INTEGRALS 225

Chapter 7
INTEGRALS

v Just as a mountaineer climbs a mountain – because it is there, so

a good mathematics student studies new material because
it is there. — JAMES B. BRISTOL v

7.1 Introduction
Differential Calculus is centred on the concept of the
derivative. The original motivation for the derivative was
the problem of defining tangent lines to the graphs of
functions and calculating the slope of such lines. Integral
Calculus is motivated by the problem of defining and
calculating the area of the region bounded by the graph of
the functions.
If a function f is differentiable in an interval I, i.e., its
derivative f ′exists at each point of I, then a natural question
arises that given f ′at each point of I, can we determine
the function? The functions that could possibly have given
function as a derivative are called anti derivatives (or G .W. Leibnitz
primitive) of the function. Further, the formula that gives (1646 -1716)
all these anti derivatives is called the indefinite integral of the function and such
process of finding anti derivatives is called integration. Such type of problems arise in
many practical situations. For instance, if we know the instantaneous velocity of an
object at any instant, then there arises a natural question, i.e., can we determine the
position of the object at any instant? There are several such practical and theoretical
situations where the process of integration is involved. The development of integral
calculus arises out of the efforts of solving the problems of the following types:
(a) the problem of finding a function whenever its derivative is given,
(b) the problem of finding the area bounded by the graph of a function under certain
conditions.
These two problems lead to the two forms of the integrals, e.g., indefinite and
definite integrals, which together constitute the Integral Calculus.

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There is a connection, known as the Fundamental Theorem of Calculus, between

indefinite integral and definite integral which makes the definite integral as a practical
tool for science and engineering. The definite integral is also used to solve many interesting
problems from various disciplines like economics, finance and probability.
In this Chapter, we shall confine ourselves to the study of indefinite and definite
integrals and their elementary properties including some techniques of integration.
7.2 Integration as an Inverse Process of Differentiation
Integration is the inverse process of differentiation. Instead of differentiating a function,
we are given the derivative of a function and asked to find its primitive, i.e., the original
function. Such a process is called integration or anti differentiation.
Let us consider the following examples:
d
We know that (sin x) = cos x ... (1)
dx
d x3
( ) = x2 ... (2)
dx 3
d x
and (e ) = e x ... (3)
dx
We observe that in (1), the function cos x is the derived function of sin x. We say
x3
that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3),and
3
ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note that
for any real number C, treated as constant function, its derivative is zero and hence, we
can write (1), (2) and (3) as follows :
d d x3 d x
(sin x + C) = cos x , ( + C) = x 2 and (e + C) = e x
dx dx 3 dx
Thus, anti derivatives (or integrals) of the above cited functions are not unique.
Actually, there exist infinitely many anti derivatives of each of these functions which
can be obtained by choosing C arbitrarily from the set of real numbers. For this reason
C is customarily referred to as arbitrary constant. In fact, C is the parameter by
varying which one gets different anti derivatives (or integrals) of the given function.
d
More generally, if there is a function F such that F (x) = f (x) , ∀ x ∈ I (interval),
dx
then for any arbitrary real number C, (also called constant of integration)
d
[ F (x) + C] = f (x), x ∈ I
dx

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 INTEGRALS 227

Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f.

Remark Functions with same derivatives differ by a constant. To show this, let g and h
be two functions having the same derivatives on an interval I.
Consider the function f = g – h defined by f (x) = g (x) – h(x), ∀ x ∈ I
df
Then = f′ = g′ – h′ giving f′ (x) = g′ (x) – h′ (x) ∀ x ∈ I
dx
or f ′ (x) = 0, ∀ x ∈ I by hypothesis,
i.e., the rate of change of f with respect to x is zero on I and hence f is constant.
In view of the above remark, it is justified to infer that the family {F + C, C ∈ R}
provides all possible anti derivatives of f.
We introduce a new symbol, namely, ∫ f (x) dx which will represent the entire
class of anti derivatives read as the indefinite integral of f with respect to x.
Symbolically, we write ∫ f (x) dx = F (x) + C .
dy
Notation Given that
dx
= f (x ) , we write y = ∫ f (x) dx .
For the sake of convenience, we mention below the following symbols/terms/phrases
with their meanings as given in the Table (7.1).

Table 7.1
Symbols/Terms/Phrases Meaning

∫ f (x) dx Integral of f with respect to x

f (x) in ∫ f (x) dx Integrand

x in ∫ f (x) dx Variable of integration

Integrate Find the integral
An integral of f A function F such that
F′(x) = f (x)
Integration The process of finding the integral
Constant of Integration Any real number C, considered as
constant function

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We already know the formulae for the derivatives of many important functions.
From these formulae, we can write down immediately the corresponding formulae
(referred to as standard formulae) for the integrals of these functions, as listed below
which will be used to find integrals of other functions.
Derivatives Integrals (Anti derivatives)

d  xn + 1  x n+1
=x ;
n
 ∫ x dx = + C , n ≠ –1
n
(i)
dx  n + 1  n +1
Particularly, we note that
d
( x) = 1 ; ∫ dx = x + C
dx
d
(ii) ( sin x ) = cos x ; ∫ cos x dx = sin x + C
dx
d
(iii) ( – cos x ) = sin x ; ∫ sin x dx = – cos x + C
dx
d
(iv) ( tan x ) = sec2 x ; ∫ sec
2
x dx = tan x + C
dx
d
(v) ( – cot x ) = cosec2 x ; ∫ cosec
2
x dx = – cot x + C
dx
d
(vi) ( sec x ) = sec x tan x ; ∫ sec x tan x dx = sec x + C
dx
d
(vii) ( – cosec x ) = cosec x cot x ; ∫ cosec x cot x dx = – cosec x + C
dx
d
(
(viii) dx sin x =
–1 1
)
; ∫
dx
= sin – 1 x + C
1 – x2 1– x 2

d
(
(ix) dx – cos x =
–1 1
; ) ∫
dx
= – cos – 1 x + C
1 – x2 1– x 2

d
(
(x) dx tan x =
–1 1
)
1 + x2
;
dx
∫ 1 + x 2 = tan
–1
x+C

d x
(e ) = e x ; ∫ e dx = e +C
x x
(xi)
dx

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d 1 1
(xii) ( log | x |) = ; ∫ x dx = log | x | +C
dx x
d  ax  ax
=a ; ∫ a dx =
x
(xiii) 
x
+C
dx  log a  log a

A Note In practice, we normally do not mention the interval over which the various
functions are defined. However, in any specific problem one has to keep it in mind.
7.2.1 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals.
(I) The process of differentiation and integration are inverses of each other in the
sense of the following results :
d
dx ∫
f (x) dx = f (x)

and ∫ f ′(x) dx = f (x) + C, where C is any arbitrary constant.

Proof Let F be any anti derivative of f, i.e.,
d
F(x ) = f (x)
dx

Then ∫ f (x) dx = F(x) + C

d d
Therefore ∫ f (x) dx = ( F (x) + C )
dx dx

d
= F (x ) = f (x)
dx
Similarly, we note that
d
f ′(x) = f (x)
dx

and hence ∫ f ′(x) dx = f (x) + C

where C is arbitrary constant called constant of integration.
(II) Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent.

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Proof Let f and g be two functions such that

d d
∫ dx ∫
f (x) dx = g (x ) dx
dx
d 
f (x) dx – ∫ g (x ) dx  = 0
dx  ∫
or

Hence ∫ f (x) dx – ∫ g (x) dx = C, where C is any real number (Why?)

or ∫ f (x) dx = ∫ g (x) dx + C
So the families of curves {∫ f (x) dx + C , C ∈ R}
1 1

and {∫ g (x) dx + C , C ∈ R} are identical.

2 2

Hence, in this sense, ∫ f (x) dx and ∫ g (x) dx are equivalent.

A Note The equivalence of the families {∫ f (x) dx + C ,C ∈ R} and 1 1

{∫ g (x) dx + C ,C ∈ R} is customarily expressed by writing ∫ f (x) dx = ∫ g (x) dx ,

2 2

without mentioning the parameter.

(III) ∫ [ f (x) + g (x)] dx = ∫ f (x) dx + ∫ g (x) dx

Proof By Property (I), we have
d 
[ f (x ) + g (x )] dx  = f (x) + g (x)
dx  ∫
... (1)
On the otherhand, we find that
d  d d
 ∫ f (x ) dx + ∫ g (x) dx  = ∫ f (x) dx + dx ∫ g (x) dx
dx dx
= f (x) + g (x) ... (2)
Thus, in view of Property (II), it follows by (1) and (2) that

∫ ( f (x) + g (x) ) dx = ∫ f (x) dx + ∫ g (x) dx .

(IV) For any real number k, ∫ k f (x ) dx = k ∫ f (x) dx

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 INTEGRALS 231

d
dx ∫
Proof By the Property (I), k f (x) dx = k f (x) .

d  d
Also
dx 
k ∫ f (x) dx  = k
dx ∫ f (x) dx = k f (x)

Therefore, using the Property (II), we have ∫k f (x ) dx = k ∫ f (x) dx .

(V) Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2, ..., fn and the real numbers, k1, k2, ..., kn giving

∫ [k1 f1 (x) + k2 f 2 (x) + ... + kn f n (x)] dx

= k1 ∫ f1 (x) dx + k 2 ∫ f 2 (x) dx + ... + kn ∫ fn (x) dx .
To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function. The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection. We illustrate it
through some examples.
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
1
(i) cos 2x (ii) 3x2 + 4x3 (iii) ,x≠0
x
Solution
(i) We look for a function whose derivative is cos 2x. Recall that
d
sin 2x = 2 cos 2x
dx
1 d d 1 
or cos 2x = (sin 2x) =  sin 2 x 
2 dx dx  2 
1
Therefore, an anti derivative of cos 2x is sin 2 x .
2
(ii) We look for a function whose derivative is 3x2 + 4x3. Note that
d 3
dx
( )
x + x 4 = 3x2 + 4x3.

Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4.

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(iii) We know that

d 1 d 1 1
(log x ) = , x > 0 and [log ( – x)] = ( – 1) = , x < 0
dx x dx –x x
d 1
Combining above, we get
dx
( log x ) = , x ≠ 0
x
1 1.
Therefore, ∫ x dx = log x is one of the anti derivatives of
x
Example 2 Find the following integrals:
3
x3 – 1 2 1
(i) ∫ 2 dx (iii) ∫ (x 2 + 2 e – ) dx
x

x
(ii)
∫ (x 3 + 1) dx x
Solution
(i) We have

x3 – 1
∫ x2 dx = ∫ x dx – ∫ x dx
–2
(by Property V)

 x1 + 1   x– 2 + 1 
= + C1 –  + C 2  ; C , C are constants of integration
 1+1   – 2 +1  1 2

x2 x– 1 x2 1
= + C1 – – C2 = + + C1 – C 2
2 –1 2 x
x2 1
= + + C , where C = C1 – C2 is another constant of integration.
2 x

A Note From now onwards, we shall write only one constant of integration in the
final answer.
(ii) We have
2 2

∫ (x 3 + 1) dx = ∫ x 3 dx + ∫ dx
2
+1
5
x3
= 2 + x + C = 3 x3 + x + C
+1 5
3

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 INTEGRALS 233

3 3
1 1
∫ + 2 e – ) dx = ∫ x 2 dx + ∫ 2 e x dx – ∫ dx
x
(iii) We have (x 2
x x
3
+1
x2
= 3 + 2 e x – log x + C
+1
2
5
2
= x 2 + 2 e x – log x + C
5
Example 3 Find the following integrals:
(i) ∫ (sin x + cos x) dx (ii) ∫ cosec x (cosec x + cot x) dx
1 – sin x
(iii) ∫ cos2 x
dx

Solution
(i) We have
∫ (sin x + cos x) dx = ∫ sin x dx + ∫ cos x dx
= – cos x + sin x + C
(ii) We have

∫ (cosec x (cosec x + cot x) dx = ∫ cosec x dx + ∫ cosec x cot x dx

2

= – cot x – cosec x + C
(iii) We have
1 – sin x 1 sin x
∫ 2
cos x
dx = ∫ 2
cos x
dx – ∫
cos 2 x
dx

= ∫ sec x dx – ∫ tan x sec x dx

2

= tan x – sec x + C
Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
d 4
(x – 6 x ) = 4x3 – 6
dx
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant.

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Given that F(0) = 3, which gives,

3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3.
Remarks
(i) We see that if F is an anti derivative of f, then so is F + C, where C is any
constant. Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ∈ R. In applications, it is often necessary to satisfy an
additional condition which then determines a specific value of C giving unique
anti derivative of the given function.
(ii) Sometimes, F is not expressible in terms of elementary functions viz., polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc. We are
therefore blocked for finding ∫ f (x) dx . For example, it is not possible to find
∫e
– x2 2
dx by inspection since we can not find a function whose derivative is e – x
(iii) When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly. For instance
y4 + 1 1
∫ y dy = + C = y5 + C
4
4 +1 5

EXERCISE 7.1
Find an anti derivative (or integral) of the following functions by the method of inspection.
1. sin 2x 2. cos 3x 3. e 2x
2 3x
4. (ax + b) 5. sin 2x – 4 e
Find the following integrals in Exercises 6 to 20:
1
∫ (4 e ∫x ∫ (ax
2
6.
3x
+ 1) dx 7. (1 – ) dx 8.
2
+ bx + c ) dx
x2
2
 1  x3 + 5x 2 – 4
∫ (2 x + e ) dx 10. ∫  x – ∫
2 x
9.  dx 11. dx
 x x2
x3 + 3x + 4 x3 − x 2 + x – 1
12. ∫ x
dx 13. ∫ x –1
dx 14. ∫ (1 – x) x dx

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 INTEGRALS 235

∫ x ( 3x + 2 x + 3) dx ∫ (2 x – 3cos x + e ) dx
2 x
15. 16.

∫ (2x – 3sin x + 5 x ) dx ∫ sec x (sec x + tan x) dx

2
17. 18.

sec 2 x 2 – 3sin x
19. ∫ cosec2 x dx 20. ∫
cos2 x
dx.

Choose the correct answer in Exercises 21 and 22.

 1 
21. The anti derivative of  x +  equals
 x
1 1 2
1 3 2 3 1 2
(A) x + 2x 2 + C (B) x + x +C
3 3 2
3 1 3 1
2 2 3 2 1 2
(C) x + 2x 2 + C (D) x + x +C
3 2 2
d 3
22. If f ( x) = 4 x3 − 4 such that f (2) = 0. Then f (x) is
dx x
1 129 1 129
(A) x + 3 − (B) x + 4 +
4 3
x 8 x 8
1 129 1 129
(C) x4 + + (D) x3 + −
x3 8 x4 8
7.3 Methods of Integration
In previous section, we discussed integrals of those functions which were readily
obtainable from derivatives of some functions. It was based on inspection, i.e., on the
search of a function F whose derivative is f which led us to the integral of f. However,
this method, which depends on inspection, is not very suitable for many functions.
Hence, we need to develop additional techniques or methods for finding the integrals
by reducing them into standard forms. Prominent among them are methods based on:
1. Integration by Substitution
2. Integration using Partial Fractions
3. Integration by Parts
7.3.1 Integration by substitution
In this section, we consider the method of integration by substitution.
The given integral ∫ f (x) dx can be transformed into another form by changing
the independent variable x to t by substituting x = g (t).

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Consider I= ∫ f (x) dx
dx
Put x = g(t) so that = g′(t).
dt
We write dx = g′(t) dt

Thus I= ∫ f ( x) dx = ∫ f ( g (t )) g′(t ) dt
This change of variable formula is one of the important tools available to us in the
name of integration by substitution. It is often important to guess what will be the useful
substitution. Usually, we make a substitution for a function whose derivative also occurs
in the integrand as illustrated in the following examples.
Example 5 Integrate the following functions w.r.t. x:
(i) sin mx (ii) 2x sin (x2 + 1)

tan 4 x sec 2 x sin (tan – 1 x)

(iii) (iv)
x 1 + x2

Solution
(i) We know that derivative of mx is m. Thus, we make the substitution
mx = t so that mdx = dt.
1 1 1
Therefore, ∫ sin mx dx = m ∫ sin t dt = – m cos t + C = – m cos mx + C
(ii) Derivative of x2 + 1 is 2x. Thus, we use the substitution x2 + 1 = t so that
2x dx = dt.
∫ 2 x sin (x + 1) dx = ∫ sin t dt = – cos t + C = – cos (x2 + 1) + C
2
Therefore,
1
1 –2 1
(iii) Derivative of x is x = . Thus, we use the substitution
2 2 x
1
x = t so that dx = dt giving dx = 2t dt.
2 x

tan 4 x sec 2 x 2t tan 4t sec 2t dt

∫ dx = ∫ = 2 ∫ tan t sec t dt
4 2
Thus,
x t
Again, we make another substitution tan t = u so that sec2 t dt = du

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 INTEGRALS 237

u5
Therefore, 2 ∫ tan 4t sec 2t dt = 2 ∫ u 4 du = 2 +C
5
2
= tan 5 t + C (since u = tan t)
5
2
= tan x + C (since t = x )
5
5
tan 4 x sec 2 x 2
Hence, ∫ x
dx =
5
tan 5 x +C

Alternatively, make the substitution tan x = t

1
(iv) Derivative of tan – 1 x = . Thus, we use the substitution
1 + x2
dx
tan–1 x = t so that = dt.
1 + x2
sin (tan – 1 x)
Therefore , ∫ 1 + x2 dx = ∫ sin t dt = – cos t + C = – cos (tan–1x) + C
Now, we discuss some important integrals involving trigonometric functions and
their standard integrals using substitution technique. These will be used later without
reference.

(i) ∫ tan x dx = log sec x + C

We have
sin x
∫ tan x dx = ∫ cos x dx
Put cos x = t so that sin x dx = – dt
dt
Then ∫ tan x dx = – ∫ t = – log t + C = – log cos x + C
or ∫ tan x dx = log sec x + C
(ii) ∫ cot x dx = log sin x + C

cos x
We have ∫ cot x dx = ∫ sin x dx

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Put sin x = t so that cos x dx = dt

dt
Then ∫ cot x dx = ∫ t = log t + C = log sin x + C
(iii) ∫ sec x dx = log sec x + tan x + C
We have
sec x (sec x + tan x)
∫ sec x dx = ∫ sec x + tan x
dx

Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt

dt
Therefore, ∫ sec x dx = ∫ = log t + C = log sec x + tan x + C
t
(iv) ∫ cosec x dx = log cosec x – cot x + C
We have
cosec x (cosec x + cot x)
∫ cosec x dx = ∫ (cosec x + cot x)
dx

Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt

dt
So ∫ cosec x dx = – ∫ t = – log | t | = – log |cosec x + cot x | + C
cosec 2 x − cot 2 x
= – log +C
cosec x − cot x
= log cosec x – cot x + C
Example 6 Find the following integrals:
sin x 1
∫ sin ∫ sin (x + a) dx ∫ 1 + tan x dx
3
(i) x cos 2 x dx (ii) (iii)

Solution
(i) We have

∫ sin x cos2 x dx = ∫ sin 2 x cos 2 x (sin x ) dx

3

= ∫ (1 – cos x) cos x (sin x ) dx

2 2

Put t = cos x so that dt = – sin x dx

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 INTEGRALS 239

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

2
Therefore,

 t3 t 5 
= – ∫ (t – t ) dt = –  –  + C
2 4

3 5
1 1
= – cos x + cos x + C
3 5
3 5

(ii) Put x + a = t. Then dx = dt. Therefore

sin x sin (t – a)
∫ sin (x + a) dx = ∫ sin t
dt

sin t cos a – cos t sin a

= ∫ sin t
dt

= cos a ∫ dt – sin a ∫ cot t dt

= (cos a ) t – (sin a) log sin t + C1 

= (cos a) (x + a ) – (sin a) log sin (x + a ) + C1 

= x cos a + a cos a – (sin a ) log sin (x + a) – C1 sin a

sin x
Hence, ∫ sin (x + a) dx = x cos a – sin a log |sin (x + a)| + C,
where, C = – C1 sin a + a cos a, is another arbitrary constant.
dx cos x dx
(iii) ∫ 1 + tan x = ∫ cos x + sin x
1 (cos x + sin x + cos x – sin x) dx
=
2 ∫ cos x + sin x
1 1 cos x – sin x
= 2 ∫ dx + 2 ∫ cos x + sin x dx
x C1 1 cos x – sin x
2 2 2 ∫ cos x + sin x
= + + dx ... (1)

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cos x – sin x
Now, consider I = ∫ dx
cos x + sin x
Put cos x + sin x = t so that (cos x – sin x) dx = dt
dt
Therefore I=∫
= log t + C2 = log cos x + sin x + C2
t
Putting it in (1), we get
dx x C1 1 C
∫ 1 + tan x = 2 + + log cos x + sin x + 2
2 2 2
x 1 C C
= + log cos x + sin x + 1 + 2
2 2 2 2
x 1  C C 
= + log cos x + sin x + C,  C = 1 + 2 
2 2  2 2 

EXERCISE 7.2
Integrate the functions in Exercises 1 to 37:

1.
2x
2.
( log x )2 3.
1
1 + x2 x x + x log x
4. sin x sin (cos x) 5. sin (ax + b) cos (ax + b)

6. ax + b 7. x x + 2 8. x 1 + 2 x 2

1 x
9. (4 x + 2) x 2 + x + 1 10. x – x 11. ,x>0
x+4

3
1
5
x2 1
12. (x – 1) 3 x 13. 14. , x > 0, m ≠ 1
(2 + 3x 3 )3 x (log x ) m

x x
15. 16. e2 x + 3 17. 2
9 – 4 x2 ex
–1
etan x e2x – 1 e2 x – e – 2 x
18. 19. 20.
1 + x2 e2 x + 1 e2 x + e – 2 x

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 INTEGRALS 241

sin – 1 x
21. tan2 (2x – 3) 22. sec2 (7 – 4x) 23.
1 – x2

2cos x – 3sin x 1 cos x

24. 6cos x + 4sin x 25. 26.
cos x (1 – tan x) 2
2
x
cos x
27. sin 2x cos 2 x 28. 29. cot x log sin x
1 + sin x

sin x sin x 1
30. 31. 32.
1 + cos x (1 + cos x ) 2
1 + cot x

33.
1
34.
tan x
35.
(1 + log x)2
1 – tan x sin x cos x x

36.
(x + 1) ( x + log x )
2

37.
(
x 3sin tan – 1 x 4 )
8
x 1+ x
Choose the correct answer in Exercises 38 and 39.

10 x 9 + 10 x log e 10 dx
38. ∫ x10 + 10 x
equals

(A) 10x – x10 + C (B) 10x + x10 + C

(C) (10x – x10)–1 + C (D) log (10x + x10) + C
dx
39. ∫ sin 2 x cos2 x equals
(A) tan x + cot x + C (B) tan x – cot x + C
(C) tan x cot x + C (D) tan x – cot 2x + C

7.3.2 Integration using trigonometric identities

When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example.

Example 7 Find (i) ∫ cos x dx (ii) ∫ sin 2 x cos 3 x dx (iii) ∫ sin x dx

2 3

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242 MATHEMATICS

Solution
(i) Recall the identity cos 2x = 2 cos2 x – 1, which gives
1 + cos 2 x
cos2 x =
2
1 1 1
Therefore, =
2 ∫ (1 + cos 2x ) dx = ∫ dx + ∫ cos 2 x dx
2 2
x 1
= + sin 2 x + C
2 4
1
(ii) Recall the identity sin x cos y = [sin (x + y) + sin (x – y)] (Why?)
2

Then =

1  1 
=
2  – 5 cos 5 x + cos x  + C

1 1
cos 5 x + cos x + C
= –
10 2
(iii) From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
3sin x – sin 3x
sin3 x =
4
3 1
∫ sin ∫ sin x dx – ∫ sin 3x dx
3
Therefore, x dx =
4 4
3 1
= – cos x + cos 3 x + C
4 12

∫ sin x dx = ∫ sin 2 x sin x dx = ∫ (1 – cos

3 2
Alternatively, x ) sin x dx
Put cos x = t so that – sin x dx = dt
t3
Therefore, 2
( )
∫ sin x dx = − ∫ 1 – t dt = – ∫ dt + ∫ t dt = – t +
3 2
3
+C

1
= – cos x +
cos3 x + C
3
Remark It can be shown using trigonometric identities that both answers are equivalent.

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 INTEGRALS 243

EXERCISE 7.3
Find the integrals of the functions in Exercises 1 to 22:
1. sin2 (2x + 5) 2. sin 3x cos 4x 3. cos 2x cos 4x cos 6x
3 3 3
4. sin (2x + 1) 5. sin x cos x 6. sin x sin 2x sin 3x
1 – cos x cos x
7. sin 4x sin 8x 8. 9.
1 + cos x 1 + cos x
sin 2 x
10. sin4 x 11. cos4 2x 12.
1 + cos x
cos 2 x – cos 2α cos x – sin x
13. 14. 15. tan3 2x sec 2x
cos x – cos α 1 + sin 2 x

4
sin 3 x + cos3 x cos 2 x + 2sin 2 x
16. tan x 17. 18.
sin 2 x cos 2 x cos 2 x
1 cos 2 x
19. 20. 21. sin – 1 (cos x)
sin x cos3 x ( cos x + sin x )2
1
22.
cos (x – a ) cos (x – b)
Choose the correct answer in Exercises 23 and 24.
sin 2 x − cos 2 x
23. ∫ sin 2 x cos 2 x dx is equal to
(A) tan x + cot x + C (B) tan x + cosec x + C
(C) – tan x + cot x + C (D) tan x + sec x + C
e x (1 + x)
24. ∫ cos2 (e x x) dx equals
(A) – cot (exx) + C (B) tan (xex) + C
(C) tan (ex) + C (D) cot (ex) + C
7.4 Integrals of Some Particular Functions
In this section, we mention below some important formulae of integrals and apply them
for integrating many other related standard integrals:
dx 1 x–a
(1) ∫ 2 2
= log +C
x –a 2 a x +a

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dx 1 a+x
(2) ∫ a 2 – x 2 = 2a log a–x
+C

dx 1 x
∫ x 2 + a 2 = a tan
–1
(3) +C
a

dx
(4) ∫ x –a2 2
= log x + x 2 – a 2 + C

dx x
(5) ∫ a2 – x2
= sin – 1
a
+C

dx
(6) ∫ x +a2 2
= log x + x 2 + a 2 + C

We now prove the above results:

1 1
(1) We have =
x –a22
(x – a ) (x + a )

1  (x + a) – (x – a)  1  1 1 
 = –
2a  (x – a) (x + a)  2a  x – a x + a 

=

dx 1  dx dx 
Therefore, ∫ x2 – a 2 = 2a  ∫ x – a – ∫ x + a 

1
= [log | (x – a)| – log | (x + a)|] + C
2a

1 x–a
= log +C
2a x+a

(2) In view of (1) above, we have

1 1  (a + x) + (a − x)  1  1 1 
=   = +
2
a –x 2
2a  (a + x) (a − x)  2a  a − x a + x 


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 INTEGRALS 245

dx 1  dx dx 
Therefore, ∫ a2 – x2 =  ∫
2a  a − x
+∫
a + x 
1
= [ − log | a − x | + log | a + x |] + C
2a
1 a+ x
= log +C
2a a−x

ANote The technique used in (1) will be explained in Section 7.5.

(3) Put x = a tan θ. Then dx = a sec2 θ dθ.
dx
Therefore, ∫ x2 + a2 =
1 1 1 x
a ∫ dθ = θ + C = tan – 1 + C
=
a a a
(4) Let x = a sec θ. Then dx = a sec θ tan θ d θ.
dx a secθ tanθ dθ
Therefore, ∫ x 2 − a 2 = ∫ a2 sec2θ − a2
= ∫ secθ dθ = log secθ + tanθ + C1

x x2
= log + – 1 + C1
a a2

= log x + x – a − log a + C1
2 2

= log x + x – a + C , where C = C1 – log |a|

2 2

(5) Let x = a sinθ. Then dx = a cosθ dθ.

dx a cosθ dθ
Therefore, ∫ a −x
2 2
= ∫ a 2 – a 2 sin 2θ
x
∫ dθ = θ + C = sin +C
–1
=
a
(6) Let x = a tan θ. Then dx = a sec2 θ dθ.
dx a sec 2θ dθ
Therefore, ∫ x2 + a2
= ∫ a 2 tan 2θ + a 2
= ∫ secθ dθ = log (secθ + tanθ) + C1

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246 MATHEMATICS

x x2
= log + + 1 + C1
a a2

2
= log x + x + a − log | a | + C1
2

2
= log x + x + a + C, where C = C1 – log |a|
2

Applying these standard formulae, we now obtain some more formulae which
are useful from applications point of view and can be applied directly to evaluate
other integrals.
dx
(7) To find the integral ∫ ax 2 + bx + c , we write
 2 b c  b   c b2 
2
2 a
ax + bx + c =  x + x + = a   x +  + – 
 a a   2a   a 4a 2  

b c b2
= t so that dx = dt and writing – 2 = ± k . We find the
2
Now, put x +
2a a 4a

1 dt  c b2 
integral reduced to the form ∫ 2 depending upon the sign of  – 2
a t ± k2  a 4a 
and hence can be evaluated.

(8) To find the integral of the type , proceeding as in (7), we

obtain the integral using the standard formulae.

px + q
(9) To find the integral of the type ∫ ax 2 + bx + c dx , where p, q, a, b, c are
constants, we are to find real numbers A, B such that
d
px + q = A (ax 2 + bx + c) + B = A (2ax + b) + B
dx
To determine A and B, we equate from both sides the coefficients of x and the
constant terms. A and B are thus obtained and hence the integral is reduced to
one of the known forms.

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 INTEGRALS 247

( px + q) dx
(10) For the evaluation of the integral of the type
ax 2 + bx + c
∫ , we proceed

as in (9) and transform the integral into known standard forms.

Let us illustrate the above methods by some examples.
Example 8 Find the following integrals:
dx dx
(i) ∫ x 2 − 16 (ii) ∫ 2x − x2
Solution
dx dx 1 x–4
(i) We have ∫ x2 − 16 = ∫ x2 – 42 =
8
log
x+4
+ C [by 7.4 (1)]

(ii)

Put x – 1 = t. Then dx = dt.

dx dt
Therefore, ∫ 2x − x 2
= ∫ 1– t 2
= sin (t ) + C
–1
[by 7.4 (5)]

= sin (x – 1) + C
–1

Example 9 Find the following integrals :

dx dx dx
(i) ∫ x2 − 6 x + 13 (ii) ∫ 3x 2 + 13x − 10 (iii) ∫ 5x2 − 2 x
Solution
(i) We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4

dx 1
So, ∫ x2 − 6 x + 13 = ∫ ( x – 3)2 + 22 dx
Let x – 3 = t. Then dx = dt
dx dt 1 t
∫ x2 − 6 x + 13 = ∫ t 2 + 22 = 2 tan +C
–1
Therefore, [by 7.4 (3)]
2
1 x–3
= tan – 1 +C
2 2

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248 MATHEMATICS

(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,

 2 13x 10 
3 x 2 + 13x – 10 = 3  x + – 
 3 3

 13   17  
2 2

= 3  x + 6  –  6   (completing the square)

    

dx 1 dx
Thus ∫ 3x 2 + 13x − 10 = ∫
3  2
13   17 
2

 x + −
  
 6  6

13
Put x + = t . Then dx = dt.
6

dx 1 dt
∫ 3x 2 + 13x − 10 3∫
Therefore, = 2
 17 
t2 −  
 6

17
t–
1 6 +C
= log 1 [by 7.4 (i)]
17 17
3× 2× t+
6 6

13 17
x+
–
1 6 6 +C
= log 1
17 13 17
x+ +
6 6

1 6x − 4
= log + C1
17 6 x + 30

1 3x − 2 1 1
= log + C1 + log
17 x+5 17 3

1 3x − 2 1 1
= log + C , where C = C1 + log
17 x+5 17 3

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 INTEGRALS 249

dx dx
(iii) We have ∫ 2
5x − 2 x
=∫
 2x 
5 x2 – 
 5 

1 dx
=
5
∫ 2 2
(completing the square)
 1 1
x–  – 
 5 5
1
Put x – = t . Then dx = dt.
5
dx 1 dt
Therefore, ∫ 5x2 − 2 x
=
5
∫ 2
1
t2 –  
5
2
1 1
= log t + t 2 –   +C [by 7.4 (4)]
5 5

1 1 2x
= log x – + x2 – +C
5 5 5

Example 10 Find the following integrals:

x+2 x+3
(i) ∫ 2 x 2 + 6 x + 5 dx (ii) ∫ 5 − 4x – x2
dx

Solution
(i) Using the formula 7.4 (9), we express

x+2= A
d
dx
( )
2 x 2 + 6 x + 5 + B = A (4 x + 6) + B
Equating the coefficients of x and the constant terms from both sides, we get
1 1
4A = 1 and 6A + B = 2 or A= and B = .
4 2
x+2 1 4x + 6 1 dx
Therefore, ∫ 2x2 + 6x + 5 = ∫ 2
4 2x + 6x + 5
dx + ∫ 2
2 2x + 6 x + 5
1 1
= I1 + I2 (say) ... (1)
4 2

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250 MATHEMATICS

In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt

dt
Therefore, I1 = ∫ t
= log t + C1

= log | 2 x + 6 x + 5 | + C1
2
... (2)

dx 1 dx
and I2 = ∫ 2x2 + 6x + 5 = 2 ∫ 5
x 2 + 3x +
2

1 dx
= ∫
2  2
3 1
2

 x + +
  
 2 2

3
Put x + = t , so that dx = dt, we get
2
1 dt 1
I2 =
2 ∫ 1
2 =
1
tan –1 2t + C 2 [by 7.4 (3)]
t2 +   2×
2 2

–1  3
= tan 2  x +  + C2 = tan ( 2 x + 3 ) + C 2
–1
... (3)
 2 
Using (2) and (3) in (1), we get
x+2 1 1
∫ 2 x 2 + 6 x + 5 dx = 4 log 2 x
2
+ 6x + 5 + tan – 1 ( 2 x + 3) + C
2

C1 C2
where, C=+
4 2
(ii) This integral is of the form given in 7.4 (10). Let us express

d
x+3= A (5 – 4 x – x 2 ) + B = A (– 4 – 2x) + B
dx
Equating the coefficients of x and the constant terms from both sides, we get
1
– 2A = 1 and – 4 A + B = 3, i.e., A = – and B = 1
2

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 INTEGRALS 251

x+3 1 ( – 4 – 2 x ) dx + dx
Therefore, ∫ dx = –
2 ∫ 5 − 4 x − x2
∫ 5 − 4 x − x2
5 − 4x − x2
1
= – I +I ... (1)
2 1 2
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt.
( – 4 − 2 x ) dx = dt
Therefore, I1 = ∫ 5 − 4x − x 2 ∫ t
= 2 t + C1

= 2 5 – 4 x – x 2 + C1 ... (2)

dx dx
Now consider I2 = ∫ 5 − 4x − x 2
=∫
9 – (x + 2) 2
Put x + 2 = t, so that dx = dt.
dt t
Therefore, I2 = ∫ 3 −t
2
= sin – 1 + C 2
2 3
[by 7.4 (5)]

x+2
= sin
–1
+ C2 ... (3)
3
Substituting (2) and (3) in (1), we obtain
x+3 x+2 C
∫ 5 – 4x – x2
= – 5 – 4x – x 2 + sin – 1
3
+ C , where C = C2 – 1
2

EXERCISE 7.4
Integrate the functions in Exercises 1 to 23.
3x 2 1 1
1. 2. 3.
x6 + 1 1 + 4x 2
(2 – x)
2
+1

1 3x x2
4. 5. 6.
9 – 25 x 2 1 + 2x4 1 − x6

x –1 x2 sec 2 x
7. 8. 9.
x2 – 1 x6 + a6 tan 2 x + 4

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252 MATHEMATICS

1 1 1
10. 11. 12.
x + 2x + 2
2 9x + 6x + 5
2
7 – 6 x – x2

1 1 1
13. 14. 15.
( x – 1)( x – 2 ) 8 + 3x – x 2
( x – a )( x – b )
4x + 1 x+2 5x − 2
16. 17. 18.
2x + x – 3
2 2
x –1 1 + 2x + 3x2

6x + 7 x+2 x+2
19. 20. 21.
( x – 5 )( x – 4 ) 4x – x 2
x + 2x + 3
2

x+3 5x + 3
22. 23. .
x – 2x − 5
2
x + 4 x + 10
2

Choose the correct answer in Exercises 24 and 25.

dx
24. ∫ x 2 + 2 x + 2 equals
(A) x tan–1 (x + 1) + C (B) tan–1 (x + 1) + C
(C) (x + 1) tan–1x + C (D) tan–1x + C
dx
25. ∫ 9x − 4x2
equals

1 –1  9 x − 8  1  8x − 9 
(A) sin  +C (B) sin –1  +C
9  8  2  9 

1 –1  9 x − 8  1  9x − 8 
(C) sin  +C (D) sin –1  +C
3  8  2  9 

7.5 Integration by Partial Fractions

Recall that a rational function is defined as the ratio of two polynomials in the form
P(x)
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(x)
Q(x)
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper. The improper rational functions can be reduced to the proper rational

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 INTEGRALS 253

P(x) P(x) P (x)

functions by long division process. Thus, if is improper, then = T(x) + 1 ,
Q(x) Q(x) Q(x)

P1 (x)
where T(x) is a polynomial in x and is a proper rational function. As we know
Q(x)
how to integrate polynomials, the integration of any rational function is reduced to the
integration of a proper rational function. The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
P(x ) P(x)
linear and quadratic factors. Assume that we want to evaluate ∫ Q(x) dx , where Q(x)
is proper rational function. It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition. After this,
the integration can be carried out easily using the already known methods. The following
Table 7.2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions.
Table 7.2
S.No. Form of the rational function Form of the partial fraction
px + q A
+
B
1. ,a≠b
(x –a) (x –b) x–a x–b

px + q A B
2. +
(x – a ) 2 x – a ( x – a )2

px 2 + qx + r A B C
3. + +
(x – a) (x – b) (x – c) x –a x –b x –c

px 2 + qx + r A B C
4. + 2
+
(x – a) 2 (x – b) x – a (x – a) x–b

px 2 + qx + r A Bx + C ,
5. + 2
(x – a) (x 2 + bx + c) x – a x + bx + c
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably.

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254 MATHEMATICS

dx
Example 11 Find ∫ (x + 1) (x + 2)
Solution The integrand is a proper rational function. Therefore, by using the form of
partial fraction [Table 7.2 (i)], we write

1 A B
= + ... (1)
(x + 1) (x + 2) x +1 x + 2
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A+B=0
and 2A + B = 1
Solving these equations, we get A =1 and B = – 1.
Thus, the integrand is given by
1 1 –1
= +
(x + 1) (x + 2) x +1 x + 2

dx dx dx
Therefore, ∫ (x + 1) (x + 2) = ∫ x + 1 – ∫ x + 2
= log x + 1 − log x + 2 + C

x +1
= log +C
x+2

Remark The equation (1) above is an identity, i.e. a statement true for all (permissible)
values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to
indicate that the statement is true only for certain values of x.
x2 + 1
Example 12 Find ∫ x 2 − 5x + 6 dx
x2 + 1
Solution Here the integrand 2 is not proper rational function, so we divide
x – 5x + 6
x2 + 1 by x2 – 5x + 6 and find that

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 INTEGRALS 255

x2 + 1 5x – 5 5x – 5
= 1+ 2 =1+
x – 5x + 6
2 x – 5x + 6 (x – 2) (x – 3)
5x – 5 A B
Let = +
(x – 2) (x – 3) x–2 x–3
So that 5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10
x2 + 1 5 10
Thus, = 1− +
x – 5x + 6
2
x –2 x –3
x2 + 1 1 dx
Therefore, ∫ x 2 – 5 x + 6 dx = ∫ dx − 5 ∫ x – 2 dx + 10∫ x – 3
= x – 5 log | x – 2 | + 10 log | x – 3 | + C.
3x − 2
Example 13 Find ∫ dx
(x + 1)2 (x + 3)
Solution The integrand is of the type as given in Table 7.2 (4). We write

3x – 2 A B C
= + +
(x + 1) (x + 3)
2
x + 1 (x + 1) 2
x+3
So that 3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2 , x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2. Solving these equations, we get
11 –5 –11 . Thus the integrand is given by
A= ,B = and C =
4 2 4
3x − 2 11 5 11
= 4 (x + 1) – –
4 (x + 3)
(x + 1) (x + 3)
2
2 (x + 1) 2

3x − 2 11 dx 5 dx 11 dx
Therefore, ∫ (x + 1)2 (x + 3) = ∫ – ∫
4 x + 1 2 (x + 1) 2
− ∫
4 x+3
11 5 11
= log x +1 + − log x + 3 + C
4 2 (x + 1) 4
11 x +1 5
= log + +C
4 x + 3 2 (x + 1)

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256 MATHEMATICS

x2
Example 14 Find ∫ (x2 + 1) (x 2 + 4) dx
x2
Solution Consider 2 and put x2 = y.
( x + 1) ( x 2 + 4)

x2 y
Then =
(x + 1) (x + 4)
2 2
(y + 1) (y + 4)

y A B
Write = +
(y + 1) (y + 4) y +1 y + 4
So that y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
1 4
A= − and B =
3 3

x2 1 4
Thus, = – +
(x + 1) (x + 4)
2 2
3 (x + 1) 3 (x + 4)
2 2

x 2 dx 1 dx 4 dx
Therefore, ∫ (x2 + 1) (x 2 + 4) = – 3 ∫ x 2 + 1 + 3 ∫ x2 + 4
1 –1 4 1 –1 x
= – tan x + × tan +C
3 3 2 2
1 –1 2 –1 x
= – tan x + tan +C
3 3 2
In the above example, the substitution was made only for the partial fraction part
and not for the integration part. Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method.
( 3 sin φ – 2 ) cos φ
Example 15 Find ∫ 5 – cos2 φ – 4 sin φ d φ
Solution Let y = sinφ
Then dy = cosφ dφ

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 INTEGRALS 257

( 3 sinφ – 2 ) cosφ (3y – 2) dy

Therefore, ∫ 5 – cos 2φ – 4 sinφ d φ = ∫ 5 – (1 – y 2 ) – 4 y
3y – 2
= ∫ y 2 – 4 y + 4 dy
3y – 2
= ∫ ( y – 2 )2 = I (say)
3y – 2 A B
Now, we write = + [by Table 7.2 (2)]
( y – 2) 2
y − 2 (y − 2) 2
Therefore, 3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4.
Therefore, the required integral is given by
3 4 dy dy
I = ∫[ + ] dy = 3 ∫ +4∫
y – 2 (y – 2) 2 y–2 (y – 2) 2

 1 
= 3 log y − 2 + 4  – +C
 y−2
4
= 3 log sin φ − 2 + +C
2 – sin φ

4
= 3 log (2 − sin φ) + + C (since, 2 – sin φ is always positive)
2 − sin φ

x 2 + x + 1 dx
Example 16 Find ∫
(x + 2) (x 2 + 1)

Solution The integrand is a proper rational function. Decompose the rational function
into partial fraction [Table 2.2(5)]. Write

x2 + x + 1 A Bx + C
= + 2
(x + 1) (x + 2)
2
x + 2 (x + 1)
Therefore, x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)

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Equating the coefficients of x2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get
3 2 1
A = , B = and C =
5 5 5
Thus, the integrand is given by
2 1
x+
x2 + x + 1 3 3 1  2x + 1 
= + 2 5 =
5 +  2 
(x + 1) (x + 2)
2
5 (x + 2) x + 1 5 (x + 2) 5  x + 1 

x2 + x + 1 3 dx 1 2x 1 1
Therefore, ∫ (x2 +1) (x + 2) dx = 5 ∫ x + 2 + 5 ∫ x2 + 1 dx + 5 ∫ x2 + 1 dx
3 1 1
= log x + 2 + log x 2 + 1 + tan –1 x + C
5 5 5

EXERCISE 7.5
Integrate the rational functions in Exercises 1 to 21.
x 1 3x – 1
1. 2. 3.
(x + 1) (x + 2) 2
x –9 (x – 1) (x – 2) (x – 3)

x 2x 1 – x2
4. 5. 2 6.
(x – 1) (x – 2) (x – 3) x + 3x + 2 x (1 – 2 x)

x x 3x + 5
7. 8. 9.
(x + 1) (x – 1)
2
(x – 1) (x + 2)
2
x – x2 − x + 1
3

2x − 3 5x x3 + x + 1
10. 11. 12.
(x – 1) (2x + 3)
2
(x + 1) (x 2 − 4) x2 − 1
2 3x – 1 1
13. 14. 15.
(1 − x) (1 + x 2 ) (x + 2) 2 x −1
4

1
16. [Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
x (x n + 1)

cos x
17. [Hint : Put sin x = t]
(1 – sin x) (2 – sin x)

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 INTEGRALS 259

(x 2 + 1) (x 2 + 2) 2x 1
18. 19. 20.
(x 2 + 3) (x 2 + 4) (x + 1) (x 2 + 3)
2
x (x 4 – 1)
1
21. x [Hint : Put ex = t]
(e – 1)
Choose the correct answer in each of the Exercises 22 and 23.
x dx
22. ∫ equals
( x − 1) ( x − 2)
( x − 1)2 ( x − 2) 2
(A) log +C (B) log +C
x−2 x −1
2
 x −1 
(C) log   +C (D) log ( x − 1) ( x − 2) + C
 x−2
dx
23. ∫ x ( x 2 + 1) equals

1 1
(A) log x − log (x +1) + C (B) log x + log (x +1) + C
2 2
2 2
1 1
(D) log x + log (x +1) + C
2
(C) − log x + log (x 2 +1) + C
2 2
7.6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions.
If u and v are any two differentiable functions of a single variable x (say). Then, by
the product rule of differentiation, we have
d dv du
(uv ) = u + v
dx dx dx
Integrating both sides, we get
dv du
uv = ∫ u dx + ∫ v dx
dx dx
dv du
or ∫ u dx dx = uv – ∫ v dx dx ... (1)
dv
Let u = f (x) and = g (x). Then
dx
du
= f ′(x) and v = ∫ g (x ) dx
dx

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Therefore, expression (1) can be rewritten as

∫ f (x) g (x) dx = f (x) ∫ g (x) dx – ∫ [ ∫ g (x) dx] f ′(x ) dx

i.e., ∫ f (x) g (x) dx = f (x) ∫ g (x ) dx – ∫ [ f ′ (x) ∫ g (x ) dx ] dx

If we take f as the first function and g as the second function, then this formula
may 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 function)
× (integral of the second function)]”
Example 17 Find ∫ x cos x dx
Solution Put f (x) = x (first function) and g (x) = cos x (second function).
Then, integration by parts gives
d
∫ x cos x dx = x ∫ cos x dx – ∫ [ (x ) ∫ cos x dx ] dx
dx

= x sin x – ∫ sin x dx = x sin x + cos x + C

Suppose, we take f (x) = cos x and g (x) = x. Then
d
∫ x cos x dx = cos x ∫ x dx – ∫ [ (cos x) ∫ x dx] dx
dx
x2 x2
= ( cos x ) + ∫ sin x dx
2 2
Thus, it shows that the integral ∫ x cos x dx is reduced to the comparatively more
complicated integral having more power of x. Therefore, the proper choice of the first
function and the second function is significant.
Remarks
(i) It is worth mentioning that integration by parts is not applicable to product of
functions in all cases. For instance, the method does not work for ∫ x sin x dx .
The reason is that there does not exist any function whose derivative is
x sin x.
(ii) Observe that while finding the integral of the second function, we did not add
any constant of integration. If we write the integral of the second function cos x

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 INTEGRALS 261

as sin x + k, where k is any constant, then

∫ x cos x dx = x (sin x + k ) − ∫ (sin x + k ) dx

= x (sin x + k ) − ∫ (sin x dx − ∫ k dx

= x (sin x + k ) − cos x – kx + C = x sin x + cos x + C

This shows that adding a constant to the integral of the second function is
superfluous so far as the final result is concerned while applying the method of
integration by parts.
(iii) Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function. However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function.
Example 18 Find ∫ log x dx
Solution To start with, we are unable to guess a function whose derivative is log x. We
take log x as the first function and the constant function 1 as the second function. Then,
the integral of the second function is x.
d
Hence, ∫ (logx.1) dx = log x ∫1 dx − ∫ [ dx (log x) ∫1 dx] dx
1
= (log x ) ⋅ x – ∫ x dx = x log x – x + C .
x
∫ x e dx
x
Example 19 Find

Solution Take first function as x and second function as ex. The integral of the second
function is ex.

∫ x e dx = x e x − ∫1 ⋅ e x dx = xex – ex + C.
x
Therefore,

x sin – 1 x
Example 20 Find ∫ 1 − x2
dx

x
Solution Let first function be sin – 1x and second function be .
1 − x2
x dx
First we find the integral of the second function, i.e., ∫ 1 − x2
.

Put t =1 – x2. Then dt = – 2x dx

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262 MATHEMATICS

x dx 1 dt
∫ ∫ = – t = − 1− x
2
Therefore, = –
1− x 2 2 t

Hence, ∫
x sin – 1 x
1− x 2
–1 2
(
dx = (sin x ) – 1 − x − ∫ ) 1
1− x 2
( – 1 − x 2 ) dx

−1 −1
= – 1 − x sin x + x + C = x – 1 − x sin x + C
2 2

Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts.

∫e
x
Example 21 Find sin x dx

Solution Take ex as the first function and sin x as second function. Then, integrating
by parts, we have

I = ∫ e x sin x dx = e x ( – cos x ) + ∫ e x cos x dx

= – ex cos x + I1 (say) ... (1)
x
Taking e and cos x as the first and second functions, respectively, in I1, we get
I1 = e x sin x – ∫ e x sin x dx
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
ex
I = ∫ e sin x dx =
x
Hence, (sin x – cos x) + C
2
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function.

∫e [ f (x) + f ′ (x )] dx
x
7.6.1 Integral of the type

∫e [ f (x ) + f ′(x )] dx = ∫ e f (x) dx + ∫ e f ′(x) dx

x x x
We have I=

= I1 + ∫ e f ′(x) dx, where I1 = ∫ e f (x) dx

x x
... (1)
x
Taking f (x) and e as the first function and second function, respectively, in I1 and

∫ f ′(x) e dx + C
x
integrating it by parts, we have I1 = f (x) ex –
Substituting I1 in (1), we get
I = e f (x) − ∫ f ′(x) e dx + ∫ e f ′(x) dx + C = ex f (x) + C
x x x

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 INTEGRALS 263

∫e [ f ( x ) + f ′( x )] dx = e x f ( x ) + C
x
Thus,

1 (x 2 + 1) e x
Example 22 Find (i) ∫ e x (tan – 1 x +
1 + x2
) dx (ii) ∫ (x + 1)2 dx
Solution
1
(i) We have I = ∫ e (tan x +
x –1
) dx
1 + x2
1
Consider f (x) = tan– 1x, then f ′(x) =
1 + x2
Thus, the given integrand is of the form ex [ f (x) + f ′(x)].
1
Therefore, I = ∫ e x (tan – 1 x + ) dx = ex tan– 1x + C
1 + x2

(x 2 + 1) e x 2
x x – 1 + 1+1)
(ii) We have I = ∫ dx = ∫ e [ ] dx
(x + 1) 2 (x + 1)2

x2 – 1 2 x –1 2
= ∫ ex [ 2
+ 2
] dx = ∫ e x [ + ] dx
(x + 1) (x +1) x + 1 (x+1) 2

x −1 2
Consider f (x ) = , then f ′(x ) =
x +1 (x + 1) 2
Thus, the given integrand is of the form ex [f (x) + f ′(x)].
x2 + 1 x x −1 x
Therefore, ∫ (x + 1)2 e dx = x + 1 e + C

EXERCISE 7.6
Integrate the functions in Exercises 1 to 22.
1. x sin x 2. x sin 3x 3. x2 ex 4. x log x
5. x log 2x 6. x2 log x 7. x sin– 1x 8. x tan–1 x

–1 –1 2
x cos−1 x
9. x cos x 10. (sin x) 11. 12. x sec2 x
1 − x2
13. tan –1x 14. x (log x)2 15. (x2 + 1) log x

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x ex x  1 + sin x 
x
16. e (sinx + cosx) 17. 18. e  1 + cos x 
(1 + x) 2  

x 1 1  (x − 3) e x
19. e  – 2  20. 21. e2x sin x
x x  (x − 1)3

– 1  2x 
22. sin  2 
 1+ x 
Choose the correct answer in Exercises 23 and 24.

∫x e
2 x3
23. dx equals

1 x3 1 x2
(A) e +C (B) e +C
3 3
1 x3 1 x2
(C) e +C (D) e +C
2 2

∫e sec x (1 + tan x ) dx equals

x
24.
(A) ex cos x + C (B) ex sec x + C
(C) ex sin x + C (D) ex tan x + C
7.6.2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :

(i)
∫ x 2 − a 2 dx (ii) ∫ x 2 + a 2 dx (iii) ∫ a 2 − x2 dx

Let I = ∫ x − a dx
2 2
(i)
Taking constant function 1 as the second function and integrating by parts, we
have
1 2x
I = x x −a −∫
2 2
x dx
2 x2 − a 2

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

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 INTEGRALS 265

dx
= x x − a − ∫ x − a dx − a ∫
2 2 2 2 2

x2 − a2
dx
= x x −a −I−a ∫
2 2 2

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

x a2
∫
2 2
or I= x 2 – a 2 dx =
x – a – log x + x 2 – a 2 + C
2 2
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
1 a2
(ii) ∫ x 2 + a 2 dx =
2
x x2 + a2 +
2
log x + x 2 + a 2 + C

(iii)

Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively.

Example 23 Find ∫ x 2 + 2 x + 5 dx

Solution Note that

∫ x 2 + 2 x + 5 dx = ∫ (x + 1)2 + 4 dx
Put x + 1 = y, so that dx = dy. Then

∫ x 2 + 2 x + 5 dx = ∫ y 2 + 22 dy

1 4
= y y2 + 4 + log y + y 2 + 4 + C [using 7.6.2 (ii)]
2 2
1
= (x + 1) x 2 + 2 x + 5 + 2 log x + 1 + x 2 + 2 x + 5 + C
2

Example 24 Find ∫ 3 − 2x − x 2 dx

Solution Note that ∫ 3 − 2 x − x2 dx = ∫ 4 − (x + 1) 2 dx

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266 MATHEMATICS

Put x + 1 = y so that dx = dy.

Thus ∫ 3 − 2x − x 2 dx = ∫ 4 − y 2 dy
1 4 y
= y 4 − y 2 + sin –1 + C [using 7.6.2 (iii)]
2 2 2
1  x +1 
= (x + 1) 3 − 2 x − x 2 + 2 sin –1  +C
2  2 

EXERCISE 7.7
Integrate the functions in Exercises 1 to 9.
1. 4 − x2 2. 1 − 4x 2 3. x2 + 4 x + 6
4. x2 + 4x + 1 5. 1 − 4x − x 2 6. x2 + 4 x − 5
x2
7. 1 + 3x − x 2 8. x + 3x
2 9. 1+
9
Choose the correct answer in Exercises 10 to 11.
10. ∫ 1 + x 2 dx is equal to

(A)
x
2
1
(
1 + x 2 + log x + 1 + x 2
2
) +C
3 3
2 2
(B) (1 + x 2 ) 2 + C (C) x (1 + x 2 ) 2 + C
3 3
x2 1
(D) 1 + x 2 + x 2 log x + 1 + x 2 + C
2 2
11. ∫ x 2 − 8 x + 7 dx is equal to
1
(A) ( x − 4) x 2 − 8 x + 7 + 9log x − 4 + x 2 − 8 x + 7 + C
2
1
(B) ( x + 4) x 2 − 8 x + 7 + 9log x + 4 + x 2 − 8 x + 7 + C
2
1
(C) ( x − 4) x 2 − 8 x + 7 − 3 2 log x − 4 + x 2 − 8 x + 7 + C
2
1 9
(D) ( x − 4) x 2 − 8x + 7 − log x − 4 + x 2 − 8 x + 7 + C
2 2

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7.7 Definite Integral

In the previous sections, we have studied about the indefinite integrals and discussed
few methods of finding them including integrals of some special functions. In this
section, we shall study what is called definite integral of a function. The definite integral
b
has a unique value. A definite integral is denoted by ∫ a f (x) dx , where a is called the
lower limit of the integral and b is called the upper limit of the integral. The definite
integral is introduced either as the limit of a sum or if it has an anti derivative F in the
interval [a, b], then its value is the difference between the values of F at the end
points, i.e., F(b) – F(a).

7.8 Fundamental Theorem of Calculus

7.8.1 Area function
b
We have defined ∫ a f ( x) dx as the area of

the region bounded by the curve y = f (x),

the ordinates x = a and x = b and x-axis. Let x
x
be a given point in [a, b]. Then ∫ a f ( x) dx
represents the area of the light shaded region
in Fig 7.1 [Here it is assumed that f (x) > 0 for
x ∈ [a, b], the assertion made below is
equally true for other functions as well].
The area of this shaded region depends upon Fig 7.1
the value of x.
In other words, the area of this shaded region is a function of x. We denote this
function of x by A(x). We call the function A(x) as Area function and is given by
x
A (x) = ∫a f ( x ) dx ... (1)
Based on this definition, the two basic fundamental theorems have been given.
However, we only state them as their proofs are beyond the scope of this text book.

7.8.2 First fundamental theorem of integral calculus

Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function. Then A′′(x) = f (x), for all x ∈ [a, b].

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268 MATHEMATICS

7.8.3 Second fundamental theorem of integral calculus

We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative.
Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
b
an anti derivative of f. Then ∫a f ( x ) dx = [F( x )] ba = F (b) – F(a).

Remarks
b
(i) In words, the Theorem 2 tells us that ∫ a f ( x) dx = (value of the anti derivative F
of f at the upper limit b – value of the same anti derivative at the lower limit a).
(ii) This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum.
(iii) The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand. This strengthens the relationship
between differentiation and integration.
b
(iv) In ∫ a f ( x) dx , the function f needs to be well defined and continuous in [a, b].
1
3
∫ − 2 x( x
2
For instance, the consideration of definite integral – 1) 2 dx is erroneous
1
2
since the function f expressed by f (x) = x( x – 1) 2 is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3].
b
Steps for calculating ∫ a f ( x) dx .
(i) Find the indefinite integral ∫ f ( x ) dx . Let this be F(x). There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
b
∫ a f ( x) dx = [F ( x) + C] a = [F(b) + C] – [F(a) + C] = F(b) – F(a) .
b

Thus, the arbitrary constant disappears in evaluating the value of the definite
integral.
b
∫ a f ( x) dx .
b
(ii) Evaluate F(b) – F(a) = [F ( x)] a , which is the value of
We now consider some examples

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Example 25 Evaluate the following integrals:

3 2 9 x
(i) ∫2 x dx (ii) ∫4 3
dx
(30 – x 2 )2
π
2 x dx
(iii) ∫1 ( x + 1) ( x + 2)
(iv) ∫ 0
4 sin 3 2 t cos 2 t dt

Solution
3 x3
(i) Let I = ∫ 2 x dx . Since ∫ x dx =
2
= F ( x) ,
2
3
Therefore, by the second fundamental theorem, we get
27 8 19
I = F (3) – F (2) = – =
3 3 3
9 x
(ii) Let I = ∫ 3
dx . We first find the anti derivative of the integrand.
4
(30 – x 2 )2
3
3 2
Put 30 – x2 = t. Then – x dx = dt or x dx = – dt
2 3

 
x 2 dt 2 1  2  1 
Thus, ∫ dx = – ∫ 2 =   =  3 
= F ( x)
3
3 t 3  t  3
(30 – x 2 ) 2  (30 – x 2 ) 
Therefore, by the second fundamental theorem of calculus, we have
9
 
2 1 
I = F(9) – F(4) =  3 
3
 (30 – x 2 ) 
4

2 1 1  2  1 1  19
− − =
3  (30 – 27) 30 – 8  3  3 22  99

= =

2 x dx
(iii) Let I = ∫
1 ( x + 1) ( x + 2)

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x –1 2
Using partial fraction, we get = +
( x + 1) ( x + 2) x + 1 x + 2

x dx
So ∫ ( x + 1) ( x + 2) = – log x + 1 + 2log x + 2 = F( x)

Therefore, by the second fundamental theorem of calculus, we have

I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
 32 
= – 3 log 3 + log 2 + 2 log 4 = log  
 27 
π
2t cos 2 t dt. Consider ∫ sin 2t cos 2t dt
3
(iv) Let I = ∫ 4
sin 3
0

1
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = du
2
1 3
∫ sin 2∫
3
So 2t cos 2t dt = u du

1 4 1 4
= [u ] = sin 2t = F (t ) say
8 8
Therefore, by the second fundamental theorem of integral calculus
π 1 π 1
I = F ( ) – F (0) = [sin 4 – sin 4 0] =
4 8 2 8

EXERCISE 7.8
Evaluate the definite integrals in Exercises 1 to 20.
1 31 2
∫ −1 ( x + 1) dx ∫ 2 x dx ∫ 1 (4 x – 5 x2 + 6 x + 9) dx
3
1. 2. 3.

π π π
5 x
∫ 0 sin 2 x dx ∫ 0 cos 2 x dx ∫ 4 e dx ∫
4 2 4
4. 5. 6. 7. tan x dx
0

π
1 dx 1 dx 3 dx
8. ∫
4
π
cosec x dx 9. ∫0 10. ∫ 0 1 + x2 11. ∫ 2 x2 − 1
6 1 – x2

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 INTEGRALS 271

π 3 x dx 1 2x + 3 1
12. ∫ 2
0
cos 2 x dx 13. ∫ 2 x2 + 1 14. ∫ 0 5 x 2 + 1 dx 15. ∫0x e
x2
dx

π
2 5x 2 π x x
∫1 ∫ ∫ 0 (sin
2
16. 17. 4
(2sec2 x + x 3 + 2) dx 18. – cos2 ) dx
x 2 + 4x + 3 0 2 2

2 6x + 3 πx 1
∫ 0 x2 + 4 dx ∫ 0 (x e + sin
x
19. 20. ) dx
4
Choose the correct answer in Exercises 21 and 22.
3 dx
21. ∫1 1 + x2
equals

π 2π π π
(A) (B) (C) (D)
3 3 6 12
2
dx
22. ∫ 0
3
4 + 9x2
equals

π π π π
(A) (B) (C) (D)
6 12 24 4

7.9 Evaluation of Definite Integrals by Substitution

In the previous sections, we have discussed several methods for finding the indefinite
integral. One of the important methods for finding the indefinite integral is the method
of substitution.
b
To evaluate ∫ a f ( x) dx , by substitution, the steps could be as follows:
1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form.
2. Integrate the new integrand with respect to the new variable without mentioning
the constant of integration.
3. Resubstitute for the new variable and write the answer in terms of the original
variable.
4. Find the values of answers obtained in (3) at the given limits of integral and find
the difference of the values at the upper and lower limits.

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272 MATHEMATICS

A Note In order to quicken this method, we can proceed as follows: After

performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept
in the new variable itself, and the limits of the integral will accordingly be changed,
so that we can perform the last step.

Let us illustrate this by examples.

1
∫ −15x x5 + 1 dx .
4
Example 26 Evaluate

Solution Put t = x5 + 1, then dt = 5x4 dx.

3 3
2 2 2
∫ 5x x + 1 dx = ∫ t = ( x5 + 1) 2
4 5
Therefore, t dt =
3 3
1
2 5 
3
1
∫ −15x x + 1 dx =  ( x + 1) 2 
4 5
Hence,
3   – 1

2 5
3 3

(
= (1 + 1) 2 – (– 1) + 1 2 
3 
5
)


2 2 
3 3
2 4 2
=  2 − 0 2  = (2 2) =
3  3 3

Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits.
Let t = x5 + 1. Then dt = 5 x4 dx.
Note that, when x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
1 2
∫ −15x x5 + 1 dx = ∫0
4
Therefore t dt
2
2  2 2 2 
3 3 3
2 4 2
=  t  =  2 – 0 2
 = (2 2) =
3   3   3 3
0

1 tan – 1 x
Example 27 Evaluate ∫0 1 + x2
dx

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1
Solution Let t = tan – 1x, then dt = dx . The new limits are, when x = 0, t = 0 and
1 + x2
π π
when x = 1, t = . Thus, as x varies from 0 to 1, t varies from 0 to .
4 4
π
π
1 tan x–1 t2  4 1  π2  π2
Therefore ∫0 1 + x2
dx = ∫ 0
4
t dt   =
2
 0 2

 16
– 0 =
 32

EXERCISE 7.9
Evaluate the integrals in Exercises 1 to 8 using substitution.
π
1x 1 –1 2x 
1. ∫ 0 x2 + 1 dx 2. ∫ 0
2
sin φ cos5 φ d φ 3. ∫ 0 sin 
1+ x
2  dx

π
2 sin x
4. ∫ 0 x x + 2 (Put x + 2 = t2) 5. ∫ 2
0 1 + cos 2 x
dx

2 dx 1 dx 2 1 1 
∫ 0 x + 4 – x2 ∫ −1 x2 + 2 x + 5 ∫ 1  x – 2 x2  e
2x
6. 7. 8. dx

Choose the correct answer in Exercises 9 and 10.

1
1 ( x − x3 ) 3
9. The value of the integral ∫1 x4
dx is
3
(A) 6 (B) 0 (C) 3 (D) 4
x
10. If f (x) = ∫ 0 t sin t dt , then f ′(x) is
(A) cosx + x sin x (B) x sinx
(C) x cosx (D) sinx + x cosx
7.10 Some Properties of Definite Integrals
We list below some important properties of definite integrals. These will be useful in
evaluating the definite integrals more easily.
b b
P0 : ∫ a f ( x) dx = ∫ a f (t ) dt
b a a
P1 : ∫a f ( x) dx = – ∫ f ( x) dx . In particular,
b ∫a f ( x ) dx = 0
b c b
P2 : ∫ a f ( x) dx = ∫ a f ( x) dx + ∫ c f ( x) dx

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b b
P3 : ∫ a f ( x) dx = ∫ a f (a + b − x) dx
a a
P4 : ∫0 f ( x) dx = ∫ f ( a − x) dx
0

(Note that P4 is a particular case of P3)

2a a a
P5 : ∫0 f ( x) dx = ∫ f ( x) dx + ∫ f (2a − x) dx
0 0

2a a
P6 : ∫0 f ( x) dx = 2 ∫ f ( x) dx, if f (2a − x) = f ( x) and
0
0 if f (2a – x) = – f (x)
a a
P7 : (i) ∫ −a f ( x) dx = 2 ∫ 0 f ( x ) dx , if f is an even function, i.e., if f (– x) = f (x).
a
(ii) ∫ −a f ( x) dx = 0 , if f is an odd function, i.e., if f (– x) = – f (x).
We give the proofs of these properties one by one.
Proof of P0 It follows directly by making the substitution x = t.
Proof of P1 Let F be anti derivative of f. Then, by the second fundamental theorem of
b a
calculus, we have ∫ a f ( x) dx = F (b) – F (a) = – [F (a) − F (b)] = −∫ b f ( x ) dx
a
Here, we observe that, if a = b, then ∫ a f ( x) dx = 0 .
Proof of P2 Let F be anti derivative of f. Then
b
∫ a f ( x) dx = F(b) – F(a) ... (1)
c
∫ a f ( x) dx = F(c) – F(a) ... (2)
b
and ∫c f ( x ) dx = F(b) – F(c) ... (3)
c b b
Adding (2) and (3), we get ∫ a f ( x) dx + ∫ c f ( x ) dx = F(b) – F( a) = ∫ f ( x) dx
a
This proves the property P2.
Proof of P3 Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a.
Therefore
b a
∫a f ( x) dx = − ∫ f ( a + b – t ) dt
b

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b
= ∫ a f (a + b – t ) dt (by P1)
b
= ∫ a f (a + b – x) dx by P 0

Proof of P4 Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now

proceed as in P3.
2a a 2a
Proof of P5 Using P2, we have ∫0 f ( x) dx = ∫ f ( x) dx + ∫
0 a
f ( x ) dx .

Let t = 2a – x in the second integral on the right hand side. Then

dt = – dx. When x = a, t = a and when x = 2a, t = 0. Also x = 2a – t.
Therefore, the second integral becomes
2a 0 a a
∫a f ( x) dx = – ∫ f (2a – t ) dt =
a ∫0 f (2a – t ) dt = ∫0 f (2a – x) dx

2a a a
Hence ∫0 f ( x) dx = ∫0 f ( x) dx + ∫ f (2a − x) dx
0
2a a a
Proof of P6 Using P5, we have ∫0 f ( x) dx = ∫ f ( x ) dx + ∫ f (2a − x) dx
0 0
... (1)
Now, if f (2a – x) = f (x), then (1) becomes
2a a a a
∫0 f ( x) dx = ∫0 f ( x) dx + ∫ f ( x) dx = 2∫ f ( x) dx,
0 0

and if f (2a – x) = – f (x), then (1) becomes

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

Proof of P7 Using P2, we have

a 0 a
∫ −a f ( x) dx = ∫ −a f ( x) dx + ∫ 0 f ( x ) dx . Then

Let t = – x in the first integral on the right hand side.

dt = – dx. When x = – a, t = a and when
x = 0, t = 0. Also x = – t.
a 0 a
Therefore ∫ −a f ( x) dx = – ∫ a f (– t ) dt + ∫ f ( x) dx
0

a a
= ∫0 f (– x) dx + ∫ f ( x) dx
0
(by P0) ... (1)

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(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
a a a a
∫ −a f ( x) dx = ∫ 0 f ( x) dx + ∫ f ( x ) dx = 2 ∫ f ( x ) dx
0 0

(ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes

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

2
Example 28 Evaluate ∫ −1 x 3 – x dx

Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that

x3 – x ≥ 0 on [1, 2]. So by P2 we write
2 0 1 2
∫ −1 x3 – x dx = ∫ −1 ( x – x) dx + ∫ – ( x 3 – x) dx + ∫ ( x3 – x ) dx
3
0 1

0 1 2
= ∫ −1 ( x 3 – x) dx + ∫ ( x – x 3 ) dx + ∫ ( x 3 – x) dx
0 1

0 1 2
 x4 x2   x2 x4   x4 x2 
=  –  +  –  + – 
4 2 – 1  2 4 0  4 2 1

1 1 1 1 1 1
= –  –  +  –  + ( 4 – 2) –  – 
4 2 2 4 4 2
1 1 1 1 1 1 3 3 11
= – + + − +2− + = − +2=
4 2 2 4 4 2 2 4 4
π

∫ sin 2
4
Example 29 Evaluate –π
x dx
4

Solution We observe that sin2 x is an even function. Therefore, by P7 (i), we get

π π

∫
4
–π
sin 2 x dx = 2 ∫ 4 sin 2 x dx
0
4

π π
(1 − cos 2 x )
= 2∫
0
4
2
dx = ∫ 0
4
(1 − cos 2 x ) dx

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π
 1 4  π 1 π π 1
=  x – sin 2 x  =  – sin  – 0 = –
 2 0 4 2 2 4 2

π x sin x
Example 30 Evaluate ∫ 0 1 + cos 2 x dx
π x sin x
Solution Let I = ∫ 0 1 + cos2 x dx . Then, by P , we have 4

π (π − x ) sin (π − x) dx
I= ∫0 1 + cos2 (π − x)
π (π − x ) sin x dx π sin x dx
= ∫0 1 + cos x
2 = π∫
0 1 + cos 2 x
−I

π sin x dx
or 2 I = π ∫0
1 + cos2 x
π π sin x dx
or I=
2 ∫ 0 1 + cos 2 x
Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1.
Therefore, (by P1) we get
–π −1 dt π 1 dt
I=
2 ∫1 1+ t 2 =
2 ∫ −1 1 + t 2
1 dt 1
= π∫0 2 (by P7,
since is even function)
1+ t 1+ t2

π  π
2
–1 1 −1
π  tan t  = π  tan –1
1 – tan 0  = π – 0 =
 4  4
=  0  
1
∫ −1 sin
5
Example 31 Evaluate x cos 4 x dx

1
∫ −1sin
5
Solution Let I = x cos 4 x dx . Let f(x) = sin5 x cos4 x. Then

f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i.e., f is an odd function.

Therefore, by P7 (ii), I = 0

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π
sin 4 x
Example 32 Evaluate ∫ 0
2
sin 4 x + cos4 x
dx

π
sin 4 x
Solution Let I = ∫ 0
2
sin 4 x + cos4 x
dx ... (1)

Then, by P4
π
π sin 4 ( − x) π
2 cos 4 x
I =∫ ∫
2 2
dx = dx ... (2)
0 π π 0 cos4 x + sin 4 x
sin 4 ( − x ) + cos4 ( − x)
2 2
Adding (1) and (2), we get
π π π
sin 4 x + cos4 x π
2I = ∫ 0
2
sin x + cos x
4 4
dx = ∫ 2 dx = [ x] 2 =
0 0 2
π
Hence I=
4
π
dx
∫
3
Example 33 Evaluate π
6
1 + tan x

π π
dx cos x dx
∫ = ∫ π3
3
Solution Let I = π
... (1)
6
1+ tan x 6
cos x + sin x

π π 
π cos  + − x  dx
3 6 
∫
3
Then, by P3 I= π
π π  π π 
6 cos  + − x  + sin  + − x 
 3 6   3 6 
π
sin x
∫
3
= π
dx ... (2)
6
sin x + cos x
Adding (1) and (2), we get
π π
π π π
2I = ∫
3
π
dx = [ x] =
3
π
− = . Hence I = π
3 6 6 12
6 6

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π
Example 34 Evaluate ∫ 0
2
log sin x dx

Solution Let I = ∫ 2
log sin x dx
0

Then, by P4
π π
π 
∫ log sin  2 − x  dx = ∫ 02 log cos x dx
I=
0
2

Adding the two values of I, we get

π
2I = ∫ ( log sin x + log cos x ) dx
0
2

π
= ∫ ( log sin x cos x + log 2 − log 2) dx (by adding and subtracting log 2)
0
2

π π

= ∫ 0
2
log sin 2 x dx − ∫
0
2
log 2 dx (Why?)

π
Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when x = ,
2
t = π.
1 π π
Therefore 2I =
2 ∫ 0
log sin t dt − log 2
2
π
2 π
=
2 ∫ 0
2
log sin t dt −
2
log 2 [by P6 as sin (π – t) = sin t)

π
π
= ∫
0
2
log sin x dx −
2
log 2 (by changing variable t to x)

π
= I − log 2
2
π
–π
Hence ∫ 0
2
log sin x dx =
2
log 2 .

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EXERCISE 7.10
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.
3
π π π
sin x sin 2 x dx
1. ∫ 0
2
cos2 x dx 2. ∫ 0
2
sin x + cos x
dx 3. ∫ 0
2
3 3
sin 2 x + cos 2 x
π
cos5 x dx 5 8
4. ∫ 0
2
sin 5 x + cos5 x
5. ∫ −5 | x + 2 | dx 6. ∫2 x − 5 dx

π
1 2
∫ 0 x (1 − x) ∫ ∫0 x
n
7. dx 8. 4
log (1 + tan x ) dx 9. 2 − x dx
0

π π

∫ 0 (2log sin x − log sin 2 x) dx ∫

2 2
10. 11. –π
sin 2 x dx
2
π
x dx
π 2π
12. ∫
0 1 + sin x
13. ∫
2
–π
sin 7 x dx 14. ∫0 cos5 x dx
2
π
sin x − cos x π a x
15. ∫ 0
2
1 + sin x cos x
dx 16. ∫ 0 log (1 + cos x) dx 17. ∫0 x + a−x
dx
4
18. ∫0 x − 1 dx
a a
19. Show that ∫0 f ( x ) g ( x ) dx = 2 ∫ f ( x) dx , if f and g are defined as f(x) = f (a – x)
0

and g(x) + g(a – x) = 4

Choose the correct answer in Exercises 20 and 21.
π

∫ + x cos x + tan 5 x + 1) dx is
2
20. The value of −π
( x3
2

(A) 0 (B) 2 (C) π (D) 1

π
 4 + 3 sin x 
21. The value of ∫ 0
2
log  dx is
 4 + 3 cos x 
3
(A) 2 (B) (C) 0 (D) –2
4

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Miscellaneous Examples

Example 35 Find ∫ cos 6 x 1 + sin 6 x dx

Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx

1
1 2
Therefore ∫ cos 6 x 1 + sin 6 x dx = 6∫
t dt
3 3
1 2 1
= × (t ) 2 + C = (1 + sin 6 x ) 2 + C
6 3 9
1
( x 4 − x) 4
Example 36 Find
∫ x5 dx
1
1 1 4
(1 − )
(x4 − x) 4 x 3 dx
Solution We have
∫ x5 dx = ∫
x4
1 3
Put 1 − 3
= 1 – x – 3 = t , so that 4 dx = dt
x x
1 5
1 5
( x 4 − x) 4 1 4 1 4 4 1 4
Therefore ∫ dx = ∫ t dt = × t 4 + C = 1 − 3  + C
x 5
3 3 5 15  x 

x 4 dx
Example 37 Find ∫
( x − 1) ( x 2 + 1)

Solution We have

x4 1
= ( x + 1) + 3
( x − 1) ( x + 1)
2
x − x + x −1
2

1
= ( x + 1) + ... (1)
( x − 1) ( x 2 + 1)

1 A Bx + C
Now express = + 2 ... (2)
( x − 1)( x + 1)
2
( x − 1) ( x + 1)

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So 1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
1 1
which give A = , B = C = – . Substituting values of A, B and C in (2), we get
2 2
1 1 1 x 1
= − − ... (3)
( x − 1) ( x + 1)
2
2( x − 1) 2 ( x + 1) 2( x + 1)
2 2

Again, substituting (3) in (1), we have

x4 1 1 x 1
= ( x + 1) + − −
( x − 1) ( x + x + 1)
2
2( x − 1) 2 ( x + 1) 2( x + 1)
2 2

Therefore
x4 x2 1 1 1
∫ ( x − 1) ( x 2 + x + 1) dx =
2
+ x + log x − 1 – log ( x 2 + 1) – tan – 1 x + C
2 4 2

 1 
Example 38 Find ∫ log (log x) + (log x)2  dx
 1 
Solution Let I = ∫ log (log x) +  dx
 (log x) 2 

1
= ∫ log (log x) dx + ∫ dx
(log x) 2
In the first integral, let us take 1 as the second function. Then integrating it by
parts, we get
1 dx
I = x log (log x ) − ∫ x dx + ∫
x log x (log x ) 2
dx dx
= x log (log x ) − ∫ +∫ ... (1)
log x (log x ) 2
dx
Again, consider ∫ log x , take 1 as the second function and integrate it by parts,
dx  x  1  1  
we have ∫ log x =  log x – ∫ x  – (log x)2  x   dx  ... (2)
 

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Putting (2) in (1), we get

x dx dx x
I = x log (log x ) − −∫ 2
+∫ 2 =
x log (log x ) − +C
log x (log x ) (log x ) log x

Example 39 Find ∫  cot x + tan x  dx

Solution We have

I= ∫ 
cot x + tan x  dx = ∫ tan x (1 + cot x ) dx
Put tan x = t2, so that sec2 x dx = 2t dt
2t dt
or dx =
1 + t4
 1  2t
Then I = ∫ t 1 + 2  dt
 t  (1 + t )
4

 1  1
(t 2 + 1) 1 + 2  dt 1 + 2  dt
dt = 2 ∫ 
t 
=2∫ 
t 
= 2∫
t4 +1  2 1  1
2
 t +  t −  + 2
 t2   t

1  1
Put t − = y, so that 1 + 2  dt = dy. Then
t  t 

 1
t − 
dy y –1  t
I = 2∫ = 2 tan – 1 + C = 2 tan +C
( 2)
2
y2 + 2 2

 t2 −1  – 1  tan x − 1 
= 2 tan – 1   + C = 2 tan  +C
 2t   2 tan x 

sin 2 x cos 2 x dx
Example 40 Find ∫ 9 – cos 4 (2 x)

sin 2 x cos 2 x
Solution Let I = ∫ dx
9 – cos 4 2 x

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Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt

1 dt 1 t 1 1 
Therefore I=–
4 ∫ 9–t 2
=–
4
sin –1   + C = − sin − 1  cos 2 2 x  + C
3 4 3 
3

∫ −1 x sin (π x) dx
2
Example 41 Evaluate

 x sin π x for − 1 ≤ x ≤ 1

Solution Here f (x) = | x sin πx | =  3
− x sin π x for 1 ≤ x ≤ 2

3 3
1
Therefore ∫ 2
−1
| x sin π x | dx = ∫ −1 x sin π x dx + ∫ 1
2
− x sin π x dx

3
1
= ∫ −1 x sin π x dx − ∫ 1
2 x sin π x dx

Integrating both integrals on righthand side, we get

3

∫ 2
−1
| x sin π x | dx =

2  1 1 3 1
= − − − = +
π  π2 π  π π2
π x dx
Example 42 Evaluate ∫ 0 a 2 cos 2 x + b2 sin 2 x
π x dx π ( π − x ) dx
Solution Let I = ∫ 0 a 2 cos 2 x + b2 sin 2 x = ∫ 0 a2 cos 2 (π − x) + b2 sin 2 (π − x) (using P4)

π dx π x dx
= π∫ −∫ 2
0 a cos x + b sin x
2 2 22 0 a cos x + b 2 sin 2 x
2

π dx
= π ∫0 −I
a cos x + b 2 sin 2 x
2 2

π dx
Thus 2I = π ∫ 0
a cos x + b2 sin 2 x
2 2

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π
π π dx π dx
or I=
2 ∫0 2 2 2 2
= ⋅2 ∫ 2 2
a cos x + b sin x 2 0 a cos x + b 2 sin 2 x
2 (using P6)

π π
 dx dx 
π ∫ 4 2 + ∫ π a2 cos2 x + b2 sin 2 x 
2
=  0 a cos2 x + b2 sin 2 x
 4


π π
 4 sec 2 x dx 2
2 cosec x dx

= π  ∫ + ∫ 2
 0 a + b tan x π a cot x + b 
2 2 2 2 2
 4

 1 dt 0 du 
2 (
= π ∫ 2 − ∫ put tan x = t and cot x = u )
 0 a +b t 1 a u +b 
2 2 2 2

1 0
π  –1 bt  π  –1 au  π  –1 b a
+ tan –1  = π
2
= ab  tan a  – ab  tan b  =  tan
 0  1 ab  a b  2ab

Miscellaneous Exercise on Chapter 7

Integrate the functions in Exercises 1 to 23.
1 1 1 a
1. 2. 3. [Hint: Put x = ]
x − x3 x+a + x+b x ax − x 2 t

1 1 1 1
4. 3 5. [Hint: = , put x = t6]
1 1 1 1 1
 1

x 2 ( x 4 + 1) 4 x2 + x3 x2 + x3 x3 1 + x 6 
 
 

5x sin x e5 log x − e4 log x

6. 7. 8.
( x + 1) ( x 2 + 9) sin ( x − a ) e3 log x − e 2 log x

cos x sin 8 − cos8 x 1

9. 10. 11.
4 − sin x 2 1 − 2sin 2 x cos 2 x cos ( x + a) cos ( x + b)

x3 ex 1
12. 13. 14.
1 − x8 (1 + e x ) (2 + e x ) ( x + 1) ( x 2 + 4)
2

15. cos3 x elog sinx 16. e3 logx (x4 + 1)– 1 17. f ′ (ax + b) [f (ax + b)]n

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286 MATHEMATICS

1 1− x 2 + sin 2 x x
18. 19. 20. e
sin x sin ( x + α )
3
1+ x 1 + cos 2 x

x2 + x + 1 –1 1− x
21. 22. tan
( x + 1) 2 ( x + 2) 1+ x

x 2 + 1  log ( x 2 + 1) − 2 log x 
23.
x4
Evaluate the definite integrals in Exercises 24 to 31.
π π
π  1 − sin x  sin x cos x cos 2 x dx
24. ∫π ex   dx 25.
 1 − cos x  ∫ 0
4
cos4 x + sin 4 x
dx 26. ∫ 0
2
cos 2 x + 4 sin 2 x
2

π π
sin x + cos x 1 dx sin x + cos x
27. ∫
3
π
sin 2 x
dx 28. ∫0 1+ x − x
29. ∫ 0
4
9 + 16 sin 2 x
dx
6
π

30. ∫ 2
sin 2 x tan −1 (sin x) dx
0

4
31. ∫ 1 [| x −1| + | x − 2 | + | x − 3 |] dx
Prove the following (Exercises 32 to 37)
3 dx 2 2 1
∫ 1 x 2 ( x + 1) = 3 + log 3 ∫ 0 x e dx = 1
x
32. 33.

1 π
2
∫ −1 x cos 4 x dx = 0
17
34. 35. ∫ 0
2 sin 3 x dx =
3
π
1 −1 π
36. ∫ 0
4
2 tan 3 x dx = 1 − log 2 37. ∫ 0 sin x dx =
2
−1

Choose the correct answers in Exercises 38 to 40

dx
38. ∫ e x + e− x is equal to

(A) tan–1 (ex) + C (B) tan–1 (e–x) + C

(C) log (ex – e–x) + C (D) log (ex + e–x) + C
cos 2 x
39. ∫ (sin x + cos x)2 dx is equal to

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 INTEGRALS 287

–1
(A) +C (B) log |sin x + cos x | + C
sin x + cos x
1
(C) log |sin x − cos x | + C (D)
(sin x + cos x) 2
b
40. If f (a + b – x) = f (x), then ∫ a x f ( x) dx is equal to

a +b b a +b b
(A)
2 ∫ a f (b − x) dx (B)
2 ∫ a f (b + x) dx
b−a b a +b b
(C)
2 ∫a f ( x ) dx (D)
2 ∫ a f ( x) dx
Summary
® Integration is the inverse process of differentiation. In the differential calculus,
we are given a function and we have to find the derivative or differential of
this function, but in the integral calculus, we are to find a function whose
differential is given. Thus, integration is a process which is the inverse of
differentiation.
d
Let F( x) = f ( x ) . Then we write ∫ f ( x ) dx = F ( x ) + C . These integrals
dx
are called indefinite integrals or general integrals, C is called constant of
integration. All these integrals differ by a constant.
® Some properties of indefinite integrals are as follows:
1. ∫ [ f ( x) + g ( x)] dx = ∫ f ( x) dx + ∫ g ( x) dx
2. For any real number k, ∫k f ( x ) dx = k ∫ f ( x) dx
More generally, if f1, f2, f3, ... , fn are functions and k1, k2, ... ,kn are real
numbers. Then

∫ [k1 f1 ( x ) + k 2 f 2 ( x) + ... + k n f n ( x )] dx

= k1 ∫ f1 ( x ) dx + k 2 ∫ f 2 ( x) dx + ... + kn ∫ f n ( x) dx
® Some standard integrals

x n +1
∫ x dx = + C , n ≠ – 1. Particularly, ∫ dx = x + C
n
(i)
n +1

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288 MATHEMATICS

(ii) ∫ cos x dx = sin x + C (iii) ∫ sin x dx = – cos x + C

∫ sec x dx = tan x + C ∫ cosec x dx = – cot x + C
2 2
(iv) (v)

(vi) ∫ sec x tan x dx = sec x + C

dx
∫ = sin − 1 x + C
(vii) ∫ cosec x cot x dx = – cosec x + C (viii)
1− x 2

dx dx
∫ = − cos − 1 x + C −1
(ix)
1− x 2 (x) ∫ 1 + x2 = tan x+C

dx −1
(xi) ∫ 1 + x2 = − cot x+C (xii) ∫ e dx = e
x x
+C

ax 1
(xiii) ∫ a dx = +C ∫ x dx = log | x | + C
x
(xiv)
log a

® Integration by partial fractions

P( x )
Recall that a rational function is ratio of two polynomials of the form ,
Q( x )
where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0. If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
P( x ) P ( x)
may divide P (x) by Q (x) so that = T ( x) + 1 , where T(x) is a
Q( x ) Q( x )
polynomial in x and degree of P1 (x) is less than the degree of Q(x). T(x)
P1 ( x )
being polynomial can be easily integrated. can be integrated by
Q( x )
P1 ( x)
expressing as the sum of partial fractions of the following type:
Q( x)
px + q A
+
B
1. = ,a≠b
( x − a) ( x − b) x − a x −b

px + q A B
2. = +
( x − a )2 x − a ( x − a) 2

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 INTEGRALS 289

px 2 + qx + r A B C
3. = + +
( x − a ) ( x − b ) ( x − c) x − a x −b x −c

px 2 + qx + r A B C
4. = + +
( x − a ) 2 ( x − b) x − a ( x − a) 2
x −b

px 2 + qx + r A Bx + C
5. = +
( x − a) ( x 2 + bx + c ) x − a x 2 + bx + c
where x2 + bx + c can not be factorised further.
® Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals. The method in which we change the variable to some
other variable is called the method of substitution. When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals. Using substitution technique, we obtain the following standard
integrals.
(i) ∫ tan x dx = log sec x + C (ii) ∫ cot x dx = log sin x + C

(iii) ∫ sec x dx = log sec x + tan x + C

(iv) ∫ cosec x dx = log cosec x − cot x + C

® Integrals of some special functions
dx 1 x−a
(i) ∫ x2 − a 2 = 2a log x+a
+C

dx 1 a+ x dx 1 x
∫ a 2 − x 2 = 2a log
−1
(ii)
a−x
+C (iii) ∫ x 2 + a 2 = a tan a
+C

dx x
dx
∫ = sin − 1 +C
(iv) ∫ x2 − a 2
= log x + x 2 − a 2 + C (v)
a −x
2 2 a

dx
(vi) ∫ x +a
2 2
= log | x + x 2 + a 2 | + C

® Integration by parts
For given functions f1 and f2, we have

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290 MATHEMATICS

, i.e., the

integral of the product of two functions = first function × integral of the

second function – integral of {differential coefficient of the first function ×
integral of the second function}. Care must be taken in choosing the first
function and the second function. Obviously, we must take that function as
the second function whose integral is well known to us.

® ∫ e x [ f ( x) + f ′( x)] dx = ∫ e x f ( x) dx + C
® Some special types of integrals
x 2 a2
(i) ∫ x 2 − a 2 dx =
2
x − a 2 − log x + x 2 − a 2 + C
2
x 2 a2
(ii) ∫ x 2 + a 2 dx =
2
x + a 2 + log x + x 2 + a 2 + C
2
x 2 a2 x
(iii) ∫ a 2 − x 2 dx = a − x2 + sin −1 + C
2 2 a
dx dx
(iv) Integrals of the types ∫ ax2 + bx + c or ∫ ax + bx + c
2
can be

transformed into standard form by expressing

 2 b c  b   c b2 
2
2 a
ax + bx + c =  x + x + = a  x +  + − 
 a a   2a   a 4a 2  

px + q dx px + q dx
(v) Integrals of the types ∫ ax2 + bx + c or ∫ ax 2 + bx + c
can be
transformed into standard form by expressing
d
px + q = A ( ax 2 + bx + c) + B = A (2ax + b) + B , where A and B are
dx
determined by comparing coefficients on both sides.
b
® We have defined ∫a f ( x) dx as the area of the region bounded by the curve
y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b. Let x be a

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 INTEGRALS 291

x
given point in [a, b]. Then ∫a f ( x ) dx represents the Area function A (x).
This concept of area function leads to the Fundamental Theorems of Integral
Calculus.
® 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 assumed to be continuous on [a, b]. Then A′ (x) = f (x) for all
x ∈ [a, b].
® Second fundamental theorem of integral calculus
Let f be a continuous function of x defined on the closed interval [a, b] and
d
let F be another function such that F( x ) = f ( x) for all x in the domain of
dx
b
∫ a f ( x) dx = [F( x) + C]a = F (b) − F (a) .
b
f, then
This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit.

—v—

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