Integral Calculus
A mathematical description of motion motivated
the creation of Calculus.
Problem of Motion:
Given x(t) find v(t) :
Differential Calculus.
Given v(t) find x(t) :
Integral Calculus.
Derivatives and integrals are operations on
functions.
One is the inverse of the other. This is the content
of the Fundamental theorem of Calculus.
Integral calculus is mainly due to the contributions from the
following well known mathematicians.(The photographs are worth
watching since these names will appear many times in the courses to
follow.)
Isaac Newton
Gottfried Leibniz
James Gregory
Pierre de Fermat
Joseph Fourier
Cauchy
Bernhard Riemann
Henri Lebesgue
Some motivations:
1. Suspension bridges
The road deck hangs on vertical cables suspended from the
main cables.
Problem : We have to find the optimal shape of the main
cable.
Mathematical description (Model):
Find the curve y = y(x) such that the derivative
of this function satisfies y' = µx. ( where
µ = t g / T ; t is density ; T is tension which can be computed.)
Solution: This is the basic problem of integral calculus and we
solve the problem by integration.
y(x) =  y′(x) dx =  μx dx = μ (x2/2) + C.
The main cable has a parabolic shape.
2. Reduction formulae are useful to compute the following:
REDUCTION FORMULAE
Reduction formula for  sinn x dx where n is a positive integer.
Let In =  sinnx dx
=  sinn-1 x.sin x. dx =  u v dx (say)
We know that  uv dx = u ( v dx) -  ( v dx ) u1 dx
In = sinn-1 x (-cos x) -  (-cos x) (n – 1) sinn-2 x. cos x dx
= - sinn-1 x cos x + ( n – 1)  sinn-2 x.cos2 x dx
= - sinn-1 x cos x + (n – 1)  sinn-2 x (1 – sin2 x) dx
= - sinn-1 x cos x + (n – 1)  sinn-2 x dx – (n – 1) In
In [1 + (n – 1)] = - sinn-1 x cos x + ( n – 1) In-2
Therefore In = 
sinn
x dx =
- sin n- 1 x cos x
+ n - 1 In-2 ...(1)
n
n
(1) is the required reduction formula.
Illustration (i): To find  sin4 x dx.
I4 = 
sin4
x dx =
- sin 3 x cos x 3
+
I2
4
4
We need to apply the result (1) again by taking n = 2
That is, I4 =
- sin3 x cos x 3 - sin x cos x 1
+
+ I0
4
4
2
2
{
I0 =  sin0 x dx =  1 dx = x
Thus I4 =  sin4 x dx =
- sin3 x cos x 3
3
- sin x cos x + x + c
4
8
8
}
Illustration (ii): To find  sin5 x dx
Solution: I5 = 
sin5
4
sin
x cos x 4
x dx =
+ I3
5
5
- sin 4 x cos x
=
+
5
But I1 =  sin1 x dx = - cos x.
2
4м
sin
x cos x 2 ьп
п
п
+ I1 п
н
э
п
5о
3
3 п
п
п
ю
Corollary : To evaluate I n =
т
p /2
sin n x dx
0
p /2
From (1) , In
=
йsin m- 1 x cos x щ
n- 1
ъ +
- к
I
кл
ъ0
n
n
ы
But cos /2 = 0 = sin 0.
Thus In
n- 1
I n-2
=
n
n-2
Now, In-2 = n - 3 I n - 4
n- 2
In = n - 1 n - 3 . I n-4 , by back substitution.
n- 2
n
Continuing the process we get:
In =
{
But I1 =
n- 1 n- 3 n- 5
..
n n- 2 n- 4
n- 1 n- 3 n- 5
..
n n- 2 n- 4
т
0
and I0 =
p /2
.
2 I
1
3
if n is odd.
1 I
0
2
if n is even.
.
sin x dx = - [cos x]0/2 = - (0 – 1) = 1
т
0
p /2
p
sin xdx = .
2
0
In =
=
{
т
p /2
sin n x dx
0
n- 1 n- 3 n- 5
..
n n- 2 n- 4
n - 1 n - 3 n - 5 ..
n n- 2 n- 4
.
2 .1
.
3
if n is odd.
1 p
.
2 2
if n is even.
Exercise : Prove the following:
(1) т
p /2
0
sin x dx = т
n
p /2
cos n x dx
0
a
a
[Hint : т f ( x)dx = т f (a - x)dx]
0
(2) I n =
т
0
n- 1
cos
x sin x n - 1
cos n x dx =
+
I n- 2
n
n
Evaluation of Integrals:
(i ) т
1
0
xn
2
(ii ) т
dx
Ґ
0
(1- x )
dx
(1 + x 2 )
n+
1
2
where n is a positive integer.
(i) We put x = sin 
Note that when x = 0,  = 0 and when x = 1,  = /2.
we get
(i ) т
1
0
xn
(1- x 2 )
dx =
т
0
1
p
2
sin n q cos qd q =
cos q
т
0
p /2
sin n x dx
(ii ) т
Ґ
0
dx
(1 + x 2 )
n+
1
2
We put x = tan 
Note that when x = 0,  = 0 and
when x  ,   /2
т
Ґ
0
dx
1
n+
2
2
=
т
1
p
2
0
(1 + x )
=
sec 2 q d q
sec 2 n q
т
0
1
p
2
cos3n- 2 d q
Exercise :
Evalute : I =
т
Ґ
dx
7
2 2
0
(1 + x )
Hint: Using above procedure, get
I=
т
1
p
2
cos7qd q
0
6 4 2
= . .
7 5 3
=
16
.
35
Reduction formula for Im,n =  sinm x cosn x dx:
Write Im,n =  (sinm-1 x) (sin x cosn x)dx
Then Im,n =
(sin
m-1
n+ 1 ь
м
cos
xп
п
п
п
x) нэ -  (m – 1) sinm-2 x cos x
п
п
оп n + 1 п
ю
n+ 1 ь
м
cos
xп
п
п
п
нэ dx
п
п
оп n + 1 п
ю
(sin m- 1 x)(cos n+ 1 x) m - 1
 sinm-2 x cosn x (1 – sin2 x) dx
=+
n+ 1
n+ 1
(sin m- 1 x)(cos n+ 1 x)
=n+ 1
+
m- 1
m- 1
I m- 2,n I m,n
n+ 1
n+ 1
(sin m- 1 x)(cos n+ 1 x) + m - 1 I
I m,n = m- 2, n
m
+
n
m+ n
Evaluation of
I m,n =
т
p /2
sin m x cos n xdx
0
p /2
I m,n
й (sin m- 1 x)(cos n+ 1 x) щ
m- 1
к
ъ
= +
I m- 2,n
к
ъ
m+ n
m+ n
л
ы0
Thus we get
I m,n
m- 1
=
I mm+ n
2, n
Changing m to m – 2 successively, we have
I m-
I m-
2, n
4, n
m- 3
=
I mm+ n- 2
m- 5
=
I mm+ n- 4
4, n
6, n
……
2
I1,n
Finally I3,n =
3+ n
I2,n =
1
I 0,n
2+ n
if m is odd
if m even
I1,n =
т
p /2
sin x cos n x dx
0
p /2
й cos n+ 1 x щ
ъ
= ккл n + 1 ы
ъ0
I 0,n =
т
p /2
1
=
n+ 1
cos n x dx
0
Im,n =  sinm x cosn x dx
м
mп
п
п
m+
=п
н
п
mп
п
п
п
оm +
1 m- 3
m- 5
2
1
.
.
....
.
n m + n- 2 m + n- 4 3+ n n + 1
p /2
1 m- 3
m- 5
1
.
.
....
.т
cos n x dx
n m+ n- 2 m+ n- 4 2+ n 0
if m is odd
if m is even
Case (i): When m is odd (and n is even or odd),
m- 1 m- 3
2
1
I m,n =
.
....
.
m + n m + n- 2 3+ n n + 1
Case (ii): When m is even and n is odd,
I m,n =
m- 1 m- 3
m- 5
1 n- 1 n- 3 2
.
.
....
.
.
...
m+ n m+ n- 2 m+ n- 4 2+ n n n- 2 3
Case (iii): When m and n are both even,
m- 1 m- 3
m- 5
1 n- 1 n- 3 1 p
I m,n =
.
.
....
.
.
... .
m+ n m+ n- 2 m+ n- 4 2+ n n n- 2 2 2
Illustrations:
(i ) т
p /2
5
4
sin x cos xdx
0
(ii ) т
p /2
(iii) т
p /2
(iv) т
p /2
4 2 1
8
. .
=
9 7 5
315
sin 7 x cos5 xdx
6
4 2 1
1
= .
. . =
12 10 8 6 120
sin 6 x cos5 xdx
5 3 1 4 2
8
=
.
. .
.
=
11 9 7 5 3
693
0
0
0
=
sin 8 x cos6 xdx
7 5
3 1 5 3 1 p
=
.
.
. . . . .
14 12 10 8 6 4 2 2
5p
=
4096
Exercise : Prove the following:
p
(i ) т sin m x cos n xdx
0
(ii ) т
2p
p /2
м
m
n
п
2
sin
x
cos
x dx, if n is even
п
п
т
=н 0
п
п
п
о0, if n is odd
sin m x cos n xdx
0
p /2
м
m
n
п
4
sin
x
cos
x dx, if both m and n are even
п
п
т
=н 0
п
п
п
о0, if m or n or both are odd
Evaluation of Integrals :
xn
dx
2 m
(1 + x )
Ґ
(i ) т
0
(ii) т
Ґ
0
x n dx
dx
2 ( m+ 1/ 2)
(1 + x )
Put x = tan ,
Sol:
(i ) т
n
Ґ
x
dx
2 m
(1 + x )
0
=
т
1/ 2 p
0
=
т
sin n q cos m q 2
sec q d q
n
cos q
1/ 2 p
sin n qcos 2m - (n+2) q d q
0
(ii ) т
Ґ
0
x n dx
(1 + x 2 )( m+ 1/ 2)
=
т
1/ 2 p
sin nqcos 2m - (n+1) qd q
0
These values can be computed.
Evaluate : I =
т
Ґ
0
x4
dx
2 4
(1 + x )
Put x = tan , dx = sec2  d
Then, I =
т
p /2
q= 0
=
т
p /2
sin 4 q cos 2 q d q
0
=
3 1 1 p
. . .
6 4 2 2
p
=
.
32
tan 4 q 2
sec qd q
8
sec q
Evaluate : I =
a
т
x ax - x 2 dx
0
Put x = a sin2 
Then dx = 2a sin  cos  d  ;  varies from 0 to /2.
Now, ax - x 2
=
=
a 2 sin 2 q cos 2 q
Therefore I =
т
p /2
т
p /2
a 2 sin 2 q - a 2 sin 4 q =
a 2 sin 2 q(1- sin 2 q)
= a sin  cos .
asin 2 q. a sin q cos q. 2a sin q cos q d q.
0
= 2a 3
0
3 1 1 p
3
4
2
sin qcos q d q =2a . . . .
6 4 2 2
p a3
=
16
Example : If n is a positive integer, show that
I=
т
2a
x n 2ax - x 2 dx
0
Solution:
I=
т
(2n + 1)! a n+ 2
=
p
n
(n + 2)!n ! 2
First we note that
2a
x n a 2 - (a - x) 2 dx
0
Now we put a – x = a cos  .
Then x = a (1 – cos ) = 2a sin2 (/2);
when x = 0,  = 0 and
when x = 2a,  =  .
I=
т
p
2n a nsin 2n (q/2) (a sin q) (a sin q) d q
0
= (2a)
n+2
т
p
т
p /2
sin 2 n+ 2 (q / 2)cos 2 (q / 2) d q
0
= (2a)
n+2
2sin 2 n+ 2 f cos 2 f d f , where f = q/2
0
= (2a)n+2 . 2 .
= (2a)
n+2
(2n + 1)(2n - 1)....1 p
(2n + 4)(2n + 2)...2 2
(2n + 1)(2n - 1)...1
. p .2. n+ 2
2 (n + 2)(n + 1)...1
(2n + 1)! a n+ 2
=
p
n
(n + 2)!n ! 2
Reduction formula for In =  tann x dx:
In =  (tann-2 x) (tan2 x) dx
=  (tann-2 x) (sec2 x – 1) dx
=  tan n-2 x sec2 x dx -  tann-2 x dx
tan n- 1 x
In =
- I n-2
n- 1
This is the reduction formula .
Evaluation of I n = = т
p /4
tan n xdx
0
p /4
йtan n- 1 x щ
ъ - I n-2
In = к
кл n - 1 ъ
ы0
=
1
- I n-2
n- 1
On changing n to n – 2 successively,
1
I n- 2 =
- I n-4 ;
n- 3
1
I n- 4 =
- I n-6 ,..
n- 5
The last expression is I1 if n is odd and I0 if n is even .
I1 =
т
p /4
tan n xdx
0
= [ log sec x]0/4 = log 2
p /4
p
dx =
4
I0 =
т
In =
1
1
1
+
n- 5
n- 1 n- 3
0
- … …..I
where I = I1 if n is odd,
I = I0 if n is even and I appears with appropriate sign
I=
т
p /4
tan 5 xdx
0
1
1
=
+
4
2
I=
т
p /4
log 2
tan 6 xdx
0
1
=
5
1
+
3
1
1
p
- .
4