Integral Calculus

UNIT 8 INTEGRAL CALCULUS

Structure
8.1 Introduction
Objectives
8.2 Antiderivatives
8.3 Basic Definitions
8.3.1 Standard Integrals
8.3.2 Algebra of Integrals
8.4 Methods of Integration
8.4.1 Integration by Substitution
8.4.2 Integration by Parts
8.5 Integration of Rational Functions
8.5.1 Some Simple Rational Functions
8.5.2 Partial Fraction Decomposition
8.6 Integration of Irrational Functions
8.7 Integration of Trigonometric Functions
8.8 Definite Integrals
8.9 Fundamental Theorem of Calculus
8.9.1 Area Function
8.9.2 First Fundamental Theorem of Integral Calculus
8.9.3 Second Fundamental Theorem of Integral Calculus
8.9.4 Evaluation of a Definite Integral by Substitution
8.10 Properties of Definite Integrals
8.11 Applications
8.12 Summary
8.13 Answers to SAQs

8.1 INTRODUCTION
In this unit, we shall introduce the notions of antiderivatives, indefinite integral and
various methods and techniques of integration. The unit will also cover definite integrals
which can be evaluated using these methods.
We know that one of the problems which motivated the concept of a derivative was a
geometrical one – that of finding a tangent to a curve at a point. The concept of
integration was also similarly motivated by a geometrical problem – that of finding the
areas of plane regions enclosed by curves. Some recently discovered Egyptian
manuscripts reveal that the formulas for finding the areas of triangles and rectangles were
known even in 1800 BC. Using these formulas, one could also find the area of any figure
bounded by straight line segments. But no method for finding the area of figures bounded
by curves had evolved till much later.
In the third century BC, Archimedes was successful in rigorously proving the formula for
the area of a circle. His solution contained the seeds of the present day integral calculus.
But it was only later, in the seventeenth century, that Newton and Leibniz were able to
generalize. Archimedes’ method and also to establish the link between differential and
integral calculus. The definition of the definite integral of a function, which we shall give 87
Calculus in this unit, was first given by Riemann in 1854. We will also acquaint you with various
application of integration.
Objectives
After studying this unit, you should be able to
• compute the antiderivative of a given function,
• define the indefinite integral of a function,
• evaluate certain standard integrals by finding the antiderivatives of the
integrals,
• compute integrals of various elementary and trigonometric functions,
• integrate rational functions of a variable by using the method of partial
fractions,
• evaluate the integrals of some specified types of irrational functions,
• define the definite integral of a given function as a limit of a sum,
• state the fundamental theorems of calculus,
• learn the different properties of definite integral,
• use the fundamental theorems to calculate the definite integral of an
integrable function, and
• use the definite integrals to evaluate areas of figures bounded by curves.

8.2 ANTIDERIVATIVES
In Unit 7, we have been occupied with the problem of finding the derivative of a given
function. Some of the important applications of the calculus lead to the inverse problem,
namely, given the derivative of a function, is it possible to find the function? This
process is called antidifferentiation and the result of antidifferentiation is called an
antiderivative. The importance of the antiderivative results partly from the fact that
scientific laws often specify the rates of change of quantities. The quantities themselves
are then found by antidifferentiation.
To get started, suppose we are given that f ′ (x) = 9, can we find f (x)? It is easy to see that
one such function f is given by f (x) = 9x, since the derivative of 9x is 9.
Before making any definite decision, consider the functions
9 x + 4, 9 x − 10, 9 x + 3
Each of these functions has 9 as its derivative. Thus, not only can f (x) be 9x, but it can
also be 9 x + 4 or 9 x − 10, 9 x + 3 . Not enough information is given to help us
determine which is the correct answer.
Let us look at each of these possible functions a bit more carefully. We notice that each
of these functions differs from another only by a constant. Therefore, we can say that if
f ′ (x) = 9, then f (x) must be of the form f (x) = 9x + c, where c is a constant. We call
9x + c the antiderivative of 9.
More generally, we have the following definition.
Definition
Suppose f is a given function. Then a function F is called an antiderivative of f, if
F ′ (x) = f (x) ∀ x .
We now state an important theorem without giving its proof.
Theorem 1
88
 If F1 and F2 are two antiderivatives of the same function, then F1 and F2 differ Integral Calculus
by a constant, that is
F1 ( x ) = F2 ( x ) + c
Remark
From above Theorem, it follows that we can find all the antiderivatives of a given
function, once we know one antiderivative of it. For instance, in the above
example, since one antiderivative of 9 is 9x, all antiderivative of 9 have the form
9x + c, where c is a constant. Let us do one example.
Example 8.1
Find all the antiderivatives of 4x.
Solution
We have to look for a function F such that F ′ (x) = 4 x . Now, an antiderivative of
4x is 2x2. Thus, by Theorem 1, all antiderivatives of 4x are given by 2x2 + c, where
c is a constant.
SAQ 1
Find all the antiderivatives of each of the following function
(i) f (x) = 10x
(ii) f (x) = 11x10
(iii) f (x) = − 5x

8.3 BASIC DEFINITIONS

We have seen, that the antiderivative of a function is not unique. More precisely, we have
seen that if a function F is an antiderivative of a function f, then F + c is also an
antiderivative of f, where c is any arbitrary constant. Now we shall introduce a notation
here : we shall use the symbol ∫ f (x) dx to denote the class of all antiderivatives of f.
We call it the indefinite integral or just the integral of f. Thus, if F (x) is an antiderivative
of f (x), then we can write ∫ f (x) dx = F (x) + c .

Here c is called the constant of integration. The function f (x) is called the integrand,
f (x) dx is called the element of integration and the symbol ∫ stands for the integral sign.

The indefinite integral ∫ f (x) dx is a class of functions which differ from one another
by constant. It is not a definite number; it is not even a definite function. We say that the
indefinite integral is unique up to an arbitrary constant.
Thus, having defined an indefinite integral, let us get acquainted with the various
techniques for evaluating integrals.

89
Calculus 8.3.1 Standard Integrals
We give below some elementary standard integrals which can be obtained directly from
our knowledge of derivatives.
Table 8.1

Sl. No. Function Integral

1 xn x n +1
+ c, n ≠ − 1
n +1

2 sin x − cos x + c
3 cos x sin x + c
4 sec2 x tan x + c
− cot x + c
2
5 cosec x
6 sec x tan x sec x + c
7 cosec x cot x − cosec x + c
8 1 sin −1 x + c or − cos −1 x + c
1 − x2

9 1 tan − 1 x + c or − cot −1 x + c
1+ x 2

10 1 sec −1 x + c or − cosec −1 x + c
x x2 − 1

11 1 ln | x | + c
x
12 ex ex + c
13 ax ax
+c
ln | a |

Now let us see how to evaluate some functions which are linear combination of the
functions listed in Table 8.1.
8.3.2 Algebra of Integrals
You are familiar with the rule for differential of sum of functions, which says
d d d
[a f (x) + b g (x)] = a [ f (x)] + b [ g (x)]
dx dx dx
There is a similar rule for integration :
Rule 1

∫ [a f (x) + b g (x)] dx = a ∫ f (x) dx + b ∫ g (x) dx

This rule follows from the two theorems.

Theorem 2
If f is an integrable function, then so is K f (x) and
∫ K f ( x ) dx = K ∫ f ( x ) dx , where K is a constant.

90 Proof
 ∫
Integral Calculus
Let f (x) dx = F (x) + c

d
Then by definition, [ F (x) + c] = f (x)
dx
d
∴ [ K {F (x) + c}] = K f (x)
dx
Again, by definition of antiderivatives, we have

∫ K f (x) dx = K [ F (x) + c] = K ∫ f (x) dx

Theorem 3
If f and g are two integrable functions, then f + g is integrable, and we have
∫ [ f ( x ) + g ( x )] dx = ∫ f ( x ) dx + ∫ g ( x ) dx .

Proof

Let ∫ f (x) dx = F (x) + c, ∫ g (x) dx = G (x) + c1

d
Then [{F (x) + c} + {G (x) + c1}] = f (x) + g (x)
dx

Thus, ∫ [f (x) + g (x) ] dx = [ F (x) + c] + [G (x) + c1 ]

= ∫ f (x) dx + ∫ g (x) dx

Rule (1) may be extended to include a finite number of functions, that is, we can
write
Rule 2

∫ [K1 f1 (x) + K 2 f 2 (x) + . . . + K n f n (x)] dx

= K1 ∫ f1 (x) dx + K 2 ∫ f 2 (x) dx + . . . + K n ∫ f n (x) dx

We can make use of Rule (2) to evaluate certain integrals which are not listed in
Table 8.1.
Example 8.2

Let us evaluate ∫ (2 + 4 x + 3 sin x + 4e x ) dx

=2 ∫ dx + 4 ∫ x dx + 3 ∫ sin x dx + 4 ∫ e x dx

= 2 x + 2 x 2 − 3 cos x + 4e x + c

Example 8.3

(1 − x) 2
Suppose we want to evaluate ∫ x x
dx

(1 − x) 2
Thus, ∫ x x
dx

91
Calculus 1 − 2x + x2
= ∫ 3
dx
x2
3 1 1
− −
= ∫ x 2 dx − ∫ 2x 2 dx + ∫ x 2 dx

1 1 3
− 2
= − 2x 2 − 4x 2 + x2 + c
3
And now some exercises for you.
SAQ 2

Write down the integrals of the following using Table 8.1 and Rule 2
5
−
(i) (a) x8 (b) x 2 (c) 4x– 2 (d) 9
2
1 ⎛ 1⎞
(ii) (a) x 2 − x − 1 (b) −3 x (c) ⎜ x − ⎟
x ⎝ x⎠

(iii) (a) e x + e − x + 4 (b) 4 cos x − 3 sin x + e x + x (c) 4 sec h 2 x + e x − 8 x

2 5 2x2 + 5
(iv) (a) + (b)
1 − x2 x x2 + 1

2
⎛ 1 ⎞
(v) (a) ax + bx + cx + d
3 2
(b) ⎜⎜ x − ⎟⎟
⎝ x⎠

sin 4 x + cos 4 x
(vi) (a) (b) (2 + x) (3 − x)
sin 2 x cos 2 x

8.4 METHODS OF INTEGRATION

We have seen in Section 8.3 that the decomposition of an integrand into the sum of a
number of integrands, with known integrals, is itself an important method of integration.
We now give two general methods of integration, namely,
(i) Integration by substitution,
(ii) Integration by parts.

The method of substitution consists in expressing the integral ∫ f ( x) dx in terms of

another simpler integral, ∫ F (t) dt , say, where the variables x and t are connected by
some suitable relation x = φ (t) .
92
The method of integration by parts enables one to express the given integral of a product Integral Calculus
of two functions in terms of another, whose integration may be simpler.
8.4.1 Integration by Substitution
Consider the following integral

∫ f ′[ g ( x)] g ′ ( x) dx . . . (1)

d
Since f [ g ( x)] = f ′[ g ( x)] g ′ ( x) (by Chain rule)
dx
∴ ∫ f ′[ g ( x)] g ′ ( x) dx = f [ g ( x)] + c

In Eq. (1), if we substitute g (x) = t

dt
Then g ′ ( x) =
dx
i.e. g ′ ( x) dx = dt
Hence f ′[ g ( x)] g ′ ( x) dx = f ′ (t) dt

∴ ∫ f ′[ g ( x)] g ′ ( x) dx = ∫ f ′ (t) dt

= f (t) + c
= f [ g ( x)] + c
Let us now illustrate this technique with examples.
Example 8.4

Find ∫ (x 2 + 1)3 2 x dx

Solution
Let t = x2 + 1
dt = 2 x dx

Therefore, ∫ (x 2 + 1)3 2 x dx

= ∫t
3
dt

t4 1
= + c = ( x 2 + 1) 4 + c , since t = x 2 + 1 .
4 4
Example 8.5

∫
4
Find x 3 e x dx .

Solution

Let t = x4

Then dt = 4 x 3 dx

1
∫ ∫
4 4
Therefore, x 3 e x dx = 4 x 3 e x dx
4
1
=
4 ∫ et dt

93
Calculus
=
1 t
4
[
e +c ]
1 ⎡ x4
= e + c ⎤ , since t = x4
4 ⎢
⎣ ⎦⎥
Some Typical Examples of Substitution

We now consider the integral ∫ f ( x) dx , where the integrand f (x) is in some

typical form and the integral can be obtained easily by the method of substitution.
Various forms of integral can be obtained easily by the method of substitution.
Various forms of integrals considered are as follows :

(a) ∫ f (ax + b) dx

To integrate f (ax + b), put ax + b = t

1
Therefore a dx = dt or dx = dt
a
1
Thus ∫ f (ax + b) dx =
a ∫ f (t) dt

which can be evaluated, once the right hand side is known, for example, to
1
find ∫
cos (ax + b) dx , we put ax + b = t and a dx = dt or dx = dt .
a
1 1
Then ∫ cos (ax + b) dx =
a ∫ cos t dt =
a
sin t + c

1
or ∫ cos (ax + b) dx =
a
sin (ax + b) + c

Similarly, we have the following results

(ax + b) n + 1
∫ (ax + b) n dx =
( n + 1) a
+ c, n ≠ − 1

1 1
∫ ax + b
dx = ln (ax + b) + c
a

1 ax + b
∫ e ax + b dx =
a
e +c

1
∫ sec 2 (ax + b) dx =
a
tan (ax + b) + c etc.

You can make direct use of the above results in solving exercises.

(b) ∫ f ( x n ) x n − 1 dx

To integrate f ( x n ) x n − 1 , we let x n = t .

Then nx n −1 dx = dt

1
and ∫ f ( x n ) x n −1 dx =
n ∫ f (t) dt

94 which can be found out once the right hand side is known.
 ∫
Integral Calculus
For example, to find x 2 sin x 3 dx , put x 3 = t ; then 3 x 2 dx = dt , that is
1
x 2 dx = dt .
3
1
Then, ∫ x 2 sin x 3 dx =
3 ∫ sin t dt

1
=− cos t + c
3
1
=− cos x 3 + c
3

(c) ∫ { f ( x )}n f ′ ( x) dx , n ≠ − 1

Putting f (x) = t; we see that f ′ ( x ) dx = dt and

t n +1
∫ {f ( x)}n f ′ ( x) dx = ∫ t n dt =
n +1
+c

{f ( x)}n + 1
= +c
n +1

For example, ∫ cos 2 x sin xdx

=− ∫t
2
dt , where t = cos x (and hence – dt = sin x dx)

1 3
Therefore, ∫ cos 2 x sin xdx = −
3
t +c

1
=− (cos x)3 + c
3
f ′( x)
(d) ∫ f ( x)
dx

Putting f (x) = t, we have f ′ ( x ) dx = dt

f ′ ( x) dt
and ∫ f (x)
dx = ∫ t
= ln | t | + c = ln | f ( x) | + c

i.e. the integral of a function in which the numerator is the differential

co-efficient of the denominator, is equal to the logarithm of the denominator
(plus a constant).
For example, applying this result, we have
sin x ( − sin x)
∫ cos x
dx = − ∫ cos x
dx = c − ln cos x

(Since f (x) = cos x, in this case.)

Therefore, ∫ tan x dx = c − ln | cos x |

Similarly, you can obtain the following integrals

∫ cot x dx = ln | sin x | + c
95
 ∫
Calculus
sec x dx = ln | (sec x + tan x) | + c

⎛ x⎞
∫ cosec x dx = ln tan ⎜ ⎟ + c
⎝2⎠
Remember that logarithm of a quantity exists only when the quantity is
positive. Thus, while making use of these formulas, make sure that the
integrand to be integrated is positive in the domain under consideration.

(e) ∫ f (a 2 ± x 2 ) dx

Under this category we now give some results obtained by putting

x = a t (and hence dx = a dt) and using some standard integrals.

dx 1 ⎛ x⎞ ⎛ d 1 ⎞⎟
Now, ∫ a +x
2 2
=
a
tan −1 ⎜ ⎟ + c
⎝a⎠
⎜Θ
⎜ dt
⎝
(tan −1 t ) =
1 + t 2 ⎟⎠

⎛ d ⎞
dx ⎛ x⎞ ⎜Θ 1 ⎟
∫
−1
= sin −1 ⎜ ⎟ + c ⎜⎜ dt (sin t ) = ⎟⎟
a2 − x2 ⎝a⎠ 1 + t2
⎝ ⎠

⎛ d ⎞
dx 1 ⎛ x⎞ ⎜Θ 1 ⎟
∫
−1
= sec −1 ⎜ ⎟ + c ⎜⎜ dt (sec t ) = ⎟
x x2 − a2 a ⎝a⎠
⎝ t t + 1 ⎟⎠
2

⎛ x + a2 + x2 ⎞ ⎛ d ⎞
= ln ⎜⎜ ⎟+c ⎜Θ ⎟
dx 1
∫ ⎜⎜ dt ln = (t + 1+ t =
2
⎜ a ⎟⎟ ⎟⎟
a2 + x2 ⎝ ⎠ ⎝ 1+ t 2 ⎠
We usually use all these integrals given under (a) – (e) directly whenever required
without actually proving them. Sometimes it may happen that two or more substitutions
have to be used in succession. We now illustrate this point with the help of the following
example.
Example 8.6

Calculate
2x
(i) ∫ 1 + x2
dx

(ii) ∫ sin 3 x cos 2 x dx

Solution
2x
(i) ∫ 1 + x2
dx

Put 1 + x2 = t
Then 2 x dx = dt
2x 1
∴ ∫ 1+ x 2
dx = ∫ t
dt

= ln t + c
= ln (1 + x2) + c
96
 ∫ ∫
Integral Calculus
(ii) sin 3 x cos 2 x dx = sin 2 x cos 2 x sin x dx

= ∫ cos 2 x (1 − cos 2 x) sin x dx

Put cos x = t
Then − sin x dx = dt

∴ ∫ sin 3 x cos 2 x dx = − ∫t (1 − t 2 ) dt
2

=− ∫ (t 2 − t 4 ) dt

⎡t3 t5 ⎤
=−⎢ − ⎥+c
⎣⎢ 3 5 ⎥⎦

t5 t3
= − +c
5 3

( cos x) 5 ( cos x) 3
= − +c
5 3
So far we have developed the method of integration by substitution, by turning the chain
rule into an integration formula. Let us do the same for the product rule. We know that
the derivative of the product of two functions f (x) and g (x) is given by
d
[ f ( x) g ( x)] = g ( x) f ′ ( x) + f ( x) g ′ ( x) ,
dx
where the dashes denote differentiation w. r. t. x. Corresponding to this formula, we have
a rule called integration by parts.
8.4.2 Integration by Parts
Let us now discuss the method of integration by parts in detail. We begin by taking two
functions f (x) and g (x). Let G (x) be an antiderivative of g (x), that is,

∫ g ( x) dx = G ( x) or G′ ( x) = g ( x)

Then, by the product rule for differentiation, we have

d
[ f ( x) G ( x)] = f ( x) G ′ ( x) + f ′ ( x) G ( x) = f ( x) g ( x) + f ′ ( x) G ( x)
dx
Integrating both sides, we get

f ( x) G ( x ) = ∫ f ( x) g ( x) dx + ∫ f ′ ( x) G ( x) dx

or ∫ f ( x) g ( x) dx = f ( x) G ( x) − ∫ f ′ ( x) G ( x) dx

Thus, ∫ f ( x) g ( x) dx = f ( x) ∫ g ( x) dx − ∫ ∫
f ′ ( x) { g ( x) dx} dx . . . (8.1)

The integration done by using the Eq. (8.1) is called integration by parts. In other words,
it can be stated as follows :
The integral of the product of two functions
= first function × integral of the second function
− integral of (differential coefficient of the first × integral of the
second). 97
Calculus We now illustrate this method through some examples.
Example 8.7
Integrate x ex with respect to x.
Solution
We use integration by parts.
Step 1

Take f ( x) = x and g ( x) = e x .

Then f ′ ( x) = 1 and ∫ e x dx = e x

Step 2
From Eq. (8.1), we have

∫ x e x dx = x e x − ∫ 1. e dx + c
x

or ∫ x e x dx = x e x − e x + c

Sometimes we need to integrate by parts more than once. We now illustrate it through the
following example.
Example 8.8

Find ∫ x 2 cos x dx

Solution

∫ x 2 cos x dx = x 2 ∫ cos x dx − ∫ 2x {∫ cos x dx} dx

= x 2 sin x − 2 ∫ x sin x dx + c1 . . . (8.2)

where c1 is a constant of integration.

Integrating ∫ x sin x dx , again by parts, we get

∫ x sin x dx = x ∫ sin x dx − ∫ 1 . {∫ sin x dx} dx

= − x cos x + ∫ cos x dx

= − x cos x + sin x + c2 . . . (8.3)

where c2 being the constant of integration. From Eqs. (8.2) and (8.3), we get

∫ x 2 cos x dx = x 2 sin x − 2 (− x cos x + sin x + c2 ) + c1 ,

= x 2 sin x + 2 x cos x − 2 sin x + c ,

where we have written c for c1 – 2c2.

We now consider some examples of integrals which occur quite frequently and can be
integrated by parts.
Example 8.9

Find ∫ e ax cos bx dx
98
Solution Integral Calculus

Step 1

Choose f (x) = e ax and g ( x ) = cos bx ; then integrating by parts gives

sin bx sin bx
∫ e ax cos bx dx = e ax
b
− ∫ ae ax
b
dx + c1 . . . (8.4)

Step 2

Integrating ∫ e ax sin bx dx by parts again, we get

(− cos bx) (− cos bx)

∫ e ax sin bx dx = e ax
b
− ∫ ae ax
b
dx + c2

1 ax a
=−
b
e cos bx +
b ∫ e ax cos bx dx + c2

Note that the second term on the right hand side is nothing but a constant
multiple of the given integral.
Step 3

Substituting the value of ∫ e ax sin bx dx , in Eq. (8.4), we have

sin bx a ⎡ 1 ax a ⎤
∫ e ax cos bx dx = e ax
b
−
b ⎢− b e cos bx + b
⎣ ∫ e ax cos bx dx + c2 ⎥ + c1
⎦

sin bx a a2
= e ax
b
+ 2 e ax cos bx − 2
b b ∫ e ax cos bx dx + c3 . . . (8.5)

a
where c3 = c1 − c2
b
Step 4
Transposing the last term from the right of Eq. (8.5) to left, we get
⎛ 2⎞
⎜1 + a ⎟ 1 ax a
⎜
⎝ b 2 ⎟⎠ ∫ e ax cos bx dx =
b
e sin bx + 2 e ax cos bx + c3
b

⎛ a2 ⎞
Dividing by ⎜1 + 2 ⎟ , we finally get
⎜ b ⎟⎠
⎝

e ax
∫ e ax cos bx dx =
a2 + b2
(b sin bx + a cos bx) + c ,

c3
where c = , as the required integral.
a + b2
2

Similarly, the integral of the type ∫ e ax sin bx dx can be obtained.

And now some exercises for you.

SAQ 3
(a) Evaluate the following integrals :
dx
(i) ∫ 9 x − 12 x + 8
2

99
Calculus x dx
(ii) ∫ x + x2 + 1
4

(iii) ∫ (tan x)5 sec 2 x dx

dx
(iv) ∫ e +1
x

cot x
(v) ∫ ln sin x
dx

1
(vi) ∫ e −1
x
dx

dx
(vii) ∫ (e + e − x ) 2
x

(viii) ∫ x sec 2 x 2 dx

(sin −1 x) 2
(ix) ∫ 1 − x2
dx

(1 + ln x) 3
(x) ∫ x
dx

(cosec 2 x)
(xi) ∫ (1 + cot x)
dx

(b) Evaluate

(i) ∫ x 2 ln x dx

(ii) ∫ x cosec2 x dx

(iii) ∫ e3 x cos 4 x dx

(iv) ∫ sin −1 x dx

(v) ∫ x tan −1 x dx

x sin −1 x
(vi) ∫ (1 − x 2 )
dx

x ex
(vii) ∫ (1 + x 2 )
dx

100
 Integral Calculus
8.5 INTEGRATION OF RATIONAL FUNCTIONS
We know, by now, that it is easy to integrate any polynomial function, that is, a function
f given by f ( x) = an x n + an −1 x n −1 + . . . + a0 . In this section, we shall see how a
rational function can be integrated.
Definition
Q ( x)
A function R is called a rational function if it is given by R ( x) = , where
P ( x)
Q (x) and P (x) are polynomials. It is defined for all x for which P (x) ≠ 0. If the
degree of Q (x) is less than the degree of P (x), we say that R (x) is a proper
rational function. Otherwise, it is called an improper rational function.
x +1 x3 + x + 5
Thus f ( x) = is a proper rational function, and g ( x) = is an
x2 + x + 2 x−2
17
improper one. But g (x) can also be written as g ( x) = ( x 2 + 3x + 6) + (by long
x−2
division).
Here we have expressed g (x), which is an improper rational function, as the sum of a
polynomial and a proper rational function. This can be done for any improper rational
function.
8.5.1 Some Simple Rational Functions
Now we shall consider some simple types of proper rational functions, like
1 1 x−m
, and 2 . We shall illustrate the method of integrating these
x − a (x − b) k
ax + bx + c
functions through some examples.
Example 8.10
1
Consider the function f ( x) = .
(x + 2) 4
Solution
To integrate this function we shall use the method of substitution.
du
Thus, if we put u = x + 2 or = 1 , and we can write
dx
1 1 u−3 1
∫ (x + 2) 4
dx = ∫ u 4
du =
−3
+c=−
3 (x + 2) 3
+ c.

Example 8.11
2x + 3
Consider the function f ( x) = .
x − 4x + 5
2

Solution
This has a quadratic polynomial in the denominator. Now
2x + 3 2x − 4 7
∫ x − 4x + 5
2
dx = ∫
x − 4x + 5
2
dx +
x − 4x + 5
2 ∫dx .

Perhaps you are wondering why we have split the integral into two parts.
The reason for this break-up is that now the integrand in the first integral on the
g ′ (x) g ′ (x)
right is of the form
g (x)
; and we know that
g (x) ∫
dx = ln | g ( x) | + c .
101
Calculus 2x − 4
Thus ∫ x − 4x + 5
2
dx = ln | x 2 − 4 x + 5| + c1 .

To evaluate the second integral on the right, we write

1 1 1
∫ x − 4x + 5
2
dx = ∫ (x − 4 x + 4) + 1
2
dx = ∫ (x − 2) 2 + 1
dx .

du
Now, if we put u = x − 2, = 1 and
dx
1 1
∫ x − 4x + 5
2
dx = ∫ u +1
2
du = tan −1 u + c2 = tan −1 ( x − 2) + c2

2x + 3
This implies, ∫ x − 4x + 5
2
dx = ln | x 2 − 4 x + 5| + 7 tan −1 (x − 2) + c .

8.5.2 Partial Fraction Decomposition

You must have studied the factorisation of polynomials. For example, we know that
x 2 − 5 x + 6 = ( x − 2) ( x − 3)

Here (x – 2) and (x – 3) are two linear factors of x 2 − 5 x + 6 .

You must have also come across polynomial like x 2 + x + 1 , which cannot be factorised
into real factors. Thus, it is not always possible to factorise a given polynomial into linear
factors. But any polynomial can, in principle, be factorised into linear and quadratic
factors. We shall not prove this statement here. It is a consequence of the fundamental
theorem of algebra. The actual factorization of a polynomial may not be very easy to
carry out. But, whenever we can factorise the denominator of a proper rational function,
we can integrate it by employing the method of partial fractions. The following examples
will illustrate this method.
Example 8.12
5x − 1
Let us evaluate ∫ x2 − 1
dx .

Solution
5x − 1
Here the integrand is a proper rational function.
x2 − 1

Its denominator x 2 − 1 can be factored into linear factors as :

x 2 − 1 = ( x − 1) ( x + 1)

5x − 1
This suggests that we can write the decomposition of into partial fraction
x2 − 1
as :
5x − 1 5x − 1 A B
= = +
x −12 (x − 1) ( x + 1) (x − 1) (x + 1)

If we multiply both sides by ( x − 1) ( x + 1) , we get

5 x − 1 = A ( x + 1) + B ( x − 1) .

That is 5 x − 1 = ( A + B) x + ( A − B)
102
 By equalling the coefficients of x, we get A + B = 5. Equating the constant terms Integral Calculus
on both sides, we get A – B = − 1.
Solving these two equations in A and B, we get A = 2 and B = 3.
5x − 1 2 3
Thus, = +
x −12 x −1 x +1

Integrating both sides of this equation, we obtain

5x − 1 2 3
∫ x −1
2
dx = ∫ x −1
dx + ∫ x +1
dx

= 2 ln | x − 1| + 3 ln | x + 1| + c
Let us go to our next example now.
Example 8.13
x
Evaluate ∫ x − 3x + 2
3
dx .

Solution
x
Take a look at the denominator of the integrand in ∫ x − 3x + 2
3
dx .

It factors into ( x − 1) 2 ( x + 2) . The linear factor (x – 1) is repeated twice in the

decomposition of x 3 − 3 x + 2 . In this case, we write

x A B C
= + +
x − 3x + 2
3 x + 2 x − 1 (x − 1) 2

From this point, we proceed as before to find A, B and C. We get

x = A (x − 1) 2 + B (x + 2) (x − 1) + C (x + 2)

1 2
We put x = 1, and x = − 2, and get C = and A = − . Then to find B, let us put
3 9
any other convenient value, say x = 0.
This gives us 0 = A – 2B + 2C
2 2
or 0=− − 2B +
9 3
2
This implies B =
9
x 2 1 2 1 1 1
Thus ∫ x − 3x + 2
3
dx = −
9 ∫ x+2
dx +
9 ∫ x −1
dx +
3 ∫ (x − 1) 2
dx

2 2 1 1
=− ln | x + 2 | + ln | x − 1| − +c
9 9 3 (x − 1)

2 x −1 1
= ln − +c
9 x+2 3 (x − 1)

In our next example, we shall consider the case when the denominator of the integrand
contains an irreducible quadratic factors (i.e. a quadratic factor which cannot be further
factored into linear factors).
Example 8.14 103
Calculus 6 x 3 − 11x 2 + 5 x − 4
Evaluate ∫ x 4 − 2 x3 + x 2 − 2 x
dx

We factorise x 4 − 2 x 3 + x 2 − 2 x as x ( x − 2) ( x 2 + 1) .

6 x 3 − 11x 2 + 5 x − 4 A B Cx + D
Then we write = + + 2
x − 2x + x − 2x
4 3 2 x ( x − 2) x +1
Thus
6 x 3 − 11x 2 + 5 x − 4 = A ( x − 2) ( x 2 + 1) + B x ( x 2 + 1) + (Cx + D) x ( x − 2)
Next, we substitute x = 0, and x = 2, to get A = 2 and B = 1. Then we put
x = 1 and x = − 1 (some convenient values) to get C = 3 and D = − 1.
6 x 3 − 11x 2 + 5 x − 4 1 1 3x − 1
Thus, ∫ x − 2x + x − 2x
4 3 2
dx = 2 ∫ x
dx + ∫ x−2
dx + ∫ x2 + 1
dx

3 2x dx
= 2 ln | x | + ln | x − 2 | +
2 ∫ x +12
dx − ∫ x +1
2

3
= 2 ln | x | + ln | x − 2 | + ln | x 2 + 1| − tan −1 x + c
2
Thus, you see, once we decompose integrand, which is a proper rational function, into
partial fractions, then the given integral can be written as the sum of some integrals of the
type discussed in previous examples.
All the functions which we integrated till now were proper rational functions. Now we
shall take up an example of an improper rational function.
Example 8.15
x3 + 2 x
Let us evaluate ∫ x2 − x − 2
dx .

Solution
Since the integrand is an improper rational function, we shall first write it as the
sum of a polynomial and a proper rational function.
x3 + 2 x 5x + 2
Now = ( x + 1) +
x −x−2
2
x −x−2
2

x3 + 2 x 5x + 2
Therefore, ∫ x −x−2
2
dx = ∫ (x + 1) dx + ∫ x −x−2
2
dx

dx dx
= ∫ x dx + ∫ dx + 4 ∫ x−2
+ ∫ x +1

x3 + 2 x x2
Hence ∫ x2 − x − 2
dx =
2
+ x + 4 ln | x − 2 | + ln | x + 1| + c

Try to do the following exercises now.

SAQ 4
Evaluate
x
(i) ∫ x − 2x − 3
2
dx

104
 3 x − 13 Integral Calculus
(ii) ∫ x 2 + 3 x − 10
dx

6 x 2 + 22 x − 23
(iii) ∫ (2 x − 1) ( x 2 + x − 6)
dx

x2 + x − 1
(iv) ∫ (x − 1) ( x 2 − x + 1)
dx

x2
(v) ∫ x2 − a2
dx

x2 + 4
(vi) ∫ x2 + 2x + 3
dx

8.6 INTEGRATION OF IRRATIONAL FUNCTIONS

The task of integrating functions gets tougher if the given function is an irrational one,
Q ( x)
that is, it is not of the form . In this section, we shall give you some tips for
P ( x)
evaluating some particular types of irrational functions. In most cases, our endeavour will
be to arrive at a rational function through an appropriate substitution. This rational
function can then be easily evaluated by using the techniques developed in Section 8.5.
Integration of Functions Containing only Fractional Powers of x
In this case, we put x = tn, where n is the lowest common multiple
(l. c. m.) of the denominators of powers of x. This substitution reduces the function
to a rational function of t.
Look at the following example.
Example 8.16
2 x1 / 2 + 3 x1 / 3
Let us evaluate ∫ 1 + x1/ 3
dx

Solution
We put x = t6, as 6 is the l. c. m. of 2 and 3. We get
2 x1 / 2 + 3x1 / 3 2t 3 + 3t 2
∫ 1+ x 1/ 3
dx = ∫ 1+ t 2
6 t 5 dt

2t 8 + 3t 7 ⎡ 6 3t − 2 ⎤
=6 ∫ dt = 6 ∫ ⎢ 2t + 3t − 2t − 3t + 2t + 3t − 2 −
5 4 3 2
⎥ dt
1 + t2 ⎢⎣ 1 + t 2 ⎥⎦

⎡2 1 2 3 2 3 3 ⎤
= 6 ⎢ t 7 + t 6 − t 5 − t 4 + t 3 + t 2 − 2t − ln (1 + t 2 ) + 2 tan −1 t ⎥ + c
⎣7 2 5 4 3 2 2 ⎦
105
Calculus 12 7 / 6 12 9
= x + 3x − x 5 / 6 − x 2 / 3 + 4 x1 / 2 + 9 x1 / 3 −12 x1 / 6 − 9 ln |1+ x1 / 3 | +12 tan −1 ( x1/ 6 ) + c
7 5 2

Integrals of the Types

(i) ∫ x 2 − a 2 dx , (ii) ∫ x 2 + a 2 dx

(iii) ∫ a 2 − x 2 dx , (iv) ∫ ax 2 + bx + c dx

(v) ∫ ( px + q ) ax 2 + bx + c dx

Now, let us evaluate the above integrals.

(i) Let I = ∫ x 2 − a 2 dx

Integrating by parts taking 1 as the second function, we have

1 2x
I=x x2 − a2 − ∫ x.
2
.
x2 − a2
dx

x2
=x x2 − a2 − ∫ x2 − a2
dx

x2 − a2 + a2
=x x2 − a2 − ∫ x2 − a2
dx

dx
=x x2 − a2 − ∫ x 2 − a 2 dx − a 2 ∫ x − a2
2

a2
=x x2 − a2 − I − log ⎛⎜ x + x 2 − a 2 ⎞⎟ + c
2 ⎝ ⎠

a2
∴ 2I = x x2 − a2 − log ⎛⎜ x + x 2 − a 2 ⎞⎟ + c
2 ⎝ ⎠

a2 ⎛ log x + x 2 − a 2 ⎞⎟ + c1 , where c1 =
x c
∴ I= x2 − a2 − ⎜
2 2 ⎝ ⎠ 2
Similarly

a2
log ⎛⎜ x + x 2 + a 2 ⎞⎟ + c , and
1
(ii) ∫ x 2 + a 2 dx =
2
x x2 + a2 +
2 ⎝ ⎠

1 a2 x
(iii) ∫ a 2 − x 2 dx =
2
x a2 − x2 +
2
sin −1 + c
a

(iv) ∫ ax 2 + bx + c dx

⎛ b c⎞
ax 2 + bx + c = a ⎜ x 2 + x + ⎟
⎝ a a⎠

⎡⎛ b ⎞
2
⎛c b 2 ⎞⎤
= a ⎢⎜ x + ⎟ + ⎜⎜ − 2 ⎟⎟⎥
⎢⎣⎝ 2a ⎠ ⎝ a 4a ⎠⎥⎦
106
 Integral Calculus
b c b2
Put x+ = t, − 2 = k 2
2a a 4a
Then the integral is reduced to any of the forms (i), (ii) or (iii).

(v) ∫ ( px + q ) ax 2 + bx + c dx

Choose constants A and B such that

⎡d ⎤
px + q = A ⎢ (ax 2 + bx + c)⎥ + B
⎣ dx ⎦
= A (2ax + b) + B
i.e. 2aA = p, Ab + B = q
Thus the integral is reduced to

A ∫ ( 2ax + b) ax 2 + bx + c dx + B ∫ ax 2 + bx + c dx

= A I1 + B I 2

I1 = ∫ ( 2ax + b) ax 2 + bx + c dx

Put ax 2 + bx + c = t
(2ax + b) dx = dt
2
i.e. I1 = (ax 2 + bx + c)3 / 2 + c2
3

Similarly, I 2 = ∫ ax 2 + bx + c dx which can be worked out as

in (iv).
∴ (v) can be determined.
Example 8.17

Evaluate ∫ 4 − x 2 dx .

Solution

∫ 4 − x 2 dx = ∫ 2 2 − x 2 dx

1 x
= x 4 − x 2 + 2 sin −1 + c .
2 2
Example 8.18

Evaluate ∫ x 2 + 2 x + 5 dx

Solution

∫ x 2 + 2 x + 5 dx = ∫ (x + 1) 2 + 4 dx

Put x + 1 = t, then dx = dt

∴ ∫ x 2 + 2 x + 5 dx = ∫ t 2 + 4 dt = ∫ t 2 + 2 2 dt

t t 2 + 4 + . 4 log ⎛⎜ t + t 2 + 4 ⎞⎟ + c
1 1
=
2 2 ⎝ ⎠

x 2 + 2 x + 5 + 2 log ⎡( x + 1) + x2 + 2x + 5 ⎤ + c
1
= ( x + 1)
2 ⎢⎣ ⎥⎦
107
Calculus Example 8.19

Evaluate ∫ x 1 + x − x 2 dx

Solution

⎡d ⎤
Let x = A ⎢ (1 + x − x 2 )⎥ + B
⎣ dx ⎦
= A (1 − 2 x) + B

1 1
∴ A=− ,B=
2 2
1
Thus ∫ x 1 + x − x 2 dx = −
2 ∫ (1 − 2 x) 1 + x − x 2 dx

1
+
2 ∫ 1 + x − x 2 dx

1 1
=− I1 + I 2
2 2

I1 = ∫ (1 − 2 x) 1 + x − x 2 dx

Put 1 + x − x2 = t

Then (1 − 2 x) dx = dt
1 3
2 2
∴ I1 = ∫ t 2 dt =
3
t + c1

3
2
= (1 + x − x 2 ) 2 + c1 . . . (1)
3
2
5 ⎛ 1⎞
I2 = ∫ 1 + x − x dx = ∫ − ⎜ x − ⎟ dx
2
4 ⎝ 2⎠

1
Put x− = t, then dx = dt
2

5
∴ I2 = ∫ 4
− t 2 dt

1 5 1 5 2t
= t − t 2 + . sin −1 + c2
2 4 2 4 5

1 5 2x − 1
= (2 x − 1) 1 + x − x 2 + sin −1 + c2
4 8 5

Hence ∫ x 1 + x − x 2 dx

3
1 1 5 2x − 1
=− (1 + x − x 2 ) 2 + ( 2 x − 1) 1 + x − x 2 + sin −1 +c
3 8 16 5
108
SAQ 5 Integral Calculus

Integrate the following functions :

(i) x2 + 4x + 6

(ii) ( x + 1) 2 x 2 + 3

(iii) 1 + 3x − x 2

8.7 INTEGRATION OF TRIGONOMETRIC

FUNCTIONS
If the integrand is a rational function of sin x or cos x or both, it can be reduced to a
x
rational function by substituting t = tan .
2
dt x 1 1 + t2
Then = sec 2 . =
dx 2 2 2
2dt
i.e. dx =
1 + t2
x
2 tan
2t 1 − t2 2
sin x = , cos x = as sin x =
1 + t2 1 + t2 1 + tan 2
x
2
x
1 − tan 2
and cos x = 2.
2 x
1 + tan
2
Example 8.20
1 + sin x
Evaluate ∫ sin x (1 + cos x)
dx

Solution
x
Let tan = t,
2
2t
1+
1 + sin x 1 + t2 2 dt
Then ∫ sin x (1 + cos x)
dx = ∫ 2t ⎡ 1 − t ⎤ 1 + t2
2
.
⎢1 + 2⎥
1 + t2 ⎣⎢ 1 + t ⎦⎥
109
Calculus 1 + t 2 + 2t
=2 ∫ 2t [1 + t 2 + 1 − t 2 ]
dt

1 1 + t 2 + 2t 1 ⎡1 ⎤
=
2 ∫ t
dt =
2 ∫ ⎢ t + t + 2⎥ dt
⎣ ⎦
1 ⎡ t2 ⎤
= ⎢log | t | + + 2 t + c⎥
2 ⎢⎣ 2 ⎥⎦
1 x 1 x x
= log | tan | + tan 2 + tan + c
2 2 4 2 2
SAQ 6
Integrate the following :
1
(i)
5 + 4 sin x
1
(ii)
2 + cos θ
1
(iii)
1 + sin x + cos x

8.8 DEFINITE INTEGRALS

We have studied indefinite integrals so far. Now, we define a definite integral and see
how it can be used to find the area under certain curves.
8.8.1 Definite Integral as the Limit of a Sum
Let f be a continuous function defined on a closed interval [a, b]. Assume that all the
values taken by the function are non-negative, i.e. the graph of the function is a curve
above the x-axis.

Figure 8.1
110
Integral Calculus

111
Consider the area of Figure 8.1. Let us find the area of this region. Integral Calculus

Let AM and BN be the ordinates for x = a and x = b. Divide MN into n equal parts of
length h each and let M 1 P1 , M 2 P2 , . . . , M n −1 Pn −1 be the ordinates at
b−a
M 1 , M 2 , . . . , M n −1 , then nh = b – a, i.e. h = .
n
Also abscissae of the point A, P1 , P2 , . . . , Pn −1 , B are
a, a + h, a + 2h, . . . , a + n −1 h, b .
∴ MA = f (a)
M 1 P1 = f ( a + h)
M 2 P2 = f ( a + 2h)
:
:
M n −1 Pn −1 = f ( a + n − 1 h)
MB = f (b)
We consider the left end points of these sub regions and construct rectangles
1, 2, 3, . . . , n as shown in Figure 8.1.
Area of the first rectangle = h f (a)
Area of the second rectangle = h f (a + h)
Area of the third rectangle = h f (a + 2h)
..................................
..................................
..................................
Area of the nth rectangle = h f (a + n − 1 h)

∴ Sum of these areas = h f (a ) + h f (a + h) + . . . + h f (a + n − 1 h)

We note that this area is approximately equal to the area of the region AMNB. Further as
the number of sub-divisions increases, the estimation becomes better. Let the
b−a
subdivisions become very large, i.e. n → ∞, then h = → 0 , which in turn implies
n
that the area of the region AMNB
= Lt h [ f (a) + f (a + h) + f (a + 2h) + . . . + f (a + n − 1 h)] . . . (1)
h→0

The expression on the R. H. S of Eq. (1) is called the definite integral of f (x) from a to b
b
and is denoted by ∫a f ( x) dx , where a is called the lower limit and b is called the upper
limit.
b
Thus, ∫a f ( x) dx = Lt [ f (a ) + f (a + 2h) + . . . + f (a + n − 1 h)] , where nh = b – a.
h→0

Cor.
b
∫a f ( x) dx = the area of the region below the curve y = f ( x) above the
x-axis and bounded by the ordinates x = a and x = b.
Remarks
For simplicity of the above concept, we have taken non-negative values of
f (x). In fact it makes sense for negative values of f (x) as well.
111
Calculus Example 8.21
b
Evaluate ∫a x 2 dx as the limit of a sum.

Solution
b
∫a x 2 dx = Lim h [ f (a ) + f (a + h) + . . . + f (a + n − 1 h)] , where nh = b – a and
h→0
2
f (x) = x .

i.e. ∫a
b
[
x 2 dx = Lim h a 2 + (a + h) 2 + (a + 2h) 2 + . . . + (a + n − 1 h) 2
h→0
]
= Lim h [(a 2
+ a 2 + . . . + a 2 ) + 2ah (1 + 2 + 3 + . . . + n − 1)
h→0

+ (12 + 2 2 + 32 + . . . + n − 1 ) h 2 ⎤
2
⎥⎦

⎡ ( n − 1) n h 2 (n − 1) n [ 2 (n − 1) + 1] ⎤
= Lim h ⎢na 2 + 2ah + ⎥
h→0 ⎢⎣ 2 6 ⎥⎦

⎡ 1 ⎤
= Lim ⎢a 2 nh + a ( nh) (nh − h) + nh (nh − h) (2nh − h)⎥
h→0 ⎣ 6 ⎦
⎡ 1 ⎤
= Lim ⎢a 2 (b − a) + a (b − a) (b − a − h) + (b − a) (b − a − h) (2b − a − h) ⎥
h→0 ⎣ 6 ⎦
1
= a 2 (b − a) + a (b − a) 2 + (b − a) 2 2 (b − a) (b − a)
6
1 1
= (b − a) [a 2 + a (b − a) + (b − a) 2 ] = (b 3 − a 3)
3 3
Example 8.22
2
Evaluate ∫0 e x dx as a limit of a sum.

Solution
2
Here b − a = 2 − 0 ∴ nh = 2, i.e. h = and f ( x) = e x
n
2
∫0 e x dx = Lim h [ f (a) + f (a + h) + . . . + f ( a + n − 1 h)]
h→0

= Lim h [e a + e a + h + e a + 2 h + . . . + e a + (n − 1) h]
h→0

= Lim h [e 0 + e h + e 2 h + . . . + e n − 1 h ] as a = 0
h→0

⎡ e nh − 1⎤
= Lim h ⎢ h ⎥ using the formula for the sum of a G. P.
h→0 ⎢⎣ e − 1 ⎥⎦
⎡ e 2 − 1⎤
= Lim h ⎢ h ⎥
h→0
⎣⎢ e − 1⎥⎦
e2 − 1
= = e2 − 1 .
eh − 1
Lim
h→0 h
112
Example 8.23 Integral Calculus
π
Evaluate ∫0 sin x dx as the limit of a sum.

Solution
Here f ( x) = sin x, a = 0, b = π
b−a π
∴ h= = , i.e. nh = π
n n
f (a) = f (0) = sin 0 = 0
f (a + h) = f (h) = sin h
f (a + 2h) = f (2h) = sin 2h
......................
f (a + n − 1 h) = sin n − 1 h

∴ ∫0
π
[
sin x dx = Lt h 0 + sin h + sin 2h + . . . + sin (n − 1 h)
h→0
]
h ⎡ h h h⎤
= Lt ⎢ 2 sin h sin + 2 sin 2h sin + . . . + 2 sin (nh − h) sin ⎥
h→0
2 sin ⎣
h 2 2 2⎦
2
1 ⎡⎛ h 3h ⎞ ⎛ 3h 5h ⎞
= Lt ⎜ cos − cos ⎟ + ⎜ cos − cos ⎟ + ... +
h→0 h ⎢⎣⎝ 2 2 ⎠ ⎝ 2 2 ⎠
sin
2
h
2
⎡ ⎛ 3h ⎞ ⎛ h ⎞⎤ ⎤
⎢cos ⎜ nh − 2 ⎟ − cos ⎜ nh − 2 ⎟⎥ ⎥
⎣ ⎝ ⎠ ⎝ ⎠⎦ ⎦

1⎡ h ⎛ h ⎞⎤
= Lt ⎢ cos − cos ⎜ nh − ⎟⎥
h→0
sin
h ⎣ 2 ⎝ 2 ⎠⎦
2
h
2
h
⎡ sin
h ⎛ h ⎞⎤ 2 =1
= 1 . Lt ⎢cos − cos ⎜ π − ⎟⎥ as nh = π and Lt
h→0 ⎣ 2 ⎝ 2 ⎠⎦ h→0 h
2
⎡ h h⎤
= Lt ⎢cos + cos ⎥ = 1 + 1 = 2 .
h→0 ⎣ 2 2⎦
SAQ 7

Evaluate the following definite integrals as a limit of a sum :

b
(i) ∫a e x dx

b
(ii) ∫a cos x dx

2
(iii) ∫1 (x 2 − 1) dx
113
Calculus
8.9 FUNDAMENTAL THEOREM OF CALCULUS
8.9.1 Area Function
b
We have defined ∫a f ( x) dx as the area of the region bounded by the curve y = f ( x) ,
x-axis and the ordinates x = a and x = b. Let x ∈ [a, b] .
x
Then ∫a f ( x) dx represents the area of the shaded region in Figure 8.2.

(Here it is assumed that f ( x) > a for x ∈ [a, b] .)

Figure 8.2
The area of this shaded region depends on x, i.e. in other words is a function of x. We
denote it by A (x)
x
∴ A ( x) = ∫a f ( x) dx

We will now state two fundamental theorems of integral calculus.

8.9.2 First Fundamental Theorem of Integral Calculus
x
Let the area function be defined by A ( x) = ∫a f ( x) dx for all x ≥ a , where the function

f is continuous on [a, b]. Then A′ ( x) = f ( x) for all x ∈ [a, b] .

(We assume it without proof.)

8.9.3 Second Fundamental Theorem of Integral Calculus
Let f be a continuous function defined on an interval [a, b] and F is an antiderivative of f,
then
b
∫a f ( x) dx = F (b) − F (a)

(We assume it without proof.)

Remarks
b
(i) ∫a f ( x) dx = (value of an antiderivative at the upper limit b) – (value of the
same antiderivative at the lower limit a).
(ii) This theorem is very useful as it gives us a method of calculating a definite
integral more easily without calculating the limit of a sum.

For convenience F (b) − F (a) is denoted by F ( x) |ba .

114
 (iii) If we consider F (x) + c to be an antiderivative value of f (x) instead of Integral Calculus
F (x), then
b
∫a f ( x) dx = [ F ( x) + c]ba = ( F (b) + c) − ( F (a) − c)

= F (b) − F (a)
Hence, there is no need to keep the integration constant c in definite
integrals.
Example 8.24
3
4
Evaluate ∫0 x2 dx

Solution
5
3 5
x2 2 2
∫ x2 dx =
5
= x
5
2

4
3
2
5
2 ⎛ 5 ⎞
⎜ 4 2 − 0⎟
∫0
4
∴ x2 dx = x 2 =
5 0 5 ⎜ ⎟
⎝ ⎠
2 5 64
= .2 =
5 5
Example 8.25
1 tan −1 x
Evaluate ∫0 1 + x2
dx

Solution
tan −1 x (tan −1 x) 2
∫ 1 + x2
dx =
2

tan −1 x t2
(Let tan −1 x = t, then ∫ 1 + x2
dx = ∫ t dt =
2
)

1 tan −1 x 1
∫0
1
∴ 2
dx = (tan −1 x) 2 0
1+ x 2

[ 1 ⎡⎛ π ⎞
]
⎤
2
1
= (tan −1 1) 2 − (tan −1 0) 2 = ⎢⎜ ⎟ − 0⎥
2 2 ⎢⎣⎝ 4 ⎠ ⎥⎦

1 π2 π2
= . =
2 16 32
Example 8.26

π
sin 2 x
Evaluate ∫ 0
2
sin x + cos 4 x
4
dx

Solution
π π
sin 2 x 2 sin x cos x
∫ 0
2
4
sin x + cos x 4
dx = ∫ 0
2
sin 4 x + cos 4 x
dx
115
Calculus π
2 tan x sec 2 x
= ∫ 0
2
tan 4 x + 1
dx

2 tan x sec 2 x
Now ∫ tan 4 x + 1
dx = tan −1 (tan 2 x)

2 tan sec 2 x dt
(Put tan 2 x = t ∴ 2 tan x sec 2 x dx = dt and ∫ 2
tan x + 1
= ∫ 2
t +1
= tan −1 t )

π π
sin 2 x
∫
−1 2
∴ 2 dx = tan (tan x) 2
0
0 sin 4 x + cos 4 x

⎛ π⎞
= tan −1 ⎜ tan 2 ⎟ − tan −1 (tan 2 0)
⎝ 2⎠
π π
= −0= .
2 2
8.9.4 Evaluation of a Definite Integral by Substitution
a
When we use the method of substitution for evaluating an integral ∫b f (x) dx , we
follow the following steps :
Step 1
Substitute x = g (y).
Step 2
Integrate the new integrand with respect to y.
Step 3
Resubstitute the value of y in terms of x in the answer.
Step 4
Find the value of the answer in Step 3 at the given limits and find the difference.
In order to quicken this method we can proceed as follows :
After performing Step 2, there is no need for Step 3. Instead the integral will be kept in
the new variable y and the limit of the integral will be accordingly changed.
Example 8.27
1
Evaluate ∫−1 5 x 4 x 5 + 1 dx .

Solution
Let t = x 5 + 1, then dt = 5 x 4 dx

When x = 1, t = 15 + 1 = 2 and when x = − 1 , t = (− 1)5 + 1 = − 1 + 1 = 0 .

1 2
∴ ∫−1 5 x 4 x 5 + 1 dx = ∫0 t dt

3
2 2 2
= t 0
3

2 ⎛ 3 ⎞
= ⎜ 2 2 − 0⎟ = 4 2 .
3 ⎜ ⎟ 3
⎝ ⎠
116
SAQ 8 Integral Calculus

(a) Evaluate
π
(i) ∫ −
2
π cos x dx
2

2 6x + 3
(ii) ∫0 x2 + 4
dx

π ⎛ 2 x x⎞
(iii) ∫0 ⎜ sin
⎝ 2
− cos 2 ⎟ dx
2⎠
(b) Evaluate
π
(i) ∫ 0
2 sin x cos x dx

π dx
(ii) ∫0 5 + 4 cos x
π
dx
(iii) ∫ 0
2
2 cos x + 4 sin x

8.10 PROPERTIES OF DEFINITE INTEGRALS

We consider below some important properties of the definite integral. These will be
useful in evaluating the definite integrals more easily.
Property 1
b a
∫a f ( x) dx = − ∫b f ( x) dx

Property 2
b c b
∫a f ( x) dx = ∫a f ( x) dx + ∫c f ( x) dx for a < c < b

Property 3
a a
∫0 f ( x) dx = ∫0 f (a − x) dx

Property 4
2a a
∫0 f ( x) dx = 2 ∫0 f ( x) dx if f (2a − x) = f ( x)

=0 if f (2a − x) = − f ( x)
Property 5
a a
∫− a f ( x) dx = 2 ∫0 f ( x) dx if f is an even function.

=0 if f is an odd function.
We give proof of these properties. 117
Calculus Property 1
Let F be an antiderivative of f.
b a
Then ∫a f ( x) dx = F (b) − F (a) = − [ F (a) − F (b)] = − ∫b f ( x) dx

Property 3
Let t=a–x
Then dt = − dx
When x = 0, t = a and when x = a, t = 0
a 0
∴ ∫0 f ( x) dx = − ∫a f (a − t) dt

a
=+ ∫0 f (a − t) dt by Property 1

a
=+ ∫0 f (a − x) dx by changing the variable t to x.

Property 4
2a a 2a
∫0 f ( x) dx = ∫0 f ( x) dx + ∫a f ( x) dx by Property 2

Put t = 2a − x in the second integral

2a 0
Then ∫a f ( x) dx = − ∫a f (2a − t) dt

a a
= ∫0 f (2a − t) dt = ∫0 f (2a − x) dx

2a a a
∴ ∫0 f ( x) dx = ∫0 f ( x) dx + ∫0 f (2a − x) dx

a
=2 ∫0 f ( x) dx or 0

according as f (2a − x) = f ( x)
or f (2a − x) = − f ( x)
Property 2 and Property 5 are left as exercises.
Example 8.28
π
sin x
Evaluate ∫ 0
2
sin x + cos x
dx

Solution
π
sin x
Let I= ∫ 0
2
sin x + cos x
dx . . . (1)

by Property 3

⎛π ⎞
π sin ⎜ − x ⎟
⎝2 ⎠
I= ∫ 0
2
⎛π ⎞ ⎛π ⎞
dx
sin ⎜ − x ⎟ + cos ⎜ − x ⎟
⎝2 ⎠ ⎝2 ⎠

118
 π Integral Calculus
cos x
= ∫ 0
2
cos x + sin x
dx . . . (2)

Adding Eqs. (1) and (2), we have

π
sin x + cos x
2I = ∫ 0
2
sin x + cos x
dx

π π
π
= ∫ 0
2 dx = x 2
0
=
2
π
∴ I=
4
Example 8.29
π
Evaluate ∫ −
4
π cos 2 x dx
4

Solution

cos 2 x is an even function.

∴ by Property 5
π π

∫ −
4
π cos 2 x dx = 2 ∫ 0
4 cos 2 x dx
4

π
1 + cos 2 x
=2 ∫ 0
4
2
dx

π
2 sin 2 x 4
= x+
2 2 0

⎛π 1 π⎞ ⎛ 1 ⎞ π 1
= ⎜ + sin ⎟ − ⎜ 0 + sin 0 ⎟ = +
⎝4 2 2⎠ ⎝ 2 ⎠ 4 2
Example 8.30
π
Evaluate ∫ 0
2 log sin x dx

Solution
π
Let I= ∫ 0
2 log sin x dx

π
⎛π ⎞
Then I= ∫ 0
2 log sin ⎜ − x ⎟ dx
⎝ 2 ⎠
π
= ∫ 0
2 log cos x dx

π
∴ 2I = ∫ 0
2 {log sin x + log cos x} dx

π
= ∫ 0
2 (log sin x cos x) dx
119
Calculus π
⎛ 2 sin x cos x ⎞
= ∫ 0
2 log ⎜
⎝ 2
⎟ dx
⎠
π
sin 2 x
= ∫ 0
2 log
2
dx

π π
= ∫ 0
2 log sin 2 x dx − ∫ 0
2 log 2 dx

π π
= ∫ 0
2 log sin 2 x dx − log 2 . x 2
0

π
π
= ∫ 0
2 log sin 2 x dx −
2
log 2 . . . (1)

π
To evaluate ∫ 0
2 log sin 2 x dx ,

π
1 π
Putting 2x = t, we have ∫ 0
2 log sin 2 x dx =
2 ∫0 log sin t dt

1 π
=
2 ∫0 log sin x dx

π
1
= .2
2 ∫ 0
2 log sin x dx , as sin ( π − x) = sin x

=I . . . (2)
From Eqs. (1) and (2), we have
π
∴ 2I = I − log 2
2
π
i.e. I =− log 2
2
SAQ 9
Evaluate
π
(i) ∫ 0
4 log (1 + tan x) dx

π
cos x
(ii) ∫ 0
2
cos x + sin x
dx

π
x dx
(iii) ∫ 0
2
sin x + cos x
π
(iv) ∫ −
2
π cos 4 x dx
2

1 log (1 + x)
(v) ∫0 1 + x2
dx

120
 Integral Calculus
8.11 APPLICATIONS
We have seen that the area below (or above) the curve y = f (x), above (or below) the
x-axis and between the ordinates x = a and x = b is represented by the definite integral
b b
∫a f ( x) dx = ∫a y dx

Likewise the area enclosed between the graph of the curve x = F (y), y-axis and the lines
y = c, y = d is given by
d d
∫c F ( y) dy = ∫c x dy

Example 8.31

Draw a rough sketch of the curve y = 3 x + 4 and find the area under the curve,
above the x-axis and between x = 0, x = 4.
Solution
y = 3x + 4
4
∴ Its domain consists of those x for which 3 x + 4 ≥ 0 , i.e. x ≥ − .
3
We construct the table of values as under
x 4 −1 0 1 2 3 4
−
3
y 0 1 2 4
7 10 13

A portion of the rough sketch of curve is shown in Figure 8.3.

4
Required area is the shaded area = ∫0 f ( x) dx .

4
= ∫0 3x + 4 dx

3
(3x + 4) 2 2 ⎛ 3 3⎞
=
4
= ⎜16 2 − 4 2 ⎟
3 0 9 ⎜ ⎟
.3 ⎝ ⎠
2

2⎡ 2 2
3 3⎤
= ⎢( 4 ) − ( 2 2 ) 2 ⎥
9 ⎢ ⎥⎦
⎣
2 3 2 112
= [ 4 − 23 ] = [64 − 8] = sq. units
9 9 9

Figure 8.3 121

Calculus Example 8.32
Make a rough sketch of the graph of the function y = 3 sin x, 0 ≤ x ≤ π and
determine the area enclosed by the curve and the x-axis.
Solution
We construct the table of values as under
X 0 π π π 2π 5π π
6 3 2 3 6
Y 0 3 3 3 3 3 3 3 0
2 2 2 2

A rough sketch of the curve is shown in Figure 8.4.

π
Required Area = ∫0 f ( x) dx

π
= ∫0 3 sin x dx

π
= [3 ( − cos x)] 0
= − 3 [cos π − cos 0o ]
= − 3 (− 1 − 1) = 6 sq. units

Figure 8.4
π
Note : Since the curve is symmetrical about the line x = .
2
∴ Required Area = 2 Area OAM
π π
=2 ∫ 0
2 f (x) dx = 2 ∫ 0
2 3 sin x dx

π
⎛ π ⎞
= − 6 cos x 2
0
= − 6 ⎜ cos − cos 0 o ⎟
⎝ 2 ⎠
= − 6 (0 − 1) = 6 sq. units
Remark
In case of symmetrical closed area, find the area of the smaller part and multiply
the result by the number of symmetrical parts.
Example 8.33
x2 y2
Find the area enclosed between the ellipse + = 1 and above the line
a2 b2
x y
+ = 1 which lies in the first quadrant.
a b
122
Solution Integral Calculus

x2 y2
The given ellipse is + =1 . . . (1)
a2 b2
x y
and the line is + =1 . . . (2)
a b
Line (2) meets the curve (1) in A (a, 0) and B (0, b). The required area is shown in
Figure 8.5.

Figure 8.5
For the ellipse
y2 x2 a2 − x2
=1− =
b2 a2 a2
b
i.e., y=± a2 − x2
a
i.e. for the first quadrant
b
y= a2 − x2
a
Shaded Area = Area OATB – Area of the triangle OAB
1 1
Area of the triangle OAB = OA . OB = ab
2 2
Area OATB = Area bounded by the ellipse, x-axis in the first quadrant.
a a b
= ∫0 y dx = ∫0 a
a 2 − x 2 dx

b ⎡ x a2 − x2 a2 x⎤
a
= ⎢ + sin −1 ⎥
a ⎢ 2 2 a⎥
⎣ ⎦ 0

b
= [0 + a 2 sin −1 1 − (0 + a 2 sin −1 0)]
2a
b ⎡ 2 π 2 ⎤ π ab
= ⎢ a 2 − a . 0⎥ = 4
2a ⎣ ⎦
π ab 1 (π − 2) ab
Required area = − ab = sq. units.
4 2 4
Example 8.34
Find the area of the region bounded by the parabola y = x 2 + 2 and the lines
y = x, x = 0, x = 3.
123
Calculus Solution
y = x is the equation of a straight line lying below the parabola and the line
x = 3 meets the parabola at (3, 11). The line y = x meets the line x = 3 at
(3, 3). The region whose area is required is shaded and shown in Figure 8.6.

Figure 8.6
Required Area = Area bounded by the parabola, x-axis and the ordinates
x = 0, x = 3 − (Area bounded by the line y = x, x-axis and the ordinates
x = 0, x = 3).
3 3
∫0 (x ∫0
2
= + 2) dx − x dx

⎛ x3 ⎞ 3 x2 3
=⎜ + 2x ⎟ −
⎜ 3 ⎟ 0 2 0
⎝ ⎠
9 21
=9+6− = sq. units
2 2
Note : Area bounded by the line y = x, x-axis and the ordinates at x = 0, and x = 3
1 1 9
is also the area of the triangle OAM = OM . AM = . 3.3 = .
2 2 2
SAQ 10
Find the area of the regions
(i) bounded by y 2 = 9 x, x = 2 and x = 4 and the x-axis in the first quadrant.

(ii) bounded by x 2 = y − 3, y = 4, y = 6 and the y-axis in the first quadrant.

x2 y2
(iii) bounded by the ellipse + =1.
16 9
(iv) bounded by the circle x 2 + y 2 = 4 , the line x = 3 y , x-axis lying in the
first quadrant.
(v) bounded by the curve x 2 = 4 y and the line x = 4 y − 2 .

(vi) enclosed between the circles x 2 + y 2 = 1 and ( x − 1) 2 + y 2 = 1 .

124
 Integral Calculus
8.12 SUMMARY
The main points covered in this unit are
• Given the derivative of a function, the process to find the function is called
antidifferentiation and the result of antidifferentiation is called an
antiderivative.
• The indefinite integral ∫ f (x) dx denotes the class of all antiderivatives of
f (x).
• (a) ∫ K f (x) dx = K ∫ f (x) dx

(b) ∫ [f (x) ± g ( x)] dx = ∫ f (x) dx ± ∫ g (x) dx

• By the method of substitution :

(a) ∫ f (x) dx = ∫ f [φ (t)] φ′ (t) dt

[f (x)]n + 1
(b) ∫ [f (x)]n f ′ ( x) dx =
n +1
, where n + 1 ≠ 0

f ′ (x)
(c) ∫ f (x)
dx = ln | f (x) | .

• By the method of integration by parts :

Integral of product of two functions = first function × integral of second
function – integral of (derivative of first function × integral of second
function).
P (x)
• A rational function f of x is given by f (x) = , where P (x) and
Q (x)
Q (x) are polynomial in x. It is called proper if the degree of P (x) is less
than the degree of Q (x). Otherwise it is called improper.
• To integrate a proper rational function, we decompose the denominator into
either linear or quadratic factors.
• Some rules to integrate irrational functions are
(a) To integrate functions containing only fractional powers of x, put
x = tn, where n is l. c. m. of denominators of powers of x.
(b) Rational functions of sin x or cos x or both can be reduced to a
x
rational function of t by substituting tan = t and then can be
2
integrated.
b
• If f is continuous on [a, b], then ∫a f ( x) dx represents the area of the
region bounded by the curve y = f (x) , x-axis and the ordinates x = a,
x = b.
n
b−a
∑
b
• ∫a f ( x) dx = Lim h
h→0
i =1
f [a + (i − 1) h] , where h =
n
.

• Fundamental theorem of calculus

x
(i) If f is continuous on [a, b] then for all x ∈ [a, b] if A ( x) ∫a f ( x) dx

then A′ ( x) = f ( x) for all x ∈ [a, b] .

125
Calculus (ii) If f is continuous function on [a, b] and F is an antiderivative of f then
b
∫a f ( x) dx = F (b) − F (a) .

• Area bounded by a curve y = f (x) , x-axis and the lines x = a, x = b is

b b
∫a f ( x) dx = ∫a y dx .

• Area bounded by a curve x = g (y), y-axis and the lines y = c, y = d is

d d
∫c g ( y ) dy = ∫c x dy .

8.13 ANSWERS TO SAQs

SAQ 1

(i) 5x 2 + c

(ii) x11 + c

− 5x2
(iii) +c
2
SAQ 2
x9
(i) (a) +c
9
2 −3/ 2
(b) − x +c
3
(c) − 4 x −1 + c
(d) 9x + c
x3 x 2
(ii) (a) − −x+c
3 2
(b) 2 x1 / 2 − 2 x 3 / 2 + c

x3 1
(c) − 2x − + c
3 x
(iii) (a) e x − e− x + 4x + c

x2
(b) 4 sin x + 3 cos x + e x + +c
2
(c) 4 tanh x + e x − 4 x 2 + c

(iv) (a) 2 sin −1 x + 5 ln | x | + c

2 ( x 2 + 1) + 3
(b) ∫ x2 + 1
dx

1
=2 ∫ dx + 3 ∫ x +12
dx

= 2 x + 3 tan −1 x + c
126
 ax 4 bx 3 cx 2 Integral Calculus
(v) (a) + + + dx + c1
4 3 2
x2
(b) − 2 x + ln | x | + c
2
sin 4 x + cos 4 x (sin 2 x + cos 2 x) − 2 sin 2 x cos 2 x
(vi) (a) ∫ sin 2 x cos 2 x
dx = ∫ sin 2 x cos 2 x
dx

1 − 2 sin 2 x cos 2 x
= ∫ sin 2 x cos 2 x
dx

1 1
= ∫ 2
sin x
dx + ∫ cos 2 x
dx − 2 ∫ dx = − cot x + tan x − 2 x + c

3 5
4 3 2
(b) 6x − x 2 + x2 − x 2 + c
3 2 5
SAQ 3
1 3x − 2
(a) (i) tan −1 +c
6 3

1 ⎛ 2x2 + 1 ⎞
(ii) tan −1 ⎜ ⎟+c
3 ⎜ 3 ⎟⎠
⎝
1
(iii) tan 6 x + c
6
(iv) x − ln | (1 + e x ) | + c
(v) ln | (ln | sin x | ) | + c

(vi) log | (e x − 1) | − x + c
1 1
(vii) − 2
+c
2 (e + 1)
x

1
(viii) tan x 2 + c
2
(ix) Put sin −1 x = t

1 (sin −1 x) 2
So
1− x 2
dx = dt and ∫ 1− x 2
dx = ∫ t 2 dt

t3 1
= + c = (sin −1 x)3 + c
3 3
1
(x) (1 + log x) 4 + c
4
(xi) − ln | (1 + cot x) | + c

x3
(b) (i) (3 ln | x | − 1) +c
9
(ii) ln | sin x) | − x cot x + c

e 3 x (4 sin 4 x + 3 cos 4 x)
(iii) +c
25
127
Calculus
(iv) x sin −1 x + 1 − x 2 + c

x2 1 x2
(v) ∫ x tan −1 x dx = ( tan −1 x) .
2
−
2 ∫ 1 + x2
dx

x2 ⎛ ⎞
1 ⎜1 − 1 ⎟ dx
=
2
tan −1 x −
2 ∫ ⎜
⎝ 1 + x 2 ⎟⎠

x2 1 1
= tan −1 x − x + tan −1 x + c
2 2 2
1 2 1
= (x + 1) tan −1 x − x + c
2 2

(vi) − φ cos φ + sin φ + c, where φ = sin −1 x

ex
(vii)
1+ x

SAQ 4
3 1
(i) ln | x − 3| + ln | x + 1| + c
4 4
(ii) 4 ln | x + 5 | − ln | x − 2 | + c

6 x 2 + 22 x − 23 6 x 2 + 22 x − 23 A B C
(iii) = = + +
(2 x − 1) ( x 2 + x − 6) (2 x − 1) ( x + 3) ( x − 2) 2 x − 1 x + 3 x − 2

6 x 2 + 22 x − 23 = A (x + 3) ( x − 2) + B (x − 2) (2 x − 1) + C (2 x − 1) ( x + 3)
x=2⇒C =3
x = − 3⇒ B = −1

1
x= ⇒ A =1
2

6 x 2 + 22 x − 23 1
∴ ∫ 2
(2 x − 1) ( x + x − 6)
dx =
2
ln | 2 x − 1| − ln | x + 3| + 3 ln | x − 2 | + c

x2 + x − 1 A Bx + C
(iv) dx = + 2
2
( x − 1) ( x − x + 1) x −1 x − x +1

∴ x 2 + x − 1 = A (x 2 − x + 1) + ( Bx + C ) (x − 1)

x =1⇒ A =1

∴ We have

x 2 + x − 1 = x 2 − x − 1 + Bx 2 + (C − B ) x − C

Thus 1=1+B
∴ B=0
Also −1=1−C
∴ C=2
128
 x2 + x − 1 dx dx Integral Calculus
∴ ∫ ( x − 1) ( x 2 − x + 1)
dx = ∫ x −1
+2 ∫ x2 − x + 1
4 2x − 1
= ln | x − 1| + tan −1 +c
3 3
x2 x2 − a2 + a2
(v) ∫ x2 − a2
dx = ∫ x2 − a2
dx

a2
= ∫ dx +
x2 − a2
dx

a2 ⎧ 1 1 ⎫
= ∫ dx +
2a ∫ ⎨ − ⎬dx
⎩x − a x + a⎭
a x−a
=x+ ln +c
2 x+a

x2 + 4 ⎧⎪ 2 x − 1 ⎫⎪
(vi) ∫ 2
x + 2x + 3
dx = ∫ ⎨1 − 2
⎪⎩
⎬dx
x + 2 x + 3 ⎪⎭
2x − 1
= ∫ dx − ∫ 2
x + 2x + 3
dx

2x − 2 3
= ∫ dx − ∫ 2
x + 2x + 3
dx + 2
x + 2x + 3
dx

3 x +1
= x − ln | x 2 + 2 x + 3| + tan −1 +c
2 2
SAQ 5
x+2
(i) x 2 + 4 x + 6 + log ⎛⎜ x + 2 + x 2 + 4 x + 6 ⎞⎟ + c
2 ⎝ ⎠
3
1 x 3 2 ⎛ 3 ⎞⎟
(ii) (2 x 2 + 3) 2 + 2x2 + 3 + log ⎜ x + x2 + +c
6 2 4 ⎜
⎝ 2 ⎟⎠

2x − 3 13 ⎛ 2x − 3 ⎞
(iii) 1 + 3x + x 2 + sin −1 ⎜⎜ ⎟⎟ + c
4 8 ⎝ 13 ⎠
SAQ 6
⎛ x ⎞
⎜ 5 tan + 4 ⎟
2 2
(i) tan −1 ⎝ ⎠ +c
3 3
2 ⎛ 1 θ⎞
(ii) tan −1 ⎜⎜ tan ⎟⎟ + c
3 ⎝ 3 2⎠

⎛ x⎞
(iii) log ⎜1 + tan ⎟ + c
⎝ 2 ⎠
SAQ 7
(i) cb − ea
(ii) sin b – sin a
4
(iii)
3 129
Calculus SAQ 8
(a) (i) 2
3π
(ii) 3 log 2 +
8
(iii) 0
2
(b) (i)
3
π
(ii)
3

1 ⎛3 + 5 ⎞
(iii) log ⎜ ⎟
5 ⎜ 2 ⎟
⎝ ⎠
SAQ 9
π
(i) log 2
8
π
(ii)
4
π
(iii) log (1 + 2 )
2 2
3π
(iv)
8
π
(v) log 2
8
SAQ 10
(i) 16 − 4 2
2
(ii) (3 3 − 1)
3
(iii) 12 π
π
(iv)
3
9
(v)
8
2π 3
(vi) −
3 2

130