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INTEGRALS
Chapter
287
7
INTEGRALS
Just as a mountaineer climbs a mountain – because it is there, so
a good mathematics student studies new material because
it is there
...
BRISTOL
7
...
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
...
If a function f is differentiable in an interval I, i
...
, 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
G
...
Leibnitz
function as a derivative are called anti derivatives (or
(1646 -1716)
primitive) of the function
...
Such type of problems arise in
many practical situations
...
e
...
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
...
g
...
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
...
In this Chapter, we shall confine ourselves to the study of indefinite and definite
integrals and their elementary properties including some techniques of integration
...
2 Integration as an Inverse Process of Differentiation
Integration is the inverse process of differentiation
...
e
...
Such a process is called integration or anti differentiation
...
(1)
...
(3)
dx
We observe that in (1), the function cos x is the derived function of sin x
...
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 :
that sin x is an anti derivative (or an integral) of cos x
...
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
...
In fact, C is the parameter by
varying which one gets different anti derivatives (or integrals) of the given function
...
Thus,
Remark Functions with same derivatives differ by a constant
...
Consider the function f = g – h defined by f (x) = g(x) – h (x), ∀ x ∈ I
df
= f′ = g′ – h′ giving f′ (x) = g′ (x) – h′ (x) ∀ x ∈ I
dx
Then
or
f ′ (x) = 0, ∀ x ∈ I by hypothesis,
i
...
, the rate of change of f with respect to x is zero on I and hence f is constant
...
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
...
dy
= f (x ) , we write y =
dx
∫ f (x) dx
...
1)
...
1
Symbols/Terms/Phrases
Meaning
∫ f (x ) dx
Integral of f with respect to x
f (x) in
x in
∫ f (x) dx
∫ f (x ) dx
Integrand
Variable of integration
Integrate
Find the integral
An integral of f
A function F such that
F′(x) = f (x)
The process of finding the integral
Integration
Constant of Integration
Any real number C, considered as
constant function
290
MATHEMATICS
We already know the formulae for the derivatives of many important functions
...
Derivatives
Integrals (Anti derivatives)
d xn +1
n
= x ;
(i)
dx n + 1
n
∫ x dx =
x n +1
+ C , n ≠ –1
n +1
Particularly, we note that
d
( x) =1 ;
dx
d
(sin x) = cos x ;
(ii)
dx
(iii)
d
( – cos x ) = sin x ;
dx
d
( tan x) = sec2 x ;
dx
d
( – cot x ) = cosec 2 x ;
(v)
dx
(iv)
∫ dx = x + C
∫ cos x dx = sin x + C
∫ sin x dx = – cos x + C
∫ sec
2
x dx = tan x + C
∫ cosec
2
x dx = – cot x + C
(vi)
d
(sec x ) = sec x tan x ;
dx
∫ sec x tan x dx = sec x + C
(vii)
d
( – cosec x) = cosec x cot x ;
dx
∫ cosec x cot x dx = – cosec x + C
d
1
–1
(viii) dx sin x =
;
1 – x2
(
)
d
1
–1
(ix) dx – cos x =
;
1 – x2
(
)
∫
∫
dx
1–x
2
dx
1–x
2
= sin – 1 x + C
= – cos
(x)
d
1
tan – 1 x =
;
dx
1 + x2
∫ 1 + x2 = tan
(xi)
d
1
– cot – 1 x =
;
dx
1 + x2
∫ 1 + x2 = – cot
(
(
)
)
dx
dx
–1
–1
x+ C
x+ C
–1
x+ C
INTEGRALS
d
1
–1
(xii) dx sec x =
;
x x2 – 1
∫x
dx
d
1
–1
(xiii) dx – cosec x =
;
x x2 – 1
)
∫x
dx
d x
( e ) = ex ;
dx
d
1
( log | x |) = ;
(xv)
dx
x
∫e
(
(
)
(xiv)
(xvi)
d ax
x
= a ;
dx log a
x
291
= sec– 1 x + C
2
x –1
= – cosec– 1 x + C
2
x –1
dx = ex + C
1
∫ x dx = log | x | +C
x
∫ a dx =
ax
+C
log a
Note In practice, we normally do not mention the interval over which the various
functions are defined
...
7
...
1 Geometrical interpretation of indefinite integral
2
Let f (x) = 2x
...
For different values of C, we get different
integrals
...
Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals
...
These together
constitute the indefinite integral
...
Clearly, for C = 0, we obtain y = x2 , a parabola with its vertex on the origin
...
For C = – 1, y = x2 – 1 is obtained by shifting the parabola
y = x2 one unit along y-axis in the negative direction
...
Some of
these have been shown in the Fig 7
...
Let us consider the intersection of all these parabolas by a line x = a
...
1,
we have taken a > 0
...
If the line x = a intersects the
parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0 , P1, P2, P–1 , P–2 etc
...
This indicates that the tangents to the
dx
2
curves at these points are parallel
...
1
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
...
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
...
7
...
2 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals
...
INTEGRALS
293
Proof Let F be any anti derivative of f, i
...
,
d
F(x) = f (x)
dx
∫ f (x ) dx = F(x) + C
Then
Therefore
d
dx
∫ f (x) dx =
=
d
( F ( x) + C )
dx
d
F (x ) = f ( x)
dx
Similarly, we note that
f ′(x) =
d
f ( x)
dx
∫ f ′(x) dx = f (x) + C
and hence
where C is arbitrary constant called constant of integration
...
Proof Let f and g be two functions such that
d
d
∫ f (x) dx = dx ∫ g (x ) dx
dx
or
d
f ( x) dx – ∫ g (x) dx = 0
dx ∫
Hence
∫ f (x ) dx – ∫ g (x) dx =
or
∫ f (x ) dx = ∫ g (x) dx + C
{∫ f (x) dx + C , C ∈ R}
{∫ g(x) dx + C , C ∈ R} are identical
...
(Why?)
294
MATHEMATICS
Note The equivalence of the families
{∫ f (x) dx + C ,C ∈ R}
1
1
and
{∫ g(x) dx + C ,C ∈ R} is customarily expressed by writing ∫ f (x ) dx = ∫ g (x) dx ,
2
2
without mentioning the parameter
...
(1)
d
∫ f (x) dx + dx ∫ g (x) dx
= f (x) + g (x)
Thus, in view of Property (II), it follows by (1) and (2) that
...
(IV) For any real number k, ∫ k f ( x) dx = k ∫ f ( x) dx
Proof By the Property (I),
Also
d
k
dx
d
k f ( x) dx = k f (x)
...
(V) Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2 ,
...
, kn giving
∫[ k1 f1 (x ) + k2 f2 (x) +
...
+ kn
∫ fn (x) dx
...
The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection
...
INTEGRALS
295
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i) cos 2x
(ii) 3x2 + 4x3
(iii)
1
,x≠0
x
Solution
(i) We look for a function whose derivative is cos 2x
...
2
(ii) We look for a function whose derivative is 3x2 + 4x3
...
dx
Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4
...
x
Example 2 Find the following integrals:
(i)
∫
x3 – 1
dx
x2
(ii)
∫
2
(x 3
3
+ 1) dx
(iii)
1
x
∫ (x 2 + 2 e – x ) dx
Solution
(i) We have
∫
x3 – 1
dx = ∫ x dx – ∫ x– 2 dx
2
x
(by Property V)
296
MATHEMATICS
x1 + 1
x– 2 + 1
+ C1 –
+ C 2 ; C , C are constants of integration
=
1
2
1 +1
– 2 +1
2
x2
x– 1
+ C1 –
– C 2 = x + 1 + C1 – C2
=
2
–1
2
x
=
x2 1
+ + C , where C = C 1 – C2 is another constant of integration
...
(ii) We have
2
2
∫ (x 3 + 1) dx = ∫ x 3 dx + ∫ dx
2
+1
5
x3
3
+ x + C = x3 + x + C
= 2
5
+1
3
3
(iii) We have
3
x
∫ (x 2 + 2 e –
1
) dx = ∫ x 2 dx + ∫ 2 e x dx –
x
1
∫ x dx
3
+1
x2
+ 2 e x – log x + C
= 3
+1
2
2
= x
5
5
2
+ 2 e x – log x + C
Example 3 Find the following integrals:
(i)
(iii)
∫ (sin x + cos x ) dx
∫
(ii) ∫ cosec x (cosec x + cot x) dx
1 – sin x
dx
cos 2 x
Solution
(i) We have
∫ (sin x + cos x ) dx = ∫ sin x dx + ∫ cos x dx
= – cos x + sin x + C
INTEGRALS
297
(ii) We have
∫ (cosec x (cosec x + cot x) dx = ∫ cosec
2
x dx + ∫ cosec x cot x dx
= – cot x – cosec x + C
(iii) We have
∫
1 – sin x
1
sin x
dx = ∫ 2 dx – ∫
dx
2
cos x
cos x
cos 2 x
2
= ∫ sec x dx – ∫ tan x sec x dx
= 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
...
Remarks
(i) We see that if F is an anti derivative of f, then so is F + C, where C is any
constant
...
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
...
, polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc
...
For example, it is not possible to find
dx by inspection since we can not find a function whose derivative is e– x
2
298
MATHEMATICS
(iii) When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly
...
2
...
Both are operations on functions
...
Both satisfy the property of linearity, i
...
,
(i)
d
d
d
[k1 f1 (x) + k2 f 2 (x )] = k1 f1 (x) + k2 f 2 (x )
dx
dx
dx
(ii)
∫[ k1
f1 ( x) + k 2 f2 ( x) ] dx = k1 ∫ f1 (x) dx + k2
∫ f2 (x) dx
Here k1 and k2 are constants
...
We have already seen that all functions are not differentiable
...
We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes
...
The derivative of a function, when it exists, is a unique function
...
However, they are unique upto an additive constant, i
...
, any
two integrals of a function differ by a constant
...
When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P
...
6
...
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
...
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
...
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
...
9
...
So is integration, as will be seen in
Section 7
...
INTEGRALS
299
10
...
2
...
EXERCISE 7
...
1
...
cos 3x
3
...
(ax + b)
5
...
∫ (4 e + 1) dx
7
...
∫ ( ax + bx + c ) dx
x
2
9
...
15
...
2
x
∫ (2 x + e ) dx
∫
x3 + 3 x + 4
10
...
x 3 + 5 x2 – 4
dx
x2
∫
1
x–
dx 11
...
x –1
∫ (1 – x)
∫ x ( 3 x + 2 x + 3) dx
2
∫ (2 x – 3sin x + 5 x ) dx
2
∫ (2 x – 3cos x + e ) dx
∫ sec x (sec x + tan x) dx
x
16
...
sec2 x
2 – 3sin x
dx
19
...
∫
dx
...
1
21
...
If
f ( x) = 4 x3 − 4 such that f (2) = 0
...
3 Methods of Integration
In previous section, we discussed integrals of those functions which were readily
obtainable from derivatives of some functions
...
e
...
However,
this method, which depends on inspection, is not very suitable for many functions
...
Prominent among them are methods based on:
1
...
Integration using Partial Fractions
3
...
3
...
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)
...
dt
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
...
Usually, we make a substitution for a function whose derivative also occurs
in the integrand as illustrated in the following examples
...
r
...
x:
(i) sin mx
(ii) 2x sin (x2 + 1)
(iii)
tan 4
x sec 2
x
x
(iv)
sin (tan – 1 x)
1 + x2
Solution
(i) We know that derivative of mx is m
...
1
1
1
Therefore,
∫ sin mx dx = m ∫ sin t dt = – m cos t + C = – m cos mx + C
INTEGRALS
301
(ii) Derivative of x2 + 1 is 2x
...
∫ 2 x sin (x
Therefore,
2
+ 1) dx = ∫ sin t dt = – cos t + C = – cos (x2 + 1) + C
1
(iii) Derivative of
x is
x = t so that
1
2 x
1 –2
1
x =
...
tan 4 x sec 2
2t tan 4 t sec2 t dt
4
2
Thus,
= 2 ∫ tan t sec t dt
∫
t
x
Again, we make another substitution tan t = u so that
sec2 t dt = du
x
dx = ∫
2 ∫ tan 4 t sec2 t dt = 2 ∫ u 4 du = 2
Therefore,
u5
+C
5
2
tan 5 t + C (since u = tan t)
5
2
5
= tan x + C (since t = x )
5
x
2
dx = tan5 x + C
5
=
Hence,
∫
tan 4 x sec 2
x
Alternatively, make the substitution tan x = t
1
(iv) Derivative of tan– 1x =
...
1 + x2
sin (tan – 1 x)
dx = ∫ sin t dt = – cos t + C = – cos (tan –1x) + C
Therefore , ∫
1 + x2
Now, we discuss some important integrals involving trigonometric functions and
their standard integrals using substitution technique
...
(i)
∫ tan x dx = log sec x + C
We have
sin x
∫ tan x dx = ∫ cos x dx
302
MATHEMATICS
Put cos x = t so that sin x dx = – dt
dt
Then
∫ tan x dx = – ∫ t = – log t + C = – log cos x + C
∫ tan x dx = log sec x + C
or
(ii)
∫ cot x dx = log
sin x + C
cos x
We have
∫ cot x dx = ∫ sin x
dx
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)
dx
sec x + tan x
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
∫ sec x dx = ∫
(iv)
∫ cosec x dx = log cosec x – cot x + C
We have
cosec x (cosec x + cot x)
dx
(cosec x + cot x)
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 x dx = ∫
= – log
cosec2 x − cot 2 x
+C
cosec x − cot x
= log cosec x – cot x + C
Example 6 Find the following integrals:
(i)
∫ sin
3
x cos 2 x dx
(ii)
sin x
∫ sin (x + a ) dx
(iii)
1
∫ 1 + tan x dx
INTEGRALS
Solution
(i) We have
∫ sin
3
x cos 2 x dx = ∫ sin 2 x cos 2 x ( sin x) dx
2
2
= ∫ (1 – cos x) cos x (sin x) dx
Put t = cos x so that dt = – sin x dx
∫ sin
Therefore,
2
x cos 2 x (sin x) dx = − ∫ (1 – t 2 ) t 2 dt
t3 t5
2
4
= – ∫ ( t – t ) dt = – – + C
3 5
= –
1
1
cos3 x + cos5 x + C
3
5
(ii) Put x + a = t
...
Therefore
sin x
∫ sin (x + a) dx = ∫
∫
=
sin (t – a)
dt
sin t
sin t cos a – cos t sin a
dt
sin t
= 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
...
(1)
cos x – sin x
dx
cos x + sin x
Put cos x + sin x = t so that (cos x – sin x) dx = dt
Now, consider I = ∫
dt
= log t + C 2 = log cos x + sin x + C 2
t
Putting it in (1), we get
I=∫
Therefore
dx
x
∫ 1 + tan x = 2 +
C1 1
C
+ 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
...
2x
1 + x2
2
...
sin x sin (cos x)
6
...
sin ( ax + b ) cos ( ax + b )
7
...
(4 x + 2) x 2 + x + 1 10
...
(x
15
...
1
x + x log x
8
...
x2
13
...
16
...
x
x+ 4
, x> 0
1
, x > 0, m ≠ 1
x (log x) m
x
ex
2
INTEGRALS
–1
etan x
18
...
e2 x – 1
e 2x + 1
20
...
tan2 (2x – 3)
22
...
24
...
27
...
2
26
...
1 + cos x
31
...
33
...
29
...
1
32
...
sin x cos x
(
x3 sin tan – 1 x4
1+ x
(1 + log x )2
x
)
8
Choose the correct answer in Exercises 38 and 39
...
39
...
3
...
2
Example 7 Find (i) ∫ cos x dx
(ii) ∫ sin 2 x cos 3 x dx
3
(iii) ∫ sin x dx
306
MATHEMATICS
Solution
(i) Recall the identity cos 2x = 2 cos2 x – 1, which gives
cos2 x =
∫ cos
Therefore,
2
x dx =
=
1 + cos 2x
2
1
1
1
∫ (1 + cos 2x) dx = 2 ∫ dx + 2 ∫ cos 2 x dx
2
x 1
+ sin 2 x + C
2 4
(ii) Recall the identity sin x cos y =
1
[sin (x + y) + sin (x – y)]
2
=
∫ sin 2 x cos 3 x dx
1
sin 5 x dx • ∫ sin x dx
2 ∫
=
Then
1
2
(Why?)
1
– 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
= –
sin3 x =
∫ sin
Therefore,
3
3sin x – sin 3 x
4
x dx =
3
1
∫ sin x dx – 4 ∫ sin 3 x dx
4
= –
Alternatively,
∫ sin
3
3
1
cos x + cos 3 x + C
4
12
x dx = ∫ sin 2 x sin x dx = ∫ (1 – cos 2 x) sin x dx
Put cos x = t so that – sin x dx = dt
Therefore,
3
∫ sin x dx = − ∫ 1 – t 2 dt = – ∫ dt + ∫ t 2 dt = – t +
(
)
t3
+C
3
1
cos 3x + C
3
Remark It can be shown using trigonometric identities that both answers are equivalent
...
3
Find the integrals of the functions in Exercises 1 to 22:
1
...
sin 3x cos 4x
3
...
sin (2x + 1)
5
...
sin x sin 2x sin 3x
7
...
sin4 x
13
...
tan4 x
19
...
1 – cos x
1 + cos x
9
...
11
...
cos x – sin x
1 + sin 2 x
15
...
sin 3 x + cos3 x
sin 2 x cos 2 x
18
...
cos 2 x
( cos x + sin x )2
cos 2 x + 2sin 2 x
cos 2 x
21
...
22
...
24
...
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
=
log
+C
(1) ∫ 2
2
x –a
2a
x +a
308
MATHEMATICS
dx
1
dx
1
(2)
∫ a 2 – x 2 = 2a log
(3)
∫ x 2 + a 2 = a tan
(4)
∫
(5)
∫
(6)
∫
dx
2
x –a
a –x
2
dx
2
x +a
x
+C
a
–1
2
2
dx
2
a+x
+C
a– x
2
2
= log x + x – a + C
= sin – 1
x
+C
a
= log x + x 2 + a 2 + C
We now prove the above results:
(1) We have
1
1
=
2
x –a
( x – a ) (x + a )
2
=
Therefore,
1 (x + a ) – (x – a ) 1 1
1
=
x – a – x + a
2a ( x – a ) ( x + a ) 2a
1
dx
dx
dx
∫ x2 – a 2 = 2 a ∫ 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 + a + x
a –x
2 a ( a + x) ( a − x)
2a
2
INTEGRALS
Therefore,
dx
∫ a 2 – x2
=
309
1 dx
dx
∫ a − x + ∫ a + x
2a
1
[− log | a − x | + log | a + x |] + C
2a
1
a+x
log
+C
=
2a
a− x
=
Note The technique used in (1) will be explained in Section 7
...
(3) Put x = a tan θ
...
dx
a sec2 θ d θ
Therefore,
∫ x2 + a 2 = ∫ a 2 tan 2 θ + a 2
1
1
1
–1 x
+C
= ∫ dθ = θ + C = tan
a
a
a
a
(4) Let x = a sec θ
...
dx
a secθ tanθ d θ
Therefore,
∫ x2 − a 2 = ∫ a 2 sec2θ − a 2
= ∫ secθ dθ = log secθ + tanθ + C1
= log
x
x2
+
– 1 + C1
a
a2
2
2
= log x + x – a − log a + C1
2
2
= log x + x – a + C , where C = C1 – log |a|
(5) Let x = a sinθ
...
Therefore,
∫
dx
a 2 − x2
=
∫
a cosθ d θ
a 2 – a 2 sin 2θ
–1
= ∫ d θ = θ + C = sin
(6) Let x = a tan θ
...
Therefore,
∫
dx
x2 + a 2
=
∫
x
+C
a
a sec2 θ dθ
a 2 tan 2 θ + a 2
= ∫ secθ dθ = log (secθ + tan θ) + C1
310
MATHEMATICS
x
x2
log
+
+ 1 + C1
=
a
a2
2
2
= log x + x + a − log |a | + C1
2
2
= log x + x + a + C , where C = C1 – 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
...
We find the
a 4a
2a
integral reduced to the form
1
dt
∫ t 2 ± k 2 depending upon the sign of
a
c b2
– 2
a 4a
and hence can be evaluated
...
(9) To find the integral of the type
px + q
∫ ax2 + bx + c dx , where
p, q, a, b, c are
constants, we are to find real numbers A, B such that
d
(ax2 + 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
...
px + q = A
INTEGRALS
(10) For the evaluation of the integral of the type
( px + q) dx
∫
ax2 + bx + c
as in (9) and transform the integral into known standard forms
...
311
, we proceed
Example 8 Find the following integrals:
(i)
dx
∫ x2 − 16
∫
(ii)
dx
2 x − x2
Solution
(i) We have
(ii)
∫
dx
dx
∫ x2 − 16 = ∫ x2 – 4 2
dx
2 x − x2
=∫
=
1
x–4
log
+ C [by 7
...
Then dx = dt
...
4 (5)]
–1
= sin (x – 1) + C
Example 9 Find the following integrals :
(i)
dx
∫ x2 − 6 x + 13
(ii)
dx
∫ 3 x2 + 13 x − 10
(iii)
dx
∫
5x 2 − 2x
Solution
(i) We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4
So,
Let
Therefore,
1
dx
∫ x2 − 6 x + 13 = ∫ ( x – 3)2 + 22 dx
x – 3 = t
...
4 (3)]
312
MATHEMATICS
(ii) The given integral is of the form 7
...
We write the denominator of the integrand,
2 13 x 10
–
3x 2 + 13 x – 10 = 3 x +
3
3
2
2
13 17
3 x + – (completing the square)
=
6 6
Thus
Put x +
1
dx
dx
= ∫
∫ 3 x2 + 13 x − 10 3 13 2 17 2
x+ −
6 6
13
= t
...
6
Therefore,
dx
∫ 3 x2 + 13 x − 10
=
1
3
∫
dt
2
17
t2 −
6
17
6 +C
log
=
1
17
17
3 × 2×
t+
6
6
1
t–
[by 7
...
Then dx = dt
...
4 (4)]
1
1
2x
log x – + x2 –
+C
5
5
5
Example 10 Find the following integrals:
(i)
x+ 2
∫ 2 x 2 + 6 x + 5 dx
(ii)
∫
x+3
5 − 4 x + x2
dx
Solution
(i) Using the formula 7
...
4
2
x+ 2
1
4x + 6
1
dx
∫ 2 x 2 + 6 x + 5 = 4 ∫ 2 x2 + 6 x + 5 dx + 2 ∫ 2 x2 + 6 x + 5
4A = 1 and 6A + B = 2 or A =
Therefore,
=
1
1
I1 + I2
4
2
(say)
...
(2)
dx
x2 + 3 x +
5
2
1
dx
∫ 3 2 1 2
2
x + +
2 2
3
= t , so that dx = dt, we get
2
I2 =
1
2
∫
dt
1
t +
2
2
=
2
1
1
2×
2
tan – 1 2 t + C 2
3
–1
–1
= tan 2 x + + C 2 = tan ( 2 x + 3 ) + C 2
2
Using (2) and (3) in (1), we get
x+ 2
1
∫ 2 x 2 + 6 x + 5 dx = 4 log 2 x
2
+ 6x + 5 +
[by 7
...
(3)
1
tan – 1 ( 2 x + 3 ) + C
2
C1 C2
+
4
2
(ii) This integral is of the form given in 7
...
Let us express
where,
C=
d
(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
x+3= A
– 2A = 1 and – 4 A + B = 3, i
...
, A = –
1
and B = 1
2
INTEGRALS
Therefore,
∫
x+3
5 − 4x − x
2
dx = –
1
2
∫
(– 4 – 2 x) dx +
5 − 4x − x
2
dx
∫
5 − 4 x − x2
1
I +I
2 1 2
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt
...
(1)
=∫
dt
t
= 2 t + C1
= 2 5 – 4x – x 2 + C1
Now consider
I2 =
∫
dx
5 − 4x − x 2
=∫
315
...
Therefore,
I2 =
∫
dt
2
3 −t
–1
= sin
2
= sin
–1
t
+ C2
3
[by 7
...
(3)
Substituting (2) and (3) in (1), we obtain
∫
x+3
5 – 4x – x
2
= – 5 – 4x – x2 + sin – 1
x+ 2
C
+ C , where C = C2 – 1
3
2
EXERCISE 7
...
1
...
7
...
5
...
1
1 + 4x 2
3x
1 + 2 x4
x2
x6 + a 6
3
...
9
...
x2 + 2 x + 2
1
13
...
2
2x + x – 3
6x + 7
19
...
( x – 5)( x – 4)
11
...
17
...
1
1
9 x + 6x + 5
12
...
8 + 3x – x2
x+2
18
...
2
( x – a )( x – b )
5x − 2
1 + 2 x + 3 x2
x+2
2
x + 2x + 3
5x + 3
x+3
2
x – 2x − 5
23
...
Choose the correct answer in Exercises 24 and 25
...
dx
∫ x2 + 2 x + 2 equals
(A) x tan–1 (x + 1) + C
(C) (x + 1) tan–1 x + C
25
...
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
...
The improper rational functions can be reduced to the proper rational
INTEGRALS
functions by long division process
...
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
...
Assume that we want to evaluate
P(x)
∫ Q(x ) dx , where
P(x)
Q(x)
is proper rational function
...
After this,
the integration can be carried out easily using the already known methods
...
2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions
...
2
S
...
Form of the rational function
Form of the partial fraction
1
...
px + q
(x – a ) 2
A
B
+
x – a ( x – a)2
3
...
px 2 + qx + r
(x – a ) 2 ( x – b )
A
B
C
+
+
2
x – a (x – a )
x–b
5
...
318
MATHEMATICS
dx
Example 11 Find
∫ (x + 1) (x + 2)
Solution The integrand is a proper rational function
...
2 (i)], we write
A
B
1
+
=
x +1 x + 2
(x + 1) ( x + 2)
...
This gives
1 = A (x + 2) + B (x + 1)
...
Thus, the integrand is given by
1
1
–1
+
=
(x + 1) ( x + 2)
x +1 x + 2
dx
dx
dx
∫ (x + 1) (x + 2) = ∫ x + 1 – ∫ x + 2
Therefore,
= log x + 1 − log x + 2 + C
= log
x +1
+C
x +2
Remark The equation (1) above is an identity, i
...
a statement true for all (permissible)
values of x
...
e
...
Example 12 Find
∫
x2 + 1
dx
x2 − 5 x + 6
x2 + 1
is not proper rational function, so we divide
x 2 – 5x + 6
x2 + 1 by x2 – 5x + 6 and find that
Solution Here the integrand
INTEGRALS
319
5x – 5
5x – 5
x2 + 1
=1 +
= 1+ 2
(x – 2) (x – 3)
x – 5x + 6
x 2 – 5x + 6
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
...
3x − 2
dx
Example 13 Find ∫
(x + 1)2 (x + 3)
Therefore,
Solution The integrand is of the type as given in Table 7
...
We write
3x – 2
A
B
C
+
+
=
2
2
(x + 1) ( x + 3)
x + 1 ( x + 1)
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 )
2, x and constant term on both sides, we get
Comparing coefficient of x
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2
...
Thus the integrand is given by
,B =
and C =
4
2
4
3x − 2
11
5
11
–
–
=
2
2
(x + 1) ( x + 3)
4 ( x + 1) 2 ( x + 1)
4 ( x + 3)
Therefore,
3x − 2
∫ (x + 1)2 (x + 3)
=
11 dx 5
dx
11 dx
∫ x + 1 – 2 ∫ (x + 1)2 − 4 ∫ x + 3
4
=
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)
320
MATHEMATICS
x2
dx
Example 14 Find ∫ 2
(x + 1) ( x2 + 4)
Solution Consider
x2
and put x2 = y
...
Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method
...
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
...
Decompose the rational function
into partial fraction [Table 2
...
Write
x2 + x + 1
A
Bx + C
+ 2
=
2
(x + 1) ( x + 2)
x + 2 ( x + 1)
Therefore,
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
322
MATHEMATICS
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
...
5
Integrate the rational functions in Exercises 1 to 21
...
x
(x + 1) ( x + 2)
4
...
2
(x –1) (x – 2) (x – 3)
x + 3x + 2
1 – x2
6
...
x
2
(x + 1) (x – 1)
9
...
2x − 3
2
(x – 1) (2x + 3)
5x
11
...
2
(1 − x) (1 + x2 )
14
...
1
[Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
x (x n + 1)
17
...
8
...
2
x
2
(x –1) (x + 2)
3x – 1
(x + 2) 2
3x + 5
x – x2 − x + 1
3
x3 + x + 1
12
...
[Hint : Put sin x = t]
1
x −1
4
INTEGRALS
18
...
2x
(x + 1) ( x2 + 3)
2
20
...
21
...
x
x dx
∫ (x − 1) (x − 2)
(A) log
equals
( x − 1) 2
+C
x−2
(B) log
( x − 2) 2
+C
x −1
2
x −1
(C) log
+C
x − 2
23
...
6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions
...
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
dx + ∫ v
dx
dx
dx
dv
du
∫ u dx dx = uv – ∫v dx dx
dv
u = f (x) and
= g (x)
...
(1)
324
MATHEMATICS
Therefore, expression (1) can be rewritten as
∫ f (x ) g (x) dx =
∫ f (x ) g (x) dx =
i
...
,
f (x) ∫ g ( x) dx – ∫ [ ∫ g (x) dx] f ′( 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
∫ x cos x dx =
cos x ∫ x dx – ∫[
= ( cos x)
Thus, it shows that the integral
d
(cos x) ∫ x dx] dx
dx
x2
x2
+ ∫ sin x
dx
2
2
∫ x cos x dx is reduced to the comparatively more
complicated integral having more power of x
...
Remarks
(i) It is worth mentioning that integration by parts is not applicable to product of
functions in all cases
...
The reason is that there does not exist any function whose derivative is
x sin x
...
If we write the integral of the second function cos x
INTEGRALS
325
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
...
However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function
...
We
take log x as the first function and the constant function 1 as the second function
...
d
Hence,
∫ (logx
...
x
Example 19 Find
∫ x e dx
x
Solution Take first function as x and second function as ex
...
∫ x e dx =
x
Therefore,
Example 20 Find
∫
x sin – 1 x
1 − x2
x ex − ∫ 1 ⋅ e x dx = xex – ex + C
...
e
...
Then dt = – 2x dx
∫
x dx
1 − x2
1 − x2
...
326
MATHEMATICS
∫
Hence,
x dx
∫
Therefore,
1− x
x sin – 1 x
1− x
= –
2
1
2
∫
dt
2
= – t = − 1− x
t
(
)
dx = (sin – 1 x) – 1 − x2 − ∫
2
1
2
( – 1 − x 2 ) dx
1− x
2
−1
= – 1− x sin x + x + C = x – 1 − x sin x + C
2
−1
Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts
...
Then, integrating
by parts, we have
I = ∫ ex sin x dx = ex ( – cos x) + ∫ e xcos x dx
= – e x cos x + I1 (say)
...
Hence,
I = ∫ e x sin x dx =
7
...
1 Integral of the type
We have
∫e
x
[ f ( x) + f ′ ( x)] dx
x
x
I = ∫ e x [ f ( x) + f ′( x)] dx = ∫ e f ( x) dx + ∫ e f ′(x) dx
x
x
= I1 + ∫ e f ′(x) dx, where I1 = ∫ e f ( x) dx
x
...
Consider f (x) = tan– 1x, then f ′(x) =
Therefore, I = ∫ ex (tan – 1 x +
(ii) We have I = ∫
1
) dx = ex tan– 1 x + C
1+ x 2
( x2 +1) ex
x2 – 1 +1+1)
dx = ∫ ex [
] dx
(x +1) 2
(x +1) 2
= ∫e x [
x2 – 1
2
x –1
2
+
] dx = ∫ e x [
+
] dx
2
2
(x +1)
( x+1)
x +1 (x +1) 2
2
x −1
, then f ′( x) =
( x + 1) 2
x +1
Thus, the given integrand is of the form ex [f (x) + f ′(x)]
...
6
Integrate the functions in Exercises 1 to 22
...
x sin x
2
...
x2 ex
4
...
x log 2x
6
...
x cos–1 x
10
...
14
...
(x2 + 1) log x
13
...
x sin– 1 x
x cos −1 x
2
8
...
x sec2 x
327
328
MATHEMATICS
x ex
(1 + x) 2
16
...
1 1
19
...
(x − 3) e x
20
...
e2x sin x
– 1 2x
22
...
23
...
1 x3
e +C
3
1 x3
e +C
2
(D)
1 x2
e +C
2
∫e
x
sec x (1 + tan x) dx equals
(A) ex cos x + C
(C) ex sin x + C
(B) ex sec x + C
(D) ex tan x + C
7
...
2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :
(i)
∫
(i)
∫
∫
Let I = ∫ x2 − a2 dx
x 2 − a 2 dx
(ii)
x 2 + a 2 dx
(iii)
a2 − x 2 dx
Taking constant function 1 as the second function and integrating by parts, we
have
I= x
x2 − a 2 − ∫
= x
x −a −∫
2
2
1
2
2x
2
x − a2
x2
2
x −a
2
x dx
dx = x
x2 − a 2 − ∫
x2 − a 2 + a 2
x2 − a 2
dx
INTEGRALS
= x
x2 − a 2 − ∫ x 2 − a 2 dx − a 2 ∫
= x
x2 − a 2 − I − a 2 ∫
x2 − a 2 − a 2 ∫
2I = x
or
329
dx
2
x − a2
dx
2
x − a2
dx
x 2 − a2
x
a2
x2 – a 2 –
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
x 2 – a2 dx =
I=
or
∫
(ii)
x 2 + a 2 dx =
1
x
2
x2 + a2 +
a2
log x + x 2 + a 2 + C
2
1
a2
x
x a 2 – x2 +
sin–1 + C
∫
2
2
a
Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a sec θ in (i), x = a tanθ in (ii) and x = a sin θ in (iii) respectively
...
Then
∫
x 2 + 2 x + 5 dx =
∫
y2 + 2 2 dy
=
=
Example 24 Find
Solution Note that
∫
∫
1
y
2
4
y2 + 4 + log y + y2 + 4 + C
2
1
(x + 1) x2 + 2x + 5 + 2 log x + 1+ x2 + 2x + 5 + C
2
3 − 2x − x 2 dx
3 − 2 x − x 2 dx = ∫ 4 − ( x + 1) 2 dx
[using 7
...
2 (ii)]
330
MATHEMATICS
Put x + 1 = y so that dx = dy
...
6
...
7
Integrate the functions in Exercises 1 to 9
...
4 − x2
2
...
x2 + 4 x + 6
4
...
1 − 4x − x2
6
...
1 + 3x − x 2
8
...
1+
x2
9
Choose the correct answer in Exercises 10 to 11
...
∫
1 + x2 dx is equal to
(A)
x
1
1+ x 2 + log x + 1 + x2
2
2
(B)
2
(1+ x 2 )2 + C
3
(D)
x2
1
1 + x 2 + x2 log x + 1 + x 2 + C
2
2
(
) +C
3
11
...
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
...
The definite integral
has a unique value
...
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
...
, F(b) – F(a)
...
7
...
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
...
To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7
...
Y
S
Q
X'
P
O a = x0 x1 x2
Y'
y=
f (x)
M
C
D
L
A
B
xr-1 xr
R
xn=b
X
Fig 7
...
,
[xr – 1, xr ],
...
, xr = a + rh and
xn = b = a + nh or n =
b −a
...
h
332
MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [xr – 1, xr ], r = 1, 2, 3, …, n
...
2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM)
...
e
...
Now we form the following sums
...
(2)
r= 0
n
Sn = h [ f ( x1 ) + f ( x2 ) +…+ f ( xn )] = h ∑ f ( xr )
and
...
In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn
...
Symbolically, we write
lim Sn = lim sn = area of the region PRSQP =
n →∞
n→∞
b
∫ a f (x)dx
...
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
...
+ f (a + ( n – 1) h ]
h →0
f ( x) dx = (b – a ) lim
n →∞
h=
1
[ f ( a) + f ( a + h ) +
...
(6)
b–a
→ 0 as n → ∞
n
The above expression (6) is known as the definition of definite integral as the limit
of sum
...
If the independent variable is denoted by
t or u instead of x, we simply write the integral as
b
∫a
b
∫ a f (t ) dt
or
b
∫a
f ( u ) du instead of
f (x) dx
...
Example 25 Find
2
∫0 (x
2
+ 1) dx as the limit of a sum
...
+ f ( a + ( n – 1) h ],
n
b–a
n
In this example, a = 0, b = 2, f (x) = x2 + 1, h =
2–0 2
=
n
n
Therefore,
2
∫ 0 (x
2
1
2
4
2 ( n – 1)
+ 1) dx = 2 lim [ f (0) + f ( ) + f ( ) +
...
+
+ 1 ]
n →∞ n
n
n
n2
= 2 lim
= 2 lim
n→ ∞
1
1
[(1 + 1 +
...
+ (2n – 2) 2 ]
14244 n
4
3
n
n -terms
1
22
[ n + 2 (12 + 22 +
...
Solution By definition
2
∫0
2
4
2n – 2
0
n + en +
...
P
...
8
Evaluate the following definite integrals as limit of sums
...
4
...
− x) dx
∫ 0 ( x + 1) dx
5
...
∫2 x
6
...
8 Fundamental Theorem of Calculus
7
...
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
...
Then
x
∫a
f (x) dx
represents the area of the light shaded region
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]
...
In other words, the area of this shaded region is a function of x
...
We call the function A(x) as Area function and is given by
A (x) =
x
∫a
f ( x ) dx
...
However, we only state them as their proofs are beyond the scope of this text book
...
8
...
Then A′(x) = f (x), for all x ∈ [a, b]
...
8
...
Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
an anti derivative of 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)
...
(iii) The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand
...
(iv) In
b
∫a
f ( x) dx , the function f needs to be well defined and continuous in [a, b]
...
3
1
∫ −2
1
–1) 2
x( x2 – 1) 2 dx is erroneous
is not defined in a portion
336
MATHEMATICS
b
∫a
Steps for calculating
f ( x) dx
...
Let this be F(x)
...
a
Thus, the arbitrary constant disappears in evaluating the value of the definite
integral
...
We now consider some examples
Example 27 Evaluate the following integrals:
3
(i)
(iii)
∫2
2
∫1
x
9
x2 dx
(ii)
∫4
(30 –
x dx
( x + 1) ( x + 2)
(iv)
∫
3
x 2 )2
π
4 sin 3 2 t
0
dx
cos 2 t dt
Solution
x3
= F ( x) ,
3
Therefore, by the second fundamental theorem, we get
3
2
(i) Let I = ∫ 2 x dx
...
We first find the anti derivative of the integrand
...
Then –
Thus,
∫
x
3
(30 – x 2 ) 2
27 8 19
– =
3 3 3
3
x dx = dt or
2
2
2 dt
dx = – ∫ 2 =
3
3 t
2
x dx = – dt
3
1 2
1
= F ( x)
t = 3
3
(30 – x 2 )
INTEGRALS
337
Therefore, by the second fundamental theorem of calculus, we have
9
2
1
I = F(9) – F(4) =
3
3
(30 – x 2 )
4
=
(iii) Let I = ∫
2
1
2
3
1
1
2 1 1 19
−
= − =
3 3 22 99
(30 – 27) 30 – 8
x dx
( x + 1) ( x + 2)
Using partial fraction, we get
x dx
∫ (x + 1) ( x + 2)
So
x
–1
2
=
+
(x + 1) (x + 2) x + 1 x + 2
= – log x + 1 + 2 log 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
(iv) Let I = ∫
π
4
sin3
0
3
2t cos2 t dt
...
9
Evaluate the definite integrals in Exercises 1 to 20
...
4
...
12
...
1
∫ −1( x + 1) dx
2
...
dx
1
– 5 x2 + 6 x + 9) dx
5 x
6
...
3
x2
∫ 2 x2 − 1
∫0 x e
π
18
...
∫ 0 5 x 2 + 1 dx
x + x3 + 2) dx
∫
π
4
tan x
0
11
...
x dx
∫ 2 x2 + 1
∫
3
10
...
x dx
x dx
2
3
...
5x 2
17
...
6x + 3
2
19
...
∫1
3
(A)
22
...
1
∫0(x e
x
+ sin
dx
equals
1 + x2
π
3
(B)
2π
3
(C)
π
6
(D)
π
12
(B)
π
12
(C)
π
24
(D)
π
4
dx
equals
4 + 9 x2
π
6
7
...
One of the important methods for finding the indefinite integral is the method
of substitution
...
Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form
...
Integrate the new integrand with respect to the new variable without mentioning
the constant of integration
...
Resubstitute for the new variable and write the answer in terms of the original
variable
...
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
...
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
...
Example 28 Evaluate
1
∫ −15 x
4
x5 + 1 dx
...
3
Therefore,
4
5
∫ 5 x x + 1 dx =
∫
t dt =
3
2 2
2
t = ( x5 + 1) 2
3
3
1
Hence,
1
∫ −1 5 x
4
3
2
x + 1 dx = ( x5 + 1) 2
3
– 1
5
=
3
3
2 5
5
(1 + 1) 2 – (– 1) + 1 2
3
=
3
3
2
4 2
2 2
2 − 0 2 = (2 2) =
3
3
3
(
)
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits
...
Then dt = 5 x4 dx
...
The new limits are, when x = 0, t = 0 and
1 + x2
π
π
when x = 1, t =
...
4
4
1
Example 29 Evaluate
1
∫0
Therefore
–1
tan x
dx =
1+ x 2
∫
π
4
t
0
π
t2 4
π2
1 π2
dt = – 0 =
2 16
2 0
32
EXERCISE 7
...
1
...
∫0
6
...
∫
π
2
0
sin φ cos 5 φ d φ 3
...
1
–1
2x
2
1+ x
π
sin x
2
0 1 + cos 2
5
...
∫ 1 x – 2x 2 e
21
x
1
9
...
If f (x) =
(B) 0
( x − x3 ) 3
dx is
x4
(C) 3
x
∫ 0 t sin t dt , then f′(x) is
(A) cosx + x sin x
(C) x cosx
(B) x sinx
(D) sinx + x cosx
dx
1
Choose the correct answer in Exercises 9 and 10
...
10 Some Properties of Definite Integrals
We list below some important properties of definite integrals
...
P0 :
b
∫a
b
P1 :
∫a
P2 :
∫a
P3 :
∫a
P4 :
∫0
b
b
a
b
f ( x) dx = ∫ f ( t ) dt
a
a
f ( x) dx = – ∫
f ( x) dx
...
e
...
f ( x) dx , if f is an even function, i
...
, if f (– x) = f (x)
...
Proof of P0 It follows directly by making the substitution x = t
...
Then, by the second fundamental theorem of
calculus,
we
have
b
∫a
a
f ( x) dx = F (b ) – F ( a ) = – [F ( a) − F ( b )] = − ∫ f ( x) dx
b
Here, we observe that, if a = b, then
a
∫a
f ( x ) dx = 0
...
Then
b
∫a
c
∫a
and
b
∫c
f ( x) dx = F(b) – F(a)
...
(2)
f ( x) dx = F(b) – F(c)
...
Proof of P3 Let t = a + b – x
...
When x = a, t = b and when x = b, t = a
...
Then dt = – dx
...
Now
proceed as in P3
...
Let
t = 2a – x in the second integral on the right hand side
...
When x = a, t = a and when x = 2a, t = 0
...
Therefore, the second integral becomes
2a
∫a
2a
Hence
∫0
0
f ( x ) dx = – ∫ f (2a – t) dt =
a
f (x ) dx =
a
∫0
f (2 a – t ) dt =
a
∫0
f (2a – x) dx
a
f (x ) dx + ∫ f (2a − x) dx
0
Proof of P6 Using P5, we have
Now, if
a
∫0
2a
∫0
f ( x ) dx = ∫
a
0
a
f ( x) dx +∫ f (2 a − x) dx
0
...
Then
t = – x in the first integral on the right hand side
...
When x = – a, t = a and when
x = 0, t = 0
...
INTEGRALS
0
a
a
∫ − a f (x) dx = – ∫ a f (–t) dt +∫ 0
Therefore
=
a
∫0
343
f ( x) dx
a
f (– x) dx + ∫ f ( x) dx
0
(by P0)
...
So by P2 we write
2
∫ −1
0
x3 – x dx =
∫ −1 ( x
=
∫ −1 ( x
0
3
1
2
– x) dx + ∫ – ( x3 – x ) dx + ∫ ( x3 – x ) dx
0
3
1
1
2
0
1
– x) dx + ∫ ( x – x3 ) dx + ∫ ( x3 – x) dx
0
1
2
x 4 x2
x 2 x4 x4 x2
= – + – + –
2 –1 2
4 0 4
2 1
4
1 1 1 1
1 1
= – – + – + (4 – 2) – –
4 2 2 4
4 2
= –
Example 31 Evaluate
∫
1 1 1 1
1 1
3 3
11
+ + − +2− + = − + 2=
4 2 2 4
4 2
2 4
4
π
4
sin 2
–π
4
x dx
Solution We observe that sin2 x is an even function
...
Then, by P , we have
4
( π − x ) sin ( π − x ) dx
1 + cos 2 ( π − x)
π
I=
=
or
∫0
∫0
π sin x dx
( π − x ) sin x dx
−I
= π∫0
2
1 + cos x
1 + cos 2 x
π
2 I = π ∫0
π
sin x dx
1 + cos 2 x
π π sin x dx
2 ∫ 0 1 + cos 2 x
Put cos x = t so that – sin x dx = dt
...
Therefore, (by P1) we get
or
I=
I=
–π
2
−1
∫1
1
= π∫0
dt
π 1 dt
2 =
1+ t
2 ∫ −1 1 + t 2
1
dt
since
is even function)
2 (by P7 ,
1+ t 2
1+ t
2
π
π
–1 1
–1
−1
= π tan t 0 = π tan 1 – tan 0 = π – 0 =
4
4
Example 33 Evaluate
Solution Let I =
1
1
∫−1 sin
∫ −1sin
5
5
x cos 4 x dx
x cos 4 x dx
...
Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i
...
, f is an odd function
...
(1)
Then, by P4
I=
π
sin 4 ( − x)
2
π
2
∫0
π
π
sin 4 ( − x) + cos 4 ( − x)
2
2
Adding (1) and (2), we get
π
2
0
2I =
Hence
∫
I=
Then, by P3
π
∫
π
cos 4 x
2
0 cos 4 x + sin 4
x
dx
...
(1)
π
3
dx =
π
6
[ x]
π
3
π
6
=
π π π
− =
...
(2)
346
MATHEMATICS
Example 36 Evaluate
Solution Let I =
∫
π
2
0
∫
π
2
log
0
sin x dx
log sin x dx
Then, by P4
I=
∫
π
2
0
π
π
log sin − x dx = ∫ 2 log cos x dx
0
2
Adding the two values of I, we get
2I =
=
=
π
2
0
∫ ( log sin x + log cos x ) dx
π
2
0
∫ ( log sin x cos x + log 2 − log 2 ) dx (by adding and subtracting log 2)
∫
π
2
log
0
π
sin 2x dx − ∫ 2 log 2 dx
0
(Why?)
Put 2x = t in the first integral
...
2I =
1
2
=
Therefore
2
2
=
π
π
∫ 0 log sin t dt − 2 log 2
∫
π
2
log
0
π
2
log
0
∫
sin t dt −
sin x dx −
π
log 2 [by P6 as sin (π – t) = sin t)
2
π
log 2 (by changing variable t to x)
2
π
= I − log 2
2
Hence
∫
π
2
log
0
sin x dx =
–π
log 2
...
11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19
...
4
...
10
...
∫ 0 x (1 − x )
∫
π
2
(2 log
0
π
∫0
π
2
0
15
...
∫0
4
2
...
dx
8
...
∫
π
2
–π
2
x dx
1 + sin x
13
...
∫ 0 log (1 + cos x ) dx
sin 7 x dx
π
sin 2 x dx
3
3
sin 2 x + cos 2 x
8
6
...
∫0 x
x − 5 dx
2
2 − x dx
π
2
sin 2
–π
2
11
...
∫0
17
...
Show that
a
∫0
a
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
...
The value of
∫
π
2
−π
2
∫
π
2
log
0
(A) 0
21
...
(1)
...
Substituting values of A, B and C in (2), we get
2
2
1
1
1
x
1
−
−
=
2
2
2
2( x − 1) 2 ( x + 1) 2( x + 1)
(x − 1) (x + 1)
Again, substituting (3) in (1), we have
...
Then integrating it by
parts, we get
= ∫ log (log x) dx + ∫
I = x log (log x ) − ∫
= x log (log x ) − ∫
Again, consider
dx
we have
1
dx
x dx + ∫
x log x
(log x) 2
dx
dx
+∫
log x
(log x)2
...
(2)
350
MATHEMATICS
Putting (2) in (1), we get
x
dx
dx
x
−∫
+∫
x log (log x ) −
+C
2
2 =
log x
(log x)
(log x)
log x
I = x log (log x) −
Example 41 Find
∫
cot x + tan x dx
Solution We have
cot x + tan x dx = ∫ tan x (1 + cot x) dx
Put tan x = t2 , so that sec2 x dx = 2t dt
I=
or
dx =
Then
∫
2t dt
1+ t 4
1 2t
dt
I = ∫ t 1 + 2
4
t (1 + t )
= 2∫
1
Put t − = y, so that
t
1
1
1 + 2 dt
1 + 2 dt
(t 2 + 1)
t
t
dt = 2 ∫
=2∫
2
t4 +1
2 1
1
t + 2
t− +2
t
t
1
1 + 2 dt = dy
...
1
...
6
...
12
...
1
3
x 2 ( x4 + 1) 4
5x
( x + 1) ( x 2 + 9)
cos x
2
4 − sin x
x3
8
1− x
15
...
sin 3 x sin ( x + α)
x+a + x+b
1
1
x2
1
+ x3
[Hint:
3
...
sin ( x − a)
sin 8 − cos8 x
1 − 2 sin 2 x cos 2 x
1
+ x3
8
...
1
cos ( x + a ) cos ( x + b)
ex
13
...
1
( x + 1) ( x 2 + 4)
16
...
f ′ (ax + b) [f (ax + b)] n
10
...
1
19
...
21
...
353
x2 + x + 1
( x + 1) 2 ( x + 2)
x2 + 1 log ( x 2 + 1) − 2 log x
1− x
23
...
1+ x
x4
Evaluate the definite integrals in Exercises 25 to 33
...
∫π
2
x 1 − sin x
e
dx 26
...
∫
π
2
sin
0
33
...
∫
π
4 sin x + cos x
0 9 + 16 sin 2 x
32
...
x
π
cos2 x dx
2
0 cos 2 x + 4 sin 2
∫0 x e
dx 27
...
sin x + cos x
dx
29
...
36
...
2
dx
3
2
∫ 1 x2 ( x + 1) = 3 + log 3
1
∫ −1
∫
x17 cos 4 x dx = 0
π
4
2
0
40
...
∫
39
...
Choose the correct answers in Exercises 41 to 44
...
dx
∫ ex + e− x
is equal to
(A) tan–1 (ex) + C
(C) log (ex – e– x) + C
cos2 x
42
...
If f (a + b – x) = f (x), then
(A)
a+b
2
b−a
2
∫ a f (x) dx
f ( x) dx is equal to
∫ a f (b − x) dx
(C)
b
∫a x
44
...
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
...
d
F( x) = f ( x)
...
These integrals
Let
dx
are called indefinite integrals or general integrals, C is called constant of
integration
...
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
...
∫[ f (x ) + g (x )] dx = ∫ f ( x) dx + ∫ g ( x) dx
2
...
, fn are functions and k1, k2,
...
Then
∫[k1 f1( x) + k2 f 2 (x) +
...
+ kn ∫ f n ( x) dx
INTEGRALS
355
Some standard integrals
x n +1
+ C , n ≠ – 1
...
If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
Recall that a rational function is ratio of two polynomials of the form
may divide P (x) by Q (x) so that
P( x)
P ( x)
= 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)
...
P1 ( x)
can be integrated by
Q( x)
356
MATHEMATICS
P1 ( x)
as the sum of partial fractions of the following type:
Q( x)
A
B
px + q
+
=
,a≠b
x− a x−b
(x − a ) (x − b )
expressing
1
...
px + q
(x − a )2
=
A
B
+
x − a ( x − a )2
3
...
px2 + qx + r
(x − a )2 ( x − b )
=
A
B
C
+
+
2
x − a ( x − a)
x−b
5
...
Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals
...
When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals
...
(i)
∫ tan x dx = log
sec x + C
(iii)
∫ sec x dx = log
sec x + tan x + C
(iv)
∫ cosec x dx = log
(ii)
∫ cot x dx = log
sin x + C
cosec x − cot x + C
Integrals of some special functions
dx
1
x−a
+C
x +a
dx
1
a+x
+C
a−x
(i)
∫ x2 − a 2 = 2 a log
(ii)
∫ a 2 − x2 = 2 a log
(iii)
dx
1
∫ x2 + a 2 = a tan
−1
x
+C
a
INTEGRALS
(iv)
(vi)
dx
∫
x2 − a 2
dx
∫
2
2
= log x + x2 − a 2 + C (v)
∫
dx
2
a −x
2
357
x
+C
a
= sin − 1
= log | x + x2 + a 2 | + C
x +a
Integration by parts
For given functions f1 and f2, we have
d
f1 ( x) ⋅ ∫ f 2 ( x) dx dx , i
...
, 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}
...
Obviously, we must take that function as
the second function whose integral is well known to us
...
px + q = A
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
...
Then
x
∫a
f ( x) dx represents the Area function A (x)
...
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]
...
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, then
b
∫a
d
F( x) = f ( x) for all x in the domain of
dx
b
f ( x) dx = [ F( x) + C] a = F ( b) − F ( a )
...
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