Notes for integral calculus. Students must read function analysis before going through this book. Read Derivative Calculus before going through this book.

1. 1
INTEGRATION BASIC
A SHORT NOTES
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

2. 2 Integration
Contents
1 Integration 3
1.1 Continuous Sampling . . . . . . . . . . . . . . . . . . . . . . . 5
1.1.1 First Order Adder . . . . . . . . . . . . . . . . . . . . 5
1.1.2 Second Order Adder . . . . . . . . . . . . . . . . . . . 6
1.2 Antiderivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.2.1 First Principle Method . . . . . . . . . . . . . . . . . . 12
1.2.2 Lower & Upper Bound . . . . . . . . . . . . . . . . . . 20
1.2.3 Integral As Summation . . . . . . . . . . . . . . . . . . 27
1.3 Direct Integration . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.3.1 Error/Residual . . . . . . . . . . . . . . . . . . . . . . 56
1.3.2 Inverse Square Law (Field Relation) . . . . . . . . . . . 57
1.3.3 Algebraic Properties . . . . . . . . . . . . . . . . . . . 58
1.3.4 Convolution . . . . . . . . . . . . . . . . . . . . . . . . 59
1.4 Integral Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
1.4.1 Constant of Integration . . . . . . . . . . . . . . . . . . 60
1.4.2 Integration By Parts . . . . . . . . . . . . . . . . . . . 61
1.4.3 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . 74
1.4.4 Integration By Substitution . . . . . . . . . . . . . . . 77

3. 3
1Integration
Integration is reverse of the derivation. It is also called anti-derivative.
It increases dimension by one. For example, on integration of straight line,
it gives area, and integration of area, it gives volume. It also increases high-
est degree by one. Successive integral of a function is called is order. For
example, integration of function f(x) as
R
f(x) dx is called plane linear in-
tegration and represents area covered under the function curve and variable
axis. Double integration,
R R
f(x, y) dx dy represents area of curve surface
and triple integration
R R R
f(x, y, z) dx dy dz represents volume. One point
may be noted here, that integral is not a simple numerical computation, but
at least it is plan area computation. In any case, if integration of a single
term function is zero, then be cautious and please recheck the rules, princi-
ples and properties you have applied as area and volume of a finite shape or
of a finite element may never be zero or negative. This problem mostly arise
when we use symmetrical limits like
I =
1
Z
−1
x dx
Here, I represents to area covered by line and x-axis. So, area shall be never
zero. Now, integrating it, we have
I =
x2
2

7. 1
−1
=
1
2
−
1
2
= 0
x
f
b
−1.00
b
−0.75
b
−0.50
b
−0.25
b
0
b
0.25
b
0.50
b
0.75
b
1.00
It is not accepted here, as area is zero. So what is wrong? It is our
assumption that when line is in negative side of x-axis, area covered between

8. 4 Integration
line and x-axis is negative. As area shall never be negative, so we apply
symmetrical limit rules or find the area separately in both sides of x-axis and
then we will sum them. First case:
I = 2 ×
1
Z
0
x dx
And second case:
I =

14. 0
Z
−1
x dx

20. +

26. 1
Z
0
x dx

32. x
F
b
−1.00
b
−0.75
b
−0.50
b
−0.25
b
0
b
0.25
b
0.50
b
0.75
b
1.00
b
F (−1) = 0.5
b
F (1) = 0.5
Again we have found that integral of x is x2
/2 which represents a parabola
but it may be rewritten as
x2
2
=
1
2
× x × x
It meant half area of the area of square of side x. Here, degree of f(x) = x
is also increased by one, i.e. F(x) = 1
2
x2
. In case of multi-term function, like
f(x) = 2x − 3 area may be positive or negative or zero as integral of this
function
I =
b
Z
a
(2x − 3) dx =
b
Z
a
2x dx −
b
Z
a
3 dx
is difference of areas covered by first term (2x) and second term (3) about
x-axis.

33. 1.1. CONTINUOUS SAMPLING 5
1.1 Continuous Sampling
In continuous sampling, values are accumulated for one step and added into
next sample. The accumulator is known as adder, scalar or integrator and
represented by
I =
x
Z
−∞
( )dx
1.1.1 First Order Adder
The integrator of a sampled value is given by
I =
x
Z
−∞
( )dx (1.1)
Equation (1.1) is equation of linear adder. The block diagram of adder
equation is given by
+
p0
x
R
−∞
( )dx
x[n] y[n]
The adder operator is represented by ℑ (say). The above figure shall also
be represented as
+
p0
ℑ
x[n] y[n]

34. 6 Integration
To find the functional equation of above block diagram, we solve the
output result for adder block. The output y[n] is ℑ times to the sum of
current output, y[n] and current input, x[n]. So,
(p0y[n] + x[n]) ℑ = y[n]
Changing the sides of similar terms and simplifying them for y[n].
x[n]ℑ = (1 − p0ℑ)y[n]
This is adder of the simple block as given above. The response of the block is
divergent if p0 > 0, unique if p0 = 1 and convergent if p0 < 0. The response
of functional equation of adder to a sample is exponential, i.e. epx
u(x).
1.1.2 Second Order Adder
The second order adders are also known as double integrals. Double integrals
are represented by
I =
x
Z
−∞


x
Z
−∞
( )dx

 dx
The block diagram of double adders is shown below, in which two simple
adders are cascaded in series.
+
p0
ℑ
x[n]
y1[n]
+
p1
ℑ y[n]
From the above figure, the output of first block is
x[n]ℑ = (1 − p0ℑ)y1[n]
The output of first block is input of the second block. The output of second
block is given by
y1[n]ℑ = (1 − p1ℑ)y[n]

35. 1.1. CONTINUOUS SAMPLING 7
Substituting the value of y1[n] in to above equation, we have
y[n] =
ℑ2
(1 − p0ℑ)(1 − p1ℑ)
x[n] (1.2)
Here ℑ is integrator. ℑ1
is meant as integrated by once. Similarly, ℑ2
is
meant as integrated by twice.
Solved Problem 1.1 In a feed-backed integrator as shown in below figure,
find the algebraic solution of this operation.
Solution
+
p
ℑ
x[n] y[n]
The algebraic solution of this block diagram is
(p y[n] + x[n])ℑ = y[n]
On simplification, we have
y[n]
x[n]
=
ℑ
1 − pℑ
We know that 1
1−pℑ
is solution of infinite long geometric series with common
ratio of pℑ where |pℑ| < 1. So, this solution can be written as
y[n]
x[n]
= 1 + pℑ + p2
ℑ2
+ p3
ℑ3
+ . . .
ℑ
This is algebraic solution of the above block diagram.

36. 8 Integration
Solved Problem 1.2 In a feed-backed integrator as shown in below figure,
find the algebraic solution of this operation. If, initially x[n] = 0 then find
the result.
Solution
+
p
ℑ
x[n] y[n]
The algebraic solution of this block diagram is
(p y[n] + x[n])ℑ = y[n]
On simplification, we have
y[n]
x[n]
=
ℑ
1 − pℑ
We know that 1
1−pℑ
is solution of infinite long geometric series with common
ratio of pℑ where |pℑ| 1. So, this solution can be written as
y[n]
x[n]
= 1 + pℑ + p2
ℑ2
+ p3
ℑ3
+ . . .
ℑ
This is algebraic solution of the above block diagram. Again,
y[n] = 1 + pℑ + p2
ℑ2
+ p3
ℑ3
+ . . .
ℑx[n]
Here, ℑ =
R
, above relation becomes
y[n] =
Z
x[n] dx + p
Z Z
x[n] dx dx + p2
Z Z Z
x[n] dx dx dx + . . .
On solving it, by replacing x[n] = 0, we have y[n] = y
y = c + p cx + p2 cx2
2
+ . . .

37. 1.1. CONTINUOUS SAMPLING 9
Or it gives
y = c
1 + px +
p2
x2
2
+ . . .
= cepx
This is required result. epx
diverges to a finite value for positive values of x.
Thus, the result is divergent.
Solved Problem 1.3 In a feed-backed integrator as shown in below figure,
find the algebraic solution of this operation. If, initially x[n] = 0 then find
the result.
Solution
+
−p
ℑ
x[n] y[n]
The algebraic solution of this block diagram is
(−p y[n] + x[n])ℑ = y[n]
On simplification, we have
y[n]
x[n]
=
ℑ
1 + pℑ
We know that 1
1+pℑ
is solution of infinite long geometric series with common
ratio of (−pℑ) where |pℑ| 1. So, this solution can be written as
y[n]
x[n]
= 1 − pℑ + p2
ℑ2
− p3
ℑ3
+ . . .
ℑ
This is algebraic solution of the above block diagram. Again,
y[n] = 1 − pℑ + p2
ℑ2
− p3
ℑ3
+ . . .
ℑx[n]
Here, ℑ =
R
, above relation becomes
y[n] =
Z
x[n] dx − p
Z Z
x[n] dx dx + p2
Z Z Z
x[n] dx dx dx − . . .

38. 10 Integration
On solving it, by replacing x[n] = 0, we have y[n] = y
y = c − p cx + p2 cx2
2
− . . .
Or it gives
y = c
1 − px +
p2
x2
2
− . . .
= ce−px
e(−p)x
converges to a finite value for positive values of x. Here, -ve sign is
part of p. Thus the result is convergent. This is required result.
Solved Problem 1.4 Assume an algebraic series operator
O = ℑ + 2pℑ2
+ 3p2
ℑ3
+ 4p3
ℑ4
+ . . .
If, f(x) = 1, then find O[f(x)]. Take, ℑ as adder operator of x.
Solution The given operator is
O = ℑ + 2pℑ2
+ 3p2
ℑ3
+ 4p3
ℑ4
+ . . .
The O[f(x)] will be given as
O[f(x)] = [ℑ + 2pℑ2
+ 3p2
ℑ3
+ 4p3
ℑ4
+ . . .]f(x)
Substituting the value of function f(x), we have
O[f(x)] = [ℑ + 2pℑ2
+ 3p2
ℑ3
+ 4p3
ℑ4
+ . . .]1
Here, ℑ =
R
, so, replacing all ℑs with
R
in right side of the above relation.
Solving all terms by using direct integral method. We shall get
O[f(x)] = x + px2
+ p2 x3
2
+ p3 x4
6
+ . . .
The right hand side of above relation, becomes exponential of natural loga-
rithmic base, ‘e’, if x is taken as common from all terms. So,
O[f(x)] =
1 + px +
p2
x2
2!
+
p3
x3
3!
+ . . .
x = epx
x
This is desired result.

39. 1.2. ANTIDERIVATIVE 11
1.2 Antiderivative
Assume a difference table of a function f(x) between lower and upper bound
levels a and b. Step size of independent variable is ∆x.
n xn y[n] ∆y[n] d[y(x)]
0 a f(a)
f(x1) − f(a) f(x1)−f(a)
∆x
1 x1 f(x1)
f(x2) − f(x1) f(x2)−f(x1)
∆x
2 x2 f(x2)
. . .
. . . . . . . . .
. . .
r − 1 xr−1 f(xr−1)
f(xr) − f(xr−1) f(xr)−f(xr−1)
∆x
r xr f(xr)
. . .
. . . . . . . . .
. . .
n b f(b)
The difference of the function is given by
∆[f(xr)] = f(xr + ∆x) − f(xr)
and derivative of the function, d[f(xr)], is given by
d[f(xr)
∆x
= lim
∆x→0
f(xr + ∆x) − f(xr)
∆x
= f
′
(xr)
The opposite process of the derivative is called anti-derivatives or integration.
From the above relation
d[f(xr)] = f
′
(xr) × ∆x
Now, taking integration in both side, we have
Z
d[f(xr)] =
Z
f
′
(xr) ∆x

40. 12 Integration
Or
f(xr) =
Z
f
′
(xr) ∆x
It shows that antiderivative of product of function derivative at a point xr
and step size, ∆x is equal to the function value at that point xr. Again, the
product of function value and step size at point xr gives area bounded by
the function and step size.
A =
Z
f(xr) ∆x
1.2.1 First Principle Method
First principle method of integration is based on the Riemann’s integration.
To understand the method of first principle of integration, the following
example is most suitable. Let
I =
a
Z
0
x dx (1.3)
The function of above integration is f(x) = x. Area between function f(x)
and x−axis withing limits from x = 0 to x = a is equal to the given integral.
Now make ‘n’ equal partitions of limit range of integration. The width of
one partition is
d =
a − 0
n
=
a
n
Take rth
partition. The lower ‘x’1
value of rth
partition is ra
n
and function
value at this lower point of the partition is
f
ra
n
=
ra
n
Area of the partition is
Ar =
ra
n
×
a
n
1
Each strip has two x-values. One lower x-value an other upper x-value. If we take
lower x-value then range of ‘r’ is from zero to ‘n-1’. If we take upper x-value then range
of ‘r’ is from ‘1’ to ‘n’.

41. 1.2. ANTIDERIVATIVE 13
Whole area of function and x-axis within limits is the sum of the areas of all
partitions. Hence
A =
n−1
X
r=0
ra
n
×
a
n
Sum of all partition is
A =
a2
n2
[0 + 1 + 2 + . . . + (n − 1)]
Sum of right hand side of above relation is
A =
a2
n2
n − 1
2
{2 × 1 + (n − 1 − 1) × 1}
On simplification
A =
a2
n2
n2
− n
2
x
f(x)
0 a
b
xr
b
f(xr)
w
x
f(x)
0 a
b
xr
b
f(xr)
Figure 1.1: Integral as area function.
It is noted from the second part of figure 1.1 that, area covered by parti-
tion approaches to the curve if partition element is fine, i.e. width of partition
is very small. As the width of partition becomes smaller, number of partition
approaches to infinity. Taking the limit of ‘n’ to infinity. So, when n → ∞
we have
A = lim
x→∞
a2
1
2
−
1
2n
=
a2
2
(1.4)
This is required answer.

42. 14 Integration
Solved Problem 1.5 Using first principle method of integration, evaluate
b
R
a
x dx.
Solution In indefinite integral, upper limit is variable itself and lower
limit is zero. Hence for a specific point other than zero or variable itself, a
constant term is added in final result. If f(x) is function whose indefinite
integral is F(x), then the function is
Z
f(x) dx = F(x) + C
Where C is boundary value under the conditions of function.
x
f(x)
b
b
f(x1)
f(x2)
b b
x1 x2
In definite integrals, C is calculated by limits itself, hence there is no need
for adding of a constant term in result. In definite integral, it is assumed
that reference point is zero. Hence area between limits x1 to x2 is given by
difference of areas calculated between limits, from 0 to x2 and from 0 to x1.
So
x2
Z
x1
f(x) dx =
x2
Z
0
f(x) dx −
x1
Z
0
f(x) dx
Using this property, the integral for function f(x) = x2
within limits x = a
to x = b will be
I =
b
Z
a
x dx =
b
Z
0
x dx −
a
Z
0
x dx
We shall compute the first integral of right hand side and it shall be used as
reference for the computation of second integral of right hand side of above
relation.

43. 1.2. ANTIDERIVATIVE 15
x
y
f(xr)
xr xr+1
b
0
b
b
Divide the limits into equal ‘n’ partitions. The width of one partition is
(b − 0)/n. For rth
partition, the lower limit is xr = rb/n and upper limit
is xr+1 = (r + 1)b/n. The width of this strip is b/n. The function value at
lower limit is f(xr) and at upper limit is f(xr+1). So for this partition, area
is
Ar = f
rb
n
×
b
n
Total area is n
X
r=0
Ar =
n
X
r=0
f
rb
n
×
b
n
Or n
X
r=0
Ar =
n
X
r=0
rb
n
×
b
n
On simplifications
Ab =
b
n
2 n
X
r=0
r
We know that the sum of ‘n’ number is given by
Sn =
1
2
(n)(n + 1) =
1
2
n2
+ n
Using this relation in above relation.
Ab =
b
n
2
×
1
2
n2
+ n
and taking limits both side
Ab = lim
n→∞
b
n
2
×
1
2
n2
+ n
=
b2
2

44. 16 Integration
This is area of function and limits from 0 to b. Now the area covered between
function and x-axis from limit x = 0 to x = a, the area is
Aa =
a2
2
x
y
b
a
b
b
Now the area of shaded region is
A = Ab − Aa =
b2
2
−
a2
2
=
b2
− a2
2
It is the desired result.
Solved Problem 1.6 Using first principle method of integration, evaluate
b
R
a
x2
dx.
Solution In indefinite integral, upper limit is variable itself and lower
limit is zero. Hence for a specific point other than zero or variable itself, a
constant term is added in final result. If f(x) is function whose indefinite
integral is F(x), then the function is
Z
f(x) dx = F(x) + C
Where C is boundary value under the conditions of function.
x
f(x)
b
b
f(x1)
f(x2)
b b
x1 x2
In definite integrals, C is calculated by limits itself, hence there is no need
for adding of a constant term in result. In definite integral, it is assumed

45. 1.2. ANTIDERIVATIVE 17
that reference point is zero. Hence area between limits x1 to x2 is given by
difference of areas calculated between limits, from 0 to x2 and from 0 to x1.
So
x2
Z
x1
f(x) dx =
x2
Z
0
f(x) dx −
x1
Z
0
f(x) dx
Using this property, the integral for function f(x) = x2
within limits x = a
to x = b will be
I =
b
Z
a
x2
dx =
b
Z
0
x2
dx −
a
Z
0
x2
dx
We shall compute the first integral of right hand side and it shall be used as
reference for the computation of second integral of right hand side of above
relation.
x
y
f(xr)
xr xr+1
b
0
b
b
Divide the limits into equal ‘n’ partitions. The width of one partition is
(b − 0)/n. For rth
partition, the lower limit is xr = rb/n and upper limit
is xr+1 = (r + 1)b/n. The width of this strip is b/n. The function value at
lower limit is f(xr) and at upper limit is f(xr+1). So for this partition, area
is
Ar = f
rb
n
×
b
n
Total area is
n
X
r=0
Ar =
n
X
r=0
f
rb
n
×
b
n
Or
n
X
r=0
Ar =
n
X
r=0
rb
n
2
×
b
n

46. 18 Integration
On simplifications
Ab =
b
n
3 n
X
r=0
r2
We know that the sum of square of ‘n’ number is given by
Sn =
1
6
(n)(n + 1)(2n + 1) =
1
6
2n3
+ 3n2
+ n
Using this relation in above relation.
Ab =
b
n
3
×
1
6
2n3
+ 3n2
+ n
and taking limits both side
Ab = lim
n→∞
b
n
3
×
1
6
2n3
+ 3n2
+ n
=
b3
3
This is area of function and limits from 0 to b. Now area covered by function
and x-axis between limits from x = 0 to x = a is given by
Aa =
a3
3
x
y
b
a
b
b
Now the area of shaded region is
A = Ab − Aa =
b3
3
−
a3
3
=
b3
− a3
3
As the answer was required.

47. 1.2. ANTIDERIVATIVE 19
Solved Problem 1.7 Using first principle method of integration, evaluate
p
R
0
ex
dx.
Solution Assuming that the given relation is
A =
p
Z
a
ex
dx
Rewriting this relation in respect of lower limit at x = 0.
I =
p
Z
a
ex
dx =
p
Z
0
ex
dx −
a
Z
0
ex
dx
First term of RHS is integrated for reference. Divide the limits into equal
‘n’ partitions. The width of one partition is (p − 0)/n. For rth
partition, the
lower limit is xr = rp/n and upper limit is xr+1 = (r + 1)p/n. The width
of this strip is p/n. The function value at lower limit is f(xr) and at upper
limit is f(xr+1). So for this partition, area is
Ar = f
hrp
n
i
×
p
n
Total area is n
X
r=0
Ar =
n
X
r=0
f
hrp
n
i
×
p
n
Or n
X
r=0
Ar =
n
X
r=0
e
rp
n ×
p
n
On simplifications
Ap =
n
X
r=0
1 +
rp
n
1!
+
rp
n
2
2!
+ . . .
#
×
p
n
Or
Ap =




n
X
r=0
1 +
p
n
n
P
r=0
r
1!
+
p2
n2
n
P
r=0
r2
2!
+ . . .



 ×
p
n

48. 20 Integration
Taking limits n → ∞, we have
Ap = p +
p2
2!
+
p3
3!
+ . . .
Ap = 1 + p +
p2
2!
+
p3
3!
+ . . . − 1 = ep
− 1
Similarly, the area covered by function and x-axis between x = 0 to x = a is
Aa = ea
− 1
Now, area between x = a to x = p is
A = Ap − Aa = ep
− 1 − (ea
− 1) = ep
− ea
Setting limits a = 0, we have ea
= e0
= 1.
A = ep
− 1
This is required answer.
Solved Problem 1.8 Using first principle method of integration, evaluate
p
R
0
sin(x) dx
Solution
x
f(x)
0
a
b
xr
b
f(xr) w
x
f(x)
0
a
b
xr
b
f(xr)
1.2.2 Lower Upper Bound
An integral of function f(x) at any point x is given by
I =
Z
f(x) dx

49. 1.2. ANTIDERIVATIVE 21
Here dx is the width of integration. An integral always gives area bounded
between function f(x) and width dx. Take two points xi and xi+1, which
are lower and upper bound limits of the element of study and corresponding
values of the function f(x) are f(xi) and f(xi+1) respectively. Assume that
f(xi+1) f(xi)
x
f(x)
b
b
f(xi)
f(xi+1)
b b
xi xi+1 a
Area of the rectangle is given like
X
dA =
n−1
X
i=0
f(xi) × (xi+1 − xi) (1.5)
Now the sum of area is known as lower bound integral as it uses the lower
value of the function of the rectangle. Similarly sum of the relation
X
dA =
n−1
X
i=0
f(xi+1) × (xi+1 − xi) (1.6)
is called upper bound integral as it uses the higher value of the function.
Here n is number of rectangles whose equal-partition width is given by dx =
(a − 0)/n.
Solved Problem 1.9 Find the lower bound integral of the relation
1
R
0
x dx.
Solution The given function is f(x) = x. Here limit points are x = 0
to x = 1. We take n = 10 rectangles, then there shall be i = n + 1, i.e. 11
points. The width of each rectangle is given by
dx =
1 − 0
10
= 0.1

50. 22 Integration
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x
f(x)
b
f(0.5)
The xi are 0.0, 0.1, . . ., 1.0. The tabular form of xi and corresponding
f(xi) = xi is given below:
i xi f(xi) f(xi) × dx
0 0.0 0.0 0.00
1 0.1 0.1 0.01
2 0.2 0.2 0.02
3 0.3 0.3 0.03
4 0.4 0.4 0.04
5 0.5 0.5 0.05
6 0.6 0.6 0.06
7 0.7 0.7 0.07
8 0.8 0.8 0.08
9 0.9 0.9 0.09
10 1.0 1.0
Total 0.45
The lower bound integral is given by
X
dA =
n−1
X
i=0
f(xi) × (xi+1 − xi)

51. 1.2. ANTIDERIVATIVE 23
The corresponding product of f(xi) and dx = xi+1 − xi is given in third
column of above table. The sum of fourth column is 0.45. This value is less
than the actual integral.
Solved Problem 1.10 Find the upper bound integral of the relation
1
R
0
x dx.
Solution The given function is f(x) = x. Here limit points are x = 0
to x = 1. We take n = 10 rectangles, then there shall be i = n + 1, i.e. 11
points. The width of each rectangle is given by
dx =
1 − 0
10
= 0.1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x
f(x)
b
f(0.5)
The xi are 0.0, 0.1, . . ., 1.0. The tabular form of xi and corresponding
f(xi) = xi is given below:

52. 24 Integration
i xi f(xi) f(xi) × dx
0 0.0 0.0
1 0.1 0.1 0.01
2 0.2 0.2 0.02
3 0.3 0.3 0.03
4 0.4 0.4 0.04
5 0.5 0.5 0.05
6 0.6 0.6 0.06
7 0.7 0.7 0.07
8 0.8 0.8 0.08
9 0.9 0.9 0.09
10 1.0 1.0 0.10
Total 0.55
The upper bound integral is given by
X
dA =
n−1
X
i=0
f(xi+1) × (xi+1 − xi)
The corresponding product of f(xi+1) and dx = xi+1 − xi is given in third
column of above table. The sum of fourth column is 0.55. This value is more
than the actual integral.
Solved Problem 1.11 Find the lower bound integral of the relation
2
R
1
x2
dx.
Solution The given function is f(x) = x2
. Here limit points are x = 1
to x = 2. We take n = 10 rectangles, then there shall be i = n + 1, i.e. 11
points. The width of each rectangle is given by
dx =
2 − 1
10
= 0.1

53. 1.2. ANTIDERIVATIVE 25
1
2
3
4
1 2
x
f(x)
b
f(1.4)
The xi are 1.0, 1.1, . . ., 2.0. The tabular form of xi and corresponding
f(xi) = x2
i is given below:
i xi f(xi) f(xi) × dx
0 1.0 1.00 0.100
1 1.1 1.21 0.121
2 1.2 1.44 0.144
3 1.3 1.69 0.169
4 1.4 1.96 0.196
5 1.5 2.25 0.225
6 1.6 2.56 0.256
7 1.7 2.89 0.289
8 1.8 3.24 0.324
9 1.9 3.61 0.361
10 2.0 4.00
Total 2.185
The lower bound integral is given by
X
dA =
n−1
X
i=0
f(xi+1) × (xi+1 − xi)
The corresponding product of f(xi+1) and dx = xi+1 − xi is given in third
column of above table. The sum of fourth column is 2.185. This value is less
than the actual integral.

54. 26 Integration
Solved Problem 1.12 Find the lower bound integral of the relation
R 2
1
x2
dx.
Here, xi are given by x0 = 1, and nine intermediate points xi = 1 + 3 ×
log(1 + 0.1 × i) where, 1 ≤ i ≤ 9 and x10 = 2.
Solution The given function is f(x) = x2
. Here limit points are x = 1
to x = 2. We take n = 10 rectangles, then there shall be i = n + 1, i.e.
11 points. The 11 points are given by x0 = 1, and nine intermediate points
xi = 1 + 3 × log(1 + 0.1 × i) where, 1 ≤ i ≤ 9 and x10 = 2.
1
2
3
4
1 2
x
f(x)
b
f(1.4384)
The tabular form of xi and corresponding f(xi) = x2
i is given below:
i xi f(xi) f(xi) × dx
0 1.0000 1.0000 0.1242
1 1.1242 1.2638 0.1433
2 1.2375 1.5315 0.1597
3 1.3418 1.8005 0.1738
4 1.4384 2.0689 0.1860
5 1.5283 2.3356 0.1964
6 1.6124 2.5997 0.2053
7 1.6913 2.8607 0.2130
8 1.7658 3.1181 0.2197
9 1.8363 3.3719 0.5521
10 2.0000 4.0000
Total 2.1735

55. 1.2. ANTIDERIVATIVE 27
The lower bound integral is given by
X
dA =
n−1
X
i=0
f(xi+1) × (xi+1 − xi)
The corresponding product of f(xi+1) and dx = xi+1 − xi is given in third
column of above table. The sum of fourth column is 2.1735. This value is
less than the actual integral.
1.2.3 Integral As Summation
Consider a function that is plotted in the following figure:
x
f(x)
a b
The limits of the plot is [a, b]. To cover the area between the function
and x-axis, we draw vertical rectangular strips of height equal to the function
value at its lower end and of equal width dx as shown in the following figure:
x
f(x)
b
b
b
b
b
a b
Area of the region covered between the function and x-axis is sum of all
rectangular strips. Consider an auxiliary rectangular strip (ith
), whose area
we want to be computed as shown in the following figure:

56. 28 Integration
x
f(x)
f(xi)
xi dxi
a b
dA = f(xi) × dxi
Where, f(xi) is height of ith
rectangular strip and dxi is width of the strip.
Total area between the function and x-axis within limit [a, b] is sum of areas
of all these rectangular strips. So,
A =
n
X
i=0
f(xi) × dxi (1.7)
Here, n is number of rectangular strips we have drawn and it is computed as
n =
b − a
dx
The values of xi is given by xi = x0 +i×dxi. The width rectangular strips in
integration is always kept constant as small as possible, therefore, dxi may
be replaced with dx everywhere in this section.
x
f(x)
b
b
b
b
b b
b
b
b
a b
As we seen from above figure, as the width of rectangular strips decrease,
they follow curves more precisely and they cover region between curve and
x-axis more correctly. Again, Consider a simple integral relation
I =
b
Z
a
f(x) dx (1.8)

57. 1.2. ANTIDERIVATIVE 29
It represents that, we have to find the sum (integrate) of product of function
value and change in x, say dx, at point x for whole range of limits [a, b].
Mathematically, relations (1.7) and (1.8) have same meaning, hence
b
Z
a
f(x) dx =
n
X
i=0
f(xi) × dxi
This shows that, integral may be written in summation form. Note that
summation and integration have the same meaning but in mathematics there
is difference between them. The summation is used in case of discrete values
while integration is used in continuous case.
Sum Methods
There are four methods of summation, commonly known as Riemann Sum-
mation with partition of equal size. Let a function f(x) is defined interval
[a, b] and is therefore divided into n parts, each equal to
∆x =
b − a
n
The points (total n + 1 numbers) for x in the partition will be
a, a + ∆x, a + 2∆x, . . . , a + (n − 1)∆x, b
Left Riemann Sum As there are n partition of the function, therefore
there shall be n + 1 points for x as shown in the following figure.
x
f(x)
b
b
b
b
b
x1 = a x2 x3 x4 x5 = b
In above figure, there are four partitions and five points of x. Each point
of x is given by
x1 = a+0∆x = a; x2 = a+∆x; x3 = a+2∆x; x4 = a+3∆x; x5 = a+4∆x = b

58. 30 Integration
When left-end point of rectangle is taken as height of the rectangle, then it
is called left hand summation. This gives integral value as
A =
n−1
X
i=0
∆x × f(a + i∆x)
Right Riemann Sum As there are n partition of the function, therefore
there shall be n + 1 points for x as shown in the following figure.
x
f(x)
b
b
b
b
b
x1 = a x2 x3 x4 x5 = b
In above figure, there are four partitions and five points of x. Each point
of x is given by
x1 = a+0∆x = a; x2 = a+∆x; x3 = a+2∆x; x4 = a+3∆x; x5 = a+4∆x = b
When right-end point of rectangle is taken as height of the rectangle, then it
is called right hand summation. This gives integral value as
A =
n−1
X
i=0
∆x × f(a + (i + 1)∆x)
Mid Riemann Sum As there are n partition of the function, therefore
there shall be n + 1 points for x as shown in the following figure.
x
f(x)
b
b
b
b
x1 = a x2 x3 x4 x5 = b

59. 1.2. ANTIDERIVATIVE 31
In above figure, there are four partitions and five points of x.
x1 = a+0∆x = a; x2 = a+∆x; x3 = a+2∆x; x4 = a+3∆x; x5 = a+4∆x = b
When mid point of rectangle is taken as height of the rectangle, then it is
called mid point summation. This gives integral value as
A =
n−1
X
i=0
∆x × f
a + (2i + 1)
∆x
2
Trapezoidal Rule As there are n partition of the function, therefore there
shall be n + 1 points for x as shown in the following figure.
x
f(x)
b
b
b
b
b
x1 = a x2 x3 x4 x5 = b
In above figure, there are four partitions and five points of x. Each point
of x is given by
x1 = a+0∆x = a; x2 = a+∆x; x3 = a+2∆x; x4 = a+3∆x; x5 = a+4∆x = b
In the trapezoidal rule, rectangular box are not made but trapezium are
constructed as shown in above figure. The summation in trapezoidal rule is
given by
A =
n−1
X
i=0
∆x
2
[f(a + i∆x) + f(a + (i + 1)∆x)]
Solved Problem 1.13 Solve the integral by summation method and direct
method. Take two different values of dx, i.e. the width of rectangular strips.
Integral is
I =
1
Z
0
x dx

60. 32 Integration
Solution The summation form of integration is
I =
n
X
i=0
f(x) × dx
Taking, dx = 0.2, we have, n = 5, and points x0 = 0, x1 = 0.2, x2 = 0.4,
x3 = 0.6 and x4 = 0.8. Now, substituting these values in sum integral
relation, we have
I = (0 + 0.2 + 0.4 + 0.6 + 0.8) × 0.2 = 0.2
Again, taking, dx = 0.1, we have, n = 10, and points x0 = 0, x1 = 0.1,
x2 = 0.2, x3 = 0.3, x4 = 0.4, x5 = 0.5, x6 = 0.6, x7 = 0.7, x8 = 0.8 and
x9 = 0.9. Now, substituting these values in sum integral relation, we have
I = (0 + 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9) × 0.2 = 0.35
Using direct method of integration, we have
I =
1
Z
0
x dx =
x2
2

64. 1
0
= 0.5
The variation in the direct method of integration and sum method of inte-
gration is due to the width of dx. As the dx moves towards zero, the result
of sum method approach to the result found from direct method.
Solved Problem 1.14 Solve the integral by summation method and direct
method. Take two different values of dx, i.e. the width of rectangular strips.
Integral is
I =
3
Z
2
2x2
dx
Solution For this problem, I have constructed a table with dx = 0.2 and
dx = 0.1 for the function 2x2
. Following table is for dx = 0.2, n = 5 and
using lower bound limit.

65. 1.2. ANTIDERIVATIVE 33
x 2x2
2x2
dx
2.00 8.00 1.60
2.20 9.68 1.94
2.40 11.52 2.30
2.60 13.52 2.70
2.80 15.68 3.14
P
2x2
dx=11.68
x
f(x)
2 3
x
f(x)
b
b
b
b
b
b
8.0
9.68
11.52
13.52
15.68
18.0
2 3
Following table is for dx = 0.1, n = 10 and using lower bound limit.
x 2x2
2x2
dx
2.00 8.00 0.80
2.10 8.82 0.88
2.20 9.68 0.97
2.30 10.58 1.06
2.40 11.52 1.15
2.50 12.50 1.25
2.60 13.52 1.35
2.70 14.58 1.46
2.80 15.68 1.57
2.90 16.82 1.68
P
2x2
dx=12.17

66. 34 Integration
Using direct method, we have
I =
3
Z
2
2x2
dx =
2
3
x3

70. 3
2
On solving it, we have I = 12.67 approximately. The difference in results
is due to comparatively large values of dx taken in summation method in
respect of direct method.
Solved Problem 1.15 Solve the integral by summation method and direct
method. Take two different values of dx, i.e. the width of rectangular strips.
Integral is
I =
3
Z
2
(x + 3) dx
Solution For this problem, I have constructed a table with dx = 0.2 and
dx = 0.1 for the function (x + 3). Following table is for dx = 0.2, n = 5 and
using lower bound limit.
x (x + 3) (x + 3) dx
2.00 5.00 1.00
2.20 5.20 1.04
2.40 5.40 1.08
2.60 5.60 1.12
2.80 5.80 1.16
P
(x + 3) dx=5.40
x
f(x)
2 3
x
f(x)
b
b
b
b
b
b
5.0
5.2
5.4
5.6
5.8
6.0
2 3

71. 1.3. DIRECT INTEGRATION 35
Following table is for dx = 0.1, n = 10 and using lower bound limit.
x (x + 3) (x + 3) dx
2.00 5.00 0.50
2.10 5.10 0.51
2.20 5.20 0.52
2.30 5.30 0.53
2.40 5.40 0.54
2.50 5.50 0.55
2.60 5.60 0.56
2.70 5.70 0.57
2.80 5.80 0.58
2.90 5.90 0.59
P
(x + 3) dx=5.45
Using direct method, we have
I =
3
Z
2
(x + 3) dx =
x2
2

75. 3
2
+ 3x|3
2
On solving it, we have I = 5.50 approximately. The difference in results
is due to comparatively large values of dx taken in summation method in
respect of direct method.
1.3 Direct Integration
Integration is reverse of differentiation. If properties of a small element of
a function is known then property of whole function can be evaluated. If a
function f(x) is to be integrated with respect to x then it would be written
as Z
f(x) dx
If the limits of integration is not defined then integration is called infinite in-
tegration. Infinite integration declares that the function is continuous within
its general values and limit exists at every points. Integration is also called
anti-differentiation.

76. 36 Integration
1 2 3 4
x
y ds
y
y =
√
x
Figure 1.2:
From figure (1.2), the integration given in equation (1.9) can be explained.
The graph shown in figure is y =
√
x. Assume a small strip of height y and
width ds as shown in figure.
I =
Z
y ds (1.9)
Here, ds is width of strip along the curve and it is given by.
ds =
s
1 +
dy
dx
2
dx
If function slop is zero or very small, then ds ≈ dx. And in this case
I =
Z
y dx
Height y of strip element, is equal to function f(x), therefore, this equation
can be written as
I =
Z
f(x) dx (1.10)
Integrals represents to area bounded by function and x-axis within the lim-
iting boundaries. Again, if we are given derivative of a function, then anti-
derivative or integration of it returns the actual function.
Z
f
′
(x) dx = f(x) (1.11)
Where f
′
(x) is first order derivative of the given function f(x). Similarly,
from different notations
Z
f(1)
(x) dx = f(x)

77. 1.3. DIRECT INTEGRATION 37
Where f(1)
(x) is first order derivative of the given function f(x). For nth
degree function derivative
Z
f(n)
(x) dx = f(n−1)
(x)
Solved Problem 1.16 Find the integral of 0.
Solution From the definition of differentiation of constant
d
dx
c = 0
Simplifying above relation
dc = 0 dx
Taking integration on both side
Z
dc =
Z
0 dx
Z
0 dx = c
Z
0 dx = c (1.12)
R
0 dx = c As we seen in previous problem, that integral of zero is a
constant value. This constant is about any given value of x. For example, if
x increases continuously, integral remain a constant function. It means, c is
parallel to the x-axis. Therefore, we can say that integral not only gives area
and volume, but it also gives the constants too. This constant is auxiliary
and independent of function (as there is zero function) so it is also known
as function constant which tells that the offset distance of the function is c
from the x-axis.
x
y
c

78. 38 Integration
Solved Problem 1.17 Find the integral of c.
Solution Let cx is a function whose differentiation with respect to x gives
constant c. Hence
d
dx
(cx) = c
Simplifying above relation
d(cx) = c dx
Taking integration on both side
Z
d(cx) =
Z
c dx
Z
c dx = cx (1.13)
If c = 1 then Z
1 dx = x (1.14)
R
c dx = cx This type of integral gives area as output where c is offset
value about x-axis, or we can say that it is y-axis value. Note that, in
derivative, coefficient of x is called line slope. Let dx is very small element
along the x-axis such that
dx = xi+1 − xi
and it is distributed in [1, 4]. If dx = 0.25 then there shall be 12 partitions
in [1, 4] with points x0 to x12. The length between points x = 1 and x = 4 is
sum length of these twelve partitions as
x
y
x0 x2 x4 x6 x8 x10 x12
x1 x3 x5 x7 x9 x11
dx1 dx12
l = dx1 + dx2 + . . . + dx12 = 0.25 + 0.25 + . . . + 0.25 (12 times)
This gives l = 3. In integral form we have
Z
c dx = cx

79. 1.3. DIRECT INTEGRATION 39
which is continuous from x = −∞ to x = ∞.
x
y
c
x0 x2 x4 x6 x8 x10 x12
x1 x3 x5 x7 x9 x11
dx1 dx12
As this gives us area, so we can get the area x ∈ [1, 4], we have
A = c × 4 − c × 1 = c × 3 = 3c
This is area which will depend on the value of c. We knew that if A = lb is
area of rectangle of length l and width b, and if width of the rectangle is 1
unit then area of that rectangle is equal to the length of the rectangle, i.e.
A = l × 1 = l. It means, integral can also be used as line integral.
1
1 2 3 4
x
y
c
If c = 1, l = 3 which is exactly the length of line segment. Therefore, this
integral is called line integral if c = 1.
1
1 2 3 4
x
y
c = 1
1 2 3 4
x
y
Here, area A is exactly equivalent to the length of line l when c = 1.

80. 40 Integration
Solved Problem 1.18 Find the integral of xn
.
Solution The differentiation of xn
is
d
dx
(xn
) = n xn−1
Simplifying above relation
d(xn
) = n xn−1
dx
Taking integration on both side
Z
d(xn
) =
Z
n xn−1
dx
Z
xn−1
dx =
xn
n
For nth
power of x, n is replaced by n + 1
Z
xn
dx =
xn+1
n + 1
Z
xn
dx =
xn+1
n + 1
(1.15)
Note that, if n = −1, then the term xn
shall be equal to 1/x and its in-
tegral can not be solved by using relation 1.15 as its result is unacceptable
mathematically.
Z
xn
dx

84. n=−1
=
xn+1
n + 1

88. n=−1
=
x0
0
=
1
0
= ∞
In other words, the integral is divergent at n = −1 irrespective the value of
x.
Solved Problem 1.19 Find the integral of 1
x
.
Solution The differentiation of ln(x) is 1/x. Hence
d
dx
ln(x) =
1
x

89. 1.3. DIRECT INTEGRATION 41
Simplifying above relation
d(ln(x)) =
1
x
dx
Taking integration on both side
Z
d(ln(x)) =
Z
1
x
dx
Z
1
x
dx = ln(x)
Z
1
x
dx = ln(x) (1.16)
Solved Problem 1.20 Find the integral of sin(x).
Solution The differentiation of cos(x) is − sin(x). Hence
d
dx
cos(x) = − sin(x)
Simplifying above relation
d[cos(x)] = − sin(x) dx
Taking integration on both side
Z
d[cos(x)] =
Z
− sin(x) dx
Z
− sin(x) dx = cos(x)
Z
sin(x) dx = − cos(x) (1.17)
Similarly Z
cos(x) dx = sin(x) (1.18)

90. 42 Integration
Solved Problem 1.21 Find the integral of tan(x).
Solution Integration of tan x can be found by using substitution method.
The integral is Z
tan x dx =
Z
sin x
cos x
dx
Now substituting cos x = t in denominator which gives − sin x dx = dt on
differentiation. Substituting these values in the above equation and its inte-
gration becomes Z
tan x dx =
Z
−
1
t
dt
On integration right hand side of above relation.
Z
tan x dx = − ln(t)
Substituting the value of t in above relation.
Z
tan(x) dx = − ln(cos x) = ln(sec x) (1.19)
Solved Problem 1.22 Find the integral of sec t.
Solution Integration of sec t can be written as
Z
sec t dt =
Z
sec t dt
Multiply and divide, right hand side of above equation by sec t + tan t
Z
sec t dt =
Z
sec t
sec t + tan t
sec t + tan t
dt
=
Z
sec2
t + sec t tan t
sec t + tan t
dt
Now substitute sec t + tan t = u and (sec t tan t + sec2
t) dt = du in above
relation
Z
sec t dt =
Z
1
u
du
= ln |u|

91. 1.3. DIRECT INTEGRATION 43
Substituting the value of u the result is
Z
sec(t) dt = ln(| sec t + tan t|) (1.20)
Solved Problem 1.23 Show that integration of sec2
x is tan x.
Solution The differentiation of tan x is
d
dx
(tan x) = sec2
x
Simplifying above relation
d(tan x) = sec2
x dx
Taking integration on both side
Z
d(tan x) =
Z
sec2
x dx
Z
sec2
x dx = tan x
Z
sec2
x dx = tan x (1.21)
Solved Problem 1.24 Show that integration of csc2
x is − cot x.
Solution The differentiation of cot x is
d
dx
(cot x) = − csc2
x
Simplifying above relation
d(cot x) = − csc2
x dx
Taking integration on both side
−
Z
d(cot x) =
Z
csc2
x dx

92. 44 Integration
Z
csc2
x dx = − cot x
Z
csc2
x dx = − cot x (1.22)
Solved Problem 1.25 Show that integration of sec x tan x is sec x.
Solution The differentiation of sec x is
d
dx
(sec x) = sec x tan x
Simplifying above relation
d(sec x) = sec x tan x dx
Taking integration on both side
Z
d(sec x) =
Z
sec x tan x dx
Z
sec x tan x dx = sec x
Z
sec x tan x dx = sec x (1.23)
Solved Problem 1.26 Find that integration of csc x cot x is − csc x.
Solution The differentiation of csc x is
d
dx
(csc x) = − csc x cot x
Simplifying above relation
d(csc x) = − csc x cot x dx
Taking integration on both side
−
Z
d(csc x) =
Z
csc x cot x dx

93. 1.3. DIRECT INTEGRATION 45
Z
csc x cot x dx = − csc x
Z
csc x cot x dx = − csc x (1.24)
Solved Problem 1.27 Find that integration of ax
is ax
ln(a)
.
Solution The differentiation of ax
is
d
dx
(ax
) = ax
ln a
Simplifying above relation
d(ax
) = ax
ln a dx
Taking integration on both side
Z
d(ax
) =
Z
ax
ln a dx
Z
ax
ln a dx = ax
Z
ax
dx =
ax
ln a
(1.25)
Solved Problem 1.28 Find the integral of ex
.
Solution The differentiation of ex
is ex
. Hence
d
dx
ex
= ex
Simplifying above relation
d(ex
) = ex
dx
Taking integration on both side
Z
d(ex
) =
Z
ex
dx

94. 46 Integration
Z
ex
dx = ex
Z
ex
dx = ex
(1.26)
Solved Problem 1.29 Find the integral of sinh x.
Solution From hyperbolic function using Euler’s theorem
sinh x =
ex
− e−x
2
Now integrating above relation with respect to x
Z
sinh x dx =
Z
ex
− e−x
2
dx
=
1
2
Z
ex
dx −
Z
e−x
dx
=
1
2
ex
−
e−x
−1
=
1
2
ex
+ e−x
Now right hand side of above is equal to cosh x, Hence
Z
sinh x dx = cosh x (1.27)
Solved Problem 1.30 Find the integral of cosh x.
Solution From hyperbolic function using Euler’s theorem
cosh x =
ex
+ e−x
2

95. 1.3. DIRECT INTEGRATION 47
Now integrating above relation with respect to x
Z
cosh x dx =
Z
ex
+ e−x
2
dx
=
1
2
Z
ex
dx +
Z
e−x
dx
=
1
2
ex
+
e−x
−1
=
1
2
ex
− e−x
Now right hand side of above is equal to sinh x, Hence
Z
cosh x dx = sinh x (1.28)
Solved Problem 1.31 Find the integral of tanh x.
Solution From hyperbolic function using Euler’s theorem
tanh x =
ex
− e−x
ex + e−x
Now integrating above relation with respect to x
Z
tanh x dx =
Z
ex
− e−x
ex + e−x
dx
Let v = ex
+ e−x
whose differentiation is dv = ex
− e−x
and above relation
becomes Z
tanh x dx =
Z
1
v
dx
On integration the relation becomes
Z
tanh x dx = log(v) + log c
= ln(ex
+ e−x
) + log c
= ln(ex
+ e−x
) + log 2 − log 2 + log c
= ln
ex
+ e−x
2
+ log 2c

96. 48 Integration
Now right hand side of above is equal to ln(cosh x), Hence
Z
tanh x dx = ln(cosh x) + c (1.29)
Solved Problem 1.32 Prove that the integration of sech2
x is tanh x.
Solution
Solved Problem 1.33 Find that integration of csch2
x is − coth x.
Solution
Solved Problem 1.34 Find that integration of sech x tan hx is − sech x.
Solution
Solved Problem 1.35 Find that integration of csch x cot hx is − csch x.
Solution
Solved Problem 1.36 Find the integral of sin−1
x.
Solution The integration of the function can be found by using integra-
tion method of product of two functions.
Z
sin−1
x dx = sin−1
x
Z
1dx −
Z
d
dx
sin−1
x
Z
1 dx
dx
= x sin−1
x −
Z
x
√
1 − x2
dx
Substitute 1 − x2
= y which gives −2x dx = dy and second part of above
relation becomes
Z
sin−1
x dx = x sin−1
x −
Z
−
1
2
√
y
dy
= x sin−1
x +
1
2
Z
y−1/2
dy
= x sin−1
x +
1
2
y−1/2+1
−1/2 + 1
= x sin−1
x +
√
y

97. 1.3. DIRECT INTEGRATION 49
Substituting the value of y, then
Z
sin−1
x dx = x sin−1
x +
√
1 − x2 (1.30)
Solved Problem 1.37 Find the integral of cos−1
x.
Solution The integration of the function can be found by using integra-
tion method of product of two functions.
Z
cos−1
x dx = cos−1
x
Z
1dx −
Z
d
dx
cos−1
x
Z
1 dx
dx
= x cos−1
x −
Z
−x
√
1 − x2
dx
= x cos−1
x +
Z
x
√
1 − x2
dx
Substitute 1 − x2
= y which gives −2x dx = dy and second part of above
relation becomes
Z
cos−1
x dx = x cos−1
x +
Z
−
1
2
√
y
dy
= x cos−1
x −
1
2
Z
y−1/2
dy
= x cos−1
x −
1
2
y−1/2+1
−1/2 + 1
= x cos−1
x −
√
y
Substituting the value of y, then
Z
cos−1
x dx = x cos−1
x −
√
1 − x2 (1.31)
Solved Problem 1.38 Find the integral of tan−1
x.
Solution The integration of the function can be found by using integra-

98. 50 Integration
tion method of product of two functions.
Z
tan−1
x dx = tan−1
x
Z
1dx −
Z
d
dx
tan−1
x
Z
1 dx
dx
= x tan−1
x −
Z
1
1 + x2
dx
Substitute 1 + x2
= y which gives 2x dx = dy and second part of above
relation becomes
Z
tan−1
x dx = x tan−1
x −
Z
1
2y
dy
= x tan−1
x −
1
2
log(y)
Substituting the value of y, then
Z
tan−1
x dx = x tan−1
x −
1
2
log(1 + x2
) (1.32)
Solved Problem 1.39 Find the integral of ln x.
Solution The integration of the function can be found by using integra-
tion method of product of two functions.
Z
ln x dx = ln x
Z
1 dx −
Z
d
dx
ln x
Z
1 dx
dx
= x ln x −
Z
1
x
× x
dx
= x ln x −
Z
1 dx
= x ln x − x
Z
ln x dx = x ln x − x (1.33)
Solved Problem 1.40 Find the integration x dy + y dx.

99. 1.3. DIRECT INTEGRATION 51
Solution Let k = xy and differentiating it with respect to x or y (Here
with respect to x)
d
dx
k =
d
dx
(xy)
Here y depends on x as in question both x adn y are differentiated. So,
dk
dx
= y
d
dx
(x) + x
d
dx
(y)
= y + x
dy
dx
Simplifying the above relation
dk = y dx + x dy
Now integrating above relation
Z
[y dx + x dy] =
Z
dk
= k
Hence right hand side is equal to xy. This is the answer.
Second Method
Z
[y dx + x dy] =
Z
d(xy)
= xy
This is the answer.
Solved Problem 1.41 Find the integration v du−u dv
v2 .
Solution Let k = u
v
and differentiating it with respect to u or v (Here
with respect to v)
d
dv
k =
d
dv
u
v
Here u depends on v as in question both u and v are differentiated. So,
dk
dv
=
v d
dv
(u) + u d
dv
(v)
v2
=
vdu
dv
+ u
v2

100. 52 Integration
Simplifying the above relation
dk =
v du + u dv
v2
Now integrating above relation
Z
v du + u dv
v2
=
Z
dk
= k
Hence right hand side is equal to u
v
. This is the answer.
Second Method
Z
v du + u dv
v2
=
Z
d
u
v
=
u
v
This is the answer.
Solved Problem 1.42 Evaluate
R
(k2
+ sin k) dk.
Solution The given integral is
I =
Z
(k2
+ sin k) dk
Using direct method for integral, the result is
I =
k3
3
− cos k + C
Solved Problem 1.43 Evaluate
R
(k + k2
+ 3k3
) dk.
Solution The given integral is
I =
Z
(k + k2
+ 3k3
) dk

101. 1.3. DIRECT INTEGRATION 53
Direct integration method gives
I =
Z
k dk +
Z
k2
dk +
Z
3k3
dk
Or
I =
k2
2
+
k3
3
+
3
4
k4
+ C
Solved Problem 1.44 Evaluate
R
(e−x
+ 2x) dx.
Solution The given integral is
I =
Z
(e−x
+ 2x) dx
Expanding right hand side, we get
I =
Z
e−x
dx +
Z
2x dx
Or
I = −e−x
+ x2
+ C
Solved Problem 1.45 A constant function is integrated with respect to x.
Find its integral output and show it geometrically.
Solution The given function is a constant function of x, therefore let
f(x) = c. Now integrating it with respect to x, we have
I =
Z
f(x) dx =
Z
c dx = cx
The integral output is an equation of line with slope magnitude c. The input
and output of this integral is shown below:
x
y
c
R
x
y
c

102. 54 Integration
x
y
c
R x
y
−c
From above figures, it is shown that the constant magnitude c becomes
the slope magnitude after integration.
Solved Problem 1.46 Evaluate
R
(sin k + cos k) dk.
Solution The given problem is
I =
Z
(sin k + cos k) dk
Note that, the trigonometric operators are dependent on variable k and inte-
gration is also performed about k. So, sin k and cos k look like constants but
they are functions actually. Integrating the given integral about k, we have
I =
Z
sin k dk +
Z
cos k dk = − cos k + c1 + sin k + c2
Or
I = − cos k + sin k + C
This is desire result.
Solved Problem 1.47 Evaluate
R
(sinh x + xn
+ tanh x) dx.
Solution The given problem is
I =
Z
(sinh x + xn
+ tanh x) dx
This is algebraic-trigonometric integral. Note that there is an algebraic term
xn
. If n = −1, then its integral shall be computed as
R 1
x
dx and if n 6= −1,
then its integral shall be as per method as applied in
R
xn
dx. Integrating
the given integral about x assuming that nneq − 1, we have
I =
Z
sinh x dx +
Z
xn
dx +
Z
tanh x dx

103. 1.3. DIRECT INTEGRATION 55
Or
I = cosh(x) +
xn+1
n + 1
+ ln (cosh(x)) + C
This is desire result.
Solved Problem 1.48 Evaluate
R 1
t
+ t2
dt.
Solution The given problem is
I =
Z
1
t
+ t2
dt
Note that, the algebraic terms are dependent on variable t and integration
is also performed about t. So, all terms are dependent on t. Integrating the
given integral about t, we have
I =
Z
1
t
dt +
Z
t2
dt
Or
I = ln t +
t3
3
+ C
This is desire result.
Solved Problem 1.49 Evaluate
R
(yt + xt2
) dt.
Solution The given problem is
I =
Z
(yt + xt2
) dt
Note that, integration is performed about t so symbols or operators which
are independent of t are considered as constants. So, y and x are constant
for this problem. Integrating the given integral about t, we have
I =
Z
yt dt +
Z
xt2
dt
Or
I = y
t2
2
+ x
t3
3
+ C
This is desire result.

104. 56 Integration
1.3.1 Error/Residual
The amount by which an observed value (O) differs from the actual value (A)
predicted by the direct method or perfect model is known as error. Errors
or residuals are not accounted in computations.
e = A − O
As the number of parts (n) increases, partition width (dx) decreases and
finally e → 0. At these states, observe value approaches to the actual value.
Solved Problem 1.50 Solve the integral by summation method and direct
method. Find the error or residual. Integral is
I =
3
Z
2
2x2
dx
Solution For this problem, I have constructed a table with dx = 0.2
function 2x2
. Following table is for dx = 0.2, n = 5 and using lower bound
limit.
x 2x2
2x2
dx
2.00 8.00 1.60
2.20 9.68 1.94
2.40 11.52 2.30
2.60 13.52 2.70
2.80 15.68 3.14
P
2x2
dx=11.68
Using direct method, we have
I =
3
Z
2
2x2
dx =
2
3
x3

108. 3
2
On solving it, we have I = 12.67 approximately. The error or residual is
difference between actual value and observed value. So, error or residual is
12.67 − 11.68 = 0.99. Error or residual may also be computed for dx = 0.1,
which may be differ from 0.99.

109. 1.3. DIRECT INTEGRATION 57
1.3.2 Inverse Square Law (Field Relation)
Inverse square law states that, value of dependent variable (y) is inversely
proportional to the square of independent variable (x). So, mathematically
y ∝
1
x2
=
k
x2
Here, k is proportional constant. Remember the gravitational field as read
in physics, which states that, gravitational field (E) between two celestial
bodies is directly proportional to the product of their masses and inversely
proportional to the square of distance between their centers (r).
E ∝
Mm
r2
Here, r is independent variable and E is dependent variable, M and m are
constant masses. In integration, we can convert a inverse cubic function into
inverse square law. From relation
Z
xn
dx =
xn+1
n + 1
Taking, n = −3, we have
Z
x−3
dx =
x−3+1
−3 + 1
=
x−2
−2
= −
1
2x2
By this method, we have find inverse square function. Similarly, we can
also convert a inverse function (k/x) into inverse square function by using
derivative.
y ∝
1
x
=
k
x
Remember the gravitational potential as read in physics, which states that,
gravitational potential (V ) between two celestial bodies is directly propor-
tional to the product of their masses and inversely proportional to the dis-
tance between their centers (r).
V ∝
Mm
r
Here, r is independent variable and V is dependent variable, M and m are
constant masses. From relation,
d
dx
xn
= nxn−1

110. 58 Integration
Taking, n = −1, we have
d
dx
x−1
= −1 × x−1−1
= −1 × x−2
=
−1
x2
These two conversion methods are very useful in physics where field rela-
tions (both electric and gravitational) are converted into potential rela-
tions (both electric and gravitational).
Solved Problem 1.51 Convert the potential function −k
x
into field function.
Here, k is constant.
Solution The potential function is −k
x
. Derivating this relation to convert
it into field function
d
dx
−
k
x
= −k
d
dx
x−1
= −k
−1 × x−2
=
k
x2
So, k/x2
is corresponding field function.
Solved Problem 1.52 Convert the field function k
x2 into potential function.
Here, k is constant.
Solution The field function is k
x2 . Integrate it to convert it into potential
function
Z
k
x2
dx = k
Z
x−2
dx
= k
x−2+1
−2 + 1
= k
x−1
−1
= −
k
x
So, −k/x is corresponding potential function.
1.3.3 Algebraic Properties
The integral of functions also follows the algebraic proprieties, such as com-
mutative, distributive etc.
Commutativity Integral of product of two functions is commutative. For
example, f(x) and g(x) are two functions of x, and their product is f(x) ×
g(x). Now, Z
f(x) × g(x) dx =
Z
g(x) × f(x) dx

111. 1.3. DIRECT INTEGRATION 59
Distributivity Integral is also distributive. If f(x), g(x) and h(x) are
three functions then
Z
f(x) × [g(x) + h(x)] dx =
Z
[f(x) × g(x) + f(x) × h(x)] dx
1.3.4 Convolution
We know that, if f(x) and g(x) are two functions of x, and F(x) and G(x) are
individual integral of these functions respectively, such that F(x) =
R
f(x)dx
and G(x) =
R
g(x)dx. Now
Z
f(x) ± g(x) dx =
Z
f(x) dx ±
Z
g(x) dx = F(x) ± G(x)
i.e. integral of sum of two functions is equal to the sum of their individ-
ual integrals. This is true for sum and difference butit is not for product,
i.e. integral of product of two function is not equal to the product of their
individual integrals. For example, take f(x) = x and g(x) = x then
F(x) =
Z
f(x)dx =
Z
x dx =
x2
2
G(x) =
Z
g(x)dx =
Z
x dx =
x2
2
So, Z
[f(x) + g(x)] dx =
Z
2x dx = x2
= F(x) + G(x)
In case of product, F(x) × G(x) = x4
4
. Now
Z
[f(x) × g(x)] dx =
Z
x2
dx =
x3
3
6= F(x) × G(x)
A question is rises about the finding of solution of this problem. There is a
method to solve this problem and this is called convolution. It is given by
x
Z
0
f(t)g(x − t) dt, 0 ≤ x ∞ (1.34)
Convolution of two function is represented by ‘*’. Note that, its result are
not simple but are complex.

112. 60 Integration
1.4 Integral Rules
Before going to discuss the rule of integration, we must remember that the
integrand should be (i) linear, (ii) have lowest powers and (iii) as possible as
simplified (i.e. as many terms as possible). If integrand is complex, it must
be simplified. For example,
I =
Z
sin2
x dx
must be simplified to
I =
Z
1 − cos 2x
2
dx
as it is simplified form, having power one, and integrand has two terms.
1.4.1 Constant of Integration
We know that The integral is a process of retrieving a function after its
differentiation. Let a function is f(x) which contains variables as well as
constant terms and it is
f(x) = ax2
+ bx + c
Now it is differentiated with respect to x and function becomes
f
′
(x) = 2ax + b
If we want to retrieve the origin function then it is to be integrated with
respect to same variable x. Hence
Z
f
′
(x) dx =
Z
(2ax + b) dx
= 2a
x2
2
+ bx
= ax2
+ bx
The function is retrieved without constant terms. This means in general dif-
ferentiation and then integration, constant terms of a function is eliminated.
This is why to find the actual function we add a constant in the result
after integration. The constant term is calculated by using boundary value
conditions (In Chapter Application of Integration).

113. 1.4. INTEGRAL RULES 61
1.4.2 Integration By Parts
If two functions f(x) and g(x) are in product, then the integration of product
function is given by
Z
f(x) · g(x) dx = f(x)
Z
g(x) dx −
Z
d
dx
f(x)
×
Z
g(x) dx
dx
(1.35)
The above statement is read as integration of product of two function
is equal to the “first function, multiply by integration of second
function minus of integration of product of differentiation of first
function and integration of second function. It is also noted that,
function whose derivative reduces its power is taken as first function.
Note In product rule or integration by parts, it is second part of right
hand side of equation (1.35) that influences the factors for assumption of
terms either as first function or as second function. This selection may be
driven such way that product of derivative of first function and integral of
second function reduced to minimal terms and powers.
Solved Problem 1.53 Find the integral of x sin(x).
Solution Applying integration by parts method, assuming x as first func-
tion and sin x second function. Now applying the part integration
Z
x sin x dx = x
Z
sin x dx −
Z
d
dx
x ×
Z
sin x dx
dx
= −x cos x −
Z
[− cos x] dx
= −x cos x + sin x Ans.
Solved Problem 1.54 Find the integral of x ln(x).

114. 62 Integration
Solution Applying integration by parts method, assuming ln x as first
function and x as second function. Now applying the part integration
Z
x ln x dx = ln x
Z
x dx −
Z
d
dx
ln x ×
Z
x dx
dx
=
x2
2
ln x −
Z
1
x
x2
2
dx
=
x2
2
ln x −
Z hx
2
i
dx
=
x2
2
ln x −
x2
4
Ans.
Solved Problem 1.55 Find the integral of sin(x) cos(x).
Solution We know that in product rule of integration, first function of
integral is derivated while second function is integrated. If derivative of first
function is equal to second function and integral of second function is equal to
first function then integration process becomes a periodic process. In other
words, there are infinite steps of integration. In these types of problems,
integral is assumed as I. For given problem
I =
Z
sin x cos x dx (1.36)
Now integrating above relation
Z
sin x cos x dx = sin x
Z
cos x dx −
Z
d
dx
sin x ×
Z
cos x dx
dx
= sin x × sin x −
Z
[cos x × sin x] dx
Second part in above relation is similar to the equation (1.36) hence
I = sin x × sin x − I
On simplification
2I = sin2
x
Or
I =
sin2
x
2

115. 1.4. INTEGRAL RULES 63
Corollary: If first and second terms are interchanged then
I = −
cos2
x
2
Solved Problem 1.56 Find the integral of sin2
(x).
Solution We know that, second part in right hand side of the part inte-
gration, first function is differentiated and second function is integrated. If
first and second function are the mutual differential or integral values then
second part of part integral becomes periodic function. In other words, there
are infinite steps of integration. To over come this problem we assume inte-
gration as I and applying integration by part method
I =
Z
sin x sin x dx (1.37)
Now integrating above relation
Z
sin x sin x dx = sin x
Z
sin x dx −
Z
d
dx
sin x ×
Z
sin x dx
dx
= − sin x × cos x +
Z
[cos x × cos x] dx
= − sin x × cos x +
Z
[1 − sin x × sin x] dx
= − sin x × cos x +
Z
1 dx −
Z
sin x × sin x dx
Second part in above relation is similar to the equation (1.37) hence
I = sin x × cos x + x − I
Applying trigonometric identity sin x cos x = sin 2x
2
2I = x −
sin 2x
2
Or
I =
2x − sin 2x
4

116. 64 Integration
Solved Problem 1.57 Find the integral of cos2
(x).
Solution We know that, second part in right hand side of the part inte-
gration, first function is differentiated and second function is integrated. If
first and second function are the mutual differential or integral values then
second part of part integral becomes periodic function. In other words, there
are infinite steps of integration. To over come this problem we assume inte-
gration as I and applying integration by part method
I =
Z
cos x cos x dx (1.38)
Now integrating above relation
Z
cos x cos x dx = cos x
Z
cos x dx −
Z
d
dx
cos x ×
Z
cos x dx
dx
= cos x × sin x −
Z
[− sin x × sin x] dx
= cos x × sin x +
Z
[1 − cos x × cos x] dx
= cos x × sin x +
Z
1 dx −
Z
cos x × cos x dx
Second part in above relation is similar to the equation (1.37) hence
I = cos x × sin x + x − I
Applying trigonometric identity sin x cos x = sin 2x
2
2I = x +
sin 2x
2
Or
I =
2x + sin 2x
4
Solved Problem 1.58 Find the integral of x2
ex
.

117. 1.4. INTEGRAL RULES 65
Solution Applying integration by parts method, assuming x2
as first
function and ex
as second function. Now applying the part integration
Z
x2
ex
dx = x2
Z
ex
dx −
Z
d
dx
x2
×
Z
ex
dx
dx
= x2
ex
−
Z
[2x ex
] dx
= x2
ex
− 2
Z
[x ex
] dx
Taking product rule of the integration in right hand side
Z
x2
ex
dx = x2
ex
− 2
x
Z
ex
dx −
Z
d
dx
x
Z
ex
dx
dx
= x2
ex
− 2
x
Z
ex
dx −
Z
(ex
) dx
= x2
ex
− 2 [xex
− ex
]
Now Z
x2
ex
dx = ex
(x2
− 2x + 2)
This is required result.
Solved Problem 1.59 Evaluate
R
k sin k dk.
Solution This integral is about variable parameter k. So, variables other
than k shall be treated as constant. Now, the integral function k sin k is
product of two terms, say k (1st
part) and sin k (2nd
part). Now, applying
product rule, we have
I =
Z
k sin k dk = k
Z
sin k dk −
Z
d
dk
k ×
Z
sin k dk
dk
Or
I = k × − cos k −
Z
[− cos k] dk
Or
I = −k cos k + sin k + C
This is desired answer.

118. 66 Integration
Solved Problem 1.60 Evaluate
R
k log k dk.
Solution This integral is about variable parameter k. So, variables other
than k shall be treated as constant. Now, the integral function k log k is
product of two terms, say log k (1st
part) and k (2nd
part). Here, log k is
taken as first part as its derivative is convergent for each |k| ≥ 1 and there is
twice integral in second part of the product rule. If it is taken as second part,
integral steps shall be increased as well as unnecessary calculations. Now,
applying product rule, we have
I =
Z
k log k dk = log k
Z
k dk −
Z
d
dk
log k ×
Z
k dk
dk
Or
I =
k2
2
× log k −
Z
1
k
×
k2
2
dk
Or
I =
k2
log k
2
−
k2
4
+ C
This is desired answer.
Integral Under Derivative or Integral Sign If a function contains either
derivative part or integral part or both, even then we can utilize product rule
of integration, to find its integral as this rule do integration of one part and
derivation of other part. Thus derivative part can be reduced by integrating
it while integral part can be reduced by differentiating it.
Solved Problem 1.61 Find the integration of function F(x) = df(x)
dx
× x.
Solution The product rule of integration is
Z
f(x) · g(x) dx = f(x)
Z
g(x) dx −
Z
d
dx
f(x)
×
Z
g(x) dx
dx
As we know that, integral of derivative of function is function itself. So,
Z
df(x)
dx
dx = f(x)
So, derivative term would be taken as second part. Now,
Z
df(x)
dx
× x dx = x
Z
df(x)
dx
dx −
Z
d
dx
x
×
Z
df(x)
dx
dx
dx

119. 1.4. INTEGRAL RULES 67
On solving it, we have
Z
df(x)
dx
× x dx = x f(x) −
Z
f(x) dx
Now, we can solve it if we know f(x).
Solved Problem 1.62 Find the integration of function F(x) =
R
f(x)dx
×x.
Solution The product rule of integration is
Z
f(x) · g(x) dx = f(x)
Z
g(x) dx −
Z
d
dx
f(x)
×
Z
g(x) dx
dx
As we know that, derivative of integral function is function itself. So,
d
dx
Z
f(x)dx
= f(x)
So, integral term would be taken as first part. Now,
Z Z
f(x) dx
× x
dx =
Z
f(x) dx
×
Z
x dx
−
Z
d
dx
Z
f(x) dx
×
Z
x dx
dx (1.39)
On solving it, we have
Z Z
f(x) dx
× x
dx =
Z
f(x) dx
×
x2
2
−
Z
f(x) ×
x2
2
dx
Now, we can solve it if we know f(x).
Periodicity Under Integral Sign
We knew that functions like sin x, cos x, ex
etc. are periodic functions under
the sign of integral and under the sign of derivative. For example,
I =
Z
ex
dx =
Z Z
ex
dx
dx = ex

120. 68 Integration
If I(n)
represents the n-times integral of a function, then
I(n)
[ex
] = ex
Similarly, if d/dx = D then D [ex
] = ex
and D(n)
[ex
] = ex
. So, if two or more
than two periodic functions are in product under the sign of integral, then
how do we identify about which is first function and which is second function,
as there are both derivative and integral in second part of product rule. The
solution of this problem is recovery of original function from second part of
product rule. See the example below:
Illustrated Example Let we have
R
e−x
sin x dx. Here, both e−x
and sin x
are periodic function. Hence, we shall take it as I. So,
I =
Z
e−x
sin x dx
We can take either of the two as first function or second function. In our
convenience, we shall take e−x
as first function. Now, on solving it
I = e−x
Z
sin x dx −
Z
d
dx
e−x
×
Z
sin x dx
dx
Here, we shall recover original I from second part of the above equation.
Now,
I = e−x
× − cos x −
Z
e−x
× −1 × − cos x
dx
Or
I = e−x
× − cos x −
Z
e−x
× −1 × − cos x
dx
I = −e−x
cos x −
Z
e−x
cos x dx
Again solving second part of above result
I = −e−x
cos x −
e−x
Z
cos x dx −
Z
d
dx
e−x
×
Z
cos x dx
dx
Or
I = −e−x
cos x − e−x
sin x +
Z
−e−x
× sin x
dx + C

121. 1.4. INTEGRAL RULES 69
We have recovered original I in above result as third term is equal to original
I. Now,
I = −e−x
cos x − e−x
sin x − I + C
Or
2I = −e−x
[cos x + sin x] + C
Or solving it, we get integral of the given integral.
I =
−e−x
[cos x + sin x]
2
+ C
This is desired result.
Solved Problem 1.63 Evaluate
R
sin 3k sin k dk.
Solution This integral has product of two periodic function, i.e. sin 3k
and sin k. There are two methods of solution of this integral. First method
is by recovering the original integral and second method is applying trigono-
metric conversion from product function to sum function. Second method is
simple and short. So,
I =
Z
sin 3k sin k dk
I =
Z
1
2
[2 sin 3k sin k] dk =
1
2
Z
[cos (3k − k) − cos (3k + k)] dk
Or
I =
1
2
Z
[cos 2k − cos 4k] dk
Or
I =
1
2
1
2
sin 2k −
1
4
sin 4k
+ C
Or
I =
1
4
sin 2k −
1
8
sin 4k + C
This is desired result.

122. 70 Integration
Solved Problem 1.64 Evaluate
R
cos x cos 3x dx.
Solution This integral has product of two periodic function, i.e. cos k and
cos 3k. There are two methods of solution of this integral. First method is by
recovering the original integral and second method is applying trigonometric
conversion from product function to sum function. Second method is simple
and short. So,
I =
Z
cos x cos 3x dk
I =
Z
1
2
[2 cos x cos 3x] dk =
1
2
Z
[cos (3k − k) + cos (3k + k)] dk
Or
I =
1
2
Z
[cos 2k + cos 4k] dk
Or
I =
1
2
1
2
sin 2k +
1
4
sin 4k
+ C
Or
I =
1
4
sin 2k +
1
8
sin 4k + C
This is desired result.
Solved Problem 1.65 Evaluate
R
ek
k2
dk.
Solution The given integral is
I =
Z
ek
k2
dk
Take k2
as first function and ek
as second function. We can not take ek
as
first function as it is periodic function and on integration of second function
k2
shall increase its degree by one. Thus second part of integral on these
assumptions, i.e.
R d
dk
ek
R
k2
dk
dk become divergent. Applying integral by
parts method taking k2
as first function, we have
I = k2
Z
ek
dk −
Z
d
dk
k2
Z
ek
dk
dk
Or
I = k2
ek
−
Z
2kek
dk

123. 1.4. INTEGRAL RULES 71
Again applying integral by part in second part of right side of above relation.
We have
I = k2
ek
−
2k
Z
ek
dk −
Z
d
dk
2k
Z
ek
dk
dk
Or
I = k2
ek
−
2kek
−
Z
2ek
dk
= k2
ek
−
2kek
− 2ek
It gives
I = k2
ek
− 2kek
+ 2ek
= (k2
− 2k + 2)ek
This is desired result. This is very important integral. It must be memorise
by the learners in standard form as given below:
Z
ek
kn
dk = kn
− nkn−1
+ n(n − 1)kn−2
− n(n − 1)(n − 2)kn−3
+ . . .
ek
Solved Problem 1.66 Evaluate
R
xπx
dx.
Solution The given integral is
I =
Z
xπx
dx
Take x as first function and πx
as second function. We have
I = x
Z
πx
dx −
Z
d
dx
x
Z
πx
dx
dx
Or
I = x
πx
ln π
−
Z
πx
ln π
dx
Or
I = x
πx
ln π
−
πx
(ln π)2
=
x
ln π
−
1
(ln π)2
πx
This is desired result.

124. 72 Integration
Solved Problem 1.67 Evaluate
R
xe2ix
dx. Here,
√
−1 = i.
Solution The given integral is
I =
Z
xe2ix
dx
Take x as first function and e2ix
as second function. We have
I = x
Z
e2ix
dx −
Z
d
dx
x
Z
e2ix
dx
dx
Or
I = x
e2ix
2i
−
Z
e2ix
2i
dx
Or
I = x
e2ix
2i
−
e2ix
4i2
= −x
e2ix
2
i +
e2ix
4
Here, i2
= −1. Again
I =
1
4
−
x
2
i
e2ix
This is desired result.
Solved Problem 1.68 Evaluate
R
cot3
k dk.
Solution The given integrand is cot3
k. To integrate it, we must simplified
it by trigonometric canonical form as
Z
cot3
k dk =
Z
cos(3k) + 3 cos k
4
dk
Or Z
cot3
k dk =
1
4
Z
cos(3k) dk +
Z
3 cos k dk
On solving right hand side and simplify it, we get
Z
cot3
k dk =
sin(3k)
3
+ 3 sin k
4
+ C
This is desired result.

125. 1.4. INTEGRAL RULES 73
Solved Problem 1.69 Evaluate
R
sec3
x dx.
Solution We have given
I =
Z
sec3
x dx =
Z
sec x × sec2
x dx
Integrating by parts, taking sec x as first function and sec2
x as second func-
tion.
I = sec x
Z
sec2
x dx −
Z
d
dx
sec x ×
Z
sec2
x dx
dx
I = sec x tan x −
Z
sec x tan2
x dx
Expanding second term of right hand side in terms of sec x.
I = sec x tan x −
Z
sec3
x dx +
Z
sec x dx
Second term is I. So,
I = sec x tan x − I + ln(| sec x + tan x|)
Or
I =
1
2
sec x tan x +
1
2
ln(| sec x + tan x|)
Solved Problem 1.70 Evaluate the integration of
R log y
y2 dy,
R log y
y3 dy,
R log y
y4 dy.
Find the pattern of result and using the pattern Find the integral of
R log y
yn dy,
where n is a non-zero integer and greater than 2.
Solution Integral is
I1 =
Z
ln y
y2
dy
Integrating by parts method
I1 = ln y
Z
1
y2
dy −
Z
d
dy
ln y ×
Z
1
y2
dy
dy
On solving it
I1 = −
ln y
y
−
1
y

126. 74 Integration
Similarly, for second relation
I2 =
Z
ln y
y3
dy
Integrating by parts method
I2 = ln y
Z
1
y3
dy −
Z
d
dy
ln y ×
Z
1
y3
dy
dy
On solving it
I2 = −
ln y
2y2
−
1
4y2
And for third relation
I3 =
Z
ln y
y4
dy
Integrating by parts method
I3 = ln y
Z
1
y4
dy −
Z
d
dy
ln y ×
Z
1
y4
dy
dy
On solving it
I3 = −
ln y
3y3
−
1
9y3
By successive comparison of I1, I2 and I3, In be
In =
Z
log y
yn
dy = −
ln y
(n − 1)y(n−1)
−
1
(n − 1)2y(n−1)
1.4.3 Chain Rule
Proof A function f of variable x is defined as f(ax). Integration of the
function is
F(x) =
Z
f(ax) dx
Changing the variable ax to y, as
d
dx
(ax) =
d
dx
y

127. 1.4. INTEGRAL RULES 75
It can be rearranged like
dx =
dy
d(ax)
dx
This dx is placed in the integral relation
F(x) =
Z
f(y)
dy
d(ax)
dx
Or
F(x) =
R
f(y) dy
d(ax)
dx
It is chain rule of integration.
Illustrated Example An integrand of a variable can be easily obtained by
direct method if we knew that it is derivative of other function. For example,
if
d
dx
sin x = cos x
Or
d[sin x] = cos x dx
Applying integration sign both side, we have
Z
d sin x =
Z
cos x dx
Left side is integral of 1 about sin x, so we have
Z
cos x dx = sin x + C
Similarly, Z
sin(x) dx = − cos x + C
Here sin(x) is function of x. Assume another function sin(ax) which has
argument ax. If it is integrated with respect to x then chain rule is applied.
In this rule, argument (ax) is assumed as single group with a as constant

128. 76 Integration
and x as variable. Its integral result about ax is divided by the derivative of
the whole group (ax). i.e.
R
sin(ax) dx =
R
[sin(ax)] dx × d(ax)
d(ax)
=
R
[sin(ax)] d(ax) × dx
d(ax)
=
R
[sin(ax)] d(ax) × 1
[d(ax)
dx ]
Or Z
sin(ax) dx =
R
[sin(ax)] d(ax)
d
dx
(ax)
Or Z
sin(ax) dx =
− cos(ax)
a
Note that this rule is valid if denominator term at right hand side is either
constant or one. If it is not constant then chain rule can not be applied. For
example, Z
sin(ax2
) dx 6=
− cos(ax2
)
a d
dx
x2
This is due that d
dx
x2
= 2x is neither constant nor one. Similarly,
Z
sin2
x dx 6=
sin2+1
x
2 + 1
×
1
d
dx
sin x
is invalid, as d
dx
sin x = cos x is neither constant nor one.
Solved Problem 1.71 Find the integral of sin(9x).
Solution We have I =
R
sin(9x) dx. Converting the base of integration,
we have
I =
R
sin(9x) d(9x)
d
dx
9x
Or simply
I = −
cos(9x)
9
This is required result.

129. 1.4. INTEGRAL RULES 77
Solved Problem 1.72 Find the integral of tan(4x).
Solution We have I =
R
tan(4x) dx. Converting the base of integration,
we have
I =
R
tan(4x) d(4x)
d
dx
4x
Or simply
I = −
loge(cos(4x))
4
This is required result.
Solved Problem 1.73 Find the integral of loge(4x).
Solution We have I =
R
ln(4x) dx. Converting the base of integration,
we have
I =
R
ln(4x) d(4x)
d
dx
4x
Or simply
I = x ln(4x) − x
This is required result.
1.4.4 Integration By Substitution
In substitution integration a part of function is replaced by another variable.
This variable reduces the function, easy to integrate. The following integra-
tion explains the substitution integration in easy way. Substitution method
is used in those relations in which differentiation of one function or part is
equal to other function or part.
Sign Problem Some time we take common of negative sign from a func-
tion to make function convenient for integration. But remember it does not
work each time. To understand this we illustrate integration of function
1/(1 − x). The integral is
Z
1
1 − x
dx = − log(1 − x)

130. 78 Integration
Now if negative sign is taken out in common from denominator then inte-
gration function becomes
−
Z
1
x − 1
dx = − log(x − 1)
The both integration have not unique solutions, hence it is prevented to
take negative sign common when integration of a function is logarithm
function.
Solved Problem 1.74 Find the integral of sin(ax).
Solution The integral is
Z
sin(ax) dx
Substituting ax = t which gives a dx = dt on differentiation and relation
becomes Z
sin(ax) dx =
Z
sin t
dt
a
And
Z
sin t
dt
a
=
1
a
Z
sin t dt
=
1
a
(− cos t)
Substituting the value of t and dt in relation
Z
sin t
dt
a
=
1
a
(− cos t)
Z
sin(ax) dx = −
1
a
cos(ax)
Solved Problem 1.75 Find the integral of log(2x).
Solution The integral is
I =
Z
log(2x) dx

131. 1.4. INTEGRAL RULES 79
Substituting 2x = t which gives 2 dx = dt and integrand becomes
I =
Z
log(t)
dt
2
And
I =
1
2
Z
log t dt
=
1
2
[t log(t) − t]
Replacing t from x according to the relation t = 2x
I =
1
2
[2x log(2x) − 2x]
Solved Problem 1.76 Find the integral of sin(x) sin(cos(x)).
Solution The integral is
I =
Z
sin(x) sin[cos(x)] dx
Substituting cos(x) = t which gives − sin(x) dx = dt and integrand becomes
I = −
Z
sin(t) dt
= cos(t)
Substituting the value of t in right hand side of above relation
I = cos[cos(x)]
Solved Problem 1.77 Find the integral of
tan(1
x )
x2 .
Solution The integral is
Z
tan 1
x
x2
dx

132. 80 Integration
Substituting 1
x
= t which gives − 1
x2 dx = dt and integrand becomes
Z
tan(1
x
)
x2
dx = −
Z
tan(t) dt
= − [− ln(cos t)]
= ln(cos t)
Substituting the value of t in right hand side of above relation
Z
tan(1
x
)
x2
dx = ln
cos
1
x
Solved Problem 1.78 Find the integral of ex
sin2
(ex
).
Solution The integral is
Z
ex
sin2
(ex
) dx
Substituting ex
= t which gives ex
dx = dt and integrand becomes
Z
ex
sin2
(ex
) dx =
Z
sin2
(t) dt
=
2t − sin 2t
4
Substituting the value of t in right hand side of above relation
Z
ex
sin2
(ex
) dx =
2ex
− sin(2ex
)
4
Solved Problem 1.79 Find the integral of ln x sin(ln x)
x
.
Solution The integral is
Z
ln x sin(ln x)
x
dx

133. 1.4. INTEGRAL RULES 81
Substituting ln(x) = t which gives 1
x
dx = dt and integrand becomes
Z
ln x sin(ln x)
x
dx =
Z
t sin(t) dt
= −t cos t + sin t
Substituting the value of t in right hand side of above relation
Z
ln x sin(ln x)
x
dx = − ln x cos(ln x) + sin(ln x)
Solved Problem 1.80 Find the integral of sin(t) ecos(t)
.
Solution The integral is
Z
sin(t) ecos(t)
dt
Substituting cos(t) = x which gives − sin t dt = dx and integrand becomes
Z
sin(t)ecos(t)
dt = −
Z
ex
dx
= −ex
Substituting the value of x in right hand side of above relation
Z
sin(t)ecos(t)
dt = −ecos t
Solved Problem 1.81 Find the integral of 2x ex2
.
Solution The integral is
Z
2x ex2
dx
Substituting x2
= t which gives 2x dx = dt and integrand becomes
Z
2x ex2
dx =
Z
et
dt
= et

134. 82 Integration
Substituting the value of t in right hand side of above relation
Z
2x ex2
dx = ex2
This is required answer.
Solved Problem 1.82 Evaluate
R
sin2
t cos3
t dt.
Solution In integration, power of a function increases by one. For exam-
ple, Z
x dx =
x2
2
Similarly, integral of line is area and integral of area is volume etc. If function
is a polynomial equation then integral increases its degree by one. So, in
this problem, integral becomes divergent as after each step, degree of either
sin t or cos t will be raised by one. It will increase the number of steps and
calculations for problem solution. Therefore, we shall first simplify it by
using trigonometric relations, so that product of functions is reduced into
sum or difference of two functions or reducible to substitution.
I =
Z
sin2
t cos3
t dt =
Z
sin2
t cos t × (1 − sin2
t) dt
Substitute sin t = k, on differentiation it gives, cos t dt = dk. Now, the above
integral becomes
I =
Z
k2
(1 − k2
) dk =
Z
(k2
− k4
)dk =
k3
3
−
k5
5
+ C
Replace k by sin t, we have
I =
sin3
t
3
−
sin5
t
5
+ C
This is desired result.
Solved Problem 1.83 Evaluate
R
cos3
t sin4
t dt.
Solution In integration, power of a function increases by one. For exam-
ple, Z
x dx =
x2
2

135. 1.4. INTEGRAL RULES 83
Similarly, integral of line is area and integral of area is volume etc. If function
is a polynomial equation then integral increases its degree by one. So, in
this problem, integral becomes divergent as after each step, degree of either
sin t or cos t will be raised by one. It will increase the number of steps and
calculations for problem solution. Therefore, we shall first simplify it by
using trigonometric relations, so that product of functions is reduced into
sum or difference of two functions or reducible to substitution.
I =
Z
cos3
t sin4
t dt =
Z
sin4
t cos t × (1 − sin2
t) dt
Substitute sin t = k, on differentiation it gives, cos t dt = dk. Now, the above
integral becomes
I =
Z
k4
(1 − k2
) dk =
Z
(k4
− k6
)dk =
k5
5
−
k7
7
+ C
Replace k by sin t, we have
I =
sin5
t
5
−
sin7
t
7
+ C
This is desired result.
Solved Problem 1.84 Evaluate
R 1
k ln k
dk.
Solution The given integral is
I =
Z
1
k ln k
dk
Substitute ln k = t. On derivating, it gives, dk/k = dt. Now, the integral
becomes
It =
Z
1
t
dt = ln t
Replacing t, we have
I = ln(ln k)
This is desired result.

136. 84 Integration
Solved Problem 1.85 Evaluate
R sin
√
k
√
k
dk.
Solution The given integral is
I =
Z
sin
√
k
√
k
dk
Substitute
√
k = t. On derivating, it gives, dk/
√
k = 2 dt. Now, the integral
becomes
It =
Z
sin t × 2 dt = −2 cos t + c
Replacing t, we have
I = −2 cos
√
k + c
This is desired result.
Solved Problem 1.86 Evaluate
R x2
1+x3 dx.
Solution The given integral is
I =
Z
x2
1 + x3
dx
Substitute 1 + x3
= t. On derivating it, we have x2
dx = dt/3. The integral
becomes
It =
1
3
×
Z
1
t
dt
This gives,
It =
1
3
ln t = ln
3
√
t
Substituting t, we have
I = ln
3
√
1 + x3
This is desired result.
Solved Problem 1.87 Evaluate
R
k e−k2
dk.
Solution The given integral is
I =
Z
k e−k2
dk

137. 1.4. INTEGRAL RULES 85
Substitute k2
= t. On derivating it, we have k dk = dt/2. The integral
becomes
It =
1
2
×
Z
e−t
dt
This gives,
It = −
1
2
e−t
=
−e−t
2
Substituting t, we have
I =
−e−k2
2
This is desired result.
Solved Problem 1.88 Evaluate
R log x
x
dx.
Solution The given integral is
I =
Z
log x
x
dx
Substitute log x = t. On derivating it, we have dx/x = dt. The integral
becomes
It =
Z
t dt
This gives,
It =
t2
2
Substituting t, we have
I =
(log x)2
2
This is desired result.
Solved Problem 1.89 Evaluate
R
xn−1
e−xn
dx.
Solution The given integral is
I =
Z
xn−1
e−xn
dx
Substitute xn
= t. On derivating it, we have x(n−1)
dx = dt/n. The integral
becomes
It =
1
n
×
Z
e−t
dt

138. 86 Integration
This gives,
It = −
1
n
e−t
=
−e−t
n
Substituting t, we have
I =
−e−xn
n
This is desired result.
Solved Problem 1.90 Evaluate
R
sin(tan t) sec2
t dt.
Solution The given integral is
I =
Z
sin(tan t) sec2
t dt
Substitute tan t = k. On derivating it, we have sec2
t dt = dk. The integral
becomes
Ik =
Z
sin(k) dk
This gives,
Ik = − cos(k)
Substituting k, we have
I = − cos (tan t)
This is desired result.
Solved Problem 1.91 Evaluate
R tan−1 x
1+x2 dx.
Solution The given integral is
Ix =
Z
tan−1
x
1 + x2
dx
Substitute tan−1
x = t. It gives dx/(1 + x2
) = dt. This simplify the integral
as
It =
Z
t dt =
t2
2
Replacing t, we have
Ix =
(tan−1
x)
2
2
This is desired result.

139. 1.4. INTEGRAL RULES 87
Solved Problem 1.92 Evaluate
R tan(log k)
k
dk.
Solution The given integral is
Ik =
Z
tan(log k)
k
dk
Substitute log k = t. It gives dk/k = dt. This simplify the integral as
It =
Z
tan t dt = ln | sec t|
Replacing t, we have
Ik = ln | sec(ln k)|
This is desired result.
Solved Problem 1.93 Evaluate
R x
x2+a2 dx.
Solution The given integral is
Ix =
Z
x
x2 + a2
dx
Substitute x2
+ a2
= t. It gives 2x dx = dt. This simplify the integral as
It =
Z
1
t
dt
2
=
1
2
ln t = ln
√
t
Replacing t, we have
Ix = ln
√
x2 + a2
This is desired result.
Solved Problem 1.94 Evaluate
R
sin x cos t dt.
Solution The integral base is t hence sin x is constant. So,
I =
Z
sin x cos t dt = sin x
Z
cos t dt = sin x × sin t
This gives I = sin x sin t. This is result.

140. 88 Integration
Solved Problem 1.95 Evaluate
R c
1+c2 dx.
Solution The integral base is t hence all terms having variable c are
constant. Hence integral is
I =
Z
c
1 + c2
dx =
c
1 + c2
Z
dx
This gives integral as
I =
c
1 + c2
x
This is desired result.
Solved Problem 1.96 Evaluate
R
tan x sin t dt.
Solution The integral base is t hence tan x is constant. So,
I = tan x
Z
sin t dt = tan x × − cos t
This gives I = − tan x cos t. This is result.
Solved Problem 1.97 Given f
′′
(x) = x2
− 4. If f
′
(0) = 0 and f(1) = 1 then
find the function f(x).
Solution We have given f
′′
(x) = x2
− 4. Now, integrating this function
with respect to x, we get
Z
f
′′
(x) dx =
Z
(x2
− 4) dx
Or
f
′
(x) =
x3
3
− 4x + C
Here, C in an integral constant. Applying the given conditions in relation
f
′
(x), we get
f
′
(0) = 0 = 0 − 0 + C ⇒ C = 0
So, the function becomes,
f
′
(x) =
x3
3
− 4x

141. 1.4. INTEGRAL RULES 89
Now, again integrating it
Z
f
′
(x) dx =
Z
x3
3
− 4x
dx
Or
f(x) =
x4
12
− 2x2
+ D
Here, D in an integral constant. Applying the given conditions in relation
f(x), we get
f(1) = 1 =
1
12
− 2 + D ⇒ D =
35
12
Now, the function becomes
f(x) =
x4
12
− 2x2
+
35
12
This is the given function.
Solved Problem 1.98 Given f
′′
(t) = sin t − t. If f
′
(π/2) = 0 and f(π/4) = 0
then find the function f(t).
Solution We have given f
′′
(t) = sin t − t. Now, integrating this function
with respect to t, we get
Z
f
′′
(t) dt =
Z
(sin t − t) dt
Or
f
′
(t) = − cos t −
t2
2
+ C
Here, C in an integral constant. Applying the given conditions in relation
f
′
(t), we get
f
′
(π/2) = 0 = − cos(π/2) −
π2
8
+ C ⇒ C =
π2
8
So, the function becomes,
f
′
(t) = − cos t −
t2
2
+
π2
8

142. 90 Integration
Now, again integrating it
Z
f
′
(t) dt =
Z
− cos t −
t2
2
+
π2
8
dt
Or
f(t) = − sin t −
t3
6
+
π2
8
t + D
Here, D in an integral constant. Applying the given conditions in relation
f(t), we get
f(π/4) = 0 =
1
12
− 2 + D ⇒ D =
1
√
2
−
11π3
384
Now, the function becomes
f(t) = − sin t −
t3
6
+
π2
8
t +
1
√
2
−
11π3
384
This is the given function.