Integral calculus vinay kumar VKR classes kota(1)

for


ABOUT THE AUTHOR
Vinay Kumar (VKR) graduated from IIT Delhi in
Mechanical Engineering.
Presently, he trains IIT aspirants at VKR Classes,
Kota, Rajasthan.


for

Second Edition

Vinay Kumar
B.Tech., IIT Delhi

McGraw Hill Education (India) Private Limited
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Integral Calculus for JEE Main & Advanced, 2/e
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PREFACE

T

his book is meant for students who aspire to join the Indian Institute of Technologies (IITs) and
various other engineering institutes through the JEE Main and Advanced examinations. The content
has been devised to cover the syllabi of JEE and other engineering entrance examinations on the topic
Integral Calculus. The book will serve as a text book as well as practice problem book for these competitive examinations.
As a tutor with more than thirteen years of teaching this topic in the coaching institutes of Kota,
I have realised the need for a comprehensive textbook in this subject.
I am grateful to McGraw-Hill Education for providing me an opportunity to translate my years of
teaching experience into a comprehensive textbook on this subject.
This book will help to develop a deep understanding of Integral Calculus through concise theory
and problem solving. The detailed table of contents will enable teachers and students to easily access
their topics of interest.
Each chapter is divided into several segments. Each segment contains theory with illustrative examples. It is followed by Concept Problems and Practice Problems, which will help students assess the
basic concepts. At the end of the theory portion, a collection of Target Problems have been given to
develop mastery over the chapter.
The problems for JEE Advanced have been clearly indicated in each chapter.
The collection of objective type questions will help in a thorough revision of the chapter. The
Review Exercises contain problems of a moderate level while the Target Exercises will assess the students’
ability to solve tougher problems. For teachers, this book could be quite helpful as it provides numerous
problems graded by difficulty level which can be given to students as assignments.
I am thankful to all teachers who have motivated me and have given their valuable recommendations. I thank my family for their whole-hearted support in writing this book. I specially thank Mr. Devendra
Kumar and Mr. S. Suman for their co-operation in bringing this book.
Suggestions for improvement are always welcomed and shall be gratefully acknowledged.
Vinay Kumar



CONTENT

About the Author

ii

Preface

v

CHAPTER 1 INDEFINITE INTEGRATION

1.1 – 1.184

1.1

Introduction

1.1

1.2

Elementary Integrals

1.4

1.3

Integration by Transformation

1.10

1.4

Integration by Substitution

1.16

1.5

Integrals Involving Sine and Cosine

1.27

1.6

Rationalization by Trigonometric Substitution

1.36

1.7

Integrals of the Form

1.40

1.8

Integrals of the Form

1.45

1.9

Integrals of the Form

1.50

1.10

Integration of Trigonometric Functions

1.55

1.11

Integration by Parts

1.65

1.12

Special Integrals

1.73

1.13

Multiple Integration by Parts

1.76

1.14

Integration by Reduction Formulae

1.81

1.15

Integration of Rational Functions Using Partial Fractions

1.88

1.16

Special Methods For Integration of Rational Functions

1.101

1.17

Integration of Irrational Functions

1.106

1.18

Integrals of the Type 

1.19

Integration of a Binomial Differential

1.118

1.20

Euler’s Substitution

1.120

1.21

Method of Undetermined Coefficients

1.124

dx
P Q

1.112


viii | Content

1.22

Non-elementary Integrals

1.127

Target Problems for JEE Advanced

1.130

Things to Remember

1.142

Objective Exercises

1.147

Review Exercises for JEE Advanced

1.160

Target Exercises for JEE Advanced

1.162

Previous Year’s Questions (JEE Advanced)

1.164

Answers

1.166

CHAPTER 2 DEFINITE INTEGRATION

2.1 – 2.204

2.1

Introduction

2.1

2.2

Definite Integral as a Limit of Sum

2.5

2.3

Rules of Definite Integration

2.12

2.4

First Fundamental Theorem of Calculus

2.19

2.5

Second Fundamental Theorem of Calculus

2.27

2.6

Integrability

2.41

2.7

Improper Integral

2.52

2.8

Substitution in Definite Integrals

2.63

2.9

Integration by parts for Definite Integrals

2.72

2.10

Reduction Formula

2.78

2.11

Evaluation of Limit of sum using Newton-leibnitz Formula

2.83

2.12

Leibnitz Rule for Differentiation of Integrals

2.91

2.13

Properties of Definite Integral

2.95

2.14

Additional Properties

2.122

2.15

Estimation of Definite Integrals

2.124

2.16

Determination of Function

2.134

2.17

Wallis’ Formula

2.138

2.18

Limit under the sign of Integral

2.143

2.19

Differentiation under the sign of Integral

2.144

2.20

Integration of Infinite Series

2.148

2.21

Approximation of Definite Integrals

2.151

Target Problems for JEE Advanced

2.154

Things to Remember

2.166

Objective Exercises

2.169

Review Exercises for JEE Advanced

2.180

Target Exercises for JEE Advanced

2.184


Content | ix

Previous Year’s Questions (JEE Advanced)

2.188

Answers

2.194

CHAPTER 3 AREA UNDER THE CURVE

3.1 – 3.86

3.1

Curve Sketching

3.1

3.2

Area of a Curvilinear Trapezoid

3.7

3.3

Area Bounded by a Function which Changes Sign

3.10

3.4

Area of a Region Between two Non-intersecting Graphs

3.13

3.5

Area of a Region Between two intersecting Graphs

3.17

3.6

Area by Horizontal Strips

3.21

3.7

Area of a Region Between Several Graphs

3.26

3.8

Determination of Parameters

3.30

3.9

Shifting of Origin

3.35

3.10

Area Bounded by a Closed Curve

3.37

3.11

Areas of Curves given by Parametric Equations

3.42

3.12

Areas of Curves given by Polar Equations

3.44

3.13

Areas of Regions given by Inequalities

3.46

Target Problems for JEE Advanced

3.51

Things to Remember

3.62

Objective Exercises

3.64

Review Exercises for JEE Advanced

3.74

Target Exercises for JEE Advanced

3.76

Previous Year’s Questions (JEE Advanced)

3.78

Answers

3.80

CHAPTER 4 DIFFERENTIAL EQUATIONS

4.1 – 4.99

4.1

Introduction

4.1

4.2

Formation of a Differential Equation

4.3

4.3

Solution of a Differential Equation

4.7

4.4

First Order and First Degree Differential Equations

4.11

4.5

Reducible to Variable Separable

4.17

4.6

Homogeneous Differential Equations

4.23

4.7

Linear Differential Equations

4.30

4.8

Solution by Inspection

4.38

4.9

First Order Higher Degree Differential Equation

4.42


x

| Content

4.10

Higher Order Differential Equation

4.46

4.11

Integral Equation

4.50

4.12

Problems in Trajectories

4.53

4.13

Applications of Differential Equation

4.55

Target Problems for JEE Advanced

4.61

Things to Remember

4.72

Objective Exercises

4.73

Review Exercises for JEE Advanced

4.84

Target Exercises for JEE Advanced

4.86

Previous Year’s Questions (JEE Advanced)

4.89

Answers

4.92


C H A P T E R

1

INDEFINITE

INTEGRATION

1.1 INTRODUCTION
Integral calculus is to find a function of a single variable
when its derivative f(x) and one of its values are
known. The process of determining the function
has two steps. The first is to find a formula that
gives us all the functions that could possibly have
f as a derivative. These functions are the so-called
antiderivatives of f, and the formula that gives them
all is called the indefinite integral of f. The second
step is to use the known function value to select
the particular antiderivative we want from the
indefinite integral.
A physicist who knows the velocity of a particle might
wish to know its position at a given time. Suppose an
engineer who can measure the variable rate at which
water is leaking from a tank wants to know the amount
leaked over a certain time period. In each case, the
problem is to find a function F whose derivative is a
known function f. If such a function F exists, it is called
an antiderivative of f.
There are two distinct ways in which we may approach
the problem of integration. In the first way we regard
integration as the reverse of differentiation; this is the
approach via the indefinite integral. In the second way
we regard integration as the limit of an algebraic
summation; this is the approach via the definite
integral. For the moment we shall consider only the
first and we begin with the formal definition.

Indefinite integration is the process which is the
inverse of differentiation, and the objective can be
stated as follows : given a function y = f(x) of a single
real variable x, there is a definite process whereby we
can find (if it exists) the function F(x) such that
dF(x)
 f(x)
dx

Definition
A function F(x) is called the antiderivative (primitive)
of the function f(x) on the interval [a, b] if at all points
of the interval F'(x) = f(x).
Find the antiderivative of the function f(x) = x2.
From the definition of an antiderivative it follows that
the function f(x)=

 x3  
 3  = x2.
 

x3
is an antiderivative, since
3

It is easy to see that if for the given function f(x) there
exists an antiderivative, then this antiderivative is not
the only one. In the foregoing example, we could take
the following functions as antiderivatives:

x3
x3
– 7 or,
 1, F(x) =
3
3
x3
generally, F(x) =
+C
3
(where C is an arbitrary constant), since

F(x) =


1.2 

INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

 x3

2
 3  C  x
On the other hand, it may be proved that functions of
x3
 C exhaust all antiderivatives of the
the form
3
function x2. This is a consequence of the following

theorem. The example shows that a function has
infinitely many antiderivatives. We are going to show
how to find all antiderivatives of a given function,
knowing one of them.

Constant Difference Theorem
If F1(x) and F2(x) are two antiderivatives of a function
f(x) on an interval [a, b], then the difference between
them is a constant.
Proof By virtue of the definition of an anti-

F1 (x) 
F2 (x) 

f(x)

...(1)
derivative we have
f(x)
for any value of x on the interval [a, b].
Let us put F1(x) – F2(x) =g(x)
...(2)
Then by (1) we have
F1 ( x ) – F2 ( x ) = f(x) – f(x) = 0
or g(x) = [F1(x) – F2(x)]' = 0
for any value of x on the interval [a, b].
But from g(x) = 0 it follows that g(x) is a constant.
Indeed, let us apply the Lagrange's theorem to the
function g(x), which, obviously, is continuous and
differentiable on the interval [a, b]. No matter what the

point x on the interval [a, b], we have, by virtue of the
Lagrange's theorem, g(x) – g(a) = (x – a) g(c) where
a < c < x.
Since g(c) = 0,
g(x) – g(a) = 0
or, g(x) = g(a)
...(3)
Thus, the function g(x) at any point x of the interval
[a, b] retains the value g(a), and this means that the
function g(x) is constant on [a, b]. Denoting the
constant g(a) by C, we get from (2) and (3),
F1(x) – F2(x) = C.
Thus, if for the function F1 and F2 there exists an interval
[a, b] such that F1'(x) = F2'(x) in [a, b], then there exists a
number C such that F1(x) = F2(x) + C in [a, b].
From this theorem it follows that if for a given function
f(x) some one antiderivative F(x) is found, then any
other antiderivative of f(x) has the form F(x) + C, where
C is an arbitrary constant.
For example, if
F '(x) = 6x2 + 2x, then
F(x) = 2x3 + x2 + C,

for some number C. There are no other antiderivatives
of 6x2 + 2x.
If F '(x) = 4x – 3 and F(1) = 3,
then
F(x) = 2x2 – 3x + C, for some number C.
Since, F(1) = 2 – 3 + C = 3, we have C = 4.
Thus F(x) = 2x2 – 3x + 4,

and there is just one function F satisfying the given
conditions.
The equation F '(x) = 4x – 3 is an example of a differential
equation, and the condition F(1) = 3 is called a
boundary condition of F.
Given F ' and a boundary condition on F, there is a
unique antiderivative F of F ' satisfying the given
boundary condition. This function F is called the
solution of the given differential equation.

Definition
If the function F(x) is an antiderivative of f(x), then the
expression F(x) + C is the indefinite integral of the
function f(x) and is denoted by the symbol f(x) dx. It
is the set of all antiderivatives of f(x). Thus, by
definition
f(x) dx = F(x) + C,
if F(x) = f(x).
Here, the function f(x) is called the integrand, f(x) dx
is the element of integration (the expression under
the integral sign), the variable x the variable of
integration, and  is the integral sign.
Thus, an indefinite integral is a family of
functions y = F(x) + C
(one antiderivative for each value of the constant C).
The symbol  for the integral was introduced by
Leibnitz. This elongated S stood for a "sum" in his
notation.
The graph of an antiderivative of a function f(x) is
called an integral curve of the function y = f(x). It is

obvious that any integral curve can be obtained by a
translation (parallel displacement) of any other integral
curve in the vertical direction.
From the geometrical point of view, an indefinite integral
is a collection (family) of curves, each of which is obtained
by translating one of the curves parallel to itself upwards
or downwards (that is, along the y-axis).


INDEFINITE INTEGRATION

Existence of Antiderivative
A natural question arises : do antiderivatives (and, hence,
indefinite integrals) exist for every function f(x) ?
The answer is no.
Let us find an antiderivative of a continuous function
 x 1
f(x) = 
2
3  x

, 0  x 1
, on the interval [0, 2].
, 1 x  2

On integrating both the formulae we get
 x2
0  x 1
 2  x  C1 ,


F(x) = 
x3
 C2 ,
3x 
1 x  2

3
To ensure that F '(1) = f(1) = 2, we first make F(x)
continuous :
F(1–) = F(1+)
1
1
+ 1 + C1 = 3 – + C2
3
2
8
3
 C1 +
= + C2
3
2
7
 C 1 = + C2
6
7
 x2
0  x 1
 2  x  6  C2 ,

Now, F(x) = 

x3
3x 
1 x  2
 C2 ,

3
We further observe that F (1+) = F (1–).
Thus, we obtain the antiderivative F(x) of the function f(x).
Let us note, however, without proof, that if a function
f(x) is continuous on an interval [a, b], then this function
has an antiderivative (and, hence, there is also an
indefinite integral).
Now, let us find an antiderivative of a discontinuous



1
function f(x) = 2 .
x
1
1
F(x) = – is an antiderivative of f(x)= 2 on (– , 0)
x
x

and on (0, ). However, it is not an an antiderivative
on [–1, 1] since the interval includes 0 where F(x) does
not exist.
We adopt the convention that when a formula for a
general indefinite integral is given, it is valid only on

an interval. Thus, we write

1

 1.3

1
+C
x
with the understanding that it is valid on the interval
(0, ) or on the interval (– , 0). This is true despite
the fact that the general antiderivative of the function
f(x) = 1/x2, x  0, is

x

2

dx  

  1  C if x  0
1
 x

.
F(x) =
  1  C if x  0
2
 x
Example 1. Prove that y = sgn x does not have an

antiderivative on any interval which contains 0.

 1 , x  0

0 , x0
Solution y = sgn (x) = 
1 , x0
Here we present two antiderivatives :
  x  c1 ,
f(x) =  x  c ,

2

x0
x0

x  c , x  0
g(x) = 
 xc , x0
f(x) is discontinuous at x = 0, if c1  c2. g(x) is continuous
at x = 0. None of these functions are differentiable at
x = 0 i.e. we are unable to ensure that f '(0) or g'(0) is 0.
Hence, y = sgn x does not have an antiderivative on
any interval which contains 0. However, the function
has an antiderivative (either f or g) on any interval
which does not contain 0.

Indefinite Integration
The finding of an antiderivative of a given function
f(x) is called indefinite integration of the function f(x).

Thus, the problem of indefinite integration is to find
the function F(x) whose derivative is the given function
f(x) i.e. given the equation

dF(x)
 f (x) , we have to
dx

find the function F(x).
The process of integration is not so simple. Although
rules may be given which cover this operation with
various types of simple functions, indefinite
integration is a tentative process, and indefinite
integrals are found by trial.
This chapter is devoted to working out methods by
means of which we can find antiderivatives (and
indefinite integrals) for certain classes of elementary
functions.


1.4



INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

Note: Though the derivative of an elementary

function is always an elementary function, the
antiderivative of the elementary function may not

prove to be representable by a finite number of
elementary functions. We shall return to this question
at the end of the chapter.

1.2

ELEMENTARY
INTEGRALS

(ii)



  f (x) dx  = kf (x)

A constant moves past the derivative symbol.
(b) We show that the derivative of f(x) +  g(x) dx
is f(x) + g(x) :

d
dx

  f (x) dx   g(x) dx 

d
d
f (x) dx 
g(x) dx = f (x) + g (x).
dx 
dx 

(c) The proof of property (c) is similar to that of
property (b).
The last two parts of theorem extend to any finite
number of functions. For instance,

=









dx = x2 ex – 2(xex – ex + C)

d
dx

(a)

Assume that f and g are functions with antiderivatives
f(x) and  g(x) dx. Then the following hold :
(a) kf(x) dx = k  f(x) dx for any constant k.
(b) (f(x) + g(x)) dx =  f(x) dx +  g(x) dx,
(c) (f(x) – g(x)) dx =  f(x) dx +  g(x) dx.
Proof
(a) It is only required to show that the derivative
of k f(x)dx is cf(x). The differentiation shows :



2 x

= x2ex – 2xex + 2ex – 2C = (x2 – 2x + 2) ex + C
where C = 2C
Wh en a rbit rary consta nts are algebra ical ly
combined with other numbers, the final algebraic
expression is just as arbitrary.
(iii) Note the following representations :



Rules of integration

d
d
k f (x) dx  k
dx 
dx

x e

  f (x)dx   f (x)


(b) d  f (x)dx  f (x ) dx
(c)

 f '(x)dx  f(x)  C

(d)

 df (x)  f (x)  C

Elementary formulae
We begin by listing a number of standard forms, that
is to say formulae for integrals which we shall be free
to quote once we have listed them. Each formula is of
the the type

 f (x)dx  F(x)

and its validity can be etablished by showing that

d F(x )  f (x )
dx

These formulae should be known and quoted, without
proof, whenever needed.
(i)

 dx = kx + C, where k is a constant

(ii)

n
 x dx =

(iii)

 x dx = ln |x| + C

(i) We have  (1  x)dx   1dx   xdx = (x + C1) +

(iv)

x
 a dx =

2
 x2

 C2  = x + x + C + C


1
2
2
 2

Here, we have two arbitrary constants when one
will suffice. This kind of problem is caused by
introducing constants of integration too soon and
can be avoided by inserting the constant of
integration in the final result, rather than in
intermediate computations.

 e dx = e + C

(vi)  sin x dx = – cos x + C
(vii)  cos x dx = sin x + C
(viii)  sec x dx = tan x + C

 (f (x) – g(x)  h(x)) dx
=  f (x) dx–  g(x) dx +  h(x) dx.
Note:

(v)

x n 1
+ C, where n ¹ 1
n 1

1

x

ax
 C , where a > 0
n a
x

2


 1.5

INDEFINITE INTEGRATION
(ix)

(x)
(xi)
(xi)

 cosec x dx = – cot x + C
 sec x tan x dx = sec x + C
 cosec x cot x dx = – cosec x + C
2



1
1  x2

dx = sin x + C

1

|x|

x2  1

1

 cot x dx = a ln sin|x|+ C

(xvi)  sec x dx = ln |sec x + tan x| + C
 x
 +C
4 2

= – ln |sec x – tan x| + C

(xvii)  cosec x dx = n |cosec x  cot x| + c

1

 x dx   n x

x
+c
2

Direct integration is such a method of computing
integrals in which they are reduced to the elementary
formulae by applying to them the principal properties
of indefinite integrals.
For example :
dx

 x  3  ln | x  3 | C
x 2x
x
x
 3 .4 dx   (3.16) dx   48 dx 

1
 x dx   n x  C
We have two cases :
(i) Let x > 0, then |x| = x and the formula attains the

 4x

dx
 nx  C
x

Differentiating, we get (nx + C)' =



1
.
x

dx
= n(–x) + C
x

5

dx =

=

x

3

48x
C

n48

4 6
2 6
x +C=
x + C.
6
3
Evaluate

 x 3  5x 2  4  7  2  dx


x
x



(ii) Let x < 0, then |x| = –x and the formula has the form



C.

In the formula and examples where integrals of this
type occurs, i.e., where the value of an integral involves
the logarithm of a function and the function may
become negative for some values of the variable of
the function, the absolute value sign enclosing the
function should be given, but it has generally been

omitted, though it is always understood to be present
and it should be supplied by the students.

= n |cosec x + cot x|
Here, we must analyse carefully the formula



1

 x dx  ln( x)  C .

Direct integration

= ln tan

form

d
1 1
 .
n(–x) =
dx
x x

Hence, both these results will be included if we write

1
(xiv)  tan x dx = ln sec |x| + C
a

= n tan

1
d
(n x) = ,
dx
x

1

Therefore when x < 0,

dx = sec–1 x + C

1
1

x
x

 x dx  ln x  C is defined for x > 0.

When x < 0, i.e. – x > 0,

We have some additional results which will be
established later :

(xv)

Since ln x is real when x > 0 and
so

–1

1
dx = tan–1 x + C
(xii) 
1  x2
(xiii)

Differentiating, we have (n(–x) + C)' =

 x 3  5x 2  4  7  2  dx


x
x

dx +  5x 2 dx –  4dx +



7
dx +
x



2

x

dx


1.6 
=

INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

x

3

dx + 5 .

+ 7. 

x

2

dx – 4 .  1 . dx

=

1
dx + 2.  x 1/ 2 dx
x

Extension of elementary formulae

 x1/ 2 
x4
x3
=
+5
– 4x + 7 ln | x | + 2  1/ 2  + C


4
3
5
x4
=
+ x3 – 4x + 7 ln | x | + 4 x + C
3
4



Evaluate


=



2 x  3x

x

5x

2 x  3x
5x

2
3 
 5x  5x  dx =

x

x

  2
 3 
       dx
5
5 




(2 / 5) x (3 / 5) x

=
+C
ln 2 / 5 ln 3 / 5
Evaluate

(i)

5

(ii)

log x
 2 4 dx

 f (x)dx  F(x) and a, b are constants, then

If

1

 f (ax  b)dx  a F(ax  b)

loge x
dx

We prove (1) by observing that, when y = ax + b,

log x
log 5
 5 e dx =  x e dx =

(ii)

2

log 4 x

dx =  2

=  2log2

x

dx =

log 2 x
2
dx



x loge 5  1
+C
log e 5  1
=  21/ 2 log2 x dx

x dx =

x 3/ 2
+C
3/ 2

2 3/ 2
x C
3

Evaluate  e x ln a  ea ln x  ea ln a dx

  2log 4 x dx 

e
x

x ln a

e

a ln x

a

e

a ln a

dx

a

=  eln a  eln x  eln a dx
=  (a x  x a  a a ) dx
=

a

x

dx +

x

a

dx +  a a dx

(ax  b) n 1
+ C, n  1
a (n  1)

(i)



(ii)

 sin(5x  2)dx   5 cos(5x  2) + C

(iii)

 sec

(iv)

 sec (ax + b) . tan (ax + b) dx

(ax + b)n dx =

1

=
(i)

...(1)

1 d
1 d
dy 1
F(y) 
F(y). = f(y). a = f(ax + b).
a dx
a dy
dx a
For example:

dx

dx

x

ax
x a 1
+
+ aa. x + C.
ln a a  1

2

(3x  5)dx 

1
tan(3x  5)  C
3

1
sec (ax + b) + C
a

(v) cos 7x + sin (2x – 6)) dx
1
= 1 sin 7x – cos (2x – 6) + C
2
7
dx
1
 ln 2x  1  C
(vi) 
2x  1 2
x
1
2x
1
1 
dx  
dx   1 
(vii) 

 dx

2x  1
2 2x  1
2
2x  1
1
1
x  ln 2x  1 + C
=
2
2
In this example, we break the function up into parts





1
whose integrals we know from the
2x  1
list of elementary integrals.

like 1 and

(viii)

dx

 25  4x

2



dx
1
dx
 
25
4  5 2
4   x2 
 x2
 4

 2


 1.7

INDEFINITE INTEGRATION

From the standard result we obtain,
 1 sec –1 x  C
,xa
a
 a
1

 x x 2  a 2 dx =  1 –1 x

– sec a  C , x  a
 a

1 1
x
1
2x
C
. tan 1  C 
tan 1
5
5
4
10
5
=
2
2

(ix)

(x)

a

=

(i)

2

dx
dx
1
bx
tan 1
 2

2 2
2
ab
a +C
a  (bx)
b x
dx

1  9x

2



dx
1
9   x2 
9




dx
3

1
 x2
9

2 / 3

1
dx
1
x
1
 sin 1  C  sin 1 3x  C
1
3  1 2
3
3
 x2
3
 3

dx is a 'convenient form' for

 x2

1

 x 2 dx and such

symbols are commonly used. Strictly, the first
symbol has no meaning save as a shorthand for
the second symbol; as the definition shows, there
can be no question here of 'dividing dx by x2'.
(ii) In the chapter of indefinite integration, the
simplification of square root functions are done
without much consideration to the sign of the
expressions. However, this must be taken
seriously in the chapter of definite integration.
For example,



=  (sin x  cos x)2 dx
=  (sin x + cos x) dx
= – cos x + sin x + C
A more elegant way of handling the situation is
illustrated below :



1  sin 2x dx =



(sin x  cos x)2 dx

=  sin x  cos x dx
= sgn (sin x + cos x) ·  (sinx + cos x) dx

= sgn (sin x + cos x) · {– cos x + sin x} + C.
(iii) We notice that the integral of
where a > 0 is

1

2 / 3

1
x
sec 1   + C. But we would
 a
a

like to know the integral of

1
x x  a2
2

.

1



dx = – sec–1 x

x x2 1

2

2 / 3
2

 5  2    

.
6
3
6

=  

Alternatively, we have
dx

x

x a
2

2

1
x
sec 1  C
a
a



Thus,
 2

dx

2 / 3

x x 1
2

 sec 1 | x |

 2


2 / 3

  
 
.
4 6 12

Determination of function
Let f be a polynomial function such
that for all real x, f(x2 + 1) = x4 + 5x2 + 3 then find the
primitive of f (x) w.r.t. x.
f (x2 + 1) = (x2 + 1)2 + 3x2 + 2
= (x2 + 1)2 + 3(x2 + 1) – 1

We replace x2 + 1 by x
 f (x) = x2 + 3x – 1
Now we integrate f(x) w.r.t. x :

 f (x)dx   (x
,

2 / 3
2

 5  2    
= 

6
3 6
This is a wrong result since the integral of a negative
function must be negative. This happened because
the antiderivative used in the calculation is wrong.

3

| x | x2  a2

dx = sec–1 x

x x2 1

2



1  sin 2x dx

1



2

2

 3x  1) dx

x
3x

xC
3
2
 3 
Given f ''(x) = cos x, f'   = e and
2
f(0) = 1, then find f (x).
f''(x) = cos x
Integrating w.r.t. x :

=


1.8



INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

f(x) = sin x + C1
 3 
Now, f '   = e
2
 e=–1+C
 C1 = e + 1
f ' (x) = sin x + e + 1
Integrating again w.r.t. x :
f (x) = – cos x + (e + 1)x + C2
f (0) = 1  1 = – 1 + C2  C2 = 2
 f (x) = (e + 1)x – cos x + 2
Example 7. A curve y = f (x) such that f ''(x) = 4x
at each point (x, y) on it and crosses the x-axis at
(–2, 0) at an angle of 45°. Find the value of f (1).
Solution f ''(x) = 4x f'(x) = 2x2 + c
f ' (–2) = tan 45° = 1 = 8 + c  c = – 7
Now
f ' (x) = 2x2 – 7
2 
 f (x) = x  7x  d
3
16
Since, f (–2) = 0 –
+ 14 + d = 0
3

26
 d=–
3
1
 f (x) = [2x3 – 21x – 26]
3
45
= – 15.
And, thus
f (1) = –
3
Example 8. Find the antiderivatives of the
function y = x + 2 which touch the curve y = x2.

Solution Since the function y = x + 2 is a derivative
of any of its antiderivatives, it follows, that the
equation for finding the abscissa of the point of
tangency has the form 2x = x + 2.
The root of this equation is x = 2. The value of the
function y = x2 at the point x = 2 is equal to 4.
Consequently, among all the antiderivatives of the
functions y = x + 2,
1
i.e. the function F(x) = x2 + 2x + C,
2
we must find that whose graph passes through the
point P(2, 4). The constant C can be found from the
condition F(2) = 4 :
1


. 4 + 2.2 + C = 4  C = – 2.
2
1
 F(x) = x2 + 2x – 2.
2
Example 9. Deduce the expansion for tan–1x from
1
= 1 – x2 + x4 – x5 + ... when x < 1,
1 x2
Solution We have
1
= 1 – x2 + x4 – x5 + .....
1 x2
Integrating both sides w.r.t. x, we have

the formula

dx
x3 x5 x7


 ...
x
–
–
3
5
7
1 x2
No constant is added since tan–1x vanishes with x.

tan–1x =



A
1.

Find an antiderivative of the function :
(i) f(x) = 1 – 4x + 9x2
(ii) f(x) = x x + x – 5

3.

 x,

(iii) f(x)  4 x  1
(iv) f(x) = (x/2 – 7)3.
2.

1
3
(3x + 4)2 and G(x)= x2 + 4x
6
2
differ by a constant by showing that they
are antiderivatives of the same function.
(b) Find the constant C such that F(x) – G(x) = C by
evaluating F(x) and G(x) at a particular value
of x.

(c) Check your answer in part (b) by simplifying
the expression F(x) – G(x) algebraically.

(a) Show that F(x) =

Let F and G be defined piecewise as

4.

5.

x 0

 x  2,

x0

 x 2,

x0
x0

and G(x) =    3
F(x) = x,
x 0
 x , x0

(a) Sh ow t h a t F a n d G h a ve t h e sa m e
derivative.
(b) Show that G(x)  F(x) + C for any constant C.

(c) Do parts (a) and (b) violate the constant
difference theorem ? Explain.
(a) Graph some representative integral curves
of the function f(x) = ex/2.
(b) Find an equation for the integral curve that
passes through the point (0, 1).
Prove that the following functions donot have an
antiderivative on any interval which contains 0.
 x  1, x  0
x0

(i) y =  x,


(ii) y =  1,



 1.9

INDEFINITE INTEGRATION
6.

7.
8.

9.

Find the function satisfying the given equation
and the boundary conditions.

(i) F'(x) = 3(x + 2)3, F(0) = 0
(ii) s"(t) = 8, s'(0) = 7, s(–1) = –3
(iii) f '(x) = x2 + 5, f(0) = –1.
If f ''(x) = 10 and f ' (1) = 6 and f (1) = 4 then find
f (–1).
Evaluate the following integrals :
e2x  1
(i)  2x . ex dx
(ii)
dx
ex
ax  b
px  q
dx
dx
(iii) e
(iv) a
Evaluate the following integrals :






 e dx
 n  e1  dx
n x

(i)

(ii)

e

 n x 2

dx



(iv) e m ln x dx
(iii)
x
10. Evaluate the following integrals :

a
e
e










mx

.b nx dx
(ii) (2x + 3x)2 dx
3x
5x
e
n 2  n x
(iii)
(iv) e
dx
x
 x dx
e
11. Evaluate the following integrals :
dx
n
(ii) (ax + b) dx
(i)
2· x
dx
(1+ x)3
(iii)
(iv)
dx
3  2x
x
(i)

12. Evaluate the following integrals :
x dx
2x  1

dx
(ii)
(i)
a  bx
x  2 dx
2
x
x4
dx
(iv)
dx
(iii)
1  x2
1  x2
13. Evaluate the following integrals :
(i)









 cos xº dx
 cot x dx

 sec (ax  b) dx
1  cos x

(iv)  1  cos x dx
2

(ii)

2

(iii)
14. Evaluate the following integrals :
dx
dx
(i)
(ii)
2
(x  1) x 2  2x
x 25x  2
dx
(iii)
(2x  1) (2x  1)2  4
15. Evaluate the following integrals :







cos  1 


x

2
dx
(ii)
x
2 
sin  1  


2

(i)



(iii)

 3  (2  3x)

1

2



1

x2 
 3  4 

dx

dx

A
16. Integrate the following functions :
1
2  5x 2
3  2x
4
1
(i) 3 sin 3  5 cos 5  1  2x 
(3x  1

(ii) (7–4x)3 +

7
2
 4cosec2 (4x  3) 
(3  7x)
16  9x2

17. Show that
I=

dx

 4sin x  5cos x 

where  = tan–1

 x 
n | tan    |  C
 2 2
41

1

5
.
4

18. Prove that



1  x 2m
1
1
1
dx  x  x 3  x 5  ... 
x 2m 1 .
2
3
5
2m

1
1 x

19. Given the continuous periodic function f(x),
x  R. Can we assert that the antiderivative of that
function is a periodic function ?
xx
+C
2
21. Find all the antiderivatives of the function
f(x) = 3/x whose graphs touch the curve y = x3.
22. In each case, find a function f satisfying the
given conditions.
(a) f(x2) = 1/x for x > 0, f(1) = 1.
(b) f(sin2 x) = cos2x for all x, f(1) = 1.
(c) f(sin x) = cos2x for all x, f(1) = 1.



20. Show that | x |dx =

 1 for 0  x  1,
(d) f(ln x) = 
, f(0) = 0.
x  1,
 x for


1.10 

INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

23. Use the following information to graph the
function f over the closed interval [– 2, 5].
(a) The graph of f is made of closed line segments
joined end to end.
(b) The graph starts at the point (– 2, 1).
(c) The derivative of f is shown below:
Y
y= f (x)

1
–2

0

1

3

5

X

–2

24. Is there a function f such that f(0) = – 2, f(1) = 1, and
f(x) = 0 for all x? If so, how many such functions
are there?
25. Find all functions f(x) such that f(x) = 2 sin 3x.
26. A function g, defined for all positive real
numbers, satisfies the following two conditions:

g(1) = 1 and g(x2) = x3 for all x > 0.
Compute g(4).
27. Find a polynomial P of degree  5 with P(0) = 1,
P(1) = 2, P(0) = P(0) = P(1) = P(1) = 0.
28. Find a function f such that f(x) = x + cos x and
such that f(0) = 1 and f (0) = 2.

1.3

INTEGRATION BY
TRANSFORMATION

Standard methods of integration
The different methods of integration all aim at reducing
a given integral to one of the fundamental or known
integrals. There are several methods of integration :
(i) Method of Transformation, i.e., it is useful to
properly transform the integrand and then take
advantage of the basic table of integration
formulae.
(ii) Method of Substitution, i.e., a change of the
independent variable helps in computing a large
number of indefinite integrals.
(iii) Integration by parts
In some cases, when the integrand is a rational
fraction it may be broken into Partial Fractions
by the rules of algebra, and then each part may be
integrated by one of the above methods.

In some cases of irrational functions, the method of

Integration by Rationalization is adopted, which is a
special case of (ii) above.
In some cases, integration by the method of Successive
Reduction is resorted to, which really falls under case
(iv) It may be noted that the classes of integrals which
are reducible to one or other of the fundamental forms
by the above processes are very limited, and that the
large majority of the expressions, under proper
restrictions, can only be integrated by the aid of infinite
series, and in some cases when the integrand involves
expressions under a radical sign containing powers of
x beyond the second, the investigation of such
integrals has necessitated the introduction of higher
classes of transcedental function such as elliptic
functions, etc.
In fact, integration is, on the whole, a more difficult
operation than differentiation. We know that
elementary functions are differentiated according to
definite rules and formulae but integration involves,
so to say, an "individual" approach to every function.
Differential calculus gives general rules for
differentiation, but integral calculus gives no such
corresponding general rules for performing the inverse
operation. In fact, so simple an integral in appearance
as



x cos xdx, or 

sin x
dx
x

cannot be worked out that is, there is no elementary
function whose derivative is x cos x, or (sin x)/x,
though the integrals exist. There is quite a large number
of integrals of these types.

Method of Transformation
In the method of transformation, we find the integrals
by manipulation i.e. by simplifying and converting the
given integrand into standard integrands. It requires
experience to find an appropriate transformation of the
integrand, thus reducing the given integral to a
standard form.
The student must not, however, take for granted that
whenever one or other of the transformations is
applicable, it furnishes the simplest method of
integration. The most suitable transformation in each
case can only be arrived at after considerable practice
and familiarity with the results introduced by such
transformations.


INDEFINITE INTEGRATION
Example 1.



Evaluate I =

Solution I =



5(x  3) 2

5x 2  30x  45
x x

= 5  x dx  30
1/ 2

dx

x x

dx

Solution

dx
3 / 2

dx

2
 5. x 3 / 2  30.2 x  45( 2x 1/ 2 )  C
3



10 3 / 2
x  60 x  90 x 1 / 2  C .
3

Evaluate

Example 2.



Solution
=



x4
dx =
x2  1



x4
dx
x2  1



x4 1 1

dx
x2  1

x

1
dx
1

Solution

=
=

1

 x6

=

dx +

x




1
dx + 2
x6

6

I=

x2  3
dx
x 6 (x 2  1)



1  x4
+ sec x tan x dx
1 x

2
dx
(x 2  1)

x4  x2  1
dx – 2
x6

cosec2

1

 x 2  1 dx

1

1
1
1
 1
dx + 2   2  4  6  dx– 2 
dx
6


x
x
x
x
1 x2

1
 1  1  1 
–1
5 +2 
 x 3x 3 5x 5  – 2 tan x + C.
5x

Evaluate I =

We have
dx
3 cos x  sin x





dx

3 cos x  sin x

dx
 3

1
2
cos x  sin x 
2
 2


1 

1
cosec  x   dx


1
3 
2
sin  x  


3
1

1

= ln tan  x  (  / 6)  + C.
2
2
1  sin 
d
Example 7. Evaluate 
cos 
Solution We break the integrand into two
summands
sin  
1  sin 
 1
 cos  d =   cos   cos   d
=

(x 6  1)  x 6
 x 6 (x 2  1) dx





Solution

(x 2  1)  2
 x 6 (x 2  1) dx

1

dx + 2
x6

=



1
2

Example 6.

x3
=
– x + tan –1 x + C
3

Example 3.

x

3/ 4

1
1
x dx +  cos x dx
2
2
1
1

= –cot x + 2 ln (sin x) + C.
2
2

2

Evaluate

1

I =  x dx +  (1 + x + x 2 + x 3 ) dx +  sec x tan x dx ,
[ (1 – xn) / (1 – x) = 1 + x + x2 + x3 + .... + xn – 1]
I = 4x1/4 + x + (x2/2) + (x3/3) + (x4/4) + sec x + C.
1  sin x
dx.
Example 5. Evaluate I = 
1  cos x
1  sin x
Solution Here I =  1  cos x dx
1
1
1  2sin x cos x
2
2
dx
=
2 1
2sin x
2
=

1
x4 1
+ 2
dx
2
x 1
x 1

=  (x 2  1) dx +

Let I =  (

1 1  x4

+ sec x tan x dx.
x3/ 4 1  x

-3/4

 45  x

x

Example 4. Evaluate (

 1.11

1
2

dx



=  (sec  – tan ) d
= ln |sec  + tan | + ln |cos | + C.
Since ln A + ln B = ln AB, the answer can be simplified
to ln (|sec  + tan | |cos |) + C.
But sec  cos  = 1 and tan  cos  = sin . The answer
becomes even simpler :


1.12 



=

1  sin 
d = ln (1 + sin ) + C.
cos 
dx
Evaluate 
sec x  cos ecx
dx
1 2sin x cos x
 sec x  cos ecx = 2  sin x  cos x dx
1 (sin x  cos x) 2  1
dx

2
sin x  cos x

= 1
2



 ( sin x + cos x) dx –

1

sin  x  

4
=

INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

1
2 2

dx

1
x 
1
ln tan    + C
( – cos x + sin x) –


2 4
2
2 2

Evaluate

sin 2x

 sin 5x sin 3x

sin 2x

 sin 5x sin 3x

dx

sin(5x  3x)
dx =  sin 5x sin 3x dx

sin 5x cos 3x  cos 5x sin 3x
dx
=
sin 5x sin 3x

=  (cot 3x – cot 5x) dx
= 1/3 ln |sin 3x| – 1/5 ln |sin 5x| + C.
sin x
Evaluate 
dx
cos 3x

sin x
 cos3 x dx =

=

sin(3x  x)

 2 cos x cos 3x

2 sin x cos x
 2 cos x cos3 x dx

Integrals of the form

 cos ax cos bx dx ,  sin ax sin bx dx
 sin ax cos bx dx , in which a b.

We use the addition formulae to change products to
sums or differences, which can be integrated easily.
For example :
1
 cos x cos 2x dx = 2  2 cos x cos 2x dx
1
=  (cos 3x  cos x) dx
2
1  sin 3x sin x 


 +C
=

1 
2  3

 sin mx sin nxdx
1
=   cos(m  n)xdx   cos(m  n)xdx 
2
 sin(m  n)x  sin(m  n)x
 2(m  n)
2(m  n)
=
x
sin
2nx


  

  2
4n 

=

 cos(m  n)x cos(m  n)x
  2(m  n)  2(m  n)
=
 cos 2nx

4n

if m 2  n 2
if

m n

1

 cos mx cos nxdx  2  cos(m  n)xdx   cos(m  n)xdx 
 sin(m  n)x sin(m  n)x
 2(m  n)  2(m  n)
=
 x  sin 2nx
2
4n

if m 2  n 2
if

m2  n 2

Evaluate  sin 8x sin 3x dx
We have sin 8x sin 3x =
and so

1
1
ln sec 3x  ln sec x + C.
6
2

m n

1

dx

 (tan 3x – tan x) dx

if

 sinmxcosnxdx  2  sin(m  n)xdx  sin(m  n)xdx

sin 3x cos x  cos 3x sin x
=
dx
2 cos x cos 3x
1
= 2

if m 2  n 2

 sin 8x sin 3x dx

=

1
(cos 5x  cos 11x) dx
2

=

1
1
sin 5x –
sin 11x + C.
10
22

1
(cos 5x – cos 11x),
2


 1.13

INDEFINITE INTEGRATION
If a question has one of the functions
like sin2x, cos2x, sin3x or cos3x, then we replace them
by
1  cos x 1  cos 2x 3sin x  sin 3x 3cos x  cos 3x
,
,
,
2
2
4
4
respectively. The idea is to first express the function in
terms of multiple angles as above and then integrate it.
Also tan2x and cot2x should be replaced by sec2x – 1

and cosec2x – 1 respectively.
For example :

therefore sin6 x =
=

x 1
=  sin 2x + C,
2 4
1
 sin2x dx = 2  (1 – cos 2x) dx
x 1
=  sin 2x + C.
2 4
1
 sin3xdx =  4 (3sin x–sin x)dx
3
1
= cos x +
cos3x + C.
4
12

1
[1 – 3 cos 2x + 3 cos2 2x – cos 3 2x]
8

Writing cos2 2x =
cos3 2x =
(1) becomes

1
15
3
1
[10 x – sin 2 x + sin 4 x  sin 6 x]  C ...(3)
32
2
2
6
If the exponents are not too big, this method works
well.


Evaluate I =  sin 3 x cos3 x dx
I=

1
(1 + cos 2x)}2
2

1
3
1
+ cos 2x + cos 4x
8
8
2

=



(

3 1
1
+ cos 2x + cos 4x) dx
8 2
8

1
8

 sin

=

1
8



=

1
32

=

1
1  3
 cos 2 x  cos 6 x  + C.
6

32  2

3

2x dx

3sin 2x  sin 6x
dx
4

 (3 sin 2x  sin 6x) dx

Evaluate

Evaluate  sin6x dx .

Since sin2 x =

1
(1 – cos 2x),
2

1
(2sin x cos x)3 dx
8 

=

3
1
1
x + sin 2x +
sin 4x + C.
8
4
32

To evaluate  sin6x dx we should first
expressed sin6 x as a sum of sines/cosines of multiples
of x and then integrate each term of the sum to obtain
the result.

...(2)

From (2) we have  sin 6 xdx

1
{1 + 2 cos 2x + cos2 2x}
4
1
1
= [1 + 2 cos 2x + (1 + cos 4x)]
4
2

 cos4 x dx =

1
(cos 6x + 3 cos 2x),
4

1
(cos 6x + 3 cos 2x)},
4
1
=
[10 – 15 cos 2x + 6 cos 4x – cos 6x]
32

=

=

1
(1 + cos 4x),
2

–

Evaluate  cos4x dx .
cos4x = {

...(1)

3

1
[1 – 3 cos 2x + (1 + cos 4x)
8
2

sin 6x =

 cos2x dx =  (1 + cos 2x) dx

1
(1 – cos 2x)3,
8

I

dx

 sin(x  )sin(x )

1
sin[(x   )  (x  )]
dx
sin(   )  sin(x   ) sin(x  )

sin(x) cos(x)  cos(x)sin(x)
dx
sin(x  )sin(x )

=

1
sin()



1
[cot(x  )  cot(x   )]dx
sin(   ) 




1.14



INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED



1
(ln sin(x  )  ln sin(x   ) ]  C .
sin(   ) 



1
sin(x  )
ln
C .

sin(   ) sin(x   )



tan 2x = tan ( x    x  )

tan(x   )  tan(x –  )
1  tan(x   )·(x –  )
or, tan 2x – tan 2x · tan (x + ) tan (x – )
= tan (x + ) + tan (x + )
 tan (x + ) tan (x + ) tan 2x
= tan 2x – tan (x + ) – tan (x – )
=

 I=



[tan 2x – tan (x + ) – tan (x – )] dx

1
ln |sec 2x| – ln sec |(x + )|
2
– ln sec |(x – )| + C
= ln | sec 2 x · cos (x + ) · cos(x – )| + C.

=

Example 17.



I=

dx

sin x cos x dx

 1  sin x  cos x

=

 sec x  tan x  1

sin xdx

Multiplying and dividing by (1+ tan x – sec x), we get
sin x(1  tan x  sec x)
=  (1  tan x)2  sec2 x dk

=



sin x(1  tan x  sec x)
dk
2 tan x

1
cos x (1 + tan x – sec x) dx

=
2
1
(cos x  sin x  1) dx
=
2
1
=  sin x  cos x  x  + C.
2





(1  sin 2 x)

1  2 sin x 1  cos 2x

w. r. t. x.
1  cos 2x
cos 2 x
Solution The given function may be written as

5cos3 x  2 sin 3 x
 (cos2 x  sin 2 x  2sin x cos x)
2sin 2 x cos2 x

+

1

2 sin x 2sin 2 x


cos2 x cos2 x 2 cos2 x

5
cosec x cot x + sec x tan x + cos x + sin x
2
+ sec2 x + 2 sec x tan x + 2 (sec2 x – 1)
5
= cosec x cot x + 3 sec x tan x + cos x + sin x
2
+ 3 sec2 x – 2.
Now integrating, we get
5
I =  cosec x cot x dx + 3  sec x tan x dx
2

=







+ cos xdx + sin xdx + 3 sec2 x dx – 2 dx

 tan x  cot x  sec x  cosec x

=

5cos x  2sin 3 x
+
2sin 2 x cos2 x



Evaluate

dx
tan x  cot x  sec x  cosec x

Solution

Integrate the function

3

+

Example 16. Evaluate tan(x –)·tan (x + ) ·tan 2xdx

Solution

Example 18.

5
= – cosec x + 3 sec x + sin x – cos x
2

+ 3 tan x – 2x + C.
cos 5x  cos 4 x
dx .
Example 19. Evaluate
1  2 cos 3x



Solution




cos 5x  cos 4 x
dx
1  2 cos 3x

sin 3x(cos 5x  cos 4x)

 sin 3x  2 cos 3x sin 3x dx

3x
3x
9x
x
cos . 2 cos cos
2
2
2
2 dx


sin 3x  sin 6 x
3x
3x
9x
x
sin cos cos cos
2
2
2
2 dx
4
9x
3x
 2 cos sin
2
2
3x
x
  2 cos cos dx   (cos 2 x  cos x) dx
2
2
sin
2x
  
 sin x   C .
 2




2 sin





