INTEGRALS 287
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
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
primitive) of the function. Further, the formula that gives
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.
Chapter
7
INTEGRALS
G .W. Leibnitz
(1646 -1716)
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288 MATHEMATICS
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:
We know that
(sin )
d
x
dx
= cos x ... (1)
3
()
3
dx
dx
= x
2
... (2)
and
()
x
d
e
dx
= e
x
... (3)
We observe that in (1), the function cos x is the derived function of sin x. We say
that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3),
3
3
x
and
e
x
are the anti derivatives (or integrals) of x
2
and e
x
, 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 :
(sin + C) cos
=
d
xx
dx
,
3
2
( + C)
3
=
dx
x
dx
and
( + C)
=
x x
d
ee
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.
More generally, if there is a function F such that
F()= ()
d
x fx
dx
,
∀
x ∈ I (interval),
then for any arbitrary real number C, (also called constant of integration)
[
]
F( )+C
d
x
dx
= f (x), x ∈ I
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INTEGRALS 289
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
Then
df
dx
= f
′
= g
′
– h
′
giving f
′
(x) = g
′
(x) – h
′
(x)
∀
x ∈ I
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( )+C
f x dx x
∫
.
Notation Given that
()
dy
fx
dx
=
, 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)
(i)
1
1
n
n
dx
x
dx n
+
=
+
;
1
C
1
n
n
x
x dx
n
+
=+
+
∫
, n ≠ –1
Particularly, we note that
()
1
d
x
dx
=
;
C
dx x
=+
∫
(ii)
()
sin cos
d
xx
dx
=
;
cos sin C
x dx x
=+
∫
(iii)
()
– cos sin
d
xx
dx
=
;
sin cos C
x dx – x
=+
∫
(iv)
()
2
tan sec
d
xx
dx
=
;
2
sec tan C
x dx x
=+
∫
(v)
()
2
– cot cosec
d
xx
dx
=
;
2
cosec cot C
x dx – x
=+
∫
(vi)
()
sec sec tan
d
x xx
dx
=
;
sec tan sec C
x x dx x
=+
∫
(vii)
()
– cosec cosec cot
d
x xx
dx
=
;
cosec cot – cosec C
x x dx x
=+
∫
(viii)
()
–1
2
1
sin
1
d
x
dx
–x
=
;
–1
2
sin C
1
dx
x
–x
=+
∫
(ix)
()
–1
2
1
– cos
1
d
x
dx
–x
=
;
–1
2
cos C
1
dx
–x
–x
=+
∫
(x)
()
–1
2
1
tan
1
d
x
dx
x
=
+
;
–1
2
tan C
1
dx
x
x
=+
+
∫
(xi)
()
–1
2
1
– cot
1
d
x
dx
x
=
+
;
–1
2
cot C
1
dx
–x
x
=+
+
∫
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INTEGRALS 291
(xii)
()
–1
2
1
sec
1
d
x
dx
x x–
=
;
–1
2
sec C
1
dx
x
x x–
=+
∫
(xiii)
()
–1
2
1
– cosec
1
d
x
dx
x x–
=
;
–1
2
cosec C
1
dx
–x
x x–
=+
∫
(xiv)
()
xx
d
ee
dx
=
;
C
xx
e dx e
=+
∫
(xv)
()
1
log | |
d
x
dx x
=
;
1
log | | C
dx x
x
=+
∫
(xvi)
x
x
da
a
dx log a
=
;
C
x
x
a
a dx
log 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 Geometrical interpretation of indefinite integral
Let f (x) = 2x. Then
2
() C
f x dx x
=+
∫
. For different values of C, we get different
integrals. But these integrals are very similar geometrically.
Thus, y = x
2
+ C, where C is arbitrary constant, represents a family of integrals. By
assigning different values to C, we get different members of the family. These together
constitute the indefinite integral. In this case, each integral represents a parabola with
its axis along y-axis.
Clearly, for C = 0, we obtain y = x
2
, a parabola with its vertex on the origin. The
curve y = x
2
+ 1 for C = 1 is obtained by shifting the parabola y = x
2
one unit along
y-axis in positive direction. For C = – 1, y = x
2
– 1 is obtained by shifting the parabola
y = x
2
one unit along y-axis in the negative direction. Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis. Some of
these have been shown in the Fig 7.1.
Let us consider the intersection of all these parabolas by a line x = a. In the Fig 7.1,
we have taken a > 0. The same is true when a < 0. If the line x = a intersects the
parabolas y = x
2
, y = x
2
+ 1, y = x
2
+ 2, y = x
2
– 1, y = x
2
– 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
etc.,
respectively, then
dy
dx
at these points equals 2a. This indicates that the tangents to the
curves at these points are parallel. Thus,
2
C
2 C F ()
x dx x x
= +=
∫
(say), implies that
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292 MATHEMATICS
the tangents to all the curves y = F
C
(x), C ∈ R, at the points of intersection of the
curves by the line x = a, (a ∈ R), are parallel.
Further, the following equation (statement)
( ) F ( ) C (say)
f x dx x y= +=
∫
,
represents a family of curves. The different values of C will correspond to different
members of this family and these members can be obtained by shifting any one of the
curves parallel to itself. This is the geometrical interpretation of indefinite integral.
7.2.2 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
f x dx
dx
∫
= f (x)
and
()
f x dx
′
∫
= f (x) + C, where C is any arbitrary constant.
Fig 7.1
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INTEGRALS 293
Proof Let F be any anti derivative of f, i.e.,
F( )
d
x
dx
= f (x)
Then
()
f x dx
∫
= F(x) + C
Therefore
()
d
f x dx
dx
∫
=
(
)
F( )+C
d
x
dx
=
F( )= ( )
d
x fx
dx
Similarly, we note that
f ′(x) =
()
d
fx
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.
Proof Let f and g be two functions such that
()
d
f x dx
dx
∫
=
()
d
g x dx
dx
∫
or
() ()
d
f x dx – g x dx
dx
∫∫
= 0
Hence
() ()
f x dx – g x dx
∫∫
= C, where C is any real number (Why?)
or
()
f x dx
∫
=
() C
g x dx
+
∫
So the families of curves
{
}
11
( ) C ,C R
f x dx
+∈
∫
and
{
}
22
( ) C ,C R
g x dx
+∈
∫
are identical.
Hence, in this sense,
() and ()
fxdx gxdx
∫∫
are equivalent.
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Note The equivalence of the families
{
}
11
( ) + C ,Cf x dx ∈
∫
R
and
{
}
22
( ) + C ,Cg x dx ∈
∫
R
is customarily expressed by writing
() = ()
f x dx g x dx
∫∫
,
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
dx
∫
= f (x) + g (x) ... (1)
On the otherhand, we find that
() + ()
d
f x dx g x dx
dx
∫∫
=
() + ()
d d
f x 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
=
∫∫
Proof By the Property (I),
() ()
d
kfxdx kfx
dx
=
∫
.
Also
()
d
k f x dx
dx
∫
=
() = ()
d
k fxdx kfx
dx
∫
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
f
1
, f
2
, ..., f
n
and the real numbers, k
1
, k
2
, ..., k
n
giving
[
]
11 2 2
() () ()
nn
kfx kf x ... kf x dx
+ ++
∫
=
11 22
() () ()
nn
k f x dx k f x dx ... k f 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.
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Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i) cos 2x (ii) 3x
2
+ 4x
3
(iii)
1
x
, x ≠ 0
Solution
(i) We look for a function whose derivative is cos 2x. Recall that
d
dx
sin 2x = 2 cos 2x
or cos 2x =
1
2
d
dx
(sin 2x) =
1
sin 2
2
d
x
dx
Therefore, an anti derivative of cos 2x is
1
sin 2
2
x
.
(ii) We look for a function whose derivative is 3x
2
+ 4x
3
. Note that
()
34
d
xx
dx
+
=3x
2
+ 4x
3
.
Therefore, an anti derivative of 3x
2
+ 4x
3
is
x
3
+ x
4
.
(iii) We know that
1
11
(log ) 0 and [log ( )] ( 1) 0
d d
x ,x –x – ,x
dx x dx –x x
=> = =<
Combining above, we get
()
1
log 0
d
x ,x
dx x
=≠
Therefore,
1
log
dx x
x
=
∫
is one of the anti derivatives of
1
x
.
Example 2 Find the following integrals:
(i)
3
2
1
x–
dx
x
∫
(ii)
2
3
( 1)
x dx
+
∫
(iii)
∫
3
2
1
( 2 –)
+
x
x e dx
x
Solution
(i) We have
3
2
2
1
–
x–
dx x dx – x dx
x
=
∫ ∫∫
(by Property V)
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296 MATHEMATICS
=
11 21
1 2
C
C
11 21
–
x x
–
–
+ +
+ +
+ +
; C
1
, C
2
are constants of integration
=
2 1
1 2
CC
21
–
xx
––
–
+
=
2
12
1
+C C
2
x
–
x
+
=
2
1
+C
2
x
x
+
, where C = C
1
– C
2
is another constant of integration.
Note From now onwards, we shall write only one constant of integration in the
final answer.
(ii) We have
2 2
3 3
( 1)
x dx x dx dx
+= +
∫ ∫∫
=
2
1
3
C
2
1
3
x
x
+
++
+
=
5
3
3
C
5
xx
++
(iii) We have
3 3
2 2
1 1
(2 ) 2
x x
x e – dx x dx e dx – dx
x x
+ =+
∫ ∫∫ ∫
=
3
1
2
2 – log + C
3
1
2
x
x
ex
+
+
+
=
5
2
2
2 – log + C
5
x
xe x+
Example 3 Find the following integrals:
(i)
(sin cos )
x x dx
+
∫
(ii)
cosec (cosec cot )
x x x dx
+
∫
(iii)
2
1 sin
cos
–x
dx
x
∫
Solution
(i) We have
(sin cos ) sin cos
x x dx x dx x dx
+=+
∫ ∫∫
=
– cos sin C
xx
++
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INTEGRALS 297
(ii) We have
2
(cosec (cosec + cot ) cosec cosec cot
x x x dx x dx x x dx
= +
∫ ∫∫
=
– cot cosec C
x– x
+
(iii) We have
2 2 2
1 sin 1 sin
cos cos cos
–x x
dx dx – dx
x xx
=
∫ ∫∫
=
2
sec tan sec
x dx – x x dx
∫∫
=
tan sec C
x– x
+
Example 4 Find the anti derivative F of f defined by f (x) = 4x
3
– 6, where F (0) = 3
Solution One anti derivative of f (x) is x
4
– 6x since
4
( 6)
d
x–x
dx
= 4x
3
– 6
Therefore, the anti derivative F is given by
F(x) = x
4
– 6x + C, where C is constant.
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) = x
4
– 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
2
–x
e dx
∫
by inspection since we can not find a function whose derivative is
2
–x
e
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298 MATHEMATICS
(iii) When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly. For instance
41
45
1
CC
41 5
y
y dy y
+
= += +
+
∫
7.2.3 Comparison between differentiation and integration
1. Both are operations on functions.
2. Both satisfy the property of linearity, i.e.,
(i)
[ ]
11 2 2 1 1 2 2
() () () ()
d d d
kfx kfx k fx k fx
dx dx dx
+= +
(ii)
[
]
11 2 2 11 22
() () () ()
k f x k f x dx k f x dx k f x dx
+ = +
∫ ∫∫
Here k
1
and k
2
are constants.
3. We have already seen that all functions are not differentiable. Similarly, all functions
are not integrable. We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes.
4. The derivative of a function, when it exists, is a unique function. The integral of
a function is not so. However, they are unique upto an additive constant, i.e., any
two integrals of a function differ by a constant.
5. When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P. When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P.
6. We can speak of the derivative at a point. We never speak of the integral at a
point, we speak of the integral of a function over an interval on which the integral
is defined as will be seen in Section 7.7.
7. The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point. Similarly, the indefinite integral of
a function represents geometrically, a family of curves placed parallel to each
other having parallel tangents at the points of intersection of the curves of the
family with the lines orthogonal (perpendicular) to the axis representing the variable
of integration.
8. The derivative is used for finding some physical quantities like the velocity of a
moving particle, when the distance traversed at any time t is known. Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known.
9. Differentiation is a process involving limits. So is integration, as will be seen in
Section 7.7.
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INTEGRALS 299
10. The process of differentiation and integration are inverses of each other as
discussed in Section 7.2.2 (i).
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
4. (ax + b)
2
5. sin 2x – 4 e
3x
Find the following integrals in Exercises 6 to 20:
6.
3
(4 + 1)
x
e dx
∫
7.
2
2
1
(1 – )
x dx
x
∫
8.
2
()
ax bx c dx
++
∫
9.
2
(2 )
x
x e dx
+
∫
10.
2
1
x – dx
x
∫
11.
32
2
54
x x–
dx
x
+
∫
12.
3
34
xx
dx
x
++
∫
13.
32
1
1
x x x–
dx
x–
−+
∫
14.
(1 )
– x x dx
∫
15.
2
( 3 2 3)
x x x dx
++
∫
16.
(2 3cos )
x
x – x e dx+
∫
17.
2
(2 3sin 5 )
x – x x dx
+
∫
18.
sec (sec tan )
x x x dx
+
∫
19.
2
2
sec
cosec
x
dx
x
∫
20.
2
2 – 3sin
cos
x
x
∫
dx.
Choose the correct answer in Exercises 21 and 22.
21. The anti derivative of
1
x
x
+
equals
(A)
11
32
1
2C
3
xx
++
(B)
2
2
3
21
C
32
xx
++
(C)
31
22
2
2C
3
xx
++
(D)
31
22
31
C
22
xx
++
22. If
3
4
3
() 4
d
fx x
dx
x
=−
such that f (2) = 0. Then f (x) is
(A)
4
3
1 129
8
x
x
+−
(B)
3
4
1 129
8
x
x
++
(C)
4
3
1 129
8
x
x
++
(D)
3
4
1 129
8
x
x
+−
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300 MATHEMATICS
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).
Consider I =
()
f x dx
∫
Put x = g(t) so that
dx
dt
= g′(t).
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 (x
2
+ 1)
(iii)
42
tan sec
xx
x
(iv)
1
2
sin (tan )
1
–
x
x+
Solution
(i) We know that derivative of mx is m. Thus, we make the substitution
mx = t so that mdx = dt.
Therefore,
1
sin sin
mx dx t dt
m
=
∫∫
= –
1
m
cos t + C = –
1
m
cos mx + C
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INTEGRALS 301
(ii) Derivative of x
2
+ 1 is 2x. Thus, we use the substitution x
2
+ 1 = t so that
2x dx = dt.
Therefore,
2
2 sin ( 1) sin
x x dx t dt
+=
∫ ∫
= – cos t + C = – cos (x
2
+ 1) + C
(iii) Derivative of
x
is
1
2
11
2
2
–
x
x
=
. Thus, we use the substitution
1
so that giving
2
x t dx dt
x
==
dx = 2t dt.
Thus,
42
42
tan sec 2 tan sec
x x t t t dt
dx
t
x
=
∫ ∫
=
42
2 tan sec
t t dt
∫
Again, we make another substitution tan t = u so that sec
2
t dt = du
Therefore,
42 4
2 tan sec 2
t t dt u du
=
∫∫
=
5
2C
5
u
+
=
5
2
tan C
5
t
+
(since u = tan t)
=
5
2
tan C (since )
5
x tx
+=
Hence,
42
tan sec
xx
dx
x
∫
=
5
2
tan C
5
x
+
Alternatively, make the substitution
tan
xt
=
(iv) Derivative of
1
2
1
tan
1
–
x
x
=
+
. Thus, we use the substitution
tan
–1
x = t so that
2
1
dx
x+
= dt.
Therefore ,
1
2
sin (tan )
sin
1
–
x
dx t dt
x
=
+
∫∫
= – cos t + C = – cos (tan
–1
x) + 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 = log sec + C
x dx x
We have
sin
tan
cos
x
x dx dx
x
=
∫∫
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302 MATHEMATICS
Put cos x = t so that sin x dx = – dt
Then
tan log C log cos C
dt
x dx – – t – x
t
= = += +
∫∫
or
tan log sec C
x dx x
=+
∫
(ii)
∫
cot = log sin + C
x dx x
We have
cos
cot
sin
x
x dx dx
x
=
∫∫
Put sin x = t so that cos x dx = dt
Then
cot
dt
x dx
t
=
∫∫
=
log C
t
+
=
log sin C
x
+
(iii)
∫
sec = log sec + tan + C
x dx x x
We have
sec (sec tan )
sec
sec + tan
xx x
x dx dx
xx
+
=
∫∫
Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt
Therefore,
sec
log + C = log sec tan C
dt
x dx t
xx
t
== ++
∫∫
(iv)
∫
cosec = log cosec – cot + C
x dx
xx
We have
cosec (cosec cot )
cosec
(cosec cot )
x xx
x dx dx
xx
+
=
+
∫∫
Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt
So
cosec – – log | | – log |cosec cot | C
dt
x dx t
xx
t
== = ++
∫∫
=
22
cosec cot
– log
C
cosec cot
xx
xx
−
+
−
=
log cosec cot C
x– x
+
Example 6 Find the following integrals:
(i)
32
sin cos
x x dx
∫
(ii)
sin
sin ( )
x
dx
xa+
∫
(iii)
1
1 tan
dx
x
+
∫
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INTEGRALS 303
Solution
(i) We have
32 22
sin cos sin cos (sin )
x x dx x x x dx
=
∫∫
=
22
(1 – cos ) cos (sin )
x x x dx
∫
Put t = cos x so that dt = – sin x dx
Therefore,
22
sin cos (sin )
xxxdx
∫
=
22
(1 – )
t t dt
−
∫
=
35
24
(–) C
35
tt
– t t dt – –
=+
∫
=
3 5
11
cos cos C
35
–x x
++
(ii) Put x + a = t. Then dx = dt. Therefore
sin sin ( )
sin ( ) sin
x t–a
dx
dt
xa t
=
+
∫∫
=
sin cos cos sin
sin
t a– t a
dt
t
∫
=
cos – sin cot
a dt a t dt
∫∫
=
1
(cos ) (sin ) log sin C
a t– a t
+
=
1
(cos ) ( ) (sin ) log sin ( ) C
axa– a xa
+ ++
=
1
cos cos (sin ) log sin ( ) C sin
x a a a– a x a – a
+ +
Hence,
sin
sin ( )
x
dx
xa
+
∫
= x cos a – sin a log |sin (x + a)| + C,
where, C = – C
1
sin a + a cos a, is another arbitrary constant.
(iii)
cos
1 tan cos sin
dx x dx
x xx
=
++
∫∫
=
1 (cos + sin + cos – sin )
2 cos sin
x x x x dx
xx+
∫
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304 MATHEMATICS
=
1 1 cos – sin
2 2 cos sin
xx
dx
dx
xx
+
+
∫∫
=
1
C 1 cos sin
2 2 2 cos sin
x x– x
dx
xx
++
+
∫
... (1)
Now, consider
cos sin
I
cos sin
x– x
dx
xx
=
+
∫
Put cos x + sin x = t so that (cos x – sin x) dx = dt
Therefore
2
I log C
dt
t
t
== +
∫
=
2
log cos sin C
xx++
Putting it in (1), we get
1 2
C C
1
+ + log cos sin
1 tan 2 2 2 2
dx x
xx
x
= ++
+
∫
=
12
CC
1
+ log cos sin
22 22
x
xx+ ++
=
12
CC
1
+ log cos sin C C
22 22
x
x x,
+ + =+
EXERCISE 7.2
Integrate the functions in Exercises 1 to 37:
1.
2
2
1
x
x
+
2.
()
2
log
x
x
3.
1
log
xx x
+
4.
sin sin (cos )
xx
5.
sin()cos()
ax b ax b
++
6.
ax b
+
7.
2
xx
+
8.
2
12
xx
+
9.
2
(4 2) 1
x xx
+ ++
10.
1
x– x
11.
4
x
x
+
, x > 0
12.
1
35
3
( 1)
x– x
13.
2
33
(2 3 )
x
x
+
14.
1
(log )
m
xx
, x > 0,
1
≠
m
15.
2
94
x
–x
16.
23
x
e
+
17.
2
x
x
e
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INTEGRALS 305
18.
1
2
1
–
tan x
e
x+
19.
2
2
1
1
x
x
e–
e
+
20.
22
22
x –x
x –x
e –e
ee
+
21. tan
2
(2x – 3) 22. sec
2
(7 – 4x) 23.
1
2
sin
1
–
x
–x
24.
2cos 3sin
6cos 4sin
x– x
xx
+
25.
2 2
1
cos (1 tan )
x– x
26.
cos
x
x
27.
sin 2 cos 2
xx
28.
cos
1 sin
x
x
+
29. cot x log sin x
30.
sin
1 cos
x
x
+
31.
()
2
sin
1 cos
x
x
+
32.
1
1 cot
x
+
33.
1
1 tan
–x
34.
tan
sin cos
x
xx
35.
(
)
2
1 log x
x
+
36.
()
2
( 1) log
xx x
x
++
37.
(
)
3 14
sin tan
1
–
xx
x
8
+
Choose the correct answer in Exercises 38 and 39.
38.
9
10
10 10 log 10
10
x
e
x
x
dx
x
+
+
∫
equals
(A) 10
x
– x
10
+ C (B) 10
x
+ x
10
+ C
(C) (10
x
– x
10
)
–1
+ C (D) log (10
x
+ x
10
) + C
39.
22
equals
sin cos
dx
xx
∫
(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)
2
cos
x dx
∫
(ii)
sin 2 cos 3
x x dx
∫
(iii)
3
sin
x dx
∫
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306 MATHEMATICS
Solution
(i) Recall the identity cos 2x = 2 cos
2
x – 1, which gives
cos
2
x =
1 cos 2
2
x
+
Therefore, =
1
(1 + cos 2 )
2
x dx
=
11
cos 2
22
dx x dx
+
=
1
sin 2 C
24
x
x
++
(ii) Recall the identity sin x cos y =
1
2
[sin (x + y) + sin (x – y)] (Why?)
Then
=
=
11
cos 5 cos C
25
– xx
++
=
11
cos 5 cos C
10 2
–xx
++
(iii) From the identity sin 3x = 3 sin x – 4 sin
3
x, we find that
sin
3
x =
3sin sin 3
4
x– x
Therefore,
3
sin
x dx
=
31
sin sin 3
44
x dx – x dx
=
31
– cos cos 3 C
4 12
xx
++
Alternatively,
3 2
sin sin sin
x dx x x dx
=
=
2
(1 – cos ) sin
x x dx
Put cos x = t so that – sin x dx = dt
Therefore,
3
sin
x dx
=
(
)
2
1
– t dt
−
=
3
2
C
3
t
– dt t dt – t
+ = ++
=
3
1
cos cos C
3
–x x
++
Remark It can be shown using trigonometric identities that both answers are equivalent.
2019-20
INTEGRALS 307
EXERCISE 7.3
Find the integrals of the functions in Exercises 1 to 22:
1. sin
2
(2x + 5) 2. sin 3x cos 4x 3. cos 2x cos 4x cos 6x
4. sin
3
(2x + 1) 5. sin
3
x cos
3
x 6. sin x sin 2x sin 3x
7. sin 4x sin 8x 8.
1 cos
1 cos
–x
x
+
9.
cos
1 cos
x
x
+
10. sin
4
x 11. cos
4
2x 12.
2
sin
1 cos
x
x
+
13.
cos 2 cos 2
cos cos
x–
x–
α
α
14.
cos sin
1 sin 2
x– x
x
+
15. tan
3
2x sec 2x
16. tan
4
x 17.
33
22
sin cos
sin cos
xx
xx
+
18.
2
2
cos 2 2sin
cos
xx
x
+
19.
3
1
sin cos
xx
20.
()
2
cos 2
cos sin
x
xx
+
21. sin
– 1
(cos x)
22.
1
cos ( ) cos ( )
x–a x–b
Choose the correct answer in Exercises 23 and 24.
23.
22
22
sin cos
is equal to
sin cos
xx
dx
xx
−
∫
(A) tan x + cot x + C (B) tan x + cosec x + C
(C) – tan x + cot x + C (D) tan x + sec x + C
24.
2
(1 )
equals
cos ( )
x
x
ex
dx
ex
+
∫
(A) – cot (ex
x
) + C (B) tan (xe
x
) + C
(C) tan (e
x
) + C (D) cot (e
x
) + 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:
(1)
∫
22
1–
= log + C
2+
–
dx x a
a xa
xa
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308 MATHEMATICS
(2)
∫
22
1+
= log + C
2–
–
dx a x
a ax
ax
(3)
∫
–1
22
1
tan C
dx x
=+
aa
x +a
(4)
∫
22
22
= log + – + C
–
dx
x xa
xa
(5)
∫
–1
22
=sin +C
–
dx x
a
ax
(6)
∫
22
22
=log + + +C
+
dx
x xa
xa
We now prove the above results:
(1) We have
22
11
( )( )
x–a x a
x –a
=
+
=
1()–()11 1
2()()2
x a x–a
–
a x–a x a a x–a x a
+
=
+ +
Therefore,
22
1
2
dx dx dx
–
a x–a x a
x –a
=
+
∫ ∫∫
=
[ ]
1
log ( )| log ( )| C
2
|x–a – |x a
a
++
=
1
log C
2
x–a
a xa
+
+
(2) In view of (1) above, we have
22
1 1 ( )( )
2 ( )( )
–
ax ax
a a xa x
ax
++−
=
+−
=
11 1
2
aa x a x
+
−+
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INTEGRALS 309
Therefore,
22
–
dx
ax
=
1
2
dx dx
a ax ax
+
−+
=
1
[ log | | log | |] C
2
ax ax
a
− −+ + +
=
1
log C
2
ax
a ax
+
+
−
Note The technique used in (1) will be explained in Section 7.5.
(3) Put x = a tan
θ. Then dx = a sec
2
θ dθ.
Therefore,
22
dx
xa
+
=
=
1
11 1
C tan C
–
x
d
a a aa
= += +
(4) Let x = a secθ. Then dx = a secθ tan θ d θ.
Therefore,
22
dx
xa
−
=
22 2
sec
tan
sec
ad
aa
−
=
1
sec
log sec + tan +C
d =
=
2
1
2
log 1 C
xx
–
a
a
++
=
22
1
log log C
x x –a a
+ −+
=
22
log +C
x x –a+
, where C = C
1
– log |a|
(5) Let x = a sin
θ. Then dx = a cosθ dθ.
Therefore,
22
dx
ax
−
=
222
cos
sin
ad
a –a
=
1
= +C=sin C
–
x
d
a
+
(6) Let x = a tan θ. Then dx = a sec
2
θ dθ.
Therefore,
22
dx
xa
+
=
2
22 2
sec
tan
ad
aa
+
=
1
sec
= log (sec tan ) C
d
++
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310 MATHEMATICS
=
2
1
2
log 1C
xx
a
a
+ ++
=
2
1
log log C
x x a |a|
2
+ +− +
=
2
log
Cx xa
2
+ ++
, where C = C
1
– log |a|
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.
(7) To find the integral
2
dx
ax bx c
++
, we write
ax
2
+ bx + c =
2
2
2
2
2
4
b c b cb
ax x a x –
aa aa
a
+ += + +
Now, put
2
b
xt
a
+=
so that dx = dt and writing
2
2
2
4
cb
–k
a
a
=±
. We find the
integral reduced to the form
22
1
dt
a
tk
±
depending upon the sign of
2
2
4
cb
–
a
a
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.
(9) To find the integral of the type
2
px q
dx
ax bx c
+
++
, where p, q, a, b, c are
constants, we are to find real numbers A, B such that
2
+ =A ( )+B=A(2 )+B
d
px q ax bx c ax 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 311
(10) For the evaluation of the integral of the type
2
()
px q dx
ax 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:
(i)
2
16
dx
x
−
(ii)
2
2
dx
xx
−
Solution
(i) We have
2 22
16 4
dx dx
x x–
=
−
=
4
log C
84
x–
x
1
+
+
[by 7.4 (1)]
(ii)
Put x – 1 = t. Then dx = dt.
Therefore,
2
2
dx
xx
−
=
2
1
dt
–t
=
1
sin ( ) C
–
t
+
[by 7.4 (5)]
=
1
sin ( – 1) C
–
x
+
Example 9 Find the following integrals :
(i)
2
6 13
dx
xx
−+
(ii)
2
3 13 10
dx
xx
+−
(iii)
2
52
dx
xx
−
Solution
(i) We have x
2
– 6x + 13 = x
2
– 6x + 3
2
– 3
2
+ 13 = (x – 3)
2
+ 4
So,
6 13
dx
xx
2
−+
=
()
2
2
1
32
dx
x– +
Let x – 3 = t. Then dx = dt
Therefore,
6 13
dx
xx
2
−+
=
1
22
1
tan C
22
2
–
dt t
t
=+
+
[by 7.4 (3)]
=
1
13
tan C
22
–
x–
+
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312 MATHEMATICS
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
2
3 13 10
x x–
+
=
2
13 10
3
33
x
x–
+
=
22
13 17
3
66
x–
+
(completing the square)
Thus
3 13 10
dx
xx
2
+−
∫
=
22
1
3
13 17
66
dx
x
+−
∫
Put
13
6
xt
+=
. Then dx = dt.
Therefore,
3 13 10
dx
xx
2
+−
∫
=
2
2
1
3
17
6
dt
t
−
∫
=
1
17
1
6
log C
17 17
32
66
t–
t
+
×× +
[by 7.4 (i)]
=
1
13 17
1
66
log
C
13 17
17
66
x–
x
+
+
++
=
1
1 64
log C
17 6 30
x
x
−
+
+
=
1
1 32 1 1
log C log
17 5 17 3
x
x
−
++
+
=
1 32
log C
17 5
x
x
−
+
+
, where C =
1
11
C log
17 3
+
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INTEGRALS 313
(iii) We have
2
2
52
5
5
dx dx
x
xx
x–
2
=
−
∫∫
=
22
1
5
11
55
dx
x– –
∫
(completing the square)
Put
1
5
x– t
=
. Then dx = dt.
Therefore,
52
dx
xx
2
−
∫
=
2
2
1
5
1
5
dt
t–
∫
=
2
2
1 1
log C
5
5
t t–
+ +
[by 7.4 (4)]
=
2
112
log
C
55
5
x
x– x –
++
Example 10 Find the following integrals:
(i)
2
2 65
x
dx
xx
2
+
++
∫
(ii)
2
3
54
x
dx
x–x
+
−
∫
Solution
(i) Using the formula 7.4 (9), we express
x + 2 =
()
2
A 2 6 5B
d
xx
dx
+++
=
A (4 6) B
x
++
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2 or A =
1
4
and B =
1
2
.
Therefore,
2
2 65
x
xx
2
+
++
∫
=
1 46 1
42
2 65 2 65
x dx
dx
xx xx
2 2
+
+
++ ++
∫ ∫
=
12
11
II
42
+
(say) ... (1)
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In I
1
, put 2x
2
+ 6x + 5 = t, so that (4x + 6) dx = dt
Therefore, I
1
=
1
log C
dt
t
t
=+
∫
=
2
1
log | 2 6 5 | C
xx
+ ++
... (2)
and I
2
=
2
2
1
5
2
2 65
3
2
dx dx
xx
xx
=
++
++
∫∫
=
22
1
2
31
22
dx
x
++
∫
Put
3
2
xt
+=
, so that dx = dt, we get
I
2
=
2
2
1
2
1
2
dt
t
+
∫
=
1
2
1
tan 2 C
1
2
2
–
t
+
×
[by 7.4 (3)]
=
1
2
3
tan 2 + C
2
–
x
+
=
(
)
1
2
tan 2 3 + C
–
x +
... (3)
Using (2) and (3) in (1), we get
()
2 1
21 1
log2 65 tan23C
4 2
2 65
–
x
dx x x x
xx
2
+
= + ++ + +
++
∫
where, C =
12
CC
42
+
(ii) This integral is of the form given in 7.4 (10). Let us express
x + 3 =
2
A (5 4 ) + B
d
– x–x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i.e., A =
1
2
–
and B = 1
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Therefore,
2
3
54
x
dx
xx
+
−−
∫
=
(
)
2 2
42
1
2
54 54
– – x dx
dx
–
xx xx
+
−− −−
∫∫
=
1
2
–
I
1
+ I
2
... (1)
In I
1
, put 5 – 4x – x
2
= t, so that (– 4 – 2x) dx = dt.
Therefore, I
1
=
(
)
2
42
54
– x dx
dt
t
xx
−
=
−−
∫∫
=
1
2C
t
+
=
2
1
25 4 C
– x–x
+
... (2)
Now consider I
2
=
2 2
54 9( 2)
dx dx
x x –x
=
−− +
∫∫
Put x + 2 = t, so that dx = dt.
Therefore, I
2
=
1
2
22
sin + C
3
3
–
dt t
t
=
−
∫
[by 7.4 (5)]
=
1
2
2
sin C
3
–
x
+
+
... (3)
Substituting (2) and (3) in (1), we obtain
21
2
32
5 – 4 – + sin C
3
54
–
xx
– xx
– x–x
++
= +
∫
, where
1
2
C
CC
2
–=
EXERCISE 7.4
Integrate the functions in Exercises 1 to 23.
1.
2
6
3
1
x
x
+
2.
2
1
14
x
+
3.
()
2
1
21
–x
+
4.
2
1
9 25
–x
5.
4
3
12
x
x
+
6.
2
6
1
x
x
−
7.
2
1
1
x–
x–
8.
2
66
x
xa
+
9.
2
2
sec
tan 4
x
x
+
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10.
2
1
22
xx
++
11.
2
1
9 65
xx
++
12.
2
1
76
– x–x
13.
( )( )
1
12
x– x–
14.
2
1
83
x–x
+
15.
( )( )
1
x–a x–b
16.
2
41
23
x
x x–
+
+
17.
2
2
1
x
x–
+
18.
2
52
12 3
x
xx
−
++
19.
( )( )
67
54
x
x– x–
+
20.
2
2
4
x
x–x
+
21.
2
2
23
x
xx
+
++
22.
2
3
25
x
x–x
+
−
23.
2
53
4 10
x
xx
+
++
.
Choose the correct answer in Exercises 24 and 25.
24.
2
equals
22
dx
xx
++
∫
(A) x tan
–1
(x + 1) + C (B) tan
–1
(x + 1) + C
(C) (x + 1) tan
–1
x + C (D) tan
–1
x + C
25.
2
equals
94
dx
xx
−
∫
(A)
–1
1 98
sin C
98
x −
+
(B)
–1
1 89
sin C
29
x −
+
(C)
–1
1 98
sin C
38
x −
+
(D)
–1
1 98
sin C
29
x −
+
7.5 Integration by Partial Fractions
Recall that a rational function is defined as the ratio of two polynomials in the form
P( )
Q( )
x
x
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(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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functions by long division process. Thus, if
P( )
Q( )
x
x
is improper, then
1
P( )
P( )
T( )
Q( ) Q( )
x
x
x
x x
=+
,
where T(x) is a polynomial in x and
1
P( )
Q( )
x
x
is a proper rational function. As we know
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
linear and quadratic factors. Assume that we want to evaluate
P( )
Q( )
x
dx
x
∫
, where
P( )
Q( )
x
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
1.
( – )( – )
px q
xa xb
+
, a ≠ b
AB
x–a x–b
+
2.
2
(–)
px q
xa
+
()
2
AB
x–a
x–a
+
3.
2
( – )( )( )
px qx r
x a x–b x–c
++
ABC
x–a x–b x–c
++
4.
2
2
(– )( )
px qx r
x a x–b
++
2
ABC
()
x–a x–b
x–a
++
5.
2
2
( – )( )
px qx r
x a x bx c
++
++
2
A B +C
x
x–a
x bx c
+
++
,
where x
2
+ 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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Example 11 Find
( 1) ( 2)
dx
xx
++
∫
Solution The integrand is a proper rational function. Therefore, by using the form of
partial fraction [Table 7.2 (i)], we write
1
( 1) ( 2)
xx
++
=
AB
12
xx
+
++
... (1)
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) ( 2)
xx
++
=
1 –1
12
xx
+
++
Therefore,
( 1) ( 2)
dx
xx
++
∫
=
12
dx dx
–
xx
++
∫∫
=
log 1 log 2 C
xx
+− + +
=
1
log C
2
x
x
+
+
+
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.
Example 12 Find
2
2
1
56
x
dx
xx
+
−+
∫
Solution Here the integrand
2
2
1
56
x
x–x
+
+
is not proper rational function, so we divide
x
2
+ 1 by x
2
– 5x + 6 and find that
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INTEGRALS 319
2
2
1
56
x
x–x
+
+
=
2
55 55
11
( 2) ( 3)
56
x– x–
x– x–
x–x
+ =+
+
Let
55
( 2) ( 3)
x–
x– x–
=
AB
23
x– x–
+
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
Thus,
2
2
1
56
x
x–x
+
+
=
5 10
1
23
x– x–
−+
Therefore,
2
2
1
56
x
dx
x–x
+
+
∫
=
1
5 10
2 3
dx
dx dx
x– x–
−+
∫∫ ∫
= x – 5 log | x – 2 | + 10 log | x – 3| + C.
Example 13 Find
2
32
( 1) ( 3)
x
dx
xx
−
++
∫
Solution The integrand is of the type as given in Table 7.2 (4). We write
2
32
( 1) ( 3)
x–
xx
++
=
2
ABC
1
3
( 1)
xx
x
++
+ +
+
So that 3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)
2
= A (x
2
+ 4x + 3) + B (x + 3) + C (x
2
+ 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
A B and C
42 4
––
,== =
. Thus the integrand is given by
2
32
( 1) ( 3)
x
xx
−
++
=
2
11 5 11
4 ( 1)
4 ( 3)
2 ( 1)
––
xx
x
+ +
+
Therefore,
2
32
( 1) ( 3)
x
xx
−
++
∫
=
2
11 5 11
4 12 43
( 1)
dx dx dx
–
xx
x
−
+
+
+
∫∫ ∫
=
11 5 11
log +1 log 3 C
4 2 ( + 1) 4
x x
x
+ − ++
=
11 +1 5
log +C
4 + 3 2 ( + 1)
x
xx
+
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320 MATHEMATICS
Example 14 Find
2
22
( 1) ( 4)
x
dx
xx
++
∫
Solution Consider
2
22
( 1) ( 4)
x
xx
++
and put x
2
= y.
Then
2
22
( 1) ( 4)
x
xx
++
=
( 1) ( 4)
y
yy
++
Write
( 1) ( 4)
y
yy
++
=
AB
14
yy
+
++
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
A =
14
and B
33
−=
Thus,
2
22
( 1) ( 4)
x
xx
++
=
2 2
14
3( 1) 3( 4)
–
xx
+
++
Therefore,
2
22
( 1) ( 4)
x dx
xx
++
∫
=
2 2
14
33
14
dx dx
–
xx
+
++
∫∫
=
1 1
1 41
tan tan C
3 32 2
– –
x
–x
+× +
=
1 1
12
tan tan C
3 32
– –
x
–x
++
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.
Example 15 Find
(
)
2
3 sin 2 cos
5 cos 4 sin
–
d
––
φφ
φ
φφ
∫
Solution Let y = sinφ
Then dy = cosφ dφ
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INTEGRALS 321
Therefore,
(
)
2
3 sin 2 cos
5 cos 4 sin
–
d
––
φφ
φ
φφ
∫
=
2
(3 – 2)
5 (1 ) 4
y dy
– –y – y
∫
=
2
32
44
y–
dy
y– y+
∫
=
()
2
32
I (say)
2
y–
y–
=
∫
Now, we write
()
2
32
2
y–
y–
=
2
AB
2
( 2)
y
y
+
−
−
[by Table 7.2 (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
I =
2
34
[+ ]
2
( 2)
dy
y–
y–
∫
=
2
3 +4
2
( 2)
dy dy
y–
y–
∫∫
=
1
3 log 2 4 C
2
y–
y
−+ +
−
=
4
3 log sin 2 C
2 sin–
φ− + +
φ
=
4
3 log (2 sin ) + C
2 sin
− φ+
−φ
(since, 2 – sin φ is always positive)
Example 16 Find
2
2
1
( 2) ( 1)
xxdx
xx
++
++
∫
Solution The integrand is a proper rational function. Decompose the rational function
into partial fraction [Table 2.2(5)]. Write
2
2
1
( 1) ( 2)
xx
xx
++
++
=
2
A B +C
2
( 1)
x
x
x
+
+
+
Therefore, x
2
+ x + 1 = A (x
2
+ 1) + (Bx + C) (x + 2)
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Equating the coefficients of x
2
, 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
32 1
A , B and C
55 5
== =
Thus, the integrand is given by
2
2
1
( 1) ( 2)
xx
xx
++
++
=
2
21
3
55
5 ( 2)
1
x
x
x
+
+
+
+
=
2
3 12 1
5 ( 2) 5
1
x
x
x
+
+
+
+
Therefore,
2
2
1
( +1) ( 2)
xx
dx
xx
++
+
∫
=
2 2
3 12 1 1
5 25 5
1
1
dx x
dx dx
x
xx
++
+
+ +
∫∫ ∫
=
2 1
3 1 1
log 2 log 1 tan C
5 5 5
–
x x x
+ + ++ +
EXERCISE 7.5
Integrate the rational functions in Exercises 1 to 21.
1.
( 1) ( 2)
x
xx
++
2.
2
1
9
x–
3.
31
( 1)( 2)( 3)
x–
x– x– x–
4.
( 1)( 2)( 3)
x
x– x– x–
5.
2
2
32
x
xx
++
6.
2
1
(1 2 )
–x
x –x
7.
2
( 1) ( – 1)
x
xx
+
8.
2
( 1) ( 2)
x
x– x
+
9.
32
35
1
x
x –x x
+
−+
10.
2
23
( 1) (2 3)
x
x– x
−
+
11.
2
5
( 1) ( 4)
x
xx
+−
12.
3
2
1
1
xx
x
++
−
13.
2
2
(1 ) (1 )
xx
−+
14.
2
31
( 2)
x–
x +
15.
4
1
1
x
−
16.
1
( 1)
n
xx
+
[Hint: multiply numerator and denominator by x
n – 1
and put x
n
= t ]
17.
cos
(1 – sin ) (2 – sin )
x
xx
[Hint : Put sin x = t]
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INTEGRALS 323
18.
22
22
( 1) ( 2)
( 3) ( 4)
xx
xx
++
++
19.
22
2
( 1) ( 3)
x
xx
++
20.
4
1
( 1)
xx –
21.
1
( 1)
x
e–
[Hint : Put e
x
= t]
Choose the correct answer in each of the Exercises 22 and 23.
22.
( 1) ( 2)
x dx
xx
−−
∫
equals
(A)
2
( 1)
log C
2
x
x
−
+
−
(B)
2
( 2)
log C
1
x
x
−
+
−
(C)
2
1
log
C
2
x
x
−
+
−
(D)
log ( 1) ( 2) C
xx
− −+
23.
2
( 1)
dx
xx
+
∫
equals
(A)
2
1
log log ( +1) + C
2
xx−
(B)
2
1
log log ( +1) + C
2
xx
+
(C)
2
1
log log ( +1) + C
2
xx
−+
(D)
2
1
log log ( +1) + C
2
xx+
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
uv
dx
=
dv du
uv
dx dx
+
Integrating both sides, we get
uv =
dv du
u dx v dx
dx dx
+
∫∫
or
dv
u dx
dx
∫
=
du
uv – v dx
dx
∫
... (1)
Let u = f (x) and
dv
dx
= g (x). Then
du
dx
= f ′(x) and v =
()
g x 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.,
() ()
fxgxdx
∫
=
() () [ () () ]
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
cos
x x dx
∫
Solution Put f (x) = x (first function) and g (x) = cos x (second function).
Then, integration by parts gives
cos
x x dx
∫
=
cos [ ( ) cos ]
d
x x dx – x x dx dx
dx
∫ ∫∫
=
sin sin
x x– xdx
∫
= x sin x + cos x + C
Suppose, we take f (x) = cos x and g (x) = x. Then
cos
x x dx
∫
=
cos [ (cos ) ]
d
x x dx – x x dx dx
dx
∫∫ ∫
=
()
2 2
cos sin
22
xx
x x dx
+
∫
Thus, it shows that the integral
cos
x 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
sin
x 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 325
as sin x + k, where k is any constant, then
cos
x x dx
∫
=
(sin ) (sin )
x xk xkdx
+− +
∫
=
(sin ) (sin
x x k x dx k dx
+− −
∫∫
=
(sin ) cos C
x x k x – kx
+− +
=
sin cos C
xx x
++
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.
Hence,
(log .1)
x dx
∫
=
log 1 [ (log ) 1 ]
d
x dx x dx dx
dx
−
∫∫ ∫
=
1
(log ) – log C
x x x dx x x – x
x
⋅ =+
∫
.
Example 19 Find
x
x e dx
∫
Solution Take first function as x and second function as e
x
. The integral of the second
function is e
x
.
Therefore,
x
x e dx
∫
=
1
xx
x e e dx
−⋅
∫
= xe
x
– e
x
+ C.
Example 20 Find
1
2
sin
1
–
xx
dx
x−
∫
Solution Let first function be sin
– 1
x and second function be
2
1
x
x
−
.
First we find the integral of the second function, i.e.,
2
1
x dx
x
−
∫
.
Put t =1 – x
2
. Then dt = – 2x dx
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326 MATHEMATICS
Therefore,
2
1
x dx
x
−
∫
=
1
2
dt
–
t
∫
=
2
–1
tx
=− −
Hence,
1
2
sin
1
–
xx
dx
x
−
∫
=
(
)
1 2 2
2
1
(sin ) 1 (1 )
1
–
x– x – x dx
x
−− −
−
∫
=
21
1 sin C
– x xx
−
− ++
=
21
1 sin C
x– x x
−
−+
Alternatively, this integral can also be worked out by making substitution sin
–1
x = θ and
then integrating by parts.
Example 21 Find
sin
x
e x dx
∫
Solution Take e
x
as the first function and sin x as second function. Then, integrating
by parts, we have
I sin ( cos ) cos
x x x
e x dx e – x e x dx
= =+
∫ ∫
= – e
x
cos x + I
1
(say) ... (1)
Taking e
x
and cos x as the first and second functions, respectively, in I
1
, we get
I
1
=
sin sin
x x
e x– e xdx
∫
Substituting the value of I
1
in (1), we get
I = – e
x
cos x + e
x
sin x – I or 2I = e
x
(sin x – cos x)
Hence, I =
sin (sin cos ) + C
2
x
x
e
e xdx x– x=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and e
x
the second function.
7.6.1 Integral of the type
[ ( ) + ( )]
x
e f x f x dx
′
∫
We have I =
[ ()+ ()]
x
e f x f x dx
′
∫
=
() + ()
x x
e f x dx e f x dx
′
∫∫
=
1 1
I () ,whereI= ()
x x
e f x dx e f x dx
′
+
∫ ∫
... (1)
Taking f (x) and e
x
as the first function and second function, respectively, in I
1
and
integrating it by parts, we have I
1
= f (x) e
x
–
() C
x
f x e dx
′
+
∫
Substituting I
1
in (1), we get
I =
() () () C
x xx
e f x f x e dx e f x dx
′ ′
− + +
∫∫
= e
x
f (x) + C
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INTEGRALS 327
Thus,
′
∫
[ () ()]
x
e f x + f x dx
=
() C
x
e f x+
Example 22 Find (i)
1
2
1
(tan )
1
x–
ex
x
+
+
∫
dx (ii)
2
2
( + 1)
( + 1)
x
xe
x
∫
dx
Solution
(i) We have I =
1
2
1
(tan )
1
x–
e x dx
x
+
+
∫
Consider f (x) = tan
– 1
x, then f ′(x) =
2
1
1
x
+
Thus, the given integrand is of the form e
x
[ f (x) + f ′(x)].
Therefore,
1
2
1
I (tan )
1
x–
e x dx
x
=+
+
∫
= e
x
tan
– 1
x + C
(ii) We have
2
2
( + 1)
I
( + 1)
x
xe
x
=
∫
dx
2
2
1 + 1+1)
[
]
( + 1)
x
x–
e
dx
x
=
∫
2
22
12
[ ]
( + 1) ( +1)
x
x–
e dx
xx
=+
∫
2
12
[+ ]
+1
( +1)
x
x–
e dx
x
x
=
∫
Consider
1
()
1
x
fx
x
−
=
+
, then
2
2
()
( 1)
fx
x
′
=
+
Thus, the given integrand is of the form e
x
[f (x) + f ′(x)].
Therefore,
2
2
11
C
1
( 1)
x x
x x
e dx e
x
x
+−
=+
+
+
∫
EXERCISE 7.6
Integrate the functions in Exercises 1 to 22.
1. x sin x 2. x sin 3x 3. x
2
e
x
4. x log x
5. x log 2x 6. x
2
log x 7. x sin
– 1
x 8. x tan
–1
x
9. x cos
–1
x 10. (sin
–1
x)
2
11.
1
2
cos
1
xx
x
−
−
12. x sec
2
x
13. tan
–1
x 14. x (log x)
2
15. (x
2
+ 1) log x
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328 MATHEMATICS
16. e
x
(sinx + cosx) 17.
2
(1 )
x
xe
x
+
18.
1 sin
1 cos
x
x
e
x
+
+
19.
2
11
–
x
e
x
x
20.
3
( 3)
( 1)
x
xe
x
−
−
21. e
2x
sin x
22.
1
2
2
sin
1
–
x
x
+
Choose the correct answer in Exercises 23 and 24.
23.
3
2 x
x e dx
∫
equals
(A)
3
1
C
3
x
e
+
(B)
2
1
C
3
x
e
+
(C)
3
1
C
2
x
e
+
(D)
2
1
C
2
x
e
+
24.
sec (1 tan )
x
e x x dx
+
∫
equals
(A) e
x
cos x + C (B) e
x
sec x + C
(C) e
x
sin x + C (D) e
x
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)
22
x a dx
−
∫
(ii)
22
x a dx
+
∫
(iii)
22
a x dx
−
∫
(i) Let
22
I
x a dx
=−
∫
Taking constant function 1 as the second function and integrating by parts, we
have
I =
22
22
12
2
x
xx a x dx
xa
−−
−
∫
=
2
22
22
x
xx a dx
xa
−−
−
∫
=
222
22
22
xaa
xx a dx
xa
−+
−−
−
∫
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INTEGRALS 329
=
22 22 2
22
dx
xxa xadxa
xa
−− − −
−
=
22 2
22
I
dx
xx a a
xa
− −−
−
or 2I =
22 2
22
dx
xx a a
xa
−−
−
or I =
22
x – a dx
=
2
22 22
– – log + – + C
2 2
x a
xa x xa
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
(ii)
2
22 22 22
1
+ = + + log + + + C
2 2
a
x a dx x x a x xa
(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
2
25
x x dx
++
Solution Note that
2
25
x x dx
++
=
2
( 1) 4
x dx
++
Put x + 1 = y, so that dx = dy. Then
2
25
x x dx
++
=
22
2
y dy
+
=
2 2
1 4
4 log 4 C
2 2
yy yy
++ + + +
[using 7.6.2 (ii)]
=
2 2
1
(1) 252log 1 25C
2
x xx x xx
+ +++ ++ +++
Example 24 Find
2
32
x x dx
−−
Solution Note that
2 2
32 4 ( 1)
x x dx x dx
− − = −+
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330 MATHEMATICS
Put x + 1 = y so that dx = dy.
Thus
2
32
x x dx
−−
∫
=
2
4 y dy−
∫
=
2 1
1 4
4 sin C
2 22
–
y
yy
−+ +
[using 7.6.2 (iii)]
=
2 1
1 1
( 1) 3 2 2 sin C
2 2
–
x
x xx
+
+ −−+ +
EXERCISE 7.7
Integrate the functions in Exercises 1 to 9.
1.
2
4
x
−
2.
2
14
x−
3.
2
46
xx
++
4.
2
41
xx
++
5.
2
14
xx
−−
6.
2
45
xx
+−
7.
2
13
xx
+−
8.
2
3
xx
+
9.
2
1
9
x
+
Choose the correct answer in Exercises 10 to 11.
10.
2
1
x dx
+
∫
is equal to
(A)
(
)
2 2
1
1 log 1 C
22
x
x xx
++ ++ +
(B)
3
2
2
2
(1 ) C
3
x
++
(C)
3
2
2
2
(1 ) C
3
xx
++
(D)
2
22 2
1
1 log 1 C
22
x
xx x x
++ ++ +
11.
2
87
x x dx
−+
∫
is equal to
(A)
2 2
1
( 4) 8 7 9log 4 8 7 C
2
x xx x xx
− −++ −+ −++
(B)
2 2
1
( 4) 8 7 9log 4 8 7 C
2
x xx x xx
+ − ++ ++ − + +
(C)
2 2
1
( 4) 8 7 3 2 log 4 8 7 C
2
x xx x xx
− − +− −+ − + +
(D)
2 2
1 9
( 4) 8 7 log 4 8 7 C
2 2
x xx x xx
− − +− −+ − + +
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INTEGRALS 331
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
has a unique value. A definite integral is denoted by
()
b
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). Here, we shall consider these two cases separately as discussed
below:
7.7.1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b]. Assume that all the
values taken by the function are non negative, so the graph of the function is a curve
above the x-axis.
The definite integral
()
b
a
f x dx
∫
is the area bounded by the curve y = f (x), the
ordinates x = a, x = b and the x-axis. To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7.2).
Divide the interval [a, b] into n equal subintervals denoted by [x
0
, x
1
], [x
1
, x
2
] ,...,
[x
r – 1
, x
r
], ..., [x
n – 1
, x
n
], where x
0
= a, x
1
= a + h, x
2
= a + 2h, ... , x
r
= a + rh and
x
n
= b = a + nh or
.
ba
n
h
−
=
We note that as n → ∞, h → 0.
Fig 7.2
O
Y
X
X
'
Y
'
Q
P
C
M
D
L
S
A
B
R
a
=
x
0
x
1
x
2
x
r
-1
x
r
x =b
n
y
f
x
=
(
)
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332 MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [x
r – 1
, x
r
], r = 1, 2, 3, …, n.
From Fig 7.2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM) ... (1)
Evidently as x
r
– x
r–1
→ 0, i.e., h → 0 all the three areas shown in (1) become
nearly equal to each other. Now we form the following sums.
s
n
= h [f(x
0
) + … + f (x
n - 1
)] =
1
0
()
n
r
r
h fx
−
=
∑
... (2)
and S
n
=
12
1
[() () ()] ()
n
n r
r
hfx fx fx h fx
=
+ +…+ =
∑
... (3)
Here, s
n
and S
n
denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [x
r–1
, x
r
] for r = 1, 2, 3, …, n, respectively.
In view of the inequality (1) for an arbitrary subinterval [x
r–1
, x
r
], we have
s
n
< area of the region PRSQP < S
n
... (4)
As
n
→∞
strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve.
Symbolically, we write
lim S
n
n→∞
=
lim
n
n
s
→∞
= area of the region PRSQP =
()
b
a
f x dx
∫
... (5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve. For
the sake of convenience, we shall take rectangles with height equal to that of the
curve at the left hand edge of each subinterval. Thus, we rewrite (5) as
()
b
a
f x dx
∫
=
0
lim [() ( ) ... ( (–1)]
h
hfa fa h fa n h
→
+ + ++ +
or
()
b
a
f x dx
∫
=
1
(–)lim [() ( ) ... ( (–1)]
n
b a fa fa h fa n h
n
→∞
+ + ++ +
... (6)
where h =
–
0
ba
as n
n
→ →∞
The above expression (6) is known as the definition of definite integral as the limit
of sum.
Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
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INTEGRALS 333
choose to represent the independent variable. If the independent variable is denoted by
t or u instead of x, we simply write the integral as
()
b
a
f t dt
or
()
b
a
f u du
instead of
()
b
a
f x dx
. Hence, the variable of integration is called a dummy variable.
Example 25 Find
2
2
0
( 1)
x dx
+
as the limit of a sum.
Solution By definition
()
b
a
f x dx
=
1
(–)lim [() ( ) ... ( (–1)],
n
b a fa fa h fa n h
n
→∞
+ + ++ +
where, h =
–
ba
n
In this example, a = 0, b = 2, f (x) = x
2
+ 1,
2–0 2
h
nn
==
Therefore,
2
2
0
( 1)
x dx
+
=
1 2 4 2 ( – 1)
2lim [(0) () () ... ( )]
n
n
ff f f
n nn n
→∞
+ + ++
=
2 2 2
2 2 2
12 4 (2 – 2)
2 lim [1 ( 1) ( 1) ... 1]
n
n
n
nn n
→∞
+ ++ +++ +
=
=
2
22 2
12
2 lim [ (1 2 ... ( – 1) ]
n
n n
n
n
2
→∞
+ +++
=
1 4 ( 1) (2 – 1)
2 lim [
]
6
n
n nn
n
n
n
2
→∞
−
+
=
1 2 ( 1) (2 – 1)
2 lim [ ]
3
n
nn
n
n n
→∞
−
+
=
21 1
2 lim [1 (1 ) ( 2 – )]
3
n
nn
→∞
+−
=
4
2 [1 ]
3
+
=
14
3
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334 MATHEMATICS
Example 26 Evaluate
2
0
x
e dx
∫
as the limit of a sum.
Solution By definition
2
0
x
e dx
∫
=
2 4 2 –2
0
1
(2 – 0) lim ...
n
nn n
n
eee e
n
→∞
++++
Using the sum to n terms of a G.P., where a = 1,
2
n
re
=
, we have
2
0
x
e dx
∫
=
2
2
1 –1
2 lim [ ]
1
n
n
n
n
e
n
e
→∞
−
=
2
2
1 –1
2 lim
–1
n
n
e
n
e
→∞
=
2
2
2 ( – 1)
–1
lim 2
2
n
n
e
e
n
→∞
⋅
= e
2
– 1 [using
0
( 1)
lim 1
h
h
e
h
→
−
=
]
EXERCISE 7.8
Evaluate the following definite integrals as limit of sums.
1.
b
a
x dx
∫
2.
5
0
( 1)
x dx
+
∫
3.
3
2
2
x dx
∫
4.
4
2
1
()
x x dx
−
∫
5.
1
1
x
e dx
−
∫
6.
4
2
0
()
x
x e dx
+
∫
7.8 Fundamental Theorem of Calculus
7.8.1 Area function
We have defined
()
b
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
be a given point in [a, b]. Then
()
x
a
f x dx
∫
represents the area of the light shaded region
Fig 7.3
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INTEGRALS 335
in Fig 7.3 [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
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
A (x) =
∫
()
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].
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
an anti derivative of f. Then
∫
()
b
a
f x dx
=
[F( )] =
b
a
x
F (b) – F(a).
Remarks
(i) In words, the Theorem 2 tells us that
()
b
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.
(iv) In
()
b
a
f x dx
∫
, the function f needs to be well defined and continuous in [a, b].
For instance, the consideration of definite integral
1
3
2
2
2
( – 1)
x x dx
−
∫
is erroneous
since the function f expressed by f (x) =
1
2
2
( – 1)
xx
is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3].
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336 MATHEMATICS
Steps for calculating
()
b
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
( ) [F ( ) C] [F( ) C] – [F( ) C] F( ) – F( )
b
b
a
a
f x dx x b a b a
= + = + +=
∫
.
Thus, the arbitrary constant disappears in evaluating the value of the definite
integral.
(ii) Evaluate F(b) – F(a) =
[F ( )]
b
a
x
, which is the value of
()
b
a
f x dx
∫
.
We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
2
x dx
∫
(ii)
9
3
4
2
2
(30 – )
x
dx
x
∫
(iii)
2
1
( 1) ( 2)
x dx
xx
++
∫
(iv)
3
4
0
sin 2 cos 2
t t dt
π
∫
Solution
(i) Let
3
2
2
I
x dx
=
∫
. Since
3
2
F( )
3
x
x dx x
==
∫
,
Therefore, by the second fundamental theorem, we get
I =
27 8 19
F(3)–F(2) –
333
==
(ii) Let
9
3
4
2
2
I
(30 – )
x
dx
x
=
∫
. We first find the anti derivative of the integrand.
Put
3
2
3
30 – . Then –
2
x t x dx dt
= =
or
2
–
3
x dx dt
=
Thus,
3 2
2
2
2
–
3
(30 – )
x dt
dx
t
x
=
∫ ∫
=
21
3
t
=
3
2
21
F( )
3
(30 – )
x
x
=
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INTEGRALS 337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
21
F(9) – F(4)
3
(30 – )
x
=
=
21 1
3 (30 – 27) 30 – 8
−
=
2 1 1 19
3 3 22 99
−=
(iii) Let
2
1
I
( 1) ( 2)
x dx
xx
=
++
∫
Using partial fraction, we get
–1 2
( 1) ( 2) 1 2
x
xx x x
=+
++ + +
So
( 1) ( 2)
x dx
xx
++
∫
=
– log 1 2 log 2 F( )
x xx
++ + =
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]
= – 3 log 3 + log 2 + 2 log 4 =
32
log
27
(iv) Let
3
4
0
I sin 2 cos 2
t t dt
π
=
∫
. Consider
3
sin 2 cos 2
t t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt =
1
2
du
So
3
sin 2 cos 2
t t dt
∫
=
3
1
2
u du
∫
=
4 4
11
[ ] sin 2 F ( ) say
88
u tt
==
Therefore, by the second fundamental theorem of integral calculus
I =
44
1 1
F ( ) – F (0) [sin – sin 0]
4 82 8
π
π
= =
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338 MATHEMATICS
EXERCISE 7.9
Evaluate the definite integrals in Exercises 1 to 20.
1.
1
1
( 1)
x dx
−
+
2.
3
2
1
dx
x
3.
2
32
1
(4 – 5 6 9)
x x x dx
++
4.
sin 2
0
4
x dx
π
∫
5.
cos 2
0
2
x dx
π
∫
6.
5
4
x
e dx
7.
4
0
tan x dx
π
8.
4
6
cosec
x dx
π
π
9.
1
0
2
1–
dx
x
10.
1
2
0
1
dx
x
+
11.
3
2
2
1
dx
x
−
12.
2
2
0
cos
x dx
π
13.
3
2
2
1
x dx
x
+
14.
1
2
0
23
51
x
dx
x
+
+
15.
21
0
x
x e dx
16.
2
2
2
1
5
43
x
xx
++
17.
23
4
0
(2sec 2)
x x dx
π
++
18.
22
0
(sin – cos )
22
xx
dx
π
19.
2
2
0
63
4
x
dx
x
+
+
20.
1
0
( sin )
4
x
x
x e dx
π
+
Choose the correct answer in Exercises 21 and 22.
21.
3
2
1
1
dx
x
+
equals
(A)
3
π
(B)
2
3
π
(C)
6
π
(D)
12
π
22.
2
3
2
0
49
dx
x
+
equals
(A)
6
π
(B)
12
π
(C)
24
π
(D)
4
π
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.
2019-20
INTEGRALS 339
To evaluate
()
b
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.
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.
Example 28 Evaluate
1
45
1
51
x x dx
−
+
∫
.
Solution Put t = x
5
+ 1, then dt = 5x
4
dx.
Therefore,
45
51
x x dx
+
∫
=
t dt
∫
=
3
2
2
3
t
=
3
5
2
2
( 1)
3
x +
Hence,
1
45
1
51
x x dx
−
+
∫
=
1
3
5
2
–1
2
( 1)
3
x
+
=
()
3
3
5 5
2
2
2
(1 1) – (– 1) 1
3
+ +
=
33
22
2
20
3
−
=
2 42
(2 2)
33
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits.
2019-20
340 MATHEMATICS
Let t = x
5
+ 1. Then dt = 5 x
4
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
Therefore
1
45
1
51
x x dx
−
+
∫
=
2
0
t dt
∫
=
2
3 33
2 22
0
22
2 –0
33
t
=
=
2 42
(2 2)
33
=
Example 29 Evaluate
–1
1
2
0
tan
1
x
dx
x
+
∫
Solution Let t = tan
– 1
x, then
2
1
1
dt dx
x
=
+
. The new limits are, when x = 0, t = 0 and
when x = 1,
4
t
π
=
. Thus, as x varies from 0 to 1, t varies from 0 to
4
π
.
Therefore
–1
1
2
0
tan
1
x
dx
x
+
∫
=
2
4
4
0
0
2
t
t dt
π
π
∫
=
2 2
1
–0
2 16 32
ππ
=
EXERCISE 7.10
Evaluate the integrals in Exercises 1 to 8 using substitution.
1.
1
2
0
1
x
dx
x
+
∫
2.
5
2
0
sin cos
d
π
φ φφ
∫
3.
1
–1
2
0
2
sin
1
x
dx
x
+
∫
4.
2
0
2
xx
+
∫
(Put x + 2 = t
2
) 5.
2
2
0
sin
1 cos
x
dx
x
π
+
∫
6.
2
2
0
4–
dx
xx
+
∫
7.
1
2
1
25
dx
xx
−
++
∫
8.
2
2
2
1
11
–
2
x
e dx
x
x
∫
Choose the correct answer in Exercises 9 and 10.
9. The value of the integral
1
3
3
1
1
4
3
()
xx
dx
x
−
∫
is
(A) 6 (B) 0 (C) 3 (D) 4
10. If f (x) =
0
sin
x
t t dt
∫
, then f ′(x) is
(A) cosx + x sin x (B) x sinx
(C) x cosx (D) sinx + x cosx
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INTEGRALS 341
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.
P
0
:
( ) ()
b b
a a
f x dx f t dt
=
∫∫
P
1
:
() – ()
b a
a b
f x dx f x dx
=
∫∫
. In particular,
() 0
a
a
f x dx
=
∫
P
2
:
() () ()
b c b
a a c
f x dx f x dx f x dx
=+
∫∫∫
P
3
:
() ( )
b b
a a
f x dx f a b x dx= +−
∫∫
P
4
:
0 0
() ( )
a a
f x dx f a x dx
=−
∫∫
(Note that P
4
is a particular case of P
3
)
P
5
:
2
0 0 0
() () (2 )
a a a
f x dx f x dx f a x dx
= +−
∫ ∫∫
P
6
:
2
0 0
() 2 () ,if (2 ) ()
a a
f x dx f x dx f a x f x= −=
∫∫
and
0 if f (2a – x) = – f (x)
P
7
: (i)
0
() 2 ()
a a
a
fxdx fxdx
−
=
∫∫
, if f is an even function, i.e., if f (– x) = f (x).
(ii)
() 0
a
a
f x dx
−
=
∫
, 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 P
0
It follows directly by making the substitution x = t.
Proof of P
1
Let F be anti derivative of f. Then, by the second fundamental theorem of
calculus, we have
( ) F( )–F( ) –[F( ) F( )] ( )
b a
a b
f x dx b a a b f x dx
= = − =−
∫ ∫
Here, we observe that, if a = b, then
() 0
a
a
f x dx
=
∫
.
Proof of P
2
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)
and
()
b
c
f x dx
∫
= F(b) – F(c) ... (3)
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342 MATHEMATICS
Adding (2) and (3), we get
( ) ( ) F( ) – F( ) ( )
c b b
a c a
f x dx f x dx b a f x dx+= =
∫∫ ∫
This proves the property P
2
.
Proof of P
3
Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a.
Therefore
()
b
a
f x dx
∫
=
( –)
a
b
f a b t dt−+
∫
=
( –)
b
a
f a b t dt
+
∫
(by P
1
)
=
( –)
b
a
fa b x+
∫
dx by P
0
Proof of P
4
Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now
proceed as in P
3
.
Proof of P
5
Using P
2
, we have
2 2
0 0
() () ()
a a a
a
f x dx f x dx 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
2
()
a
a
f x dx
∫
=
0
– (2 – )
a
f a t dt
∫
=
0
(2 – )
a
f a t dt
∫
=
0
(2 – )
a
f a x dx
∫
Hence
2
0
()
a
f x dx
∫
=
0 0
( ) (2 )
a a
f x dx f a x dx
+−
∫∫
Proof of P
6
Using P
5
, we have
2
0 0 0
( ) ( ) (2 )
a a a
fxdx fxdx f a xdx
= +−
∫ ∫∫
... (1)
Now, if f (2a – x) = f (x), then (1) becomes
2
0
()
a
f x dx
∫
=
0 0 0
() () 2 () ,
a a a
fxdx fxdx fxdx
+=
∫∫ ∫
and if f(2a – x) = – f (x), then (1) becomes
2
0
()
a
f x dx
∫
=
0 0
() () 0
a a
f x dx f x dx
−=
∫∫
Proof of P
7
Using P
2
, we have
()
a
a
f x dx
−
∫
=
0
0
() ()
a
a
f x dx 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.
2019-20
INTEGRALS 343
Therefore
()
a
a
f x dx
−
∫
=
0
0
– (– ) ( )
a
a
f t dt f x dx+
∫∫
=
0 0
(– ) ( )
a a
f xdx fxdx
+
∫∫
(by P
0
) ... (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0 0 0
() () () 2 ()
a a a a
a
f x dx f x dx f x dx f x dx
−
=+=
∫ ∫∫ ∫
(ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes
0 0
() () () 0
a a a
a
f x dx f x dx f x dx
−
=− + =
∫ ∫∫
Example 30 Evaluate
2
3
1
–
x x dx
−
∫
Solution We note that x
3
– x ≥ 0 on [– 1, 0] and x
3
– x ≤ 0 on [0, 1] and that
x
3
– x ≥ 0 on [1, 2]. So by P
2
we write
2
3
1
–
x x dx
−
∫
=
0 1 2
3 3 3
1 0 1
( –) –( –) ( –)
x x dx x x dx x x dx
−
++
∫∫ ∫
=
0 1 2
3 3 3
1 0 1
(–) (–) (–)
x x dx x x dx x x dx
−
++
∫ ∫∫
=
0 1 2
42 24 42
–1
0 1
–
––
42 24 42
xx xx xx
++
=
()
11 11 11
– – – 4–2 – –
42 24 42
++
=
1111 11
– 2
4224 42
++−+−+
=
3 3 11
2
24 4
−+=
Example 31 Evaluate
24
–
4
sin
x dx
π
π
∫
Solution We observe that sin
2
x is an even function. Therefore, by P
7
(i), we get
24
–
4
sin
x dx
π
π
∫
=
2
4
0
2 sin
x dx
π
∫
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344 MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x
dx
π
−
=
4
0
(1 cos 2 )
x dx
π
−
=
4
0
1
– sin 2
2
xx
π
=
1
1
– sin – 0 –
42 2 42
π ππ
=
Example 32 Evaluate
2
0
sin
1 cos
x x
dx
x
π
+
Solution Let I =
2
0
sin
1 cos
x x
dx
x
π
+
. Then, by P
4
, we have
I =
2
0
()sin()
1 cos ( )
x x dx
x
π
π− π−
+ π−
=
2
0
( ) sin
1 cos
x x dx
x
π
π−
+
=
2
0
sin
I
1 cos
x dx
x
π
π −
+
or 2 I =
π
π
sin
cos
x dx
x1
2
0
+
∫
or I =
2
0
sin
2
1 cos
x dx
x
π
π
+
Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1.
Therefore, (by P
1
) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
=
1
2
1
2
1
dt
t
−
π
+
=
1
2
0
1
dt
t
π
+
(by P
7
,
2
1
since
1
t
+
is even function)
=
2
1
–1 –1 1
0
tan tan 1 – tan 0 – 0
44
t
−
ππ
π =π =π =
Example 33 Evaluate
1
54
1
sin cos
x x dx
−
Solution Let I =
1
54
1
sin cos
x x dx
−
. Let f(x) = sin
5
x cos
4
x. Then
f (– x) = sin
5
(– x) cos
4
(– x) = – sin
5
x cos
4
x = – f (x), i.e., f is an odd function.
Therefore, by P
7
(ii), I = 0
2019-20
INTEGRALS 345
Example 34 Evaluate
4
2
44
0
sin
sin cos
x
dx
xx
π
+
∫
Solution Let I =
4
2
44
0
sin
sin cos
x
dx
xx
π
+
∫
... (1)
Then, by P
4
I =
4
2
0
44
sin ( )
2
sin ( ) cos ( )
2 2
x
dx
x x
π
π
−
ππ
−+ −
∫
=
4
2
44
0
cos
cos sin
x
dx
xx
π
+
∫
... (2)
Adding (1) and (2), we get
2I =
44
2
22
44
0 00
sin cos
[]
2
sin cos
xx
dx dx x
xx
π
ππ
+ π
= ==
+
∫∫
Hence I =
4
π
Example 35 Evaluate
3
6
1 tan
dx
x
π
π
+
∫
Solution Let I =
3 3
6 6
cos
1 tan cos sin
x dx
dx
x xx
π π
π π
=
+ +
∫∫
... (1)
Then, by P
3
I =
3
6
cos
36
cos sin
36 36
x dx
x x
π
π
ππ
+−
π π ππ
+− + +−
∫
=
3
6
sin
sin cos
x
dx
xx
π
π
+
∫
... (2)
Adding (1) and (2), we get
2I =
[]
3
3
6
6
366
dx x
π
π
π
π
πππ
= =−=
∫
. Hence
I
12
π
=
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346 MATHEMATICS
Example 36 Evaluate
2
0
log sin
x dx
π
∫
Solution Let I =
2
0
log sin
x dx
π
∫
Then, by P
4
I =
2 2
0 0
log sin log cos
2
x dx x dx
π π
π
−=
∫ ∫
Adding the two values of I, we get
2I =
( )
2
0
log sin log cos
x x dx
π
+
∫
=
( )
2
0
log sin cos log 2 log 2
xx dx
π
+−
∫
(by adding and subtracting log2)
=
2 2
0 0
log sin 2 log 2
x dx dx
π π
−
∫∫
(Why?)
Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when
2
x
π
=
,
t = π.
Therefore 2I =
0
1
log sin log 2
2 2
t dt
π
π
−
∫
=
2
0
2
log sin log 2
2 2
t dt
π
π
−
∫
[by P
6
as sin (π – t) = sin t)
=
2
0
log sin log 2
2
x dx
π
π
−
∫
(by changing variable t to x)
=
I log 2
2
π
−
Hence
2
0
log sin
x dx
π
∫
=
–
log 2
2
π
.
2019-20
INTEGRALS 347
EXERCISE 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.
1.
2
2
0
cos
x dx
π
∫
2.
2
0
sin
sin cos
x
dx
xx
π
+
∫
3.
3
2
2
33
0
22
sin
sin cos
x dx
xx
π
+
∫
4.
5
2
55
0
cos
sin cos
x dx
xx
π
+
∫
5.
5
5
| 2|
x dx
−
+
∫
6.
8
2
5
x dx
−
∫
7.
1
0
(1 )
n
x x dx−
∫
8.
4
0
log (1 tan )
x dx
π
+
∫
9.
2
0
2x x dx−
∫
10.
2
0
(2log sin log sin 2 )
x x dx
π
−
∫
11.
22
–
2
sin
x dx
π
π
∫
12.
0
1 sin
x dx
x
π
+
∫
13.
72
–
2
sin
x dx
π
π
∫
14.
2
5
0
cos
x dx
π
∫
15.
2
0
sin cos
1 sin cos
xx
dx
xx
π
−
+
∫
16.
0
log (1 cos )
x dx
π
+
∫
17.
0
a
x
dx
x ax
+−
∫
18.
4
0
1
x dx
−
∫
19. Show that
0 0
() () 2 ()
a a
fxgxdx fxdx=
∫∫
, if f and g are defined as f(x) = f(a – x)
and g(x) + g(a – x) = 4
Choose the correct answer in Exercises 20 and 21.
20. The value of
3 52
2
( cos tan 1)
x x x x dx
π
−π
+ ++
∫
is
(A) 0 (B) 2 (C) π (D) 1
21. The value of
2
0
4 3 sin
log
4 3 cos
x
dx
x
π
+
+
∫
is
(A) 2 (B)
3
4
(C) 0 (D) –2
2019-20
348 MATHEMATICS
Miscellaneous Examples
Example 37 Find
cos 6 1 sin 6
x x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
1
2
1
cos 6 1 sin 6
6
x x dx t dt
+=
∫ ∫
=
3 3
2 2
12 1
( ) C = (1 sin 6 ) C
63 9
t x
×+ + +
Example 38 Find
1
4
4
5
()
xx
dx
x
−
∫
Solution We have
1
1
4
4
4
3
54
1
(1 )
()xx
x
dx
dx
x x
−
−
=
∫∫
Put
–3
34
1 3
1 1– ,s
o that
x t dx dt
x x
−= = =
Therefore
1
1
4
4
4
5
() 1
3
xx
dx t dt
x
−
=
∫∫
=
5
5
4
4
3
14 4 1
C= 1 C
3 5 15
t
x
×+ − +
Example 39 Find
4
2
( 1) ( 1)
x dx
xx
−+
∫
Solution We have
4
2
( 1) ( 1)
x
xx
−+
=
32
1
( 1)
1
x
xxx
++
− +−
=
2
1
( 1)
( 1) ( 1)
x
xx
++
−+
... (1)
Now express
2
1
( 1) ( 1)
xx
−+
=
2
A BC
( 1)
( 1)
x
x
x
+
+
−
+
... (2)
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INTEGRALS 349
So 1 = A (x
2
+ 1) + (Bx + C) (x – 1)
= (A + B) x
2
+ (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give
11
A ,B C –
22
= ==
. Substituting values of A, B and C in (2), we get
2
1
( 1) ( 1)
xx
−+
=
2 2
11 1
2( 1) 2
( 1) 2( 1)
x
x
xx
−−
−
++
... (3)
Again, substituting (3) in (1), we have
4
2
( 1) ( 1)
x
x xx
− ++
=
2 2
11 1
( 1)
2( 1) 2
( 1) 2( 1)
x
x
x
xx
++ − −
−
++
Therefore
4 2
2 –1
2
1 1 1
log 1 – log ( 1) – tan C
22 4 2
( 1) ( 1)
x x
dx x x x x
x xx
= ++ − + +
− ++
∫
Example 40 Find
2
1
log (log )
(log )
x dx
x
+
∫
Solution Let
2
1
I log (log )
(log )
x dx
x
= +
∫
=
2
1
log (log )
(log )
x dx dx
x
+
∫∫
In the first integral, let us take 1 as the second function. Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x x x dx
xx
x
− +
∫∫
=
2
log (log )
log
(log )
dx dx
xx
x
x
−+
∫∫
... (1)
Again, consider
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have
2
11
––
log log
(log )
dx x
x
dx
xx x
x
=
∫∫
... (2)
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350 MATHEMATICS
Putting (2) in (1), we get
2 2
I log (log )
log
(log ) (log )
x dx dx
xx
x
xx
= −− +
∫∫
=
log (log ) C
log
x
xx
x
−+
Example 41 Find
cot tanx x dx
+
∫
Solution We have
I =
cot tan
x x dx
+
∫
tan (1 cot )
x x dx
=+
∫
Put tan x = t
2
, so that sec
2
x dx = 2t dt
or dx =
4
2
1
t dt
t
+
Then I =
24
12
1
(1 )
t
t dt
tt
+
+
∫
=
2
22
4 2
2
2
11
11
( 1)
2 =2 =2
1
1
1
2
dt
dt
t
tt
dt
t
t
t
t
t
+ +
+
+
+
−+
∫∫∫
Put
1
t
t
−
= y, so that
2
1
1
t
+
dt = dy. Then
I =
()
–1 –1
2
2
1
2 2tan C= 2tan C
2 2
2
t
dy y
t
y
−
= + +
+
∫
=
2
–1 –1
1 tan 1
2 tan C = 2 tan C
2 2 tan
t x
t x
− −
+ +
Example 42 Find
4
sin 2 cos 2
9 – cos (2 )
x x dx
x
∫
Solution Let
4
sin 2 cos 2
I
9–cos 2
xx
dx
x
=
∫
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INTEGRALS 351
Put cos
2
(2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1 12
2
1 1 11
I– – sin C sin cos 2 C
4 43 43
9–
dt t
x
t
−
= = + =− +
Example 43 Evaluate
3
2
1
sin ( )
x x dx
−
π
Solution Here f (x) = | x sin πx | =
sin for 1 1
3
sin for 1
2
xx x
xx x
π −≤ ≤
− π ≤≤
Therefore
3
2
1
| sin |
x x dx
−
π
=
3
1
2
11
sin sin
x x dx x x dx
−
π +− π
=
3
1
2
11
sin sin
x x dx x x dx
−
π− π
Integrating both integrals on righthand side, we get
3
2
1
| sin |
x x dx
−
π
=
=
2 2
2 11 31
−− − = +
π ππ
π π
Example 44 Evaluate
2 2 22
0
cos sin
x dx
a xb x
π
+
Solution Let I =
2 2 22 22 22
0 0
()
cos sin cos ( ) sin ( )
x dx x dx
a xb x a x b x
π π
π−
=
+ π− + π−
(using P
4
)
=
2 2 22 2 2 22
0 0
cos sin cos sin
dx x dx
a xb x a xb x
π π
π −
+ +
=
2 2 22
0
I
cos sin
dx
a xb x
π
π −
+
Thus 2I =
2 2 22
0
cos sin
dx
a xb x
π
π
+
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352 MATHEMATICS
or I =
2
2 2 22 2 2 22
0 0
2
2 2
cos sin cos sin
dx dx
a xb x a xb x
π
π
π π
=⋅
+ +
∫ ∫
(using P
6
)
=
2
4
2 2 22 2 2 22
0
4
cos sin cos sin
π
π
π
π +
+ +
∫ ∫
dx dx
a xb x a xb x
=
2 2
2
4
222 22 2
0
4
sec cosec
tan cot
π
π
π
π +
+ +
∫∫
x dx x dx
a b x a xb
=
( )
0
1
222 222
0
1
tan t cot
π
−
==
+ +
∫∫
dt du
put x and x u
a bt au b
=
1 0
–1 –1
0 1
tan – tan
ππ
bt au
ab a ab b
=
–1 –1
tan tan
π
+
ba
ab a b
=
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24.
1.
3
1
xx
−
2.
1
xa xb
++ +
3.
2
1
x ax x
−
[Hint: Put x =
a
t
]
4.
3
24
4
1
( 1)
xx+
5.
1
1
3
2
1
xx
+
[Hint:
11
11
3
2
36
11
1
xx
xx
=
+
+
, put x = t
6
]
6.
2
5
( 1) ( 9)
x
xx
++
7.
sin
sin ( )
x
xa
−
8.
5 log 4 log
3 log 2 log
xx
xx
ee
ee
−
−
9.
2
cos
4 sin
x
x
−
10.
88
22
sin cos
1 2sin cos
x
xx
−
−
11.
1
cos ( ) cos ( )
xa xb
++
12.
3
8
1
x
x
−
13.
(1 ) ( 2 )
x
xx
e
ee
++
14.
22
1
( 1) ( 4)
xx
++
15. cos
3
x e
log sinx
16. e
3 logx
(x
4
+ 1)
– 1
17. f ′ (ax + b) [f (ax + b)]
n
18.
3
1
sin sin ( )
xx
+α
19.
1 1
1 1
sin cos
sin cos
xx
xx
− −
− −
−
+
, x ∈ [0, 1]
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INTEGRALS 353
20.
1
1
x
x
−
+
21.
2 sin 2
1 cos 2
x
x
e
x
+
+
22.
2
2
1
( 1) ( 2)
xx
xx
++
++
23.
–1
1
tan
1
x
x
−
+
24.
2 2
4
1 log ( 1) 2 log
xx x
x
+ +−
Evaluate the definite integrals in Exercises 25 to 33.
25.
2
1 sin
1 cos
π
π
−
−
∫
x
x
e dx
x
26.
4
44
0
sin cos
cos sin
xx
dx
xx
π
+
∫
27.
2
2
22
0
cos
cos 4 sin
x dx
xx
π
+
∫
28.
3
6
sin cos
sin 2
xx
dx
x
π
π
+
∫
29.
1
0
1
dx
xx
+−
∫
30.
4
0
sin cos
9 16 sin2
xx
dx
x
π
+
+
∫
31.
1
2
0
sin 2 tan (sin )
x x dx
π
−
∫
32.
0
tan
sec tan
xx
dx
xx
π
+
∫
33.
4
1
[| 1| | 2 | | 3 |]
x x x dx
−+ − + −
∫
Prove the following (Exercises 34 to 39)
34.
3
2
1
22
log
33
( 1)
dx
xx
=+
+
∫
35.
1
0
1
x
x e dx
=
∫
36.
1
17 4
1
cos 0
x x dx
−
=
∫
37.
3
2
0
2
sin
3
x dx
π
=
∫
38.
3
4
0
2 tan 1 log 2
x dx
π
=−
∫
39.
1
1
0
sin 1
2
x dx
−
π
=−
∫
40. Evaluate
1
23
0
x
e dx
−
∫
as a limit of a sum.
Choose the correct answers in Exercises 41 to 44.
41.
xx
dx
ee
−
+
∫
is equal to
(A) tan
–1
(e
x
) + C (B) tan
–1
(e
–x
) + C
(C) log (e
x
– e
–x
) + C (D) log (e
x
+ e
–x
) + C
42.
2
cos 2
(sin cos )
x
dx
xx
+
∫
is equal to
(A)
–1
C
sin cos
xx
+
+
(B)
log |sin cos | C
xx
++
(C)
log |sin cos | C
xx
−+
(D)
2
1
(sin cos )
xx
+
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354 MATHEMATICS
43. If f (a + b – x) = f (x), then
()
b
a
x f x dx
∫
is equal to
(A)
()
2
b
a
ab
f b x dx
+
−
∫
(B)
()
2
b
a
ab
f b x dx
+
+
∫
(C)
()
2
b
a
ba
f x dx
−
∫
(D)
()
2
b
a
ab
f x dx
+
∫
44. The value of
1
1
2
0
21
tan
1
x
dx
xx
−
−
+−
∫
is
(A) 1 (B) 0 (C) –1 (D)
4
π
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.
Let
F() ()
d
x fx
dx
=
. Then we write
() F() C
f x dx x
=+
∫
. These integrals
are called indefinite integrals or general integrals, C is called constant of
integration. All these integrals differ by a constant.
From the geometric point of view, an indefinite integral is collection of family
of curves, each of which is obtained by translating one of the curves parallel
to itself upwards or downwards along the y-axis.
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,
() ()
kfxdx k fxdx
=
∫∫
More generally, if f
1
, f
2
, f
3
, ... , f
n
are functions and k
1
, k
2
, ... ,k
n
are real
numbers. Then
1 1 22
[ () () ... ()]
nn
k f x kf x kf x dx
+ ++
∫
=
11 22
() () ... ()
nn
k f x dx k f x dx k f x dx
+ ++
∫∫ ∫
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INTEGRALS 355
Some standard integrals
(i)
1
C
1
n
n
x
x dx
n
+
=+
+
∫
, n ≠ – 1. Particularly,
C
dx x
=+
∫
(ii)
cos sin C
x dx x
=+
∫
(iii)
sin – cos C
x dx x
=+
∫
(iv)
2
sec tan C
x dx x
=+
∫
(v)
2
cosec – cot C
x dx x
=+
∫
(vi)
sec tan sec C
x x dx x
=+
∫
(vii)
cosec cot – cosec C
x x dx x
=+
∫
(viii)
1
2
sin C
1
dx
x
x
−
=+
−
∫
(ix)
1
2
cos C
1
dx
x
x
−
=− +
−
∫
(x)
1
2
tan C
1
dx
x
x
−
=+
+
∫
(xi)
1
2
cot C
1
dx
x
x
−
=− +
+
∫
(xii)
C
xx
e dx e
=+
∫
(xiii)
C
log
x
x
a
a dx
a
=+
∫
(xiv)
1
2
sec C
1
dx
x
xx
−
=+
−
∫
(xv)
1
2
cosec C
1
dx
x
xx
−
=− +
−
∫
(xvi)
1
log | | C
dx x
x
=+
∫
Integration by partial fractions
Recall that a rational function is ratio of two polynomials of the form
P( )
Q( )
x
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
may divide P (x) by Q (x) so that
1
P( )
P( )
T( )
Q( ) Q( )
x
x
x
x x
=+
, where T(x) is a
polynomial in x and degree of P
1
(x) is less than the degree of Q(x). T(x)
being polynomial can be easily integrated.
1
P( )
Q( )
x
x
can be integrated by
2019-20
356 MATHEMATICS
expressing
1
P( )
Q( )
x
x
as the sum of partial fractions of the following type:
1.
( )( )
px q
xaxb
+
−−
=
AB
xa xb
+
−−
, a ≠ b
2.
2
()
px q
xa
+
−
=
2
AB
()
xa
xa
+
−
−
3.
2
( )( )( )
px qx r
xaxbxc
++
−−−
=
ABC
xa xb xc
++
−−−
4.
2
2
( )( )
px qx r
xa xb
++
−−
=
2
ABC
()
xa xb
xa
++
− −
−
5.
2
2
( )( )
px qx r
x a x bx c
++
− ++
=
2
A B +C
x
xa
x bx c
+
−
++
where x
2
+ 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 log sec C
x dx x
=+
∫
(ii)
cot log sin C
x dx x
=+
∫
(iii)
sec log sec tan C
x dx x x
= ++
∫
(iv)
cosec log cosec cot C
x dx
xx
= −+
∫
Integrals of some special functions
(i)
22
1
log C
2
dx x a
a xa
xa
−
= +
+
−
∫
(ii)
22
1
log C
2
dx a x
a ax
ax
+
= +
−
−
∫
(iii)
1
22
1
tan C
dx x
aa
xa
−
=+
+
∫
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INTEGRALS 357
(iv)
22
22
log C
dx
x xa
xa
= + −+
−
(v)
1
22
sin C
dx x
a
ax
−
=+
−
(vi)
22
22
log | |C
dx
x xa
xa
= + ++
+
Integration by parts
For given functions f
1
and f
2
, we have
, 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.
[ () ()] () C
x x
e f x f x dx e f x dx
′
+= +
Some special types of integrals
(i)
2
22 22 22
log C
2
2
xa
xadx xa x xa
− = −− + − +
(ii)
2
22 22 22
log
C
2 2
xa
xadx xa x xa
+ = ++ + + +
(iii)
2
22 22 1
sin C
2 2
x ax
axdx ax
a
−
− = −+ +
(iv) Integrals of the types
2
2
or
dx dx
ax bx c
ax bx c
++
++
can be
transformed into standard form by expressing
ax
2
+ bx + c =
2
2
2
2
2
4
b c b cb
ax x a x
aa a a
a
++= + +−
(v) Integrals of the types
2
2
or
px q dx px q dx
ax bx c
ax bx c
+ +
++
++
can be
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358 MATHEMATICS
transformed into standard form by expressing
2
A ( ) B A (2 ) B
d
px q ax bx c ax b
dx
+= + + += + +
, where A and B are
determined by comparing coefficients on both sides.
We have defined
()
b
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
given point in [a, b]. Then
()
x
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
Let the area function be defined by A(x) =
()
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
let F be another function such that
F() ()
d
x fx
dx
=
for all x in the domain of
f, then
[]
() F() C F() F()
b
b
a
a
f x dx x b a
= += −
∫
.
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.
—
—
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