing through (2, 3, 4) + 1 z____ –7 x – 3 y_____ and parallel to the line _____ = = . 3 4 5 27. Find the equation of a line passing through (– 3, 2, – 4) and parallel to the vector 2 i + 3 j + 7k. y – 2 x – 1 28. Find the point where the line _____ = _____ = 2 – 3 z_____ +3 meets the plane 2x – 2y – z = 7. 4 29. Find the equation of the line through 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to x = 2y = 3z.. 30. Prove that the lines 4x = 3y = – z and 3x = y = – 4z are perpendicular. 31. If the lines x = az + b, y = cz + d and x = a¢z + b¢, y = c¢z + d ¢ are perpendicular, prove that a ◊ a¢ + c ◊ c¢ = – 1. 32. Find the distance of the point (1, – 2, 3) from the plane x – y + z = 5 measured parallel to the y z x line __ = __ = ___ . 2 3 – 6 33. Find the foot of the perpendicular from the point – 1 z____ –7 x – 6 y_____ (1, 2, 3) to the line _____ = = and also 3 2 3 find the length of the perpendicular. 34. Find the image of a point (1, 6. 3) in the line –7 x y – 1 z____ __ = _____ = . 1 2 3 35. Find the equation of the line which can be drawn from the point (1, – 1, 0) to intersect the – 1 z____ y z+1 –3 x – 2 y_____ x – 4 __ lines _____ = = and _____ = = _____ . 2 3 4 4 2 5 orthogonally. 36. Find the equation of the line through (2, – 3, 1) parallel to the plane 2x – y + z = 6 so as to meet y z–2 x – 2 ___ the line _____ = = ____ at right angle. 2 – 3 – 1 Shortest distance between two lines 37. Find the shortest distance between the lines r = 3i + 8 j + 3k + l (3i – j + k) and r = – 3i – 7 j + 6k + m (– 3i + 2 j + 4k). 38. Find the shortest distance between the lines
– 3 z____ –4 x – 2 y_____ L1: _____ = = 3 4 5
and
+ 1 z_____ +2 x – 2 y_____ L2: _____ = = . 2 –1 3
39. Prove that the shortest distance between any two opposite edges of a tetrahedron formed by the planes __ y + z =__0, x + z = 0, x + y = 0 and x + y + z = ÷3 a is a ÷2 .
6.12 Integral Calculus, 3D Geometry & Vector Booster 40. If 2d be the shortest distance between the y z x __z 1 lines __ + __ c = 1, x = 0 and __ – c = __ y = 0, prove a b 1 1 1 1 that __ 2 + __ 2 + __ 2 = ___ 2 . a b c d Plane 41. Find the equation of the plane through the points P (1, 1, 1), Q (3, – 1, 2) and R (–3, 5, – 4). 42. Show that the four points (1, 2, 3), (2, 1, 4) (3, – 1, 3) and (2, 2, 6) are coplanar. 43. Find the intercepts of the plane 2x – 4y + 5z = 20 on the co-ordinate axes. 44. Find the equation of the plane passes through the points (3, 0, 0), (0, 4, 0) and (0, 0, 5). 45. Find the equations of the plane through the points (0, 4, – 3), (6, – 4, 3) other than the plane through the origin, which cut off from the axes intercepts, whose sum is zero, 46. A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point (p, q, r). Find the equation of the plane. 47. A variable plane cuts the co-ordinate axes in A, B, C and is of constant distance 3p from the origin. Find the locus of the centroid of the triangle ABC. x y __z 48. The plane __ a + __ + c = 1 meets the co-ordinate axes b at the points A, B, C respectively. Find the area of the D ABC. 49. A variable plane cuts the co-ordinate axes in A, B, C and is of constant distance p from the origin. Find the locus of the centroid of the tetrahedron OABC. 50. If a variable plane forms a tetrahedron of constant volume 64 k3 with the co-ordinate planes, find the locus of the centroid of the tetrahedron. 51. Find the angle between the planes x – 2y + 2z + 2014 = 0 and 2x + y + 2z + 2015 = 0. 52. Prove that the planes 3x – 2y + 3z + 10 = 0 and 2x – 3y – 4z + 7 = 0 are perpendicular. 53. Find the equation of the plane passing through (1, 2, 3) and the perpendicular to the planes 2x + 3y + 4z + 7 = 0, 3x + 4y + 5z + 10 = 0 x – 2 54. Find the angle between the line _____ = 3 y + 1 z – 3 _____ = ____ and the plane 3x + 4y + z + 5 = 0. – 1 – 2 – 4 z____ –5 x – 3 y_____ 55. Prove that the line _____ = = is parallel 2 3 4 to the plane 4x + 4y – 5z + 2 = 0. 56. Find the equation of a plane passing through (2, 3, – 1) and parallel to x – 2y + 3z + 10 = 0. 57. Find the equation of the plane passing through the points (2, 3, – 4), (1, – 1, 3) and parallel to x-axis. 58. Find the equation of the plane passing through the points (1, 2, 3), (4, 3, 1) and parallel to y-axis
59. Find the equation of the plane passing through the points (3, 4, 5), (– 2, 3, – 1) and parallel to z-axis. 60. Find the equation of the plane passing through (1, 2, 3) and parallel to xy-plane. 61. Find the equation of a plane parallel to yz-plane and passes through the point (2, 3, 4). 62. Find the equation of the plane passing through (3, 4, 5) and parallel to zx-plane. 63. Find the equation of the plane passing through the points (2, – 1, 0), (3, – 4, – 5) and parallel to the line 2x = 3y = 4z. 64. Find the equation of the plane passing through (2, 3, 4) and parallel to the lines x = 2y = 3z and 2x = 5y = z. 65. Find the equation of the plane passing through the intersection of the planes 2x + 5y – 5z = 6 and 2x + 7y – 8z = 7 and the point (– 1, 4, 3). 66. Find the equation of the plane passing through the point (2, – 1, 1) and the line 4x – 3y + 5 = 0 = y – 2z – 5. 67. Find the equation of the plane passing through the intersection of the planes 2x + 3y + 10z = 8 2x – 3y + 7z = 2 and the perpendicular to 3x – 2y + 4z = 5. 68. Find the equation of the plane passing through the line x + y + z + 3 = 0, 2x – y + 3z + 1 = 0 and x y __z parallel to __ = __ = . 1 2 3 69. Find the length of the perpendicular from a point P (1, 2, 3) to the plane 5x + 4y – 3z – 1 = 0. 70. Find the equation of the line through (– 1, 3, 2), perpendicular to the plane x + 2y + 2z = 3 and also find the co-ordinates of its foot. 71. Find the distance of the point (1, 2, 3) from the plane x + y + z = 11 measured parallel + 2 z____ –7 x + 1 y_____ to _____ = = . 1 – 2 2 72. Find the locus of a point, the sum of the squares of whose distances from the planes x + y + z = 0, x – z = 0, x – 2y + z = 0 is 9. 73. Two system of the co-ordinate axes have the same origin. If a plane cuts them at distances a, b, c and p, q, r respectively from the origin. Prove that 1 1 1 1 1 1 __2 + __ 2 + __ 2 = __ 2 + __ 2 + __ 2 . a b c p q r 74. Find the distance of the point (4, 1, 1) from the lines x + y + z – 4 = 0 = x – 2y – 2z – 4. 75. Find the distance between the planes x + 2y – 2z + 1 = 0 and 2x + 4y – 4z + 5 = 0 76. Find the co-ordinates of the point of intersection y – 2 z + 3 x – 1 of the line _____ = _____ = _____ with the plane 2 – 3 4 2x – 2y – z = 7.
3D-Co-ordinate Geometry
77. Find the equation of the plane passing through (1, 2, 0) – 1 z____ –2 x + 3 y_____ which contains the line _____ = = . 3 4 – 2 y – 2 z – 3 x – 1 78. Prove that the lines _____ = _____ = ____ and 2 3 4 y – 3 –4 x – 2 _____ z____ _____ = = are coplanar and also find the 3 4 5 equation of the plane where they lie. 79. Find the equation of the plane passing through the line of intersection of the planes 4x – 5y – 4z = 4 and 2x + y + 2z = 8 at the point (1, 2, 3). 80. Find the eqution of the plane bisecting the acute angle between the planes x + 2y + 2z – 3 = 0 and 3x + 4y + 12z + 1 = 0. 81. Find the image of the point (2, – 3, 4) with respect to the palne 4x + 2y – 4z + 3 = 0. 82. Find the equation of the image of the y – 2 z + 3 x – 1 line _____ = _____ = _____ in the plane 3x – 3y 9 –1 – 3 + 10z – 26. 83. Find the equation of the plane containning the y + 6 z + 1 x – 1 line _____ = _____ = _____ and is parallel to the 3 4 2 – 1 z_____ +3 x – 4 y_____ line _____ = = . 2 – 3 5 84. Find the equation of the plane containing the lines – 1 z_____ – 4 z____ +3 –6 x + 1 y_____ x – 2 y_____ _____ = = and _____ = = . 7 3 1 3 5 5
1. The shortest distance of the point (a, b, c) from the x-axis is ______
______
(a) ÷a 2 + b2
(b) ÷ a2 + c2
(c) ÷b 2 + c2
(d) ÷ a2 + b2 + c2
______
__________
2. The equation of the x-axis is x y __z x (a) __ = __ = (b) __ = 1 0 0 0 x y __z x (c) __ = __ = (d) __ = 0 0 1 1
y __ = 1 y __ = 0
(
z __ 0
5. The foot of perpendicular from (a, b, c) on the z-axis is (a) (a, 0, 0) (b) (0, b, 0) (c) (0, 0, 0) (d) (0, 0, c). 6. The angle between any two diagonals of a cube is __
(b) tan– 1 (÷2 ) __ 2 (c) cot– 1(÷2 ) (d) cos– 1 __ 3 7. The angle between a diagonal of a unit cube and a diagonal of a face is __ 2 2 – 1 __ –1 __ (a) cos (b) cos 3 3
( ) 1 ( __ ÷ 3 )
(c) cos– 1
3 ÷ (a) ___ 2 __
)
1 1 __ 1 3. If the direction cosines of a line are __ c , __ c , c , then c is (a) 1 (b) ± 2__ (c) ± 3 (d) ± ÷3 4. The number of lines, through the origin makes equal angles with the axes, is (a) 1 (b) 2 (c) 4 (d) 0
( )
( ÷ ) ( )
1 (d) cos– 1 __ 3
8. A line makes an angle a, b and g with the axes, the value of sin2a + sin2b + sin2g is (a) 1 (b) 2 (c) – 1 (d) 0. 9. A line makes an angle a, b and g with the axes, then the value of cos (2a) + cos (2b) + cos (2g) is (a) 1 (b) 2 (c) –1 (d) 0. 10. The shortest distance from the origin to the p l a n e 2x + 3y + 6z – 21 = 0 is (a) 2 (b) 1 (c) 3 (d) 4 11. A plane meets the co-ordinate axes at A, B and C, respectively. If the centroid of the triangle is (2, 2, 2), the equation of the plane is (a) x + y + z = 5 (b) x + y + z = 6 (c) x + y + z = 9 (d) x + y + z = 4 12. The locus of the first degree equation in x, y, z reprsents a (a) stline (b) plane (c) sphere (d) conicoid 13. If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1 (k) with x-axis, then k is __
z __ 1
__
(a) tan– 1 (2÷2 )
__
(Mixed Problems)
6.13
2 (b) __ 7
2 ÷ (c) ___ (d) 1 3 14. The distance between the parallel lines 2x – 3y + 6z + 5 = 0 and 6x – 9y + 18z + 21 = 0 is 3 1 (a) __ (b) __ 7 7 2 4 (c) __ (d) __ . 7 7 15. The equation of the plane passing through (1, 2, 3) and parallel to 2x + y + 2z + 10 = 0 is
6.14 Integral Calculus, 3D Geometry & Vector Booster (a) 2x + y + 2z – 10 = 0 (b) 2x + y + 2z – 12 = 0 (c) 2x + y + 2z + 12 = 0 (d) 2x + y + 2z + 10 = 0 16. The angle between the planes 2x – y + z = 6 and x + y + 2z = 7 is p p (a) __ (b) __ 4 3 p p __ __ (c) (d) 2 6 17. The line x = 1, y = 2 is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) none of these 18. The line y = 2 and z = 3 is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) none of these 19. The equation of a line passing through (2, 3, 4) and perpendicular to the plane 2x + 3y – z = 5 is – 3 z____ –4 x – 2 y_____ (a) _____ = = 2 1 3 + 3 z____ –4 x – 2 y_____ (b) _____ = = 2 1 3 + 3 z____ –4 x – 2 y_____ (c) _____ = = 2 1 3 – 3 z____ –4 x – 2 y_____ (d) _____ = = 2 4 5 y – 3 z – 4 x – 2 20. The line _____ = _____ = ____ is parallel to the 3 4 5 plane (a) 3x + 4y + 5z = 7 (b) 2x + 3y + 5z = 10 (c) 2x + y – 2z = 0 (d) 2x + 3y + 2z = 5 21. The angle between the plane 3x + 6y – 2z + 5 = 0 + 1 z____ –3 x – 2 y_____ and the line _____ = = is 2 – 1 2 8 4 (a) cos– 1 ___ (b) cos– 1 ___ 21 21 8 4 (c) sin– 1 ___ (d) sin– 1 ___ . 21 21 22. The co-ordinates of the point of intersection of the + 1 z_____ +3 x – 6 y_____ line _____ = = and the plane x + y – z = 3 –1 0 4 is (a) (2, 1, 0) (b) (1, 2, – 6) (c) (5, 1, 2) (d) (5, – 1, 1) x – 1 23. The shortest distance between the lines _____ = 2 y_____ – 2 z____ – 4 z____ –3 –5 x_____ – 2 y_____ = and = = is 3 4 3 4 5 1 1__ (a) __ (b) ___ 6 ÷6 1__ 1 (c) ___ (d) __ 3 3 ÷
( ) ( )
( ) ( )
24. The plane passing through (3, 2, 0) and the line – 6 z____ –4 x – 3 y_____ _____ = = is 1 4 5 (a) x – y + z = 1 (b) x + y + z = 5 (c) x + 2y – z = 1 (d) 2x – y + z = 5. ___›
25. If the projection of PQ on OX, OY and___ OZ are › respectively 12, 3 and 4, the magnitude of PQ is (a) 169 (b) 13 (c) 12 (d) 144 26. The distance of the plane passing through (1, 1, 1) and – 1 z____ –1 x – 1 y_____ perpendicular to the line _____ = = from 3 0 4 the origin is 3 4 (a) __ (b) __ 4 3 7 (c) __ (d) 1 5 27. A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the centroid D (x, y, z) of DABC satisfies the rela1 1 1 tion __ 2 + __ 2 + __ 2 = k, then k is x y z 28.
29.
30.
31.
32.
(a) 1 (c) 9
(b) 3 (d) 4 y – 3 x – 2 x – 1 4 – z The lines _____ = _____ = _____ and _____ = 1 1 k k y – 4 z____ –5 _____ = are coplanar if k is 2 1 (a) 1 (b) 2 (c) 0 (d) 4 The equation of the obtuse angle bisector between the planes 3x – 2y + 6z + 8 = 0 and 2x – y + 2z + 3 = 0 is (a) 5x – y – 4z = 3 (b) 5x + y – 4z = 5 (c) 4x + y – 5z = 5 (d) 7x + 3y – 9z The plane x – y – z = 4 is rotated through an angle 90° about its line of intersection with the plane x + y + 2z = 4. Then the new position of the plane is (a) 5x + 3y + 2z = 4 (b) 5x + y + 4z = 20 (c) 4x + y + 5z = 20 (d) 4x + 5y + z = 20 The value of m for which the straight lines 3x – 2y + z + 3 = 0 = 4x – 3y + 4z + 1 are parallel to the plane 2x – y + m z = 2 is (a) 6 (b) 8 (c) – 2 (d) – 11 A straight line is given by r = (1 + t) iˆ + 3t + (1 – t) , where t in R. If this line lies in the plane x + y + c z = d, the value of (c + d – 5) is
3D-Co-ordinate Geometry
(a) 8 (c) 4
(b) 9 (d) 2
– 1 z_____ +1 x – 2 y_____ 33. Let L1: _____ = = 1 0 2 – 1 ___ z x – 3 y_____ and L2: _____ = = 1 1 – 1 and let p be the plane which contains the line L1 and is parallel to L2. The distance of the plane p from the origin is __ 2 1 (a) __ (b) __ 7 7 __ (c) ÷6 (d) none of these 34. If the planes x = cy + bz, y = az + bx and z = bx + ay pass through the same line, the value of a2 + b2 + c2 + 2ab + 2 is (a) 2 (b) 1 (c) 3 (d) 4 35. The equation of the plane passing through (0, 2, 4) – 1 2____ –z x + 3 y_____ and containing the line _____ = = is 3 4 2 (a) x – 2y + 4z – 12 = 0 (b) 5x + y + 9z – 38 = 0 (c) 10x – 12y – 9z + 60 = 0 (d) 7x + 5y – 3z + 2 = 0 36. The two lines x = ay + b, z = cy + d and x = a¢y + b¢, z = c¢y + d¢ will be perpendicular if and only if (a) aa¢ + bb¢ + cc¢ + 1 = 0 (b) aa¢ + bb¢ + cc¢ = 0 (c) (a + a¢) + (b + b¢) + (c + c¢) = 0 (d) a a¢ + c c¢ + 1 = 0 1 – y _____ z + 2 x – 2 _____ 37. If the line _____ = = lies in the plane 3 2 6 x + 3y – a z + b = 0. Then the value of (a + b + 1) is (a) 2 (b) 1 (c) 3 (d) 4 y – y z z – 3 x x – 1 2_____ 38. If the lines __ = __ = __ ; _____ , = ____ and 1 2 3 3 1 4 – 1 z____ –2 x + k y_____ _____ = = are concurrent, then 3 2 h
÷
(a) h = – 2, k = – 6 (b) h = 6, k = 2 (c) h = 1/2, k = 2 (d) h = 2, k = 1/2 39. If the four points P (2 – x, 2, 2), Q(2, 2 – y, 2) R (2, 2, 2 – z) and S (1, 1, 1) are coplanar, then 1 __ 1 __ 1 (a) __ x + y + z = 1 (b) x + y + z = 1 1 1 1 (c) _____ + _____ + ____ = 1 1–x 1–y 1–z 1 1 1 (d) _____ + _____ + _____ = 2 1+x 1+y 1+z
6.15
40. The equation of the right bisector plane of the line segment joining the points (2, 3, 4) and (6, 7, 8) is (a) x + y + z = 15 (b) x + 2y + z = 15 (c) 3x + 2y + z = 15 (d) x + 2y + 5z = 15 41. The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2 + y2 + z2 + 4x – 2y – 6z = 155 is (a) 39 (b) 26 (c) 11 (d) 13 42. The intersection of the spheres x2 + y2 + z2 + 7z – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane (a) 2x – y – z = 1 (b) x – 2y – z = 1 (c) x – y – 2z = 1 (d) x – y – z = 1 43. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius (a) 3 (b) 1__ (c) 2 (d) ÷ 2 44. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the mid-point of the line joining the centres of the spheres x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8, then a is (a) – 1 (b) 1 (c) – 2 (d) 2 45. The intersection of the spheres x2 + y2 + z2 + 7x – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane. Then (a) 2x – y – z = 1 (b) x – 2y – z = 1 (c) x – y – 2z = 1 (d) x – y – z = 1.
(Problems for JEE-Advanced)
1. Show that the straight lines whose direction cosines are given by the equation al + bm + cn = 0 and ul2 + vm2 + wn2 = 0 are perpendicular if a2 (u + v) + b2 (u + w) + c2 (u + v) = 0 a2 b2 __2 and parallel if __ u + __ v + w = 0. 2. If P be any point on the plane l x + m y + n z = p and Q be a point on the line OP such that OP ◊ OQ, find the locus of the point Q. y x __ __z 3. A point P moves on a plane __ a + b + c = 1. A plane through P and perpendicular to OP meets the co-ordinate axes at A, B and C. If the plane through A, B and C and parallel to the planes x = 0, y = 0, z = 0 intersect at Q, find the locus of Q. 4. A plane passes through a fixed point (a, b, c). Show that the locus of the foot of the perpendicular to it from the origin is the sphere x2 + y2 + z2 – ax – by – cz = 0.
6.16 Integral Calculus, 3D Geometry & Vector Booster
5. The plane x – y – z = 4 is rotated through an angle of 90° about the line of intersection with the plane x + y + 2z = 4. Find its equation in the new position. 6. If the planes x – c y – b z = 0, c y – y + a z = 0 and b x + a y – z = 0 passing through a straight line, find the value of a2 + b2 + c2 + 2 abc. 7. Find the co-ordinates of those points on the + 2 z____ –3 x – 1 y_____ line _____ = = which are at a distance 2 3 6 of 3 units from the point (1, – 2, 3). 8. Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the + 2 6____ –z x – 2 y_____ line _____ = = . 2 3 6 9. Find the equation of the plane passing through (1, 2, 0) – 1 2____ –z x + 3 y_____ which contains the line _____ = = . 3 4 2 10. Find the equation of the projection of the + 1 z____ –3 x – 1 y_____ line _____ = = on the plane x + 2y + z 2 – 1 4 = 9. 11. Find the shortest distance between the lines 2x + y + z – 1 = 0 = 3x + y + 2z – 2 and x = y = z. 12. Find the distance of the point of intersection of the + 3 z_____ + 1 + 1 x – 4 y_____ x – 1 y_____ lines _____ = = and _____ = = 7 1 – 4 2 – 3 z + 10 ______ from (1, – 4, 7). 8
19. If the equation of the plane through the straight y + 2 z x – 3 line _____ = _____ = __ and perpendicular to 2 – 3 6 x – y + z + 2 = 0 is ax + by + cz + 4 = 0, find the value of a + b + c + 2. 20. Find the equation of the plane through (2, 3, 4) and also which is at the maximum distance from the origin. 21. A sphere of radius r passes through the origin and meets the axes in A, B, C. Find the locus of the centroid of DABC lies on the sphere. 22. A plane passes through a fixed point (f, g, h) and cuts the axes in A, B, C. Show that the locus of the centre of the sphere OABC. 23. A sphere of constant radius r passes through the origin and cuts the axes in A, B, C. Find the locus of the foot of the perpendicular from O to the plane ABC. 24. Find the equation of a sphere circumscribing the tetrahedron whose faces are x = 0, y = 0, z = 0 y __z x __ and __ a + b + c = 1 25. Find the equation of a sphere for which the circle x2 + y2 + z2 + 7y – 2z + 2 = 0 and 2x + 3y + 4z = 8 is a great circle.
(Tougher Problems for Jee-Advanced)
13. If the lines x = a y + b, z = c y + d and x = a1y + b1, z = c1y + d1 are perpendicular such that a a1 + b b1 + k = 0, find the value of (k2 + 4). 14. Find the distance of the point(1, 2, 3) from the plane x + y + z = 11 measured parallel – 12 z____ –7 x + 1 y______ to _____ = = . 1 – 2 2 15. If a variable plane forms a tetrahedron of constant volume 64k2 with the co-ordinate planes, find the locus of the centroid of the tetrahedron. y z 16. If 2d be the shortest distance between the lines __ + __ b c x __z 1 1 __ __ = 1, x = 0 and __ a – c = 1, y = c, prove that a2 + b2 1 1 + __ 2 = __ 2 . c d
3.
17. If the foot of the perpendicular from the origin to a plane is P (a, b, c), find the equation of the plane. 18. Find the equation of a plane which passes through y – 2 x – 1 the point of intersection of lines _____ = _____ = 3 1 y – 1 z–3 z–2 x–3 ____ and _____ = _____ = ____ and at the greatest 2 1 2 3 distance from the point (0, 0, 0).
4.
5.
6.
1. Show that the straight lines whose direction cosines are given by al + b m + c n = 0 and f m n + g n l + h l m = 0 are f g h (i) perpendicular if __ a + __ + __ = 0 b c __
2.
___
___
(ii) parallel if ÷ af + ÷ bg + ÷ ch = 0. If the edges of a rectangular parallelopiped are a, b and c, show that the angles between the four a2 b2 __ c2 diagonals are given by cos–1 __2 + __ 2 + 2 . a b c A variable plane passes through the fixed point (a, b, c) and meets the axes of reference in P, Q, R. Show that the locus of the point of intersection of the plane through P, Q, R parallel to the co-ordinate a b __c planes is __ x + __ y + z = 1. A variable plane makes intercepts on the co-ordinate axes, the sum of whose squares is constant and equal to k2. Find the locus of the foot of the perpendicular from the origin to the plane. Find the incentre of the tetrahedron formed by the planes x = 0, y = 0, z = 0 and x + y + z = a. If the image of the point (3, 5, 7) in the plane 2x + y + z = 16 is (a, b, c), find the value of a2 + b2 + c2 + 18.
(
)
3D-Co-ordinate Geometry
x – 1 7. If q be the acute angle between the lines _____ = a y_____ + 1 __z y – 3 –1 x + 1 _____ z____ = c and _____ = = c ,where a > b > c a b b and a, b and c are the zeroes of x3 + x2 – 4x – 4 = 0, find q. 8. If the planes x = cy + bz, y = az + cx and z = bx + ay meet in a line, show that the line of intersection y z x of these planes is _______ __ 2 = _______ __ 2 = ______ __ 2 1 – a 1 – b 1 – c ÷ ÷ ÷ 9. Let the plane x – y – z = 2 is rotated through 90° about its line, of intersection with the plane x + 2y + z = 2. Find the equation of the plane in the new position. y x __ __z 10. A point P moves on a plane __ a + b + c = 1. A plane through P and perpendicular to OP meets the co-ordinate axes at A, B and C. If the planes through A, B and C parallel to the planes x = 0, y = 0 and z = 0 intersect at Q, find the locus of Q.
Integer Type Questions
1. If the angle between the lines whose direction cosines are given by 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0 a –1 __ is cos , find the value of (a + b). b 2. A line makes angles a, b, g, d with the four diagonals of a cube, find the value of 3(cos2a + cos2b + cos2g + cos2d).
( )
3. If the image of the point P (1, 6, 3) in the –2 x y – 1 z____ line __ = _____ = is Q (a, b, g), find the vaue 1 2 3 of (a + b + g). 4. If P(a, b, c) be any point on the plane 3x + 2y + z = 7, find the least value of 2(a2 + b2 + c2). 5. Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the + 2 6____ –z x – 2 y_____ line _____ = = . 2 3 6 6. Find the shortest distance between the lines x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4 and the z-axis. 7. Find the number of planes that are equidistant from four non-coplanar points. 8. If the distance between the plane x – 2y + z = d – 2 x – 1 y_____ and the plane containing the lines _____ = = 2 3 __ y – 3 z – 3 z – 4 x – 2 ____ and _____ = _____ = ____ is ÷6 , find the 4 3 4 5 value of |d|.
6.17
9. If the angle between two faces of a regular tetrahedron be q, find the value of (3cosq + 2). 10. Find the number of lines, which are equally inclined to the axes. – 3 z____ –4 x – 2 y_____ 11. If the lines _____ = = and 1 1 – k – 4 z____ – 5 x – 1 y_____ _____ = = are coplanar, find the value 2 1 k of (k + 5). 9 __ – z y_____ – 3 x_____ – 2 2 12. If the straight lines = = _____ and 1 2 k y – 4 z x–1 _____ = _____ = __ intersect, find the value of k. 2 1 k 13. If the planes x = cy + bz, y = cx + az and z = bx + ay pass through a straight line, find the value of a2 + b2 + c + 2abc + 2. –2 z–k x – 4 y_____ 14. If the line _____ = = ____ lies in the plane 1 1 2 2x – 4y + z = 7, find the value of k. 15. If the plane ax – by + cz = 0 contains the – 2d z____ –c x – a y______ b line _____ = = c , find the value of __ . a b d
( )
Comprehensive Link Passages Passage I A line makes angles a, b, g with the co-ordinate axes, then cos2a + cos2b + cos2g = 1. 1. If a + b = 90°, then g is (a) 0 (b) 60° (c) 90° (d) 120° 2. If a – b = 90°, then g is (a) 30° (b) 60° (c) 120° (d) 90° 3. The direction cosines of a line equally inclined to the axes are 1 1 1 1 1 1 (a) __ , __ , __ (b) – __ , – __ , – __ , 3 3 3 3 3 3 1__ ___ 1 1__ 1__ ___ 1 1__ (c) ___ , __ , ___ (d) – ___ , – __ , ___ 3 ÷ 3 ÷ 3 3 ÷ 3 ÷ 3 ÷ ÷ Passage II A line makes angles a, b, g, d with the four diagonals of a unit cube. 1. cos2a + cos2b + cos2g + cos2d is (a) 2/3 (b) 4/3 (c) 8/3 (d) 10/3 2 2 2 2. sin a + sin b + sin g + cos2d is (a) 1/3 (b) 4/3 (c) 8/3 (d) 15/3
6.18 Integral Calculus, 3D Geometry & Vector Booster
3. cos (2a) + cos (2b) + cos (2g) + cos (2d) is (a) 1/3 (b) 4/3 (c) – 8/3 (d) – 4/3
Passage III
x1 1 x V = __ 2 6 x3 x4
y1 y2 y3 y4
z1 z2 z3 z4
1__ (a) ___ 3 ÷
1__ (b) ___ 2 ÷
2 (c) __ 3
1__ (d) ___ ÷6
÷
1. The volume of a tetrahedron with vertices (0, 0, 0), (2, 0, 0), (0, 3, 0) and (0, 0, 4) is (a) 1 (b) 4 (c) 6 (d) 2 2. The volume of the terahedron with vertices (1, 2, 3), (2, 0, 0), (0, 4, 0) and (0, 0, 3) is (a) 2 (b) 6 (c) 4 (d) 8 3. The volume of tetrahedron formed by the planes x + y = 0, y + z = 0, z + x = 0 and x + y + z = 1 is (a) 2/3 (b) 4/3 (c) 8/3 (d) 10/3 4. The volume of the tetrahedron formed by the y y y x x __ __z __ __ __ __z planes __ a + b + c = 1, a + b = 0, b + c = 0 x z and __ a + __ c = 0 is 1 (a) __ abc 3 2 __ (c) abc 3
4 (b) __ abc 3 10 ___ (d) abc 3
( )
__
( )
are perpendicular or, if a1a2 + b1b2 + c1c2 = 0 and parallel if a1 __ b1 __ c1 __ a2 = b2 = c2
1. The number of lines that are equally inclined with the axes is (a) 8 (b) 6 (c) 4 (d) 3 2. If the lines L1 : x = ay + b, z = cy + d and L2 : x = x¢y + b¢, z = c¢y + d ¢ are perpendicular, then (a) aa¢ + cc¢ = 0 (b) aa¢ + cc¢ = –1 (c) aa¢ – cc¢ = 0 (d) aa¢ – cc¢ = 1 – 14 z____ –3 1 – x 7y 3. Let L1: _____ = _______ = 3 2 25
If L1 meets L2 at right angles, the value of (11p – 64) is (a) 5 (b) 6 (c) 7 (d) 8. Passage VI x – x1 y_____ – y1 z_____ – z1 Two lines L1: _____ = = c a1 b1 1 x – x2 y_____ – y2 z_____ – z2 and L2: _____ = = c are coplanar if a2 b2 2
( ) __
(c) tan (2÷2 ) (d) cos ____ 2÷2 3 2. The angle between one diagonal of a unit cube and a diagonal of a face is __ (a) cot–1(÷2 ) (b) tan–1 ___ 1__ 3 ÷
(c) cos–1 __ 1 3
x – x2 y_____ – y2 z_____ – z2 and L2: _____ = y = c a2 2 2
– 5 6____ –z 7 – 7x y_____ and L2: ______ = = . 3p 1 5
1. The angle between two diagonals of a unit cube is 1 (a) tan–1(3) (b) tan–1 __ 3 –1
x – x1 y_____ – y1 z_____ – z1 Let L1: _____ = = c a1 b1 1
a1a2 + b1b2 + c1c2 __________ __________ . Then cos(q) = ______________________ 2 ÷a 1 + b21 + c 21 ÷a 22 + b 22 + c22
Passage V
1 1 1 1
Passage IV _› Let r 1 = a1 + b1 + c1 _› and r 2 = a2 + b2 + c2 .
3. If q be the angle between the vector a = 2i – j + k and the z-axis, then sinq is
__
The volume of the tetrahedron ABCD with vertices A(x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) is
–1
( ) ( ÷ )
__
2 (d) sin–1 __ 3
x2 – x1 a1 a2
y2 – y1 b1 b2
z2 – z1 c1 = 0 c2
y – 2 ____ y–1 z–3 x – 1 _____ x – 4 _____ 1. If two lines _____ = = and _____ = = z 2 3 4 2 5 are coplanar, the point of intersection is (a) (1, 1, 1) (c) (–1, –1, –1)
(b) (1, –1, –1) (d) (1, 1, –1)
6.19
3D-Co-ordinate Geometry
y + 1 ____ y – k __z z–1 x – 3 ____ x – 1 _____ 2. If the lines _____ = = and _____ = = 2 3 4 1 2 1 are coplanar, the value of k is (a) 3/2 (b) 9/2 (c) – 2/9 (d) – 3/2 y – 3 4 – z x – 2 x – 1 3. The lines _____ = _____ = ____ and _____ = 1 1 k k y_____ –4 = z – 5 are coplanar, if 2 (a) k = 0 (b) k = 1 (c) k = 2 (d) k = – 3
Passage VII – 2 z____ –3 x – 1 y_____ Let L1: _____ = = 2 3 4 – 4 z____ –5 x – 2 y_____ and L2: _____ = = 3 4 5
Let A = (2, 3, 1), P : 2x + y + z = 6 – 2 z____ –3 x – 1 y_____ and L: _____ = = 2 1 4 1. The foot of perpendicular from the point A to the line L is 37 43 11 43 ___ 11 37 ___ (a) ___ , ___ , (b) ___ , ___ , 21 21 21 21 21 21
( ) 43 11 37 ___ (c) ( – ___ , ___ , – 21 21 21 )
( ) 11 43 37 (d) ( ___ , – ___ , ___ 21 21 21 )
2. The image of the point A (2, 3, 1) w.r.t. the plane 2x + y + z is 2 7 __ 1 2 7 __ 1 (a) __ , __ , (b) – __ , __ , – 3 3 3 3 3 3 2 1 7 2 7 1 (c) __ , – __ , – __ (d) – __ , – __ , __ 3 3 3 3 3 3
( (
Matrix Match (For Jee-Advanced Examination Only)
Column I
)
)
( (
) )
Column II
(A)
S cos a
(P)
(B)
S sin2a
(Q) –1
(C)
S cos(2a)
(R) 2
(D)
S (cos2a + sin2a)
(S)
2
3
1
2. Match the following Columns: A line makes angles a, b, g, d with the four diagonals of a cube. Column I
( ) 7 4 (d) ( __ , __ , 11 ) 5 5
Passage VIII
3. The image of the line L with respect to the plane P is – 5 3z –8 3x – 1 3y (a) ______ = ______ = _____ 4 2 3 – 4 3z –5 3x – 1 3y (b) ______ = ______ = _____ 4 2 3 – 5 8_____ – 3z 3x – 1 3y (c) ______ = ______ = 4 2 1 – 3y 3z –8 3x – 1 5______ (d) ______ = = _____ 4 3 5
1. Match the following Columns: A line makes angles a, b, g with the co-ordinate axes.
and A = (2, 3, 5). 1. The shortest distance between the lines L1 and L2 is 1__ 1__ (a) ___ (b) ___ 3 ÷2 ÷ 1__ 1 (c) ___ (d) __ 2 ÷6 2. The equation of the plane passing through A and containing the line L1 is (a) 2x – z + 1 = 0 (b) 2x – y – z + 2 = 0 (c) 2x – y + 3z + 2 = 0 (d) 3x – y + 5z + 10 = 0 3. The point A = (2, 3, 5). is shifted to the line L2 by a distance 1. Then the co-ordinates of the new position of A is 7 11 __ 7 4 11 4 (a) __ , ___ , (b) __ , __ , ___ 5 5 5 5 5 5
( ) 7 11 (c) ( __ , ___ , 4 ) 5 5
Column II
(A)
S cos a
(P) 4
(B)
S sin2a
(Q) 4/3
(C)
S cos (2a)
(R) 8/3
(D)
S (cos2a + sin2a)
(S) – 4/3
2
3. Match the following columns: Column I
Column II
(A) The angle between any two (P) diagonals of a cube is
( ÷ ) __
2 cos–1 __ 3
( )
(B)
The angle between one (Q) sin–1 __ 1 diagonal of a unit cube and 3 a diagonal of a face is
(C)
The angle between a (R) diagonal of a cube and a diagonal of a face intersecting it
cot–1 ___ 1__ 3 ÷
(D) The angle between the (S) diagonal of the faces of the cube through the same (T) vertex is
cot–1( ÷2 ) 1 –1 __ cos 3
( ) __
( )
6.20 Integral Calculus, 3D Geometry & Vector Booster
4. Match the following Columns: Column I (A) The direction cosines of two lines are connected with l + m + n = 0 and l2 + m2 + n2 = 0, the angle between them is (B) The angle between the planes 3x – 4y + 5z = 0 and 2x – y – 2z = 5 is (C) The angle between the planes 2x – y + z = 6 and x + y + 2z = 7 is (D) The angle between the –2 +3 x – 1 y_____ line _____ = = _____ and 2 1 – 2 the plane x + y + 4 = 0 is
Column II (P) 60°
point (2, 6, 3) to the line –2 x y – 1 z____ __ = _____ = is 2 2 3 (C) The point of intersection of (R) (1, 3, 5) the lines –1 z–6 x – 2 y_____ _____ = = ____ 1 – 2 1 + 3 z_____ +3 x + 3 y_____ and _____ = = is 7 1 – 6
(Q) 45°
(R) 90°
(S)
(D) If the lines (S) (– 9, – 14, – 19) y + 2 z+3 x+1 _____ = _____ = _____ 2 3 4 – 2 z____ –1 x – 3 y_____ _____ and = = 3 4 5 are
120°
5. Match the following Columns: Column I y – 2 (A) x – 4 = _____ = If the line _____ 1 1 z____ –k lies on the plane 2 2x – 4y + z = 7, then k is y – 3 x – 2 (B) = _____ = The lines _____ 1 1 y – 4 4 – z x – 1 ____ and _____ = _____ = 2 k k z–5 ____ are coplanar, then k is 1 (C) If the lines + 1 z____ –1 x – 1 y_____ _____ = = 2 3 4 – k __z x – 3 y____ and _____ = = intersect, 1 2 1 then k is (D) The shortest distance between the lines – 3 z____ –6 x – 2 y_____ _____ = = 3 4 5 – 2 z____ –1 x – 5 y_____ and _____ = = is 1 2 3
(B) The co-ordinates of the foot (Q) (5, 10, of the perpendicular from the – 6)
Column II (P) – 3
(Q) 7
Column I
(R) 0 (S) 9/2 (T) 5/2
6. Match the following Columns: Column I
7. Match the following Columns: Let P (0, 3, – 2), Q (3, 7, –1) and R (1, – 3, –1) be three given points. Let L1 be the line passing through P and Q and L be the line through R and parallel to _› 2 the vector v = + .
Column II
(A) The co-ordinates of the point (P) (2, 3, 5) of intersection of the line + 2 z____ –2 x – 1 y_____ _____ = = 1 3 2 and the plane 3x + 4y + 5z = 15 is
Column II __
(A)
The perpendicular distance (P) of P from L2 is
7÷3
(B)
The shortest distance (Q) between L1 and L2 is
2
(C)
Area of the DPQR is
6
(D)
The distance from (0, 0, 0) (S) to the plane PQR is
(R)
19 _____ ____ 147 ÷
8. Match the following Columns: Column I
Column II
(A) The shortest distance from the (P) point (3, 4, 5) to the x-axis is
5
(B) The shortest distance from the (Q) point (3, 2, 4) to y-axis is
÷41
(C) If the plane 2x – 3y + 6z = 11 (R) makes an angle sin–1(k) with the x-axis, then 7k is
9
(D) A straight line (S) _ › r = (1 + t) + 5t + (1 – t) where t Œ R. If this line lies in (T) the plane x + y + cz = d, the value of (c + d) is
2
___
4
3D-Co-ordinate Geometry
9. Match the following Columns: Column I
Column II
(A) The volume of the tetrahedron, (P) whose vertices are (0, 0, 0), (2, 0, 0), (0, 3, 0) and (0, 0, 4) is
6
(B) The volume of the tetrahedron (Q) whose vertices are (0, 1, 2), (3, 0, 1), (4, 3, 6) and (2, 3, 2) is
4
(C) The volume of the tetrahedron (R) formed by the planes whose equations are x + y = 0, y + z = 0, z + x = 0 and x + y + z = 1 is
20
(D) The volume of the tetrahedron (S) formed by the planes whose equations are y z x y __ (T) + __ = 0, __ + __ = 0, 2 3 3 4 y z z x x __ + __ = 0 and __ + __ + __ = 1 4 2 2 3 4 is
16
2/3
Column II 2
2
2
(A)
If 2x + 3y + 4z + 2kxy (P) + 4yz = 0 represents two planes, the value of k2 is
2
(B)
If x2 + 5y2 + 5z2 + 2kxy (Q) + 4xz = 0 represents two planes, the value of 5k 2 is
5
(C)
If x2 + y2 + 4z2 – 2kyz = 0 (R) represents two planes, the value of (k2 + 2) is
10
(D)
If 9x2 + y2 + z2 – 6kxy = (S) 0 represents two planes, the (T) value of (k2 + 7) is
8 1
Questions asked in Previous Years’ JEE-Advanced Examinations
10. Match the following Columns: Column I
–2 x – 4 y_____ 1. The value of k, for which the line _____ = = 1 1 z____ –k lies in the plane 2x – 4y + z = 7, is 2 (a) 7 (b) 6 (c) no real value (d) – 7 [IIT-JEE, 2003] 2. Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1). [IIT-JEE, 2003]
6.21
3. If P be the point (2, 1, 6), find a point Q such that PQ is perpendicular to the plane in x + y – 2z = 3 and the mid-point of PQ lies on it. [IIT-JEE, 2003] y + 1 ____ y – k __z z–1 x – 3 ____ x – 1 _____ 4. If the lines _____ = = and _____ = = 2 3 4 1 2 1 intersect at a point, the value of k is (a) 3/2 (b) 9/2 (c) 2/9 (d) – 3/2 [IIT-JEE, 2004] 5. A parallelopiped T has one of its face as ABCD. The face opposite to ABCD is A¢B ¢C ¢D ¢. The parallelopiped T is compressed to another parallelopiped S with the face ABCD remaining the same but A¢B ¢C ¢D ¢ changes to A≤B≤C≤D≤. If the volume of S is 90% of T, show that the locus of A≤ is a plane. [IIT-JEE, 2004] 6. p1 and p2 are two planes passing through the origin. L1 and L2 are two lines passing through the origin such that L1 lies on p1 not on p2 and L2 lies on p2 not on p1. Show that there exist three points A, B, C whose permutation A¢, B¢, C ¢ can be choosen such that (a) A is on L1, B on p1 but not on L1 and C not on p1. (b) A¢ is on L2, B¢ on p2 but not on L2 and C ¢ not on p2. [IIT-JEE, 2004] 7. A plane p passes through the point (1, 1, 1) and is parallel to the vectors b = (1, 0, –1) and c = (–1, 1, 0). If p meets the axes in A, B and C, find the volume of the tetrahedron OABC. [IIT-JEE, 2004]
8. Find the equation of the planes passing through the lines 2x – y + z = 3, 3x + y + z = 5 and which at a 1 distance of ___ __ from the point (2, 1, –1). ÷6 [IIT-JEE, 2005]
9. A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If
the centroid G (x, y, z) of DABC satisfies the rela1 1 1 tion __ 2 + __ 2 + __ 2 = k, the value of k is x y z (a) 9 (b) 3 (c) 1 (d) 1/3 [IIT-JEE, 2005] 10. A plane p is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4 and passes through the point (1, – 2, 1). The distance of p from the point (1, 2, 2) is (a) 0 __ (b) 1 __ (c) ÷2 (d) 2÷2 [IIT-JEE, 2006]
6.22 Integral Calculus, 3D Geometry & Vector Booster 11. A point (a, b, g) lies on the plane x + y + z = 2 Let a = a i + b j + g k and k × (k × a) = 0, then g equals .......... [IIT-JEE, 2006] 12. A line L is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance L from the origin equals .......... [IIT-JEE, 2006] 13. A plane passes through (1, 2, 3) and is perpendicular to two planes x = 0 and y = 0. The distance of the plane from the point (0, –1, 0) equals .......... [IIT-JEE, 2006] 14. Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5. Statement I: The parametric equations of the line of intersection of given planes are x = 3 14t, y = 1 + 2t, and z = 15t. Statement II: The vector 14i + 2j + 15k is parallel to the given planes. [IIT-JEE, 2007] 15. Consider the following linear equations: ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0. Match the conditions/expressions in Column I with statements in Column II. Column I
Column II
(A) a + b + c π 0 (p) the equations represent planes meeting at a and a2 + b2 + c2 single point. = ab + bc + ca, (B) a + b + c = 0 (q) the equations represent the line x = y = z and a2 + b2 + c2 π ab + bc + ca, (C) a + b + c π 0 (r) the equations represent the identical planes and a2 + b2 + c2 π ab + bc + ca, (D) a + b + c = 0 (s) the equations represent the whole of the threeand a2 + b2 + c2 dimensional space. = ab + bc + ca, [IIT-JEE, 2007] 16.
Consider the planes P1 : x – y + z = 1, P2 : x + y – z = –1, P3 : x – 3y + 3z = 2. Let L1, L2 and L3 be the lines of intersection of the planes P2 and P3, P3 and P1 and P1 and P2, respectively. Statement I: At least two of the lines L1, L2 and L3 are non-parallel.
Statement II: The three planes do not have a common point. [IIT-JEE, 2008] 17. Consider the lines y + 2 z_____ +1 x + 1 _____ L1: _____ = = 3 1 2 + 2 _____ z–3 x – 2 y_____ L2: _____ = = 1 2 3 (i) The unit vector perpendicular to both L1 and L2 is i + 7j + 7k – i + 7j + 5k ___ __ (a) ___________ (b) ___________ ÷99 5÷3 – i + 7j + 3k – 7i + 7j + k __ ___ (c) ___________ (d) ___________ 5÷3 ÷99 (ii) The shortest distance between L1 and L2 is 17 __ (a) 0 (b) ___ 3 ÷ 17__ 41__ ____ ____ (c) (d) 5÷3 5÷3
(iii) The distance of the point (1, 1, 1) from the plane passing through the point (–1, – 2, –1) and whose normal is perpendicular to both the lines L1 and L2 is 7 2 ___ ___ (a) ____ (b) ____ ÷75 ÷75 13 23 ___ ___ (c) ____ (d) ____ ÷75 ÷75 [IIT-JEE, 2008] 18. A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the co-ordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals __ (a) 1 __ (b) ÷ 2 (c) ÷3 (d) 2 [IIT-JEE, 2009] 19. Let P (3, 2, 6) be a point in space and Q be a point on the line r = (i – j + 2k) + l (– 3i + j + 5k). Then the value of l for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is (a) 1/4 (b) –1/4 (c) 1/8 (d) –1/8 [IIT-JEE, 2009] 20. The equation of the plane containing the y z x straight line __ = __ = __ and perpendicular to 2 3 4 y z x the plane containing the straight lines __ = __ = __ 3 4 2 x y __z and __ = __ = is 4 2 3 (a) x + 2y – 2z = 0 (b) 3x + 2y – 2z = 0 (c) x – 2y + z = 0 (d) 5x + 2y – 4z = 0 [IIT-JEE, 2010]
3D-Co-ordinate Geometry
21. If the distance of the point P (1, – 2, 1) from the plane x + 2y – 2z = a, where a > 0, is 5, the foot of the perpendicular from P to the plane is
( ) 10 1 2 ___ (c) ( __ , __ , 3 3 3)
( ) 2 1 5 (d) ( __ , – __ , __ 3 3 2)
8 4 __ 7 (a) __ , __ , – 3 3 3
4 4 1 (b) __ , – __ , __ 3 3 3
[IIT-JEE, 2010] 22. If the distance between the plane x – 2y + z = d – 2 x – 1 y_____ and the plane containing the lines _____ = = 2 3 __ y – 3 z – 3 z – 4 x – 2 ____ and _____ = _____ = ____ is ÷ 6 , then |d| 4 3 4 1 is... [IIT-JEE, 2010] 24. Let a, b, c be three real numbers satisfying
[ ]
1 9 7 [a b c] 8 2 7 = [0 0 0] ... (E). 737 If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, the value of 7a + b + c is (a) 0 (b) 12 (c) 7 (d) 6. [IIT-JEE, 2011] 25. The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R(1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, the length of the line segment PS is __ 1__ (a) ___ (b) ÷ 2 2 ÷ __ (c) 2 (d) 2÷2 [IIT-JEE, 2012] 26. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and 2__ x – y + z = 3 and at a distance ___ from the point 3 ÷ (3, 1, –1) is
(a) 5x – 11y + z = 17
(c) x + y + z = ÷3
__
___ __ (b) ÷ 2x + y = 3÷2 – __ __ (d) x – ÷y = 1 – ÷ 2
1
[IIT-JEE, 2012] + 1 __z x – 1 y_____ x+1 27. If the straight lines _____ = = and _____ = 2 2 k 5 y + 1 z _____ = __ are coplanar, the planes containing these 2 k two lines is (are) (a) y + 2z = –1 (b) y + z = –1 (c) y – z = –1 (d) y – 2z = –1 [IIT-JEE, 2012]
6.23
28. The perpendiculars are drawn from points on the + 1 __z x + 2 y_____ line _____ = = to the plane x + y + z = 3. 2 –1 3 The feet of the perpendiculars lie on the line –2 –2 x y – 1 z____ x y – 1 z____ (a) __ = _____ = (b) __ = _____ = 8 –13 2 3 5 – 5 –2 –2 x y – 1 z____ x y – 1 z____ (c) __ = _____ = (d) __ = _____ = – 7 – 7 4 3 2 5 [IIT-JEE, 2013] 29. A line L passing through the origin is perpendicular to the lines L1 : (3 + t)i + (–1 + 2t) j + (4 + 2t) k and L2 : (3 + 2s) i + (3 + 2s) j + (2 + s) k, where – • < t, s 0 Hence, the obtuse angle bisector 3x – 2y + 6z + 8 2x – y + 2z + 3 _________ ________ _______________ = ______________ 4 + 1 + 4 ÷ ÷9 + 4 + 36 3x – 2y + 6z + 8 ______________ 2x – y + 2z + 3 fi _______________ = 7 3 fi
5x – y – 4z = 3
30. The equation of any plane passing through the intersection of the given planes
(x – y – z – 4) + l (x + y + 2z – 4) = 0
and
(l + 1)x + (l – 1) y + (2l – 1) x = 4l + 4
and
4a – 3b + 4c = 0
Solving, we get a b __ c fi __ = __ = 5 8 1 Since the line is parallel to the plane, so
10 – 8 + m = 0
fi m = – 2 32. Given striaght line can written as – 0 z____ –1 x – 1 y_____ _____ = = 1 3 –1 Since the lies in the plane, so 1 + 3 – c = 0 and 1 + 0 + c = d fi c = 4 and d = 5 Hence, the value of (c + d – 5) = 4 33. Let the equation of the plane be
a(x – 2) + b(y – 1) + c(z + 1) = 0
Since the plane contains the line L1 and parallel to L2, so a+b+c=0 and
a + 0 + 2c = 0
a b c fi __ = ___ = ___ 2 –1 –1 Hence, the equation of the plane is
2(x – 2) – (y – 1) + (z + 1) = 0
fi
2x – 4 – y + 1 + z + 1 = 0
fi
2x – y + z = 2
Hence, the distance from the origin
÷
__
2 2 2 = _________ ________ = ___ __ = __ 3 4 + 1 + 1 ÷6 ÷
34. Since the three planes pass through the same line, so 1 – c – b c a = 0 –1 b a –1 fi
|
2
|
(1 – a ) + c(– c – ab) – b(ac + b) = 0
which is perpendicular with x – y – z = 4. So,
fi
(1 – a2) – c2 – abc – abc – b2 = 0
fi
a2 + b2 + c2 + 2abc = 1
fi
a2 + b2 + c2 + 2abc + 2 = 1 + 2 = 3
fi
(l + 1) – (l – 1) – (2l – 1) = 0 3 l = __ 2
35. The equation of any plane through (0, 2, 4) is a(x – 0) + b(y – 2) + c(z – 4) = 0 which containing the line – 1 2____ –z x + 3 y_____ _____ = = 3 4 2 So,
3a + 4b – 2c = 0
and
– 3a – b – 2c = 0
Solving, we get a b c ____ = ___ = __ –10 12 9 Hence, the equation of the plane is
–10x + 12(y – 2) + 9(z – 4) = 0
fi
–10x + 12y – 24 + 9z – 36 = 0
fi
10x – 12y – 9z + 60 = 0
36. Ans. (d) x–2 37. Since the line _____ = 3 x + 3y – az + b = 0,
1 – y z_____ +2 _____ = lies in the plane 1 2 so
|
3D-Co-ordinate Geometry
6.41
|
– y x 0 x 0 – z = 0 x – 1 –1 –1 fi
x(0 – z) + y(– x + xz – z) = 0
fi
– xz – xy + xyz – yz = 0
fi
xy + yz + zx = xyz
1 __ 1 __ 1 fi __ x + y + z = 1 40. The equation of the line joining the points (2, 3, 4) and (6, 7, 8) is y–3 z–4 x–2 _____ = _____ = _____ 7–3 8–4 6–2 –3 x – 2 y_____ fi _____ = = 4 4 –3 x – 2 y_____ fi _____ = = 1 1
z–4 ____ 4 z–4 ____ 1
3 – 3 – 2a = 0
and
2 + 3 + 2a + b = 0
The mid-point of the line segment joining the points (2, 3, 4) and (6, 7, 8) is (4, 5, 6). Hence, the equation of the right bisector of the plane 1(x – 4) + 1(y – 5) + 1(z – 6) = 0
So,
a = 0, b = – 5
fi
Comprehensive Link Passages
Hence, the value of
(a + b + 7) = 2
x + y + z = 15
Clearly, l = 3m + 1, 2l = 2 – m
Passage I 1. Given,
cos2a + cos2b + cos2g = 1
Solving, we get
fi
cos2a + cos2 (90° – a) + cos2g = 1
fi
cos2a + sin2a + cos2g = 1
Hence, the point of intersection is (1, 2, 3).
fi
1 + cos2g = 1
–2 1 + k 2_____ – 1 3_____ Therefore, _____ = = 3 2 h 1 + k __ 1 1 fi _____ = = __ 3 2 h 3 1 fi h = 2, k = __ – 1 = __ 2 2
fi
cos2g = 0
fi
g = 90°
fi
cos2a + cos2(90° + a) + cos2g = 1
39. Given four points are
fi
cos2a + sin2a + cos2g = 1
P(2 – x, 2, 2), Q(2, 2 – y, 2)
fi
1 + cos2g = 1
R(2, 2, 2, – z) and S(1, 1, 1)
fi
g = 90°
Now,
PQ = (x, – y, 0)
PR = (x, 0, – z)
cos2a + cos2a + cos2a = 1
and
PS = (x – 1, –1, –1)
fi fi
3 cos2a = 1
fi
1 cos2a = __ 3
38. Let the point of intersection is (l, 2l, 3l).
l = 1, m = 0
It is given that, the vectors P, Q, R and S are coplanar, so the vectors PQ, PR, PS are coplanar.
2. cos2a + cos2b + cos2g = 1
3. cos2a + cos2b + cos2g = 1
6.42 Integral Calculus, 3D Geometry & Vector Booster 1 cos a = ___ __ 3 ÷
fi
(
)
1 1__ ___ 1 Hence, the direction cosines are ___ __ , ___ , __ 3 ÷3 ÷3 ÷ Passage II 2
0 0 0 1 1 1 –1 = __ 6 1 –1 1 –1 1 1
1 1 –1 1 __ = – 1 – 1 1 6 –1 1 1
= 2(cos2a + cos2b + cos2g + cos2d) – 4
4 = 2 ◊ __ 3 8 __ = – 4 3 4 = – __ 3
0 0 –1 1 2 0 1 __ = – 6 0 2 1
4 2 = __ = __ cu. units. 6 3
2
2
2
1. cos a + cos b + cos g + cos d = 4/3
2. sin2a + sin2b + sin2g + sin2d
= 4 – (cos2a + cos2b + cos2g + cos2d)
4 = 4 – __ 3 8 __ = 3 3. cos(2a) + cos(2b) + cos(2g) + cos(2d)
Passage III
0 12 = 60 0
0 0 3 0
0 0 0 4
1 1 1 1
2 0 0 1 0 3 0 – = 6 0 0 4
|
|
1 = – __ × 24 = |– 4| = 4 cu units. 6
(
) (
0 1 a 1 = __ 1 6 a 1 – a
0 1 b 1 – b 1 b
2. The volume of a tetrahedron
1 1 2 = __ 6 0 0
2 0 4 0
3 0 0 3
1 1 1 1
Ï0 0 1 2 1 Ô __ = Ì 4 0 1 – 2 4 6 Ô 0 Ó0 3 1 1 = __ [12 – 2{2(0 – 3) 6 1 = __ [12 + 12] 6 =4
1 1 1 1
Ê C1 Æ C1 + C3 ˆ ÁË C Æ C + C ˜¯ 2 2 3
4. On solving, we get the vertices of a tetrahedron as 1 1 __ 1 1 1 1 1 __ 1 __ 1 (0, 0, 0), __ a , __ , – c , __ a , – __ , __ and – __ a , b , c b b c respectively. Volume of a tetrahedron
1. Volume of a tetrahedron
3. On solving, we get the vertices of a tetrahedron as (0, 0, 0, 0), (1, 1, –1), (1, –1, 1), (–1, 1, 1). Volume of a tetrahedron
3 1¸ Ô 0 1˝ 4 1 Ô˛ – 4(0 – 0)}]
1 a 1 a 1 – a
1 b 1 – b 1 b
)
0 1 – c 1 c 1 c
1 1 1 1
1 c 1 c 1 c
–
1 = – __ 6
1 1 –1 1 __ = – abc 1 – 1 1 6 –1 1 1
4 = __ abc 6
2 = __ abc 3
(
)
3D-Co-ordinate Geometry
÷
_____
________
Passage IV
6.43
2 ___ 1 sin q = ÷ 1 – cos2q = 1 – __ = __ 3 ÷3 Passage V
1. Ans. (a)
x–b 2. L1: _____ = a
x – b¢ L2: _____ = a¢
y z–d __ = ____ c 1 y z – d¢ __ = _____ 1 c¢
Since L1 and L2 are perpendicular, so sum of the product of their sirection ratios is zero.
Thus,
a ◊ a¢ + 1 ◊ 1 + c ◊ c¢ = 0
fi
aa¢ + cc¢ = –1.
– 14 z____ –3 1 – x 7y L1: _____ = _______ = 3 2 25
1. OP = i + j + k
BN = i + j – k OP ◊ OL = 1 + 1 __
3. Given
x–1 _____ = – 3
__
÷ 3 ◊ ÷3 cos q = 1 + 1
y–2 z–3 _____ = ____ 2 25/7
– 5 6____ –z 7 – 7x y_____ L2: ______ = = . 3p 1 5
1 cos q = __ 3
and
1 q = cos–1 __ 3
y – 5 z____ –6 x–1 _____ = _____ = – 3p/7 1 – 5
Hence, the angle between any two diagonals is __ 1 q = cos–1 __ = tan–1 (2÷2 ) 3
Since L1 meets L2 at right angles, so
( ) ( )
2. OP = i + j + k
( )
3p 25 (– 3) – ___ + ___ – 10 = 0 7 7
9p 25 fi ___ = 10 – ___ 7 7
OL = i + j OP ◊ OL = 1 + 1 ÷ 3 ◊ ÷2 cos q = 1 + 1
9p fi ___ = 7
÷ 3 ◊ ÷2 cos q = 2
fi
÷ 3 cos q = ÷ 2
Passage VI
2 cos q = __ 3
__
__
__
__
__
__
÷
__
__
Hence, the angle between a diagonal and a diagonal of a face is
( ÷ ) __
__ 2 q = cos–1 __ = cot–1 (÷2 ) 3
3. a ◊ k = 2
|a| |i| cos q = 2 __
÷ 6 cos q = 2
÷
– 2 z____ –3 x – 1 y_____ _____ = = = l 2 3 4
Let two lines meet at P. So,
2l + 1 = 5m + 4, 3l + 2 = 2l + 1
and
4l + 3 = m
fi
3l – 3m = –1, 4l – m = – 3
l = –1 = m
Hence, the point of intersection is (– 1, – 1, – 1). 2. Since both the lines are coplaner, so
__
2__ 2 cos q = ___ = __ 3 ÷6
p=5
–1 x – 4 y_____ and _____ = = z = m 2 5
( ÷ )
2 q = cos–1 __ 3
1. Let
45 ___ 7
3–1 k +1 0 –1 2 3 4 =0 1 2 1
6.44 Integral Calculus, 3D Geometry & Vector Booster
fi
2 k + 1 –1 2 3 4 =0 1 2 1
x–2 1 2
y–3 z–5 1 2 =0 3 4
fi
2(3 – 8) – (k + 1)(2 – 4) – (4 – 3) = 0
fi
– 2(x – 2) + (z – 5) = 0
fi
–10 + 2(k + 1) – 1 = 0
fi
– 2x + 4 + z – 5 = 0
fi
2(k + 1) = 11
fi
– 2x + z – 1 = 0
fi
2x – z + 1 = 0
9 11 k = ___ – 1 = __ 2 2 3. Since the lines are coplaner, so
fi
3. Ans. (c)
Passage VIII Let
A = (2, 3, 1), P : 2x + y + z = 6
1– 2 4 – 3 5– 4 1 1 –k = 0 k 2 1
and
– 2 z____ –3 x – 1 y_____ L : _____ = = 2 1 4
fi
–1 1 1 1 1 –k = 0 k 2 1
1. Any point on the line is
M(2l + 1, l + 2, 4l + 3)
The direction ratios of AM are (2l – 1, l – 1, 4l + 2)
fi
(2 – k) – (1 + k2) – (1 + 2k) = 0
fi
k2 + 3k = 0
Clearly, AM is perpendicular to the line L. So,
fi
k(k + 3) = 0
2(2l – 1) + (l – 1) + 4(4l + 2) = 0
fi
k = 0, – 3
fi
21l + 5 = 0
fi
5 l = – ___ 21
Passage VII
1. The shortest distance between two lines
(r2 – r1) . (u × v) = ______________ |(u × v)|
Now,
(u × v)
i j k = 2 3 4 3 4 5
= – i + 2j – k
and
r2 – r1 = i + 2j + 2k
(
Let
and
–1 – 3 g_____ a – 2 b_____ 8 fi _____ = = = – __ 2 1 1 3 fi
16 8 8 a = 2 – ___ , b = 3 – __ , g = 1 – __ 2 3 3
fi
10 5 1 a = ___ , b = __ , g = – __ 3 3 3
(
)
z–3 ____ 4
10 1 __ 5 Hence the required image is ___ , __ , – . 3 3 3
z–5 ____ 3
Also, let A = (2, 3, 5).
2. Hence, the image of the point (2, 3, 1) w.r.t. the plane 2x + y + z = 6 is
–1 – 3 g_____ 2(4 + 3 + 1) a – 2 b_____ _____ = = = – __________ 2 1 1 (4 + 1 + 1)
(– 1 + 4 – 2) 1 ________ = __________ = ___ __ 1 + 4 + 1 ÷6 ÷ –2 x – 1 y_____ L1 : _____ = = 2 3 –4 x – 2 y_____ L2 : _____ = = 3 4
)
43 11 37 ___ – ___ , ___ , – 21 21 21
Hence, the shortest distance
Hence, the foot of the perpendicular is the
2. Hence, the equation of the plane passing through A and containing the line L1 is
3. Given plane is P : 2x + y + z = 6
– 2 z____ –3 x – 1 y_____ and the line is L : _____ = = 2 1 4 Let the image of the point A(2, 3, 1) w.r.t. to the given plane is Q(a, b, g ).
6.45
3D-Co-ordinate Geometry
S cos(2a) = cos(2a) + cos(2b) + cos(2g)
So, the image of the given line w.r.t. to the given –g – b z____ x – a y_____ plane is _____ = = 2 1 4
–1 – 3 g_____ a – 2 b_____ Let _____ = = = g 2 1 1
(D)
S (cos a + sin2a) = S 1 = 1 + 1 + 1 = 3
Let M is the mid-point of A and Q, i.e.
2. (A)
S cos2a = cos2a + cos2b + cos2g + cos2d
(
)
+1 + 3 g_____ a + 2 b_____ M = _____ , , 2 2 2
(
)
Clearly, the point M lies on the plane
(
) (
)
2l + 4 l+6 l+2 2 ______ + _____ + _____ = 0 2 2 2
fi
l l 2l + 4 + __ + 3 + __ + 1 = 0 2 2
fi
3l = – 8
fi
8 l = – __ 3
So,
a = 2l + 2, b = l + 3, g = l + 1
fi
16 8 8 a = 2 – ___ , b = 3 – __ , g = 1 – __ 3 3 3
fi
10 – 5 1 a = – ___ , b = __ , g = ___ 3 3 3
– 7 3z –1 3x + 2 3y fi ______ = ______ = _____ 2 1 4 10 5 7 x + ___ y – __ z + __ 3 3 3 fi ______ = _____ = _____ 2 1 4 – 7 3z +5 3x + 10 3y fi _______ = ______ = ______ 2 1 4
Match Matrix 1. (A)
(B)
2
(B)
= 4/3
S sin2a = sin2a + sin2b + sin2g + sin2d 4 8 = 4 – __ = __ 3 3
(C)
S cos(2a)
= cos(2a) + cos(2b) + cos(2g) + cos(2d) = 2(cos2a + cos2b + cos2g + cos2d) – 4
( )
8 4 4 = 2 __ – 4 = __ – 4 = – __ 3 3 3 (D)
S (cos2a + sin2a) = S 1
=1+1+1+1
=4
3. (A) Let two diagonals of a cube are ai + aj + ak and – ai + aj + ak Let q be the angle between them a ◊ – a + __a ◊ a + a ◊ a __ cos q = _______________ = a÷3 ◊ a÷3
a2 ___ 2 = 3a
1 __ 3
( )
1 fi q = cos–1 __ 3 (B) Let the diagonal of the cube be i + j + k and the diagonal of the face be i + j. Let q be the angle between them,
÷
__
1 __+ 1__ _____ 2 2 cos(q) = _____ = __ __ = __ 3 3 ÷ 2 ÷3 ÷ 2 ÷
( ÷ ) __
__ 2 fi (q) = cos–1 __ = cot–1 (÷2 ) 3
(C) Let the diagonal of the cube be i + j + k and the diagonal of the face intersecting it is j + k. Let q be the angle between them,
( ÷ ) __
S cos a = cos a + cos b + cos g = 1 2
= 2 – 3 = –1 2
10 5 7 x + ___ y – __ z + __ 3 3 3 ______ _____ _____ = = 2 1 4
Hence, the required image of the line w.r.t. to the plane is
2
= 2(cos2a + cos2b + cos2g) – 3
2l + 4 l________ + 3 + 3 l_____ +2 = ______ , , 2 2 2
(C)
2
S sin2a = sin2a + sin2b + sin2g
= 3 – (cos2a + cos2b + cos2g)
=3–1=2
__ 2 q = cos __ = cot–1 (÷2 ) 3 –1
(D) Let the diagonal of the faces of the same vertex are i + j, j + k. Let q be the angle between them. Then
6.46 Integral Calculus, 3D Geometry & Vector Booster
0 + __1 +__ 0 __ 1 cos q = ________ = 2 2 ÷ 2 ÷
1– 2 4 – 3 5– 4 1 1 –k = 0 k 2 1
fi
–1 1 1 1 1 –k = 0 k 2 1
fi
–1 0 0 1 2 1– k = 0 k k + 2 k +1
1 = ___ __ 3 ÷
fi
1__ q = cot–1 ___ 3 ÷
l + m + n = 0, l2 = m2 + n2
4. (A) Given
fi
– (m + n) = m2 + n2
fi
– 2m n = 0
fi
m=0=n
fi 2(k + 1) + (k – 1)(k + 2) = 0
When m = 0, then l = – n
fi 2k + 2 + k2 + k – 2 = 0
l m n 1 fi __ = __ = ___ = ___ __ 1 0 –1 ÷ 2 When n = 0, then l = – m l fi __ = 1
m ___ = –1
n __ = 0
1 ___ __ 2 ÷
fi k2 + 3k = 0 fi k = 0, – 3. (C) Since the given lines are intersect, so they are coplanar. Thus,
Let q be the angle between them. Then
cos (q) = l1l2 + m1m2 + n1n2
p q = __ 3
fi
(B) Let q be the angle between them. Then
6 + 4 – _________ 10 __________ cos q = _______________________ = 0 ÷9 + 16 + 25 ÷4 + 1 + 4
fi
p q = __ 2
(C) Let q be the angle between them. Then
3 1 2 – __1 +__ 2 __ cos (q) = ________ = = __ 6 2 ÷6 ÷ 6 p fi q = __ 3 (D) Let q be the angle between them. Then
fi
2 + __1 +__ 0 ___ 1 sin q = ________ = __ ÷9 ÷ 2 ÷2 p __ q = 4
5. (A) Since the line lies in the plane, so the point (4, 2, k) lies in the plane.
Thus,
8–8+k=7
k=7
fi
2 k + 1 –1 2 3 4 =0 1 2 1
1 1__ 1 = ___ __ ◊ ___ + 0 + 0 = __ 2 2 ÷2 ÷
3 – 1 k + 1 –1 2 3 4 =0 1 2 1
(B) Since the given lines are coplanar so
fi 2(3 – 8) – (k + 1)(2 – 4) – (4 – 3) = 0 fi –10 + 2(k + 1) – 1 = 0 fi 2k – 9 = 0 9 fi k = __ 2 (D) Given lines are – 3 z____ –6 x – 2 y_____ _____ = = 3 4 5 – 2 z____ –1 x – 5 y_____ and _____ = = 1 2 3 Hence, the shortest distance (r2 – r1) ◊ (u × v) = ______________ |(u × v)|
=0
where, (r2 – r1) =
(5 – 2, 2 – 3, 1 – 6) = (3, –1, – 5)
i j k (u × v) = 3 4 5 = 2i – 4j + 2k 1 2 3 _________
__
fi |u × v| = ÷ 4 + 16 + 4 = 2÷6
3D-Co-ordinate Geometry
6. (A) Let any point on the given line be
(l + 1, 3l – 2, 2 – 2l) Clearly, this point lies on the plane so, 3(l + 1) + 4(3l – 2) + 5(2 – 2l) = 15 fi 5l + 5 = 15 fi 5l = 10 fi l = 2 Hence, the point of intersection is (3, 4, – 2). y – 1 z____ – 2 x (B) Any point on the line __ = _____ = can 2 2 3 be considered as M(2l, 2l + 1, 3l + 2). Let the point P be P(2, 6, 3). The direction ratios of PM are (2l – 2, 2l – 5, 3l – 1). Clearly PM is perpendicular to the given line L. So, 2(2l – 2) + 2(2l – 5) + 3(3l – 1) = 0 fi 17l – 17 = 0 fi l – 1 = 0 fi l = 1 Hence, the foot of the perpendicular is (2, 3, 5). (C) Any point on the first line be (l + 2, 1 – 2l, l + 6). Any point on the second line be (7m – 3, – 6 – 3m, m – 3) Clearly, both the points are same. So, l + 2 = 7m – 3, l + 6 = m – 3 l – 7m = – 5, l – m = – 9 Solving we get 29 2 l = – ___ , m = – __ 3 3 Hence, the point of intersection is
(
)
23 61 ___ 11 ___ , ___ , – 3 3 3 (D) Clearly, 2l – 3m = 4, 3l – 4m = 4 Solving we get l = – 2, m = – 4 Hence, the point of intersection is (– 5, – 8, – 9) 7. Given P(0, 3, – 2), Q(3, 7, – 1) and R(1, – 3, – 1). +2 x y – 3 z_____ Clearly, L1 : __ = _____ = 3 4 1 + 3 z_____ +1 x – 1 y_____ and L2 : _____ = = 1 0 1 The equation of the plane PQR is
x 3 1
y–3 z+2 4 1 =0 –6 1
fi
10x – 2(y – 3) – 22(z + 2) = 0
fi
10x – 2y – 22z = 38
fi
5x – y – 11z = 19
(A)
6.47
Let the foot of the perpendicular on L2 is M. Thus, M = (l + 1, – 3, l – 1) The direction ratios of PM are (l + 1, – 6, l + 1) Since PM is perpendicular to L2, so
1(l + 1) + 0 + 1(l + 1) = 0
fi l = – 2 Hence, the foot of the perpendicular is (–1, – 3, – 3) Thus, the length of the perpendicular _________
___
=÷ 1 + 36 + 1 =÷ 38 (B) The shortest distance between L1 and L2
(r2 – r1) ◊ (u × v) = ______________ |(u × v)|
4 + 12 – 4 = _________ = 2, 6 where (r2 – r1) = (1, – 6, 1) = i – 6j + k i j k 3 4 1 = 4i – 2j – 4k fi u × v = 1 0 1 __________
fi |u × v| = ÷16 + 4 + 16 = 6
(C) Area of DPQR
1 = __ |PQ × PR| 2
i j k 1 __ = 3 4 1 2 1 –6 1
1 = __ |10i – 2j – 22k| 2 ____ __ ÷588 1 ____________ __ _____ = ÷ 100 + 4 + 484 = = 7÷3 2 2 (D) Hence, the distance from the origin to the plane 5x – y – 11z = 19
19 = ____________ ___________ = ÷25 + 1 + 121
19 _____ ____ 147 ÷
8. (A) The shortest distance from the given point to the x-axis ______
___
=÷ 4 2 + 52 = ÷41
6.48 Integral Calculus, 3D Geometry & Vector Booster
(B) The shortest distance from the given point to ______ the y-axis = ÷ 3 2 + 42 = 5
(C) Let q be the angle between them. Then 2 sin q = __ 7 2 –1 __ fi q = sin 7 2 __ Clearly, k = 7
( )
fi
fi
r = (1 + t)i + 5t j + (1 – t)k
= (i + k) + t(i + 5j – k)
g f =0 c
abc + 2fgh – af 2 – bg2 – ch2 = 0 2x2 + 3y2 + 4z2 + 2kxy + 4yz = 0
represents two planes if
h b f
(A) The given equation
7k = 2
(D) Given line is
a h g
2 k 0 k 3 2 =0 0 2 4
fi 2(12 – 4) – k(4k) = 0
Since the given line lies in the plane
fi 16 – 4k2 = 0
c = d so, 1 + 0 + 1 + 5 – c = 0
fi k2 – 4 = 0 fi k2 = 4
fi 1 + c = d c=6 Thus,
c = 6, d = 7
Hence, the value of c + d is 13 9. (A) Hence, the volume of the tetrahedron
1 1 = __ abc = __ × 2 × 3 × 4 = 4 6 6
(B) Let the points A, B, C and D are (0, 1, 2), (3, 0, 1), (4, 3, 6) and (2, 3, 2), respectively.
Now,
AB = (3, –1, –1)
AC = (4, 2, 4)
AD = (2, 2, 0)
fi 21 – k(5k) = 0 21 fi k2 = ___ 5
3 –1 –1 1 __ = 4 2 4 6 2 2 0
|
1 k 0 k 5 2 =0 0 2 5
fi 5k2 = 21
Hence, the volume of the tetrahedron 1 = __ [AB, AC, AD] 6
(B) x2 + 5y2 + 2kxy + 4xz = 0 represents a pair of planes if
(C) x2 + y2 + 4z2 – 2k yz = 0 represents a pair of planes if
1 0 –K
0 –K 1 0 =0 0 4
fi 4 – k2 = 0
|
fi k2 = 4
1 = __ (– 24 – 8 – 4) 6
fi k2 + 2 = 6
= 6 cu. units.
(C) Volume of a tetrahedron = 2/3 2 (D) Volume of a tetrahedron = __ abc 3 2 __ = × 2 × 3 × 4 3 = 16 10. As we know that the homogeneous equation of 2nd degree ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0 represents two planes if
(D) 9x2 + y2 + z2 – 6k xz = 0 represents a pair of straight lines if
9 0 0 1 0 – 3k
0 – 3k = 0 1
fi 1 – 9k2 = 0 fi 9k2 = 1
6.49
3D-Co-ordinate Geometry
1. We have, al + bm + cn = 0 (bm + cn) fi l = – ________ a Now, ul2 + vm2 + wn2 = 0. bm + cn 2 fi u _______ + vm2 + wn2 = 0 a
(
)
fi
u(bm + cn)2 + va2 m2 + wa2 n2 = 0
fi
u(b2 m2 + c2 n2 + 2bcmn) + va2 m2 + wa2 n2 = 0
(ub2 + va2) m2 + 2ubcmn + (uc2 + wa2) n2 = 0 m 2 m fi (ub2 + va2) __ n + 2ubc __ n + (uc2 + wa2) = 0 ...(i) m1 m2 ___ ___ Let its roots are n and n . 1 2 fi
( )
( )
(a) If two straight lines are parallel, so the Eq. (i) will provide us equal roots. Thus, D = 0 fi (2ubc)2 – 4(b2u + a2v)(c2u + a2w) = 0 fi (ubc)2 – (b2u + a2v)(c2u + a2w) = 0 fi (b2u + a2v)(c2u + a2w) – (ubc)2 = 0 fi a2b2 u w + a2c2uv + a4vw = 0 fi b2uw + c2uv + a2vw = 0 2
2
2. Let P = (a, b, g) and Q = (x1, y1, z1) Thus, the direction ratios of OP are (a, b, g) and the direction ratios of OQ are (x1, y1, z1). Since O, Q, P are collinear, we have a __ x1 =
Similarly, eliminating n, we get l1l2 m1m2 ________ 2 = ________ 2 2 wb + vc uc + wa2
g __ z 1
...(i)
As P(a, b, g) lies on the plane lx + my + nz = p so
la + mb + ng = p
fi
k(lx1 + my1 + nz1) = p
Given
...(ii)
2
OP ◊ OQ = p
____________
___________
fi ÷a 2 + b2 + g 2 ◊ ÷x 21 + y 21 + z21 = p2 ______________
___________
fi ÷k 2 (x21 + y 21 + z21) ◊ ÷x 21 + y 21 + z 21 = p2 _____________
fi ÷k 2(x21 + y 21 + z21) 2 = p2 fi
k(x21 + y 21 + z 21) = p2
...(iii)
On dividing Eq. (ii) and Eq. (iii), we get lx1 + my1 + nz1 __ 1 _____________ = p (x21 + y21 + z21) fi
p(lx1 + my1 + nz1) = (x21 + y 21 + z 21)
Hence, the locus of the point Q is p(lx + my + nz) = (x2 + y2 + z2) x y __z 3. Given plane is __ a + __ + c = 1 b
2
a b c __ __ fi __ u + v + w = 0 (b) When two straight lines are perpendicular. So, uc2 + wa2 the product of the roots = ________ ub2 + va2 2 m1 ___ m2 uc + wa2 ________ fi ___ ◊ = n1 n2 ub2 + va2 m1m2 n1n2 fi ________ 2 = ________ 2 2 2 uc + wa ub + va
b __ y = 1
Let
...(i)
P = (a, b, g)
b __g a __ Then __ a + b + c = 1
...(ii)
___________
So,
OP = ÷ a 2 + b 2 + g 2
Therefore, the direction ratios of OP are g b a ___________ ___________ ____________ , ____________ , ____________ ___________ 2 2 2 2 2 2 2 ÷a + b + g ÷a + b + g ÷a + b 2 + g 2
l1l2 m1m2 n1n2 Thus, ________ 2 2 = ________ 2 2 = ________ wb + vc uc + wa ub2 + va2
The equation of the normal through P and normal to OP is b a __________ ___________ x + ____________ ___________ y 2 2 2 2 ÷a + b + g ÷a + b2 + g 2
Two lines are perpendicular, if
___________ g + ____________ ___________ z = ÷a 2 + b 2 + g 2 2 2 2 ÷a + b + g
fi (wb2 + vc2) + (uc2 + wa2) + (ub2 + va2) = 0
fi
ax + by + g z = ÷ a 2 + b 2 + g 2
fi a2 (u + v) + b2 (u + w) + c2 (u + v) = 0
a 2 + b 2 + g 2 Clearly, A = ____________ , 0, 0 a
l1l2 + m1m2 + n1n2 = 0
Hence, the result.
___________
(
)
...(iii)
6.50 Integral Calculus, 3D Geometry & Vector Booster
(
) )
a 2 + b 2 + g 2 B = 0, ___________ , 0 g
(
2
2
2
a + b + g C = 0, 0, ____________ g
Let
Q = (p, q, r)
Then
a 2 + b 2 + g 2 a 2 + b 2 + g 2 p = ___________ , q = ___________ a b 2
and
2
2
a + b + g r = ___________ g
...(iv)
a2 + b2 + g 2 1 1 1 Now, __2 + __ 2 + __ 2 = ______________ p q r (a 2 + b 2 + g 2)2 1 1 1 1 fi __2 + __ 2 + __ 2 = _____________ 2 p q r (a + b2 + g 2)
...(v)
From Since From From
2 2 + b2 + g 2 __ + b2 + g 2 q a__________ p a___________ __ = , = a aa b bb 2 2 2 r a + b + g __c = ___________ cg
...(iv) ...(v) ...(vi)
ka 2 + kb2 + kg 2 – k(aa + bb + cg) = 0 fi a 2 + b 2 + g 2 – (aa + bb + cg ) = 0 Hence, the locus of the foot of the perpendicular P is x2 + y2 + z2 – (ax + by + cz) = 0. 5. Given planes are x – y – z = 4 ...(i) and
From Eq. (iv), we get
Eq. (ii) and (iii), we get ka a + kbb + kcg + D = 0 (a, b, g) lies in the plane (i), so Aa + Bb + Cg + D = 0 Eqs. (iii) and (v), we get ka 2 + kb 2 + kg 2 + D = 0 Eqs. (iv) and (vi), we get
x + y + 2z = 4
...(ii)
Since the required planes pass through the line of intersection of the planes (i) and (ii). Thus, its equation will be (x – y – z – 4) + l (x + y + 2z – 4) = 0 fi
(1 + l) x + (1 – l) y + (2 – l) z – 4 – 4l = 0 ...(iii) Therefore, Since Eq. (i) and (iii) are mutually perpendicular, 2 2 2 2 2 2 a + b + g a + b + g ____________ + ____________ so, aa bb (1 + l) – (1 – l) – (2 – l) = 0 2 2 2 a + b + g 2 + ___________ fi l = __ cg 3 p q __r 2 = __ a + __ + c = 1 Substituting the values of l = __ in Eq. (iii), we get 3 b 5x + y + 4z = 20. 1 1 1 1 ___ __ ___________ fi ___ 2 which is the required equation of the plane. ap + bq + cr = a 2 + b 2 + g 6. Given planes are 1 1 1 fi = __ 2 + __ 2 + __ 2 (from Eq. v) x – cy – bz = 0 ...(i) p q r cx – y + az = 0 ...(ii) Hence, the locus of Q(p, q, r) is bx + ay – z = 0 ...(iii) 1 1 1 1 1 1 ___ ___ __ __ __ __ ax + + cz = 2 + 2 + 2 The equation of any plane passing through the line by x y z of intersection of the planes (i) and (ii) may be taken 4. Let the equation of the plane be as (x – cy – bz) + l(cx – y + az) = 0 Ax + By + Cz + D = 0 ...(i) fi (1 + lc)x – (c + l)y + (al – b)z = 0
which is passing through (a, b, c). So,
Aa + Bb + Cc + D = 0
...(ii)
Let P(a, b, g) be the foot of the perpendicular from the origin to the plane (i). The direction ratios of OP are (a, b, g). Therefore, (a, b, g) be the direction ratios of the same line OP. So, g a b __ 1 __ = __ = = __ (say) ...(iii) A B C k
...(iv)
If planes (iii) and (iv) are the same, so the equations (iii) and (iv) are identical. (1 + lc) (c + l) (al – b) _______ = – ______ = _______ a –1 b (1 + lc) (c + l) Thus, _______ = – ______ a b fi
(a + bc) l = – _______ (ac + b)
...(v)
3D-Co-ordinate Geometry
(c + l) (al – b) – ______ = _______ a –1 (ab + c) fi l = – _______ (1 – a2) From Eqs (v) and (vi), we get
So,
and
3a + 4b – 2c = 0
(ac + b) (ab + c) – _______ = – _______ (a + bc) (1 – a2)
and – 4a – b + 2c = 0 Solving, we get a b c _____ = _____ = _______ 8 – 1 8 – 6 – 3 + 16 a b c fi __ = ___ = ___ 7 12 13 Hence, the equation of the plane is
fi
a – a3 + bc – a2bc = a2bc + ac2 + ab2 + bc
fi fi
3
2
2
...(vi)
2
a + ab + ac + 2a bc – a = 0 2
2
2
a + b + c + 2abc = 1
7. Any point on the given line is
P : (2l + 1, 3l – 2, 6l + 3)
Let
Q : (1, – 2, 3)
Given
PQ = 3
fi
PQ2 = 9
fi
(4l2 + 9l2 + 36l2) = 9
49l2 = 9 9 3 2 fi l2 = ___ = __ 7 49 3 fi l = ± __ . 7 Hence, the points are fi
( )
( ) ( ) ) ( ( ) 17 11 = ( ___ , 5, 17 ) and ( – ___ , – 9, – 11 ). 3 3
7 7 7 ± 2 __ + 1, ± 3 __ – 2, ± 6 __ + 3 3 3 3
8. The equation of any line through (1, 0, – 3) and + 2 6____ –z x – 2 y_____ parallel to the line _____ = = is 2 3 6 y z+3 x – 1 __ _____ = = _____ . 2 3 – 6 Any point on the line (i) is (2l + 1, 3l, – 6 – 3l) which is lies in the plane x – y – z = 9. So,
(2l + 1) – (3l) – (– 6 – 3l) = 9
fi
2l + 7 = 9
fi
2l = 2 fi l = 1
Thus, the point is (3, 3, – 9). Hence, the required distance
________________________
=÷ (3 – 1)2 + (3 – 0)2 + (– 9 + 3)2
=÷ 4 + 9 + 36
=÷ 49 = 7.
_________ ___
9. The equation of any plane passing through (1, 2, 0) is a(x – 1) + b(y – 2) + cz = 0 which contains the – 1 2____ –z x + 3 y_____ line _____ = = . 3 4 2
6.51
7(x – 1) + 2(y – 2) + 13z = 0
fi 7x + 2y + 13z = 11 10. Let the given line AB be + 1 z____ –2 x – 1 y_____ _____ = = ...(i) 2 –1 4 and the given plane is x + 2y + z = 9 ...(ii) Let DC be the projection of AB on plane (ii). Clearly the plane ABCD is perpendicular to the plane (ii). Equation of any plane through AB is a(x – 1) + b(y + 1) + c(z – 3) = 0 ...(iii) where 2a – b + 4c = 0 ...(iv) since the plane (iii) is perpendicular to the plane (ii), so a + 2b + c = 0 ...(v) Solving Eqs (iv) and (v), we get a b c ___ = __ = __ – 9 2 3 Putting these values of a, b and c in Eq. (iii), we get 9(x – 1) – 2(y + 1) – 5(z – 3) = 0 fi 9x – 2y – 5z + 4 = 0 ...(vi) Since the projection DC of AB on the plane (ii) is the line of intersection of the plane ABCD and the plane (ii). Thus, the equation of DC will be 9x – 2y – 5z + 4 = 0 fi x + 2y + z – 9 = 0 ...(vii) Let l, m, n be the direction ratios of the line of intersection of the planes (i) and (ii). So, 9l – 2m – 5n = 0 and l + 2m + n = 0 Solving, we get l m n __ = ____ = ___ 8 –14 20 l m n fi __ = ___ = ___ 4 – 7 10 Let the line of intersection of the planes (i) and (ii) meet xy-plane at P(a, b, 0). So, P lies on the planes (i) and (ii) Therefore, 9a – 2b + 4 = 0 and a + 2b – 9 = 0
6.52 Integral Calculus, 3D Geometry & Vector Booster 17 1 a = __ , b = ___ 2 4 Hence, the equation of DC is 17 1 x – __ y – ___ z 2 4 _____ ______ = = ___ – 7 4 10 – 17 ___ z 2x – 1 4y fi ______ = _______ = 8 – 28 10 – 17 __z 2x – 1 4y fi ______ = _______ = 4 –14 5 11. Any plane passing through the first line is (2x + y + z – 1) + l(3x + y + 2z – 2) = 0 ...(i) which is parallel to the 2nd line x = y = z. So, (2 + 3l) ◊ 1 + (1 + l) ◊ 1 + (1 + 2l) ◊ 1 = 0 fi
(2 + 3l) + (1 + l) + (1 + 2l) = 0
fi
6l + 4 = 0 2 fi l = – __ 3 2 __ Put l = – in Eq. (i), we get 3 2 (2x + y + z – 1) – __ (3x + y + 2z – 2) = 0 3 fi y – z + 1 = 0 ...(ii) 1__ Thus, distance from (0, 0, 0) to the plane (ii) = ___ . 2 ÷ 12. We have, 4 + l = 1 + 2m fi
l – 2m = – 3
Also,
– 3 – 4l = –1 – 3m
fi
4l – 3m = – 2
fi aa1 + cc1 + 1 = 0 fi k=1 Now, k2 + 4 = 1 + 4 = 5 The equation of any line through (1, 2, 3) parallel to the given line is – 2 z____ –3 x – 1 y_____ _____ = = . 1 – 2 2 Any point on the above line can be considered as (l + 1, 2 – 2l, 2l + 3) which is also a common point of the plane. Thus, l + 1 + 2 – 2l + 2l + 3 = 11 14.
So,
fi l=5 Therefore, the point is (6, – 8, 13). Hence, the required distance
_________________________
= ÷(6 – 1)2 + (– 8 – 2)2 + (13 – 3)2
= ÷25 + 100 + 100
=÷ 225
_____________ ____
= 15 15. Let the equation of the plane be y __z x __ __ a + b + c = 1
...(i) ...(ii)
From Eq. (i) and (ii), we get l = 2, m = 1 Thus, the point of intersection
= (4 + 1, – 3 – 4, –1 + 7)
= (5, – 7, 6)
Hence, the required distance from (5, – 7, 6) to (1, – 4, 7) _________ = ÷16 + 9 + 1
___
= ÷26 .
13. Let
L1 : x = ay + b, z = cy + d y z–d x – b __ fi _____ = = ____ a c 1 and L2 : x = a1y + b1, z = c1y + d1 x – b1 fi _____ = a1
y z – d1 __ = _____ c1 1 Since L1 and L 2 are perpendicular, so
aa1 + 1.1 + c ◊ c1 = 0
It is given that the = 64k3 0 0 1 a 0 fi __ 6 0 b 0 0
volume of the tetrahedron OABC 0 0 0 c
1 1 = 64k3 1 1
1 fi __ (abc) = 64k3 6 fi (abc) = 384 k3 Let the centroid be G(a, b, g). a b c Clearly, __ = a, __ = b, __ = g 4 4 4 fi a = 4a, b = 4b, c = 4g Therefore, 64(a b g) = 384k3 fi
(a b g) = 6k3
Hence, the locus of G(a, b, g) is xyz = 6k3
3D-Co-ordinate Geometry
y z x z 16. Given lines __ + __c = 1, x = 0 and __ a – __c = 1, y = 0 b can be written as z x y – b ___ L1 : __ = _____ = – c 0 b y z x – a __ and L2 : _____ = = __c a 0 Here, and Now,
r1 = bj r2 = ai u = bj – ck v = ai + ck (r2 – r1) = ai – bj i
and
k
u × v = 0 b –c a 0 c
fi
j
= bci – acj – abk _________________ (bc)2 + (ac)2 + (ab)2
|(u × v)| = ÷
Given shortest distance = 2d (r2 – r1) ◊ (u × v) fi ______________ = 2d |(u × v)| abc + abc _________________ fi __________________ = 2d ÷(bc) 2 + (ac)2 + (ab)2 2abc _________________ fi __________________ = 2d 2 + (ac)2 + (ab)2 ÷ (bc) abc _________________ fi __________________ = d 2 + (ac)2 + (ab)2 ÷ (bc) (abc)2 fi __________________ = d2 2 2 2 [(bc) + (ac) + (ab) ] ((bc)2 + (ac)2 + (ab)2) 1 fi ___2 = ___________________ d (abc)2 1 fi ___2 = d
1 __ 2 + a
1 __ 2 + b
1 __ 2 c
17. Let O be the origin and the direction ratios of OP are (a, b, c) Hence, the equation of the plane is
6.53
3l – 2 l_____ + 1 2l +1 fi ______ = = ______ 1 2 3 fi
l=1
Thus, the point of intersection of the two lines is P(4, 3, 5). Therefore, the equation of the plane perpendicular to OP, where O is (0, 0, 0) and passing through P is 4x + 3y + 5z = 50. 19. The equation of the plane through the line be A(x – 1) + B(y + 2) + C(z – 0) = 0, then
2A – 3B + 5C = 0
and
A–B+C=0
Solving, we get C A B __ __ = __ = 2 3 1 Hence, the equation of the plane is
2(x – 1) + 3(y + 2) + z = 0
fi
2x + 3y + z + 4 = 0
Thus, a = 2, b = – 3, c = 1 Hence, the value of a + b + c + 2.
=2–3+1+2
= 2.
20. The direction ratios of the normal to the plane is (2, 3, 4). The equation of any plane through (2, 3, 4) and the maximum distance from the origin is
2(x – 2) + 3(y – 3) + 4(z – 4) = 0
fi
2x + 3y + 4z = 29.
21. Let O be the origin (0, 0, 0). Let the co-ordinates of the points A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Thus, the equation of the sphere passing through O, A, B, and C is
x2 + y2 + z2 – ax – by – cz = 0
But the radius of the sphere is given by r.
÷
___________
So,
a2 + b2 + c2 r = ___________ 4
fi (a2 + b2 + c2) = 4r 2 ...(i) Let (a, b, g ) be the centroid of the DABC fi ax + by + cz = a2 + b2 + c2 a + 0 + 0 __ a a = ________ = 18. Let a point (3l + 1, l + 2, 2l + 3) of the first line Then 3 3 lies on the 2nd line. Thus, b c Similarly, b = __ = g = __ 3 3 3l + 1 – 3 l________ + 2 – 1 2l +3–2 _________ = = _________ Putting the values of a, b and c in Eq. (i), we get 1 2 3 9(a 2 + b 2 + g 2) = 4r 2
a(x – a) + b(y – b) + c(z – c) = 0
6.54 Integral Calculus, 3D Geometry & Vector Booster Hence, the locus of (a, b, g ) is
2
2
2
9(x + y + z ) = 4r
2
22. Let the co-ordinates of A, B and C be (a, 0, 0), (0, b, 0) and (0, 0, c), respectively Hence, the equation of the plane is y __z x __ __ a + b + c = 1 Since the given plane passes through (f, g, h), we get f __ g __ h __ ...(i) a + b + c = 1 The equation of the sphere passing through 0, A, B and C is x2 + y2 + z2 – ax – by – cz = 0 Let (a, b, g ) be its centre. a b c Thus, a = __ , b = __ , g = __ 2 2 2 fi a = 2a, b = 2b, c = 2g Putting the values of a, b and c in (i), we get f g h ___ + ___ + ___ = 1 2a 2b 2g f g h fi __ + __ + __ = 2 a b g Hence, the locus of the centre (a, b, g ) is f g __ h __x + __ y + z = 2 23. Let O be the origin (0, 0, 0). Let the co-ordinates of the points A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Then the equation of the plane ABC is y __z x __ __ ...(i) a + b + c = 1 Also, the equation of the sphere passing through O, A, B, and C is
x2 + y2 + z2 – ax – by – cz = 0
÷
2
2
2
So,
a +b +c r = __________ 4
fi
(a2 + b2 + c2) = 4r 2
...(ii)
Let (a, b, g ) be the foot of the perpendicular from the origin O to the plane (i) Now, the equation of the line through O and the perpendicular to the plane (i) is y z x ___ = ___ = ___ = r (say) 1/a 1/b 1/c Thus, the co-ordinates of any point on the line are r r __r __ a , __ , c ...(iii) b If it is the foot of the perpendicular from O to the plane (i), its co-ordinates will satisfy the equation of the plane.
(
)
(
)
Then from Eq. (iii), we get
and
and
1/a r a = __ a = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/b r b = __ = ____________ b 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/c r g = __ c = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
Now,
a 2 + b2 + g 2
1 1 1 __ 2 + __2 + __2 a b c = _____________ 2 = 1 1 1 __ 2 + __ 2 + __ 2 a b c
= aa = bb = cg
Thus,
a2 + b2 + g 2 a2 + b 2 + g 2 a = ___________ , b = ___________ a b
and
a 2 + b 2 + g 2 c = ____________ g
( (
) ) (
1 ____________ 1 1 1 __ __ 2 + 2 + __ 2 a b c
)
Putting the values of a, b and c in (ii), we get
(
)
1 1 1 (a 2 + b 2 + g 2) ___ 2 + ___ 2 + ___ 2 = 4r 2 a b g
Hence, the locus of the foot of the perpendicular (a, b, g ) is 1 1 1 (x2 + y2 + z2) __ 2 + __ 2 + __ 2 = 4r2 x y z 24. Given plane is y __z x __ __ a + b + c = 1 Let O be the origin. Clearly, the plane intersects the axes at (a, 0, 0), (0, b, 0) and (0, 0, c), respectively. Hence, the equation of the sphere passing through O, A, B and C is
(
But the radius of the sphere is given by r. __________
1 __r __ 1 __r __ 1 __r Hence, __ a ◊ a + b ◊ b + c ◊ c = 1 1 fi r = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
)
x2 + y2 + z2 – ax – by – cz = 0
25. The equation of any sphere through the given circle is x2 + y2 + z2 + 7y – 2z + 2 + l (2x + 3y + 4z – 8) =0 fi x2 + y2 + z2 + 2ll + (7 + 3l) y + (4l – 2) z + (2 – 8l) = 0
3D-Co-ordinate Geometry __ ___
The centre of the sphere is
(
)
7 + 3l 4l –2 – l, – ______ , – ______ 2 2 If the given circle is a great circle of the sphere, so the centre of the sphere lies on the plane
(
6.55
)
Thus,
7 + 3l 2(– l) + 3 – ______ + 4(1 – 2l) = 8 2
fi
– 4l – 21 – 9l + 8 – 16l = 16
fi
– 29l = 29
fi
l = –1
fi (af + bg ± 2 ÷af ÷bg ) = ch __
___
__
___
__
___
___
fi (÷af ± ÷ bg ) 2 = (÷ch ) 2 ___
fi (÷af ± ÷ bg ) = ± (÷ch ) ___
fi (÷af ± ÷ bg + ÷ ch ) = 0 which is the required condition. 2. Let OA = a, OB = b and OC = c
Hence, the equation of the sphere is x2 + y2 + z2 + 7y – 2z + 2 – 2x – 3y – 4z + 8 = 0 fi x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0
1. Given al + bm + cn = 0 and fmn + gnl + hlm = 0 Eliminating n, we get
(
)
(
)
al + bm al + bm = gl – _______ + hlm = 0 f m – _______ c c – afm – bfm2 – agl2 – bglm + chlm = 0 l l 2 m + bf = 0 fi ag __ m + (af + bg – ch) __ l1 l2 m . Let its roots are ___ m and ___ 1 2 Now, l1 ___ l2 bf ___ ___ m1 ◊ m2 = ag fi
( )
( )
l2l2 m1m2 ag fi ___ = _____ bf l2l2 m1m2 ____ n1n2 fi ___ = _____ ag = bf ch
Similarly, we can find the angle between other pairs of diagonals and we have six such pairs out of these four digonals and all these angles are given by
± a2 ± b2 + c2 cos q = ___________ (a2 + b2 + c2)
fi
a2 ± b2 ± c2 q = cos– 1 __________ a2 + b2 + c2
l1l2 = m1m2 + n1n2 = 0
f fi __ a +
g h __ + __ = 0 b c (ii) If the lines are parallel, so l1 m1 __ n1 __ = ___ m = n l2 2 2 So, the roots of the Eq. (i) have equal roots, i.e. D = 0 fi (af + bg – ch)2 = 4abfg __
(
___
fi (af + bg – ch) = ± 2÷af ÷bg
)
3. The equation of any plane passing through P, Q, R is
x __ + a
(i) If the lines are perpendicular, so
– a2 + b2 + c2 = ___________ (a2 + b2 + c2)
l2l2 m1m2 ____ n1n2 fi ____ = _____ = (f/a) (g/b) (h/c)
Now, OP, CN, AM and BL are four diagonals. Let q be the angle between the diagonals OP and AM. The direction ratios of OP and AM are (a, b, c) and (– a, b, c) – a◊a + b ◊ b__________ + c ◊ c __________ Thus, cos q = _______________________ 2 2 2 2 + c ÷ a + b2 + c2 ÷ a + b
where
y __ + b
z __ = 1 g
...(i)
P = (a, 0, 0), Q = (0, b, 0), R = (0, 0, g)
which is passing through (a, b, c) a b __c Thus, __ + __ + = 1 a b g The equation of any plane passing through Q (0, b, 0) and parallel to yz-plane is x = a The equation of any plane passing through R (0, 0, g) and parallel to -plane is y = b The Equation of any plane passing through P (a, 0, 0) and parallel to xy-plane is z = b Hence, the locus of the point (a, b, g) is x __ b __c __ a + y + z = 1
6.56 Integral Calculus, 3D Geometry & Vector Booster
4. Let the equation of the plane be y __z x __ __ a + b + c = 1 It is given that,
Putting the values of a, b and c in Eq. (ii), we get ...(i)
a2 + b2 + c2 = k2
...(ii)
Let P (a, b, g) be the foot of the perpendicular drawn from the origin O to the plane (i). Now the equation of the line through (0, 0, 0) and the perpendicular to the plane (i) is y z x ___ = ___ = ___ 1/a 1/b 1/c So, the co-ordinates of any point on the line are r r __r __ a , __ , c ...(iii) b
(
)
It is the foot of the perpendicular from O to the plane (i), its co-ordinates will satisfy the equation of the plane. 1 __r __ 1 __r __ 1 __r Hence, __ a ◊ a + b ◊ b + c ◊ c = 1 1 1 1 2 + __ 2 = 1 fi r __ 2 + __ a b c
(
)
and
1/a r a = __ a = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/b r b = __ = ____________ b 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
1/c r g = __ c = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
(
)
Now,
a2 + b2 + g2
1 1 1 __ 2 + __ 2 + __ 2 a b c = _____________ 2 1 1 1 __ 2 + __ 2 + __ 2 a b c
( ( (
)
Hence, the locus of P (a, b, g ) is
(
)
1 1 1 (x2 + y2 + z2) __ 2 + __ 2 + __ 2 = k2 x y z
5. Clearly, the planes x = 0, y = 0 and z = 0 meet in (0, 0, 0) Hence, the incentre lies on the perpendicular from (0, 0, 0) to the plane x + y + z = a and divides it in the ratio 3 : 1, i.e. 3 from the vertex (0, 0, 0) and 1 from the plane x + y + z = a. The equation of the perpendicular from (0, 0, 0) to the plane x + y + z x y __z = a is __ = __ = = l (say) 1 1 1 Any point on the perpendicular is (l, l, l) a If it lies on the plane x + y + z = a, so, l = __ 3 Thus, the perpendicular from (0, 0, 0) meets the plane a a __ a x + y + z = a in (l + l + l), i.e. __ , __ , . 3 3 3 Also, the incentre divides the join of (0, 0, 0) and a a __ a __ , __ , in the ratio 3 : 1. 3 3 3
(
1 r = ____________ 1 1 1 __ 2 + __ 2 + __ 2 a b c
Then from (iii)
)
(
)
(
(
1 1 1 (a2 + b2 + g 2 ) ___ 2 + __ 2 + __ 2 = k2 a b g
1 = ____________ = aa = bg = cg 1 1 1 __ __ 2 + 2 + __ 2 a b c
)
)
Let (a, b, g ) be its incentre. Then a 3 ◊ __ + 1.0 3 a a = ________ = __ 3+1 4 a Similarly, b = __ = g 4 a a __ a Thus, the required incentre is __ , __ , . 4 4 4 6. The equation of any line passing through (3, 5, 7) and parallel to the normal to the plane is
(
)
– 5 z____ –7 x – 3 y_____ _____ = = ...(i) 2 1 1 and the equation of any point on the line (i) is
)
)
M (2l + 3, l + 5, l + 7)
which lies on the plane 2x + y + z = 16. So, fi fi Thus,
(2l + 3) + (l + 5) + (l + 7) = 16 4l + 15 = 16 1 l = __ 4 1 1 1 M = __ + 3, __ + 5, __ + 7 2 2 2
( 15 7 11 ___ = ( __ , ___ , 2 2 2)
)
Thus,
a2 + b 2 + g 2 a2 + b2 + g2 a = __________ , b = __________ a b
and
a2 + b 2 + g2 c = __________ g
Let Q (a, b, c) is the image of P (3, 5, 7). Clearly, M is the mid-point of P and Q.
3D-Co-ordinate Geometry
a + 3 __ 15 7 b + 5 ___ 11 c + 7 ___ Thus, _____ = , _____ = , _____ = 2 2 2 2 2 2 fi a = 4, b = 6, c = 8 Hence, the value of a2 + b2 + c2 + 18 = 16 + 36 + 64 + 18 = 134. 7. Given lines are + 1 __z – 3 z____ –1 x – 2 y_____ x + 1 y_____ _____ = = c and _____ = c = a a b b ab + bc + ca Therefore, cos q = ____________ 2 ...(i) a + b2 + c2 Since a, b and c are the roots of x3 + x2 – 4x – 4 = 0, so a + b + c = – 1 ab + bc + ca = 4 and
abc = 4
Now,
a2 + b2 + c2
= (a + b + c)2 – 2 (ab + bc + ca)
=1+8=9
From Eq. (i), we get 4 cos q = – __ 9 Hence, the acute angle between the lines is
( )
Thus,
6.57
_____________
(ac + b) = ÷ (1 – a2) (1 – c2) _____________
Similarly, (bc + a) = ÷ (1 – b2) (1 – c2) l m n Hence, _______ _____ = _______ _____ = ______ _____ 2 2 ÷1 – b ÷ 1 – c2 ÷ 1 – a Since the three given planes pass through the origin, so the line of intersection of the planes also pass through the origin. Thefore, the equation of the line is y z x __ = __ m = __ n l y z x _____ fi _______ = _______ _____ = ______ _____ 2 2 ÷ 1 – b ÷ 1 – c2 ÷ 1 – a 9. Given planes are x – y – z = 2 ...(i) and x + 2y + z = 2 ...(ii) The equation of any plane passing through the line of intersection of planes (i) and (ii) can be written as (x – y – z – 2) + l (x + 2y + z – 2) = 0 ...(iii) The direction of the cosines of the normal to the plane (iii) are 1+l 2l – 1 l–1 ___________ ___________ ___________ _____________ , ____________ , _____________ 2 2 2 – 4l + 3 ÷6l – 4l + 3 ÷ 6l – 4l + 3 ÷ 6l
fi
(1 – a2) + (– c2 – abc) + (– abc – b2) = 0
The direction cosines of the normal to the plane (i) are 1__ ___ 1 1 ___ – __ , – ___ __ 3 ÷3 ÷3 ÷ Since the angle between the planes (i) and (ii) is 90°, so, (1 + l) (2l + 1) (l – 1) __ __ __ cos (90°) = ______ – _______ – ______ p÷3 p÷3 p÷3 1 where p = ____________ ___________ 2 – 4l + 3 ÷ 6l
fi
1 – a2 – b2 – c2 – 2abc = 0
fi
q = cos–1 __ 4 9 8. Since the planes x = cy + bz, y = az + cx and z = bx + ay meet in a line, so – c – b 1 c – 1 a b a – 1
fi
fi
| |
1(1 – a2) + c (– c – ab) – b (ca + b) = 0
2
2
2
a + b + c + 2abc = 1
Let l, m, n be the direction cosines of the line of intersection. Therefore, l – c m – b n = 0 and cl – m + an = 0 l m n Thus, ______ = ______ = _____ ac + b bc + a 1 – c2 Now, (ac + b)2 = a2c2 + b2 + 2ab
= a2c2 + b2 + 1 – a2 – b2 – c2 = 1 – a2 – c2 + a2c2 = (1 – a2) (1 – c2)
(1 + l) – (2l + 1) – (l – 1) = 0
3 l = __ 2 Hence, the required equation of the plane is 3 (x – y – z – 2) + __ (x + 2y + z – 2) = 0 2 fi 2 (x – y – z – 2) + 3 (x + 2y + z – 2) = 0 fi
fi
5x + 4y + z = 10
10. Given plane is y __z x __ __ a + b + c = 1 Let P (m, n, p) be a point on (i), so p m __ n __ __ a + b + c = 1
...(i)
...(ii)
6.58 Integral Calculus, 3D Geometry & Vector Booster __________
i.e
OP = ÷ m 2 + n2 + p2
1 1 1 1 fi ___ + ___ + __ = __________ aa bb cg m2 + n2 + p2
The direction cosines of OP are
(
1 = ___ 2 + a
1 ___ 2 + b
1 ___ 2 g
p m n ____________ __________ , ___________ __________ , ____________ __________ 2 + n2 + p2 ÷m 2 + n2 + p ÷ m 2 + n2 + p2 ÷ m
The equation of the plane through P and normal to OP is n y m x __________ ____________ + ____________ __________ 2 + n2 + p2 ÷m 2 + n2 + p2 ÷ m
1 1 1 1 1 1 ___ ___ __ __ __ ___ a x + b y + c z = x2 + y2 + z2
)
Hence, the required locus of Q is
Integer Type Questions
__________ 1. Let q be angle between the lines. p z 2 __________ + ____________ = ÷ m + n2 + p2 1 \ cos q = __ ÷m 2 + n2 + p2 3
fi
m x + n y + p z = (m2 + n2 + p2)
Thus,
(m2 + n2 + p2) A = ____________ 0, 0 m ,
(m2 + n2 + p2) B = 0, ____________ 0 n ,
and
(m2 + n2 + p2) C = 0, 0, ____________ p
( (
) )
(
)
( )
1 q = cos– 1 __ 3 Hence, (a + b) = 1 + 3 = 4 2. Four diagonals of a cube are OP, AM, CL, BN fi
Let the point Q be (a, b, g) Thus,
2 m2 + n2 + p2 m__________ + n2 + p2 a = __________ , b = m n
and
m2 + n2 + p2 g = ___________ p
...(iii)
Now, 1 1 1 ___2 + ___ 2 + ___ 2 = a b g
(m2 + n2 + p2) _____________ (m2 + n2 + p2)2
1 1__ ___ 1 Direction cosines of BN are ___ __ , – ___ , __ 3 ÷3 ÷3 ÷
1 = ____________ 2 (m + n2 + p2) 2
2
2
(m + n + p ) From Eqs. (iii), m = _____________ a 2 + n2 + p2) m (m _____________ fi __ = a aa 2 2 2 n (m + n + p ) Similarly, __ = _____________ b bb 2 p (m + n2 + p2) ____________ and __ = c cg 2 (m2 + n2 + p2) (m + n2 + p2) Now, ____________ + _____________ aa bb
1 1__ ___ 1 Direction cosines of OP are ___ __ , ___ , __ 3 ÷ 3 ÷3 ÷ 1 1 1__ __ , ___ __ , ___ Direction cosines of AM are – ___ 3 ÷ 3 ÷ 3 ÷ 1 1__ ___ 1 Direction cosines of CL are ___ __ , ___ , – __ 3 ÷ 3 ÷ 3 ÷
(m2 + n2 + p2) __ p m n __ + _____________ = a + __ + c = 1 cg b
Let l, m, n be the direction cosines of a line which is inclined at angles a, b, g, d. l + m__+ n cos a = ________ 3 ÷ – l + m__ + n l + m __– n Similarly, cos b = _________ , cos g = ________ 3 ÷3 ÷ l – m__+ n and cos d = ________ 3 ÷ Thus,
Thus,
cos2a + cos2b + cos2g + cos2d 1 = __ [(l + m + n)2 + (– 1 + m + n)2 3 + (l + m + n)2 + (l – m + n)2˚ 1 = __ [4 (l2 + m2 + n2)] 3 4 = __ 3
3D-Co-ordinate Geometry
Thus, 3(cos2a + cos2b + cos2g + cos2d) = 4 3. The image of P (1, 6, 3) w.r.t. the line –2 x y – 1 z____ __ = _____ = is Q (1, 0, 7) 1 2 3 It is given that Q (a, b, g) = Q (1, 0, 7) Thus,
a+b+g=8
4. Clearly, 3a + 2b + c = 7 Thus, the minimum value of (a2 + b2 + c2) is fi fi
|
|
3.0 +__________ 2.0 + 0 – 7 2 = ______________ 2 + 12 ÷ 3 + 22 49 7 = ___ = __ 14 2 7 (a2 + b2 + c2) ≥ __ 2 2 2 2 2 (a + b + c ) ≥ 7
Hence, the least value of (a2 + b2 + c2) is 7.
5. The equation of any line passing through (1, 0, – 3) + 2 6____ –z x – 2 y_____ and parallel to _____ = = 2 3 6 y z+3 x – 1 __ is _____ = = _____ ...(i) 2 3 – 6 Any point on (i) can be considered as P (2l + 1, 3l, – 34 – 6l) which is also a point of the plane 2l + 1 – 3l + 3 + 6l = 9 fi
5l + 4 = 9
fi
l=1
= ÷4 + 9 + 36
= ÷49
Shortest Distance = =
Distance of any point on z-axis 2 ____ __ = 2 ÷ 1 2 7. Let the points be A, B, C and D. The number of planes which have three points on one side and the fourth point on the othe side is 4. The number of planes which have two points on each side of the plane is 3. Thus, the number of planes is 7. 8. Let the equation of the any plane be ax + by + cz + d = 0 Since the plane (i) containing the lines, so 2a + 3b + 4c = 0 and 3a + 4b + 4c = 0. a b c Therefore, _______ = _______ = _____ 15 – 16 12 – 10 8 – 9 a b fi ___ = __ = – 1 2 a b fi __ = ___ = 1 – 2 Also,
a (x – 1) + b (y – 2) + c (z – 3) = 0
fi
(x – 1) – 2 (y – 2) + (z – 3) = 0
fi |d| = 6. 9. We shall find the angle between the faces OAB and OAC. The angle between the faces OAB and OAC is the angle between the normal _› to the faces. The _› normal to the_ face_ OAB is a × b and the normal to › › face OAC is a × c .
_________ ___
=7 6. The equation of any plane containing the given line is (x + y + 2z – 3) + l (2x + 3y + 4z – 4) = 0 fi (1 + l) x + (1 + 3l) y + (2 + 4l) z – (3 + 4l) = 0 ...(i) y z x If the plane is parallel to z-axis, i.e __ = __ = __ , 0 0 1 the normal to the plane (i) is perpendicular to z-axis (1 + l (0) + (1 + 3l) (0) + (2 + 4l) (1) = 0 fi
1 l = – __ 2
...(i)
c ___ – 1 c __ 1
| |
________________________
= ÷(3 – 1)2 + (3 – 0)2 + (– 9 + 3)2
From Eqs. (i) and (ii), we get 1 (x + y + 2z – 3) – __ (2x + 3y + 4z – 4) = 0 2 fi y + 2 = 0 ...(iii)
fi x – 2y + z = 0 __ Again distance between the plane is ÷ 6 . __ d –__0 fi _____ = ÷6 ÷6
Thus, the point P is (3, 3, – 9). Hence, the required distance
6.59
...(ii)
_›
Thus,
_›
_›
_›
(a × b ) ◊ |(a × c ) cos q = _______________ _› _› _› _› |(a × b )| |(a × c )| › _› _ a ◊ (b ×
_›
_›
(a × c )) = ______________ {1 ◊ 1 ◊ sin(60°)}2
6.60 Integral Calculus, 3D Geometry & Vector Booster _› _ _ _› › › a ◊ {(b ◊ c ) a –
› _ _› _ ›
(a ◊ b ) c } = ____________________ 3/4
(a ◊ a ) (b ◊ c ) – (a ◊ b ) (a ◊ c ) = _______________________ 3/4
› _ _ › _ › _ ›
› _ _ › _ › _ ›
1(1 ◊ 1 ◊ cos (60°)) – (1 ◊ 1 ◊ cos (60°))2 = _____________________________ 3/4 1 1 __ – __ 2 4 __ 1 = _____ = 3/4 3
Hence, the value of (3 cos q + 2) = 1 + 2 = 3. 10. Clearly, l + ± m = ± n l 2 + m2 + n2 = 1 1__ fi l = m = n = ± ___ 3 ÷ Hence, 8 possible direction cosines and 4 lines are possible. – 3 z____ –4 x – 2 y_____ Since the lines _____ = = 1 1 – k – 4 z____ –5 x – 1 y_____ and _____ = = are coplanar, so 2 1 k
fi 11.
|
|
y2 – y1 z2 – z1 x 2 – x1 a c1 b1 1 a2 c2 b2
| | | | | |
–4 – 3 5 1 – 2 4 fi = 0 – k 1 1 2 k 1 – 1 1 1 fi 1 1 – k = 0 k 2 1
0 0 –1 fi 1 1 = 0 – k 2 k 2+k 1+k fi fi fi
– {2 (1 + k) + (k – 1) (k + 2)} = 0 2
– {2 + 2k + k + k – 2} = 0 k2 + 3k = 0
fi k = 0, – 3 Hence, the value of (k + 5) is 5 or 2. 9 __ – z – 3 2_____ x_____ – 2 y_____ 12. Let = = = l 1 2 k –4 x – 1 y_____ and _____ = = 2 k Thus, and
z __ = m 1
9 x + l + 2, y = 2l + 3, z = __ – k l 2 x = m + 1, y = 2m + 4, z = m
Solving, we get 1 l = 1, m = __ and k = 4 2 Hence, the value of k is 4. 13. Given planes are x = cy + bz y = cx + az z = bx + ay fi x – cy – bz = 0 cx – y + az = 0 bx + ay – z = 0
| |
Since these three planes pass through a line, so – c – b 1 c – 1 a = 0 b a – 1 fi fi fi fi
1 (1 – a2) + c (– c – ab) – b (ca + b) = 0 (1 – a2) – (c2 + abc) – (abc + b2) = 0 1 – a2 – b2 – c2 – 2abc = 0 a2 + b2 + c2 + 2abc = 1
a2 + b2 + c2 + 2abc + 2 = 1 + 2 = 3 – 2 z____ –k x – 4 y_____ 14. Since the line _____ = = lies in the plane 1 1 2 2x – 4y + z = 7 So, the point (4, 2, k) lies in the plane Thus, 2(4) – 4(2) + = 7 fi 8–8+k=7 fi k=7 15. Given plane is a x – b y + c z = 0 ....(i) and the line is – 2d z____ –c x – a y______ _____ = = c ...(ii) a b fi
Since the plane (i), contains the line (ii), so,
a (a) – b (b) + c (c) = 0
a (a) – b (2d) + c (c) = 0
fi
a2 + c2 = b2
a2 + c2 = 2bd
Clearly, b2 = 2bd
b = __ = 2 d
Questions asked in Past Iit-JEE Examinations
1. Since the line lies on the plane, so, the point (4, 2, k) lies on the plane.
Thus,
8–8+k=7
fi
k = 7.
3D-Co-ordinate Geometry
2. The equation of the plane passing through (2, 1, 0), and (4, 1, 1) is
|
|
– 1 z x – 2 y 1 = 0 3 – 1 0 1 2 fi fi
(x – 2) (– 1 – 0) – (y – 1) (3 – 2) + z (0 + 2) = 0 x + y – 2z = 3
3. The equation of the line PQ is
– 1 z____ –6 x – 2 y_____ _____ = = 1 2 – 2 Any point on the above line is
Q (l + 2, l + 1, 6 – 2l)
6.61
Let V1, the volume of the new box = r ◊ (b × c) 90 9 It is given that V1 = ____ V2 = ___ V2. 100 10 Thus, V1 = (r – 0 ◊ 9a) ◊ (b × c) fi r lies on a plane passing through 0 ◊ 9 a, and perpendicular to (b × c). 6. We take A = A¢ as the origin and B = B¢ = any point other than A on the line intersection of p1 and p2. Now, consider C = C¢ = any point niether on p1 nor p2. Thus, in this case, both the conditions of (a) and (b) are fullfilled.
The mid-point of PQ is
(
)
l l __ + 2, __ + 1, 6 – l 2 2 Since the mid-point of PQ lies on the plane, so,
(
) ( __ l2 + 1 ) – 2 (6 – l) = 3
l __ + 2 + 2
fi
3l = 12
fi
l = 4.
Thus, the co-ordinates of Q are (6, 5, – 2). 4. Since the given lines intersect, so they are coplanar.
|
|
+ 1 – 1 2 k 2 4 = 0 3 1 2 1 fi
– 10 – (k + 1) (2 – 4) – (4 – 3) = 0
fi
– 10 + 2 (k + 1) – 1 = 0
fi
2k = 9
fi
9 k = __ . 2
Similarly if we take, A = non-origin point on L1 B = non-origin point on the line of intersection of p1 and p2 and C = non-origin point on L2. If we take A = C≤, B = B¢, C = A¢, both the conditions of (a) and (b) are fulfilled.
5. Let AA¢ = a, AB = b, AD = d Volume of the parallelopiped T,
V = a ◊ (b × c)
Let
AA≤ = r
7. As the plane p parallel to b = (1, 0, – 1) and c = (– 1, 1, 0) normal to the plane is given by
| |
i j k b × c 1 0 = i + j + k – 1 –1 1 1
\ The equation of the plane ABC is
1 ◊ (x – 1) + 1 ◊ (y – 1) + 1 ◊ (z – 1) = 0
fi
x+y+z–3=0
6.62 Integral Calculus, 3D Geometry & Vector Booster fi
x+y+z=3 x y __z fi __ + __ + = 1. 3 3 3 This planes meets the axes in A (3, 0, 0), B (0, 3, 0), C (0, 0, 3). Thus, volume of the tetrahedron OABC 1 = __ [OA OB OC] 6 1 = __ [3i 3j 2k] 6 27 = ___ [i j k] 6 9 = __ 2 8. The equation of the plane passing through the line of intersection 2x – y + z = 3, 3x + y + z = 5 is (2x – y + z – 3) + l (3x + y + z – 5) = 0 ...(i) fi (2 + 3l) x + (l – 1) y + (l + 1) z = (5l + 3) = 0 Also, it is given that, the distance from (2, 1, – 1) to 1 the above plane is ___ __ . ÷6
|
|
2 (2 + 3l) + (l – 1) – (l + 1) – (5l + 3) _________________________ ____________________________________ ÷(2 + 3l)2 + (l – 1)2 + (l + 1)2 1__ = ___ ÷6 l–1 1 _____________ fi _______________ = ___ __ 2 ÷6 + 12l + 6 ÷ 11l
|
|
fi 6 (l – 1)2 = 11l2 + 12l + 6 fi 5l2 + 24l = 0 24 fi l = 0, – ___ . 5 24 Putting the values of l = 0, – ___ in Eq. (i), we get 5 the required equations of the planes are 2x – y + z = 3 or 9.
62x + 29y + 19z – 105 = 0
Equation of any plane passing through A, B, C X is __ a + Thus, centroid
Y Z __ + __ = 1 b c = G (x, y, z)
a b __ c = __ , __ , . 3 3 3
| ÷
(
)
Also, it is given that OM = 1
|
0 + 0 + 0 – 1 fi ____________ __________ = 1 1 1 1 __ 2 + __ 2 + __ 2 a b c 1 1 1 fi __2 + __ 2 + __ 2 = 1 a b c 1 1 1 fi ____ + ____ + ___ = 1 9 x2 9 y2 9 z2 1 1 1 fi __2 + __ 2 + __ 2 = 9 x y z 10. The equation of any plane passing through (1, – 2, 1) is
a (x – 1) + b (y + 2) + c (z – 1) = 0 ...(i)
The plane (i) is perpendicular to the planes So, and
2x – 2y + z = 0 and x – y + 2z = 4 2a – 2b + c = 0 a – a + 2c = 0
a b c Thus, ______ = _____ = ______ – 4 + 1 1 + 2 – 2 + 2 a b c fi ___ = __ = __ – 3 3 0 a b c fi __ = ___ = __ 1 – 1 0 Therefore, the equation of the plane is (x – 1) – (y + 2) = 0 fi x – y – 3 = 0. Thus, the distance of p from the point (1, 2, 2)
|
|
__ 1 – _____ 2–3 4 is ________ = ___ __ = 2÷2 . 1 + 1 ÷2 ÷
11. Given
k × (k × a) = 0
fi
(k ◊ a) k – (k ◊ k) a = 0
fi
g k – a = 0
fi
g k – (a i + b j + g k) = 0
fi
– (a i + b j) = 0
fi
– (a i + b j) = 0 ◊ i + 0 ◊ j
fi
a = 0, b = 0
Since the point (a, b, g) lies on the plane x + y + z = 2, so a+b+g=2
3D-Co-ordinate Geometry
fi
0+0+g=2
fi
g=2
12. The equation of any line passing through (0, 1, 0) and perpendicular to x + 2y + 2z = 0 is x y–1 __ = _____ = 1 2
z __ . 2
6.63
So 3 x – 6 y = 15 and 2 x + y = 5. On solving, we get x = 3 and y = – 1 Hence, the equation of the line is + 1 ___ z x – 3 y_____ _____ = = = t (say) 14 2 15 The parametric equations of the line are x = 14t + 3, y = 2t – 1, z = 15t. 15. The given system of equations ax + by + cz = 0, bx + cy + az = 0, and cx + ay + bz = 0 can be written as
( ) ( ) ( )
a b c x b c a y = c ab z Clearly
OA = 1,
2 2 ________ AM = _________ = __ 1 + 4 + 4 3 ÷
Thus,
4 5 OM 2 = OA2 – AM2 = 1 – __ = __ 9 9
fi
| |
__
÷5 OM = ___ . 3
0 0 0
AX = O a b c Now, |A| = b c a c ab
= – (a3 + b3 + c3 – 3abc)
= – (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(a) Let a + b + c = 0
13. The equation of any plane passing through (1, 2, 3) is
Then (a2 + b2 + c2 – ab – bc – ca) = 0
1 fi __ ((a – b)2 + (a – b)2 + (a – b)2) = 0 2 fi (a – b) = 0 = (b – c) = (c – a)
a (x – 1) + b (y – 2) + c (z – 3) = 0
which is perpendicular to x = 0 and y = 0 So,
a = 0, b = 0.
Thus, the equation of the plane is
c (z – 3) = 0
fi
(z – 3) = 0
Thus, the required distance from (0, – 1, 0) to the plane (z – 3) = 0 is
| |
0 –__3 _____ = 3. 1 ÷ 14. Let a, b, c be the direction ratios of the line of intersection of 3x – 6y – 2z = 15 and 2x + y – 2z = 5. Then
3a – 6b – 2c = 0
and
2a + b – 2c = 0
a b c Thus, ______ = ______ = ______ 12 + 2 – 4 + 6 3 + 12 a b c fi ___ = __ = ___ 14 2 15 Clearly, the vector 14i + 2 j + 15k is parrallel to the line of intersections of the plane. Now, we shall find the equation of the line put z = 0 in the given planes.
fi a = b = c Thus, the three system of equations are identical planes. i.e x + y + z = 0. (b) Suppose a + b + c = 0 and a2 + b2 + c2 π ab + bc + ca In this case, at least one of the following is true. a2 π bc, b2 π ca, c2 = ab Consider b2 π ac We can write the first two equations are
a x + b y = (a + b) z
and b x + c y = (b + c) z Eliminating y, we get (ac – b2) x = (ac – b2) z fi x = z Putting this in the first two equations, we get ax + b y + c x = 0 fi by = – (a + c) x
6.64 Integral Calculus, 3D Geometry & Vector Booster by = b x fi x = y Thus, x = y = z (c) In this case |A| π 0. So, the system of equations provide us a trivial solution. Thus, x = 0 = y = z. (d) Given a + b + c = 0, a2 + b2 + c2 = ab + bc + ca 1 fi a + b + c = 0, __ [(a – b)2 + (b – c)2] 2 + (c – a)2) = 0 fi a + b + c = 0, a = b = c fi a = 0 = b = 0. Thus, the system of equations represent the whole three-dimensional space. 16. Let l, m, n be the direction ratios of the line L1. Then l+m–n=0 and
l – 3m + 3n = 0 l m n Thus, _____ = ______ = ______ 3 – 3 –1 – 3 – 3 – 1 l m n fi __ = __ = __ 0 1 1 Thus, the direction ratios of L1 are (0, 1, 1) Similarly, the direction ratios of L2 and L3 are (0, 1, 1) and (0, 1, 1), respectively. Therefore, the lines L1, L2 and L3 are parallel. So, Statement I is false. Put z = 0 in P2: x + y – z = – 1, P3: x – 3y + 3z = 2 3 1 we get, x = – __ , y = – __ 4 4
(
)
1 3 Thus, any point on L1 is – __ , – __ , 0 . 4 4 So, the equation of the line L1 is + 3/4 __z x + 1/4 y______ ______ = = 0 1 1 which is parallel to the plane P1. 1 3 Since – __ , – __ , 0 does not lie on P1 , so no point 4 4 of L1 lies on P1.
(
)
Therefore, the three planes do not have a common point. Thus, the statement II is true. 17. (i) The unit vector perpendicular to both L1 and L2 u×v is ______ |u × v|
| |
Now,
i j k u × v = 1 1 2 = – i – 7j + 5k 1 2 3
Thus,
– i – 7j + 5k u×v __ ______ = __________ . |u × v| 5÷3
(ii) The shortest distance between L1 and L2
(r2 ◊ r1) ◊ (u × v) = _____________ |(u × v)|
Here, and fi
r2 = – i – 2j – k r1 = 2i – 2j + 3k r2 – r1 = 3i + 4k
|
|
– 3 +__20 17 Shortest distance = _______ = ____ __ . 5÷3 5÷3 (iii) The equation of any plane passing through (– 1, – 2, – 1) is Thus,
a (x + 1) + b (y + 2) + c (z + 1) = 0
...(i)
which is perpendiculat to the lines + 2 z_____ +1 x + 1 y_____ L1: _____ = = 3 1 2 + 2 z____ –3 x – 2 y_____ and L2: _____ = = 1 2 3 So,
3a + b + 2c = 0
fi
a + 2b + 3c = 0
a b c Therefore, _____ = _____ = _____ 3–4 2–9 6–1 a b ___ c __ fi = __ = . 1 7 – 5 Hence, the equation of the plane is
(x + 1) + 7 (y + 2) – 5 (z + 1) = 0
fi
x + 7y – 5z + 10 = 0.
Therefore, the required distance
|
|
1 +__________ 7 – 5 + 10 13 = ____________ = ____ __ . 5÷3 ÷1 + 49 + 25
18. Let the equation of the line passing through P (2, – 1, 2) and makes equal angles with the co-ordinate axes is + 1 z____ –2 x – 2 y_____ _____ = = 1 1 1 ___ ___ ___ __ __ __ 3 3 3 ÷ ÷ ÷ + 1 z____ –2 x – 2 y_____ fi _____ = = . 1 1 1 Any point on the above line is
Q (l + 2, + l – 1, l + 2)
Since the point Q lies on the plane, so fi
2 (l + 2) + (l – 1) + (l + 2) = 9
3D-Co-ordinate Geometry
fi
4l = 9 – 5 = 4
21. Given
fi
l = 1.
1 – ________ 4–2–a fi ___________ = 5 1 + 4 + 4 ÷
Thus, the point Q is (3, 0, 3). Therefore, the length of PQ
______________________ __ (3 – 2)2 (0 + 1)2 + (3 – 2)2 = ÷3
= ÷
19. The cartesian form of the given line is
Let
PQ = 5
|
|
| |
a+5 fi _____ = 5 3 fi fi
+ 1 x_____ x – 1 y_____ –2 _____ = = – 3 1 5
6.65
a + 5 = ± 15 a = – 5 ± 15 = 10, – 15
Since a is positive, so a = 10. The equation of the line PQ is
Q = (1 – 3l, l – 1, 5l + 2)
The direction ratios of PQ is ( –2 – 3l, – l – 3, 5l – 4) Since PQ is parallel to the given plane. So,
(– 2 – 3l) – 4 (l – 3) + 3 (5l – 4) = 0.
fi
1 l = __ 4
+ 2 z____ –1 x – 1 y_____ _____ = = . 1 2 – 2
1 Hence, the value of l is __ . 4 20. Let plane 1: ax + by + cz = 0 contains the line x y __z __ = __ = 2 3 4 Thus, 2a + 3b + 4c = 0 ...(i) Let plane 2: a; x + b¢y + c¢z = 0 is perpendicular to the plane containing the lines x __ = 3
y __ = 4
z x __ and __ = 2 4
y __ = 2
Let
Q = (l + 1, 2l – 2, 1 – 2l)
Also,
PQ = 5
fi
l + 1 + 2 (2l – 2) – 2 (1 – 2l) = a = 10
fi
9l = 10 + 5 15 5 fi l = ___ = __ 9 3 8 4 __ 7 Thus the point Q is __ , __ , – 3 3 3
(
z __ 3
)
Thus,
3a¢ + 4b¢ + 2c¢ = 0
fi
22. Let the equation of the any plane be
4a¢ + 2b¢ + 2c¢ = 0
ax + by + cz + d = 0 Since Eqs. (i) containing the lines, so
a¢ b¢ c¢ Hence, ______ = _____ = ______ 12 – 4 8 – 9 6 – 16
a¢ b¢ c¢ fi __ = ___ = ____ 8 – 1 – 10 Also, Plane 1 is perpendicular to Plane 2 So,
a ◊ a¢ + bb¢ c ◊ c¢ = 0
fi
8a – b – 10c = 0
From Eqs. (i) and (ii), we get a b c _______ = _______ = _______ – 30 + 4 32 + 20 –2 – 24 a b c fi ____ = ___ = ____ – 26 52 – 26 a b c fi __ = ___ = __ 1 – 2 1 Hence, the euation of the required plane is x – 2y + z = 0
...(i)
2a + 3b + 4c = 0
and ...(ii)
3a + 4b + 5c = 0. a b c Therefore, _______ = _______ = _____ 15 – 16 12 – 10 8 – 9 a b c fi ___ = __ = ___ – 1 2 – 1 a b c fi __ = ___ = __ 1 – 2 1 Also, a (x – 1) + b (y – 2) + c (z – 3) = 0
fi
(x – 1) – 2 (y – 2) + (z – 3) = 0
fi
x – 2y + z = 0
__
Again distance between the plane is ÷ 6 .
| |
__ d –__0 fi _____ = ÷ 6 ÷6
fi
|d| = 6.
6.66 Integral Calculus, 3D Geometry & Vector Booster 23. Let
and
– 1 z_____ +1 x – 2 y_____ L1: _____ = = 1 – 2 1 8 x – __ y + 3 –1 3 _____ z____ _____ L2: = = 2 – 1 1
Let the line be y z __ = __ ...(i) b c intersects the lines L1 and L2. So, the shortest distances between Eq. (i) and L1, and Eq. (i) and L2 are zero. For PQ and L1, x __ a =
| |
24. Given
1 9 7 [a b c] 8 2 7 = [0 0 0] 737
fi and
a + 8b + 7c = 0 9a + 2b + 3c = 0 a+b+c=0
Solving, we get b = 6 a, c = – 7 a Since the point P lies on the plane, so 2a + b + c = 1 fi 2a + 6a – 7a = 1 fi a=1 Thus, b = 6 and c = – 7 Now the value of 7a + b + c =7+6–7 = 6. 25.
| |
j i k a b c = (b + 2c) i + (c – a) j – (2a + b) k 1 – 2 1 Since PQ is perpendicular to L1 So,
2 (b + 2c) + (c – a) + (2a + b) = 0
fi
a + 3b + 5c = 2
| |
For PQ and L2
j i k a b c = (b + c) i – (a – 2c) j – (a + 2b) k 2 – 1 1 Since PQ is perpendicular to L2, 8 __ So, (b + c) + 3 (a – 2c) – (a + 2b) k 3 fi 3a + b – 5c = 0 ...(ii) Solving Eq. (i) and (ii), we get a b c _______ = ______ = _____ – 15 – 5 15 + 5 1 – 9
Hence, the required equation of the line be y z x __ = ___ = __ 5 – 5 2 Thus the point of intersections P and Q are
(
(
is (1, 4, 1). 1 13 __ , ___ 3 3
)
The direction ratios of PT is (2, 2, – 1) The angle between OR and PT is 9 2 +___ 8 –__1 ____ 1 cos q = ________ = __ = ___ __ 18 ÷ 9 9÷2 ÷2 ÷ p fi q = __ 4 and the length of PT
÷( ) ( ) ( ) _______________
a b c fi __ = ___ = __ 5 – 5 2
22 22 12 = __ + __ + __ = 1 3 3 3
\
p cos __ = 4
)
10 10 8 P = (5, – 5, 2), Q = ___ , – ___ , __ 3 3 3 Therefore, d 2 = PQ2 = 6.
The direction ratios of QR 4 The co-rdinates of P = __ , 3
PS 1 fi PS fi ___ ( ) ___ 1 ÷2 p TS 1 sin ( __ ) = ___ fi TS = ___ 4 1 ÷2 __
__
1 PS = TS = ___ __ . 2 ÷ 26. The equation of the required plane is P: (x + 2y + 3z – 2) + l (x – y + z – 3) = 0 Thus,
3D-Co-ordinate Geometry
fi (1+ l) x + (2 – l) y + (3 + l) z – (2 + 3l) = 0
Thus,
(2l – 2) + (– l – 1) + 3l = 3
2 Its distance from (3, 1, – 1) is ___ __ . 3 ÷
fi
4l = 6 3 l = __ 2
|
|
3(1 + l) + (2 – l) – (3 + l) – (2 + 3l) ________________________ fi _________________________________ + 1)2 + (2 – l)2 + (3 + l)2 ÷ (l
fi
(
)
5 9 So, the point P is 1, – __ , __ . 2 2 The foot of the perpendicular from the point (– 2, – 1, 2 __ = ___ 0) on the plane is the point Q (0, 1, 2). 3 ÷ 7 5 The direction ratio of PQ = 1, – __ , __ = (2, – 7, 5) 2 2
(
(– 2l)2 4 fi ____________ = __ 3l2 + 4l + 14 3
)
Hence, the equation of the line is
fi 4l + 14 = 0
–2 x y – 1 z____ __ = _____ = . – 7 2 5
7 fi l = – __ . 2 Hence, the required equation of the plane is
| |
29. The common perpendicular is along
z 17 5 11 – __ x + ___ y – __ + ___ = 0 2 2 2 2
i j k 1 2 2 = 2i + 3j – 2k 22 1
fi
– 5x + 11y – z + 17 = 0
Let
fi
5x – 11y + z = 17
2l – 3 – 3l + 1 2l –4 So, ______ = _______ = ______ 1 2 2
| |
27. Since given lines are coplanar, so we get 2 k 2 5 2 k = 0 2 0 0 fi fi
6.67
M = (2l, – 3l, 2l)
fi
l=1
Thus,
M = (2, – 3, 2)
Let the required point be P.
k2 – 4 = 0
Given,
k = ± 2
___
PM = ÷17
The equation of the plane containing these two lines is
fi (3 + 2s – 2)2 + (3 + 2s + 3)2 + (2 + s – 2)2 = 17
+ 1 z x – 1 y 2 = 0 k 2 k 2 5
fi (s + 2) (9s + 10) = 0 10 fi s = –2, – ___ 9 8 7 7 __ Thus, P = (–1, –1, 0) or __ , __ , 9 9 9 30. Given lines are – y z x – 5 _____ L1: _____ = = ___ 0 a – 3 – 2
|
|
fi (x – 1) (k2 – 4) – (y + 1) (2k – 10) + z (4 – 5k) = 0 fi – (y + 1) (2k – 10) + z (4 – 5k) = 0 fi – (y + 1) (± 4 – 10) + (4 – ± 10) = 0 Taking positive sign, we get
( + 1) – z = 0
fi y – z = – 1 Taking negative sign, we get
– (y + 1) (– 14) + 14z = 0
(y + 1) + z = 0
fi
fi
y + z = – 1
28. Any point P on the given line is (2l – 2, – l – 1, 3l) The point P lies on the given plane for some l
fi 9s2 + 28s + 20 = 0
(
and
x–a L2: _____ = 0
|
)
y z ___ = _____ – 1 2 – a
Since L1 and L2 are coplanar, so
|
5 – a 0 0 3 = 0 – a 0 – 2 2–a 0 –1 fi
(5 – a) (3 – a) (2 – a) – 2) = 0
fi
(a – 5) (a 2 – 5a + 4) = 0
fi
(a – 5) (a – 1) (a – 4) = 0
fi
a = 1, 4, 5.
6.68 Integral Calculus, 3D Geometry & Vector Booster 31. Ans. (A) y z+ 3 ___ = _____ – 1 1
x– 1 L1: _____ = 2
y + 3 z_____ +3 x – 4 _____ L2: _____ = = 1 1 2
| |
i j k Normal of plane P : n = 7 1 2 3 5 – 6
= – 16i + 48j + 32k
fi
n = i – 3 j – 2 k
The point of intersection of L1 and L2 are
2l + 1 = m + 4, – l = m – 3
1 = 3m – 2
m=1 Thus, the point of intersection is (5, – 2, – 1). The equation of the plane passing through (5, – 2, – 1) is a (x – 5) + b (y + 2) + c (z + 1) = 0 ...(i) Also, ax + by + cz = d is perpendicular with P1: 7x + y + 2z = 3, P2: 3x + 5y – 6z = 4 Thus,
7a + b + 2c = 0
and
3a + 5b – 6c = 0 a b c Therefore, _______ = ______ = ______ – 6 – 10 6 + 42 35 – 3 a b c fi ____ = ___ = ___ – 16 48 32 a b c fi __ = ___ = ___ 1 – 3 – 2 Now, from Eq. (i), we get
(x – 5) – 3 (y + 2) – 2 (z + 1) = 0.
fi
x – 3y – 2z = 13
Thus, a = 1, b = – 3, c = – 2, d = 13 32.
x L1: __ = 1 x L2: __ = 1
y z–1 __ = ____ = r, Q (r, r, 1) 1 0 y z+1 ___ = _____ = s, Q (s, – s, – 1,) – 1 0
fi
(l – 1) (l + 1) = 0 ( l = r)
fi
l = – 1, 1
For l = 1, the point P and Q are concides Hence, the value of l = – 1. 33. Let the required plane be x + z + ly – 1 = 0
| |
l______ –1 Now, _______ = 1 2 ÷ l + 2 fi
(l – 1)2 = l2 + 2
fi
– 2l + 1 = 2 1 l = – __ 2
fi Thus,
P3: 2x – y + 2z – 2 = 0
Distance of P3 from (a, b, g ) is 2.
|
|
2a – b + 2g – 2 ________ fi ____________ = 2 4 + 1 + 4 ÷ fi
2a – b + 2g – 2 = ± 6
fi
2a – b + 2g + 4 = 0 and 2a – b + 2g – 8 = 0
34. Line L will be parallel to the line of intersection of P1 and P2. Let a, b and c be the direction ratios of line L. Thus,
a + 2b – c = 0 and 2a – b + c = 0
fi a : b : c :: 1 : – 3, : – 5 The equation of the line L is – 0 z____ –0 x – 0 y_____ fi _____ = = 1 – 3 – 5 Again foot of the perpendicular from origin to plane 1 1 1 P1 is – __ , – __ , __ 6 3 6
(
)
The equation of the projection of line L on plane P1 is
1 1 1 x + __ y + __ z – __ 6 3 6 _____ = _____ = _____ = k 1 – 3 – 5
(
) (
)
5 2 1 1 1 Clearly points 0, __ , – __ and – __ , – __ , __ 6 3 6 3 6
PQ = (l + r) i + (l – r) j + (l – 1) k
satisfy the line of projection, i.e. M.
and
l – r + l – r = 0, as PQ is ^ t0 L1
fi
2l = 2r
fi
l=r
PR = (l – s) i + (l + s) j + (l + 1) k
35. Points O, P, Q, R, S are (0, 0, 0), (3, 0, 0), (3, 3, 0), 3 3 (0, 3, 0), __ , __ , 0 respectively. 2 2 1 The angle between OQ nad OS is cos– 1 ___ __ . 3 ÷
and l–s–l–s=0fis=0 Since PQ ^ PR, so (l – r) (l – s) + (l – r) (l + s) + (l – 1) (l + 1) = 0
(
)
( )
The equation of the plane containing the points O, Q and S is x–y=0
3D-Co-ordinate Geometry
The perpendicular distance from P (3, 0, 0) to the plane x – y = 0 is
| |
3 –__0 fi _____ = 2 ÷
3 ___ __ 2 ÷
line RS x y–3 __ = _____ = 1 – 1
x – y + z = 3 is obtained by – 1 z____ – 1 + 7 – 3) – 7 – 2 (3 x – 3 y_____ _____ = = = _______________ 1 – 1 1 3 i.e.
and the perpendicular distance from O (0, 0, 0) to the
÷
___
z 15 __ is ___ . 2 2
36. The mirrior image of (3, 1, 7) w.r.t. the plane
6.69
P = (– 1, 5, 3)
The equation of the plane passing through line and
| |
P = (– 1, 5, 3) is x y z _› n = – 1 5 3 = 0] 1 21 fi
x – 4y + 7z = 0
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