Transcript reduction formulae - EngineeringDuniya.com

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 de rivative
of this function satisfies y' = µ x. ( w h ere
µ = t g / T ; t is density ; T is tension w hich can b e com puted.)
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
n
(1) is the required reduction formula.
Illustration (i): To find  sin4 x dx.
x cos x
+
n- 1
n
I n-2 ...(1)
I4 = 
sin4
x dx =
- sin
3
x co s x
+
4
3
I2
4
We need to apply the result (1) again by taking n = 2
- sin
That is, I4 =
3
x cos x
+
4
3 - sin x cos x
{
4
2
I0 =  sin0 x dx =  1 dx = x
Thus I4 =  sin4 x dx =
3
- sin x cos x
4
-
3
8
sin x cos x +
3
8
x + c
+
1
2
I0
}
Illustration (ii): To find  sin5 x dx
Solution: I5 = 
sin5
4
x dx =
- sin x co s x
5
4
=
- sin x cos x
5
But I1 =  sin1 x dx = - cos x.
+
4
5
I3
2
4 ìïï - sin x cos x 2 ü
ïï
+ í
+ I1 ý
5 ïîï
3
3 ïïþ
Corollary : To evaluate I n =
ò
p /2
n
sin x d x
0
p /2
From (1) , In
ésin m - 1 x cos x ù
ú
- ê
êë
ú
n
û0
=
But cos /2 = 0 = sin 0.
Thus In
=
n- 1
n
I
n-2
+
n- 1
n
I
n-2
Now, In-2 = n - 3 I n - 4
n- 2
In =
n- 1
n- 3
n
n- 2
. I n-4 , by back substitution.
Continuing the process we get:
n- 1 n- 3 n- 5
In =
{
But I1 =
n- 1 n- 3
n
n- 5
sin x dx =
0
ò
p /2
0
..
.
1
2
I1
if n is odd.
I0
if n is even.
- [cos x]0/2 = - (0 – 1) = 1
sin xdx
0
2
3
n- 4
n- 2
p /2
and I0 =
.
n- 2 n- 4
n
ò
..
=
p
2
.
In =
ò
p /2
n
sin x d x
0
n- 1 n- 3
=
{
n
n- 2
n- 5
..
.
n- 4
n - 1 n - 3 n - 5 ..
n- 2 n- 4
n
2
.1
if n is odd.
p
if n is even.
3
.
1
2
.
2
Exercise : Prove the following:
(1) ò
p /2
0
sin x dx = ò
[H int : ò
(2) I n =
p /2
n
ò
0
a
f ( x ) dx =
0
n
n
cos x dx
cos x dx =
ò
cos
a
f ( a - x )dx ]
0
n- 1
x sin x
n
+
n- 1
n
I n-
2
Evaluation of Integrals:
(i ) ò
x
1
( ii ) ò
dx
2
0
¥
n
(1 - x )
dx
0
2
n+
(1 + x )
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 ) ò
0
1
x
n
1
dx
2
(1 - x )
=
ò
2
0
p
n
sin q cos q d q
cos q
=
ò
p /2
n
sin x d x
0
( ii ) ò
¥
dx
0
2
n+
1
2
(1 + x )
We put x = tan 
Note that when x = 0,  = 0 and
when x  ,   /2
ò
¥
0
1
dx
2
(1 + x )
n+
1
=
ò
2
0
2
p
2
sec q d q
sec
1
=
ò
2
0
2n
q
p
cos
3 n- 2
dq
E xercise :
ò
E valute : I =
¥
dx
7
0
2
(1 + x ) 2
H int: U sing above procedure, get
1
ò
I =
=
p
2
0
6
.
7
=
7
cos q d q
16
35
4
5
.
.
2
3
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
= -
ìï cos n + 1 x ü
ïï
ï
m-2 x cos x
x) í ý -  (m – 1) sin
ïîï
n + 1 ïïþ
(sin
m- 1
x )(cos
n+ 1
x)
n+ 1
= -
(sin
m- 1
x )(cos
n+ 1
+
m- 1
ìï cos n + 1 x ü
ïï
ïí
ý dx
ïîï
n + 1 ïïþ
 sinm-2 x cosn x (1 – sin2 x) dx
n+ 1
n+ 1
x)
+
m- 1
n+ 1
I m-
2 ,n
-
m- 1
n+ 1
I m ,n
I m ,n = -
Evaluation of
(sin
m- 1
x )(cos
n+ 1
x)
m+ n
m+ n
I m ,n =
ò
p /2
m- 1
+
m
I m-
2 ,n
n
sin x cos xdx
0
p /2
I m ,n
é (sin m - 1 x )(cos n + 1 x ) ù
ú
= êêë
ú
m+ n
û0
Thus we get
I m ,n =
m- 1
m + n
I m-
2 ,n
+
m- 1
m+ n
I m-
2 ,n
Changing m to m – 2 successively, we have
I m-
I m-
2 ,n
4 ,n
m- 3
=
m + n- 2
m- 5
=
m + n- 4
I m-
I m-
4 ,n
6 ,n
……
Finally I3,n =
I2,n =
2
3+ n
I 1, n
1
2+ n
I 0 ,n
if m is odd
if m even
I 1, n =
ò
p /2
n
sin x cos x dx
0
p /2
é cos n + 1 x ù
ú
= êêë
n+ 1 ú
û0
I 0 ,n =
ò
p /2
1
=
n+ 1
n
co s x d x
0
Im,n =  sinm x cosn x dx
ìï
ï
ï
ï
= í
ï
ï
ï
ïïî
m- 1
.
m- 3
.
m- 5
....
m + n m + n- 2 m + n- 4
m- 1
.
m- 3
.
m- 5
m + n m + n- 2 m + n- 4
2
1
.
if m is odd
3+ n n+ 1
....
1
2+ n
.ò
0
p /2
n
cos x dx
if m is even
Case (i): When m is odd (and n is even or odd),
I m ,n =
m- 1
.
m- 3
m + n m + n- 2
....
2
.
1
3+ n n+ 1
Case (ii): When m is even and n is odd,
I m ,n =
m- 1
.
m- 3
.
m- 5
m + n m + n- 2 m + n- 4
....
1
n- 1 n- 3 2
.
...
2+ n n n- 2 3
.
Case (iii): When m and n are both even,
I m ,n
m- 1
m- 3
m- 5
1
n- 1 n- 3 1 p
=
.
.
....
.
.
... .
m + n m + n- 2 m + n- 4
2+ n n n- 2 2 2
Illustrations:
(i ) ò
p /2
5
4
4
=
sin x co s xd x
9
0
( ii ) ò
p /2
7
5
sin x co s xd x
6
=
0
( iii ) ò
p /2
6
5
sin x co s xd x
=
p /2
8
6
sin x cos xdx
0
5
4
.
8
=
.
315
2
1
.
10
8
6
5
3
1
4
14
=
1
12
7
=
.
7
.
11
0
( iv ) ò
2
.
.
9
5
.
12
5p
4096
7
.
3
10
.
=
120
.
5
.
1
8
1
2
=
3
.
5
6
8
693
.
3 1 p
. .
4 2 2
Exercise : Prove the following:
p
( i ) ò sin
m
n
x co s xd x
0
( ii ) ò
2p
sin
m
p /2
ìï
m
n
sin x co s x d x , if n is even
ïï 2 ò
0
= í
ï
ïïî 0 , if n is o d d
n
x co s xd x
0
p /2
ìï
m
n
sin x cos x dx, if both m and n are even
ïï 4 ò
0
= í
ï
ïïî 0, if m or n or both are odd
Evaluation of Integrals :
¥
(i ) ò
x
n
2
(1 + x )
0
m
( ii ) ò
dx
n
¥
x dx
2
(1 + x )
0
(m+ 1/ 2)
Put x = tan ,
Sol:
(i ) ò
¥
x
n
2
(1 + x )
0
m
dx
ò
=
=
( ii ) ò
0
¥
1/ 2p
n
n
2
sec q d q
cos q
0
ò
m
sin q cos q
1/ 2p
n
sin q cos
2m - (n+ 2)
q dq
0
n
x dx
2
(1 + x )
(m+ 1/ 2)
=
ò
1/ 2p
n
sin q cos
2m - (n+ 1)
qd q
0
These values can be computed.
dx
E valuate : I =
ò
¥
0
x
4
2
(1 + x )
4
dx
Put x = tan , dx = sec2  d
ò
T hen , I =
p /2
q= 0
=
=
=
ò
p /2
4
4
tan q
8
sec q
2
sin q co s q d q
0
3 1 1 p
. . .
6 4 2 2
p
32
.
2
sec q d q
E valuate : I =
ò
a
2
x ax - x dx
0
Put x = a sin2 
Then dx = 2a sin  cos  d  ;  varies from 0 to /2.
N ow , ax - x
=
2
2
2
2
2
4
a sin q - a sin q =
=
2
T herefore I =
= 2a
3
ò
a sin
2
2
= a sin  cos .
2
a sin q cos q
p /2
2
a sin q (1 - sin q )
2
q . a sin q cos q . 2a sin q cos q d q .
0
ò
p /2
4
2
sin q co s q d q
0
3 1 1 p
3
= 2a . . . .
6 4 2 2
=
pa
16
3
Example : If n is a positive integer, show that
I =
ò
2a
x
2
=
2ax - x dx
ò
n+ 2
( n + 2) ! n ! 2
0
Solution:
I =
n
(2 n + 1) ! a
First we note that
2a
x
n
2
2
a - ( a - x ) 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,  =  .
n
p
I =
ò
p
n
n
2n
2 a sin
( q /2) (a sin q ) (a sin q ) d q
0
= (2a)
= (2a)
= (2a)
n+ 2
n+ 2
n+ 2
ò
p
sin
2 n+ 2
2
( q / 2 )cos ( q / 2 ) d q
0
ò
p /2
2 sin
2 n+ 2
2
f cos f
d f , w here f = q /2
0
.2.
(2 n + 1)(2 n - 1)....1 p
(2 n + 4)(2 n + 2)...2 2
= (2a)
n+ 2
(2 n + 1)(2 n - 1)...1
. p .2.
2
=
(2 n + 1) ! a
n+ 2
n+ 2
( n + 2) ! n ! 2
n
p
( n + 2)( n + 1)...1
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
In =
tan
n- 1
n- 1
x
- I n-2
This is the reduction formula .
E valuation of I n = = ò
p /4
n
tan xdx
0
p /4
étan n - 1 x ù
ú
In = ê
êë n - 1 ú
û0
1
=
n- 1
- I n-2
- I n-2
On changing n to n – 2 successively,
I n-
2
=
1
n- 3
- I n-4 ;
I n-
4
=
1
n- 5
- I n-6 ,..
The last expression is I1 if n is odd and I0 if n is even .
I1 =
ò
p /4
n
tan xdx
0
= [ log sec x]0/4 = log 2
I0 =
In =
ò
p /4
dx =
4
0
1
n- 1
p
-
1
+
n- 3
1
- … …..I
n- 5
where I = I1 if n is odd,
I = I0 if n is even and I appears with appropriate sign
I=
ò
=
p /4
5
tan xdx
0
1
1
-
4
I=
=
ò
1
5
+
log
2
2
p /4
6
tan xd x
0
-
1
3
+
1
1
-
p
4
.