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

Isaac Newton Gottfried

Leibniz

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.)

James Gregory Pierre de Fermat

Joseph Fourier Cauchy

Bernhard Riemann Henri Lebesgue

The road deck hangs on vertical cables suspended from the main cables.

Problem : We have to find the optimal shape of the main cable.

Some motivations: 1. Suspension bridges

Mathematical description (Model):

Find the curve y = y(x) such that the derivative

of this function satisfies y' = µx. ( where

/ ; is density ; is tension which can be computed.)µ g T Tt t=

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 =n-2I ...(1)

1

sin cos

1n x x nn n

-- -+

(1) is the required reduction formula.

Illustration (i): To find sin4 x dx.

I4 = sin4 x dx = 2

3sin cosI

4

34

x x-+

We need to apply the result (1) again by taking n = 2

That is, I4 = { }0

3sin cos 3 sin cos 14 4 2 2x x x x

I- -

+ +

I0 = sin0 x dx = 1 dx = x

Thus I4 = sin4 x dx =

3sin cos 3 3sin x cos x + x + c

4

8 8x x-

-

Illustration (ii): To find sin5 x dx

Solution: I5 = sin5 x dx =4

3

sin cos 4 I

5 5x x-

+

4 2

1

sin cos 4 sin cos 25 5 3 3x x x x

Iì ü- -ï ïï ï= + +í ýï ïï ïî þ

But I1 = sin1 x dx = - cos x.

Corollary : To evaluate/ 2

0sin dxn

nI xp

=ò

From (1) , In = / 21

n - 2

0

sin cos 1

I m x x nn n

p-é ù -ê ú- +ê úë û

But cos /2 = 0 = sin 0.

n - 2

1 I

nn-

Thus In =

Now, In-2 = n - 4

3I

2nn

--

In =1nn-

n-4

3. I , by back substitution.

2nn

--

Continuing the process we get:

In =

1nn- 3

2nn

--

5..

4nn--

2.3{ I1 if n is odd.

1nn- 3

2nn

--

5..

4nn--

1.2

I0 if n is even.

But I1 =/ 2

0sin dx =x

p

òand I0 =

/ 20

0sin xdx

p

ò .2p

=

- [cos x]0/2 = - (0 – 1) = 1

=

1nn- 3

2nn

--

5..

4nn--

2.3{

.1 if n is odd.

1nn- 3

2nn

--

5..

4nn--

1.2 if n is even..

2p

/ 2

0sin dxn

nI xp

=ò

Exercise : Prove the following:

1n

2

cos sin 1(2) = cos x dx =

n

n n

x x nI I

n n

-

-

-+ò

/ 2 / 2

0 0(1) sin dx = cos dxn nx x

p p

ò ò

0 0[Hint : ( ) ( ) ]

a a

f x dx f a x dx= -ò ò

Evaluation of Integrals:

1

20( )

(1 )

nxi dx

x-ò 1

02 2

( )

(1 )n

dxii

x

¥

++

ò

where n is a positive integer.

(i) We put x = sin

Note that when x = 0, = 0 and when x = 1, = /2.

we get

1

20( )

(1 )

nxi dx

x-ò

/ 2

0 sin dxn x

p=ò

1

2

0

sin coscos

n dp q q qq

=ò

10

2 2

( )

(1 )n

dxii

x

¥

++

ò

We put x = tan

Note that when x = 0, = 0 and

when x , /2

10

2 2(1 )n

dx

x

¥

++

ò1 22

20

sec sec n

dp q qq

=ò

13 22

0cos n d

pq-=ò

70

2 2

:

(1 )

dxEvalute I

x

¥=

+ò

172

0cosI d

pq q=ò

2.3

4.5

67

=

16.

35=

:Exercise

Hint: Using above procedure, get

Reduction formula for Im,n = sinm x cosn x dx:

Write Im,n = (sinm-1 x) (sin x cosn x)dx

Then Im,n =

1cos

dx1

n xn

+ì üï ïï ï-í ýï ï+ï ïî þ

1m-1 cos

(sin x)1

n xn

+ì üï ïï ï-í ýï ï+ï ïî þ

1 1(sin )(cos )1

m nx xn

- +

=-+

11

mn

-+

+

- (m – 1) sinm-2 x cos x

sinm-2 x cosn x (1 – sin2 x) dx

1 1(sin )(cos )1

m nx xn

- +

=-+

2, ,

1 11 1m n m n

m mI I

n n-

- -+ -

+ +

1 1

,

(sin )(cos )m n

m n

x xI

m n

- +

=-+

2,

1m n

mI

m n -

-+

+

Evaluation of/ 2

,0

sin cosm nm nI x xdx

p=ò

/ 21 1

, 2,

0

(sin )(cos ) 1m n

m n m n

x x mI I

m n m n

p- +

-

é ù -ê ú= - +ê ú+ +ë û

, 2,

1m n m n

mI I

m n -

-=

+

Thus we get

Changing m to m – 2 successively, we have

2, 4,

32m n m n

mI I

m n- -

-=

+ -

4, 6,

54m n m n

mI I

m n- -

-=

+ -

……

Finally I3,n = if m is odd 1,

23 nIn+

I2,n = if m even0,

12 nIn+

/ 2n

1,0

sin x cos x dx nIp

=ò/ 21

0

cos1

n xn

p+é ùê ú= -ê ú+ë û

11n

=+

/ 2n

0,0

cos x dx nIp

=ò

Im,n = sinm x cosn x dx

/ 2n

0

1 3 5 2 1. . .... . if m is odd

2 4 3 11 3 5 1

. . .... . cos x dx if m is even2 4 2

m m mm n m n m n n nm m mm n m n m n n

p

ì - - -ïïï + + - + - + +ïï=íï - - -ïïï + + - + - +ïî ò

Case (i): When m is odd (and n is even or odd),

,

1 3 2 1. .... .

2 3 1m n

m mI

m n m n n n- -

=+ + - + +

Case (ii): When m is even and n is odd,

,

1 3 5 1 1 3 2. . .... . . ...

2 4 2 2 3m n

m m m n nI

m n m n m n n n n- - - - -

=+ + - + - + -

Case (iii): When m and n are both even,

,

1 3 5 1 1 3 1. . .... . . ... .

2 4 2 2 2 2m n

m m m n nI

m n m n m n n n np- - - - -

=+ + - + - + -

Illustrations:

/ 25 4

0( ) sin cosi x xdx

p

ò4

.9

=

/ 27 5

0( ) sin cosii x xdx

p

ò 6.

12=

/ 26 5

0( ) sin cosiii x xdx

p

ò 511

=

/ 28 6

0( ) sin cosiv x xdx

p

ò 714

=

2.

715

8315

=

410

2.8

1.6

1120

=

3.9

1.7

4.5

2.3

8693

=

5.12

3.10

1.8

5.6

3.4

1.2

.2p

54096p

=

Exercise : Prove the following:

0( ) sin cosm ni x xdx

p

ò/ 2

02 sin cos dx, if n is even

0, if n is odd

m nx xpìïïï=íïïïî

ò

2

0( ) sin cosm nii x xdx

p

ò/ 2

04 sin cos dx, if both m and n are even

0, if m or n or both are odd

m nx xpìïïï=íïïïî

ò

Evaluation of Integrals :

20( )

(1 )

n

m

xi dx

x

¥

+ò 2 ( 1/ 2)0( )

(1 )

n

m

x dxii dx

x

¥

++ò

Put x = tan ,

Sol:

20( )

(1 )

n

m

xi dx

x

¥

+ò1/ 2

2

0

sin cossec

cos

n m

n dp q q

q qq

=ò1/ 2

2m - (n+2)

0sin cos n d

pq q q=ò

These values can be computed.

2 ( 1/ 2)0( )

(1 )

n

m

x dxii

x

¥

++ò1/ 2

2m - (n+1)

0sin cos n d

pq q q=ò

4

2 40:

(1 )x

Evaluate I dxx

¥=

+ò

Put x = tan , dx = sec2 d

4/ 22

80

tan, sec

secThen I d

p

q

qq q

q==ò

/ 24 2

0sin cos d

pq q q=ò

3 1 1 . . .6 4 2 2

p=

.32p

=

2

0:

a

Evaluate I x ax x dx= -ò

Put x = a sin2

Then dx = 2a sin cos d ; varies from 0 to /2.

2,Now ax x- 2 2 2 4sin sina aq q= - 2 2 2sin (1 sin )a q q= -

2 2 2sin cosa q q= = a sin cos .

/ 22

0Therefore I = sin . a sin cos . 2a sin cos .a d

pq q q q q qò

/ 23 4 2

0= 2a sin cos d

pq q qò 3 3 1 1

=2a . . . .6 4 2 2

p 3

16ap

=

Example : If n is a positive integer, show that

22

02

anI x ax x dx= -ò

2(2 1)!( 2)! ! 2

n

n

n an n

p++

=+

Solution: First we note that

22 2

0( )

anI x a a x dx= - -ò

Now we put a – x = a cos .

Then x = a (1 – cos ) = 2a sin2 (/2);

when x = 0, = 0 and

when x = 2a, = .

2n

02 sin ( /2) (a sin ) (a sin ) n nI a d

pq q q q=ò

n+2 2 2 2

0= (2a) sin ( / 2)cos ( / 2)n d

pq q q+ò

/ 2n+2 2 2 2

0= (2a) 2sin cos , where = /2n d

pff ff q+ò

n+2 (2 1)(2 1)....1= (2a) . 2 .

(2 4)(2 2)...2 2n nn n

p+ -+ +

n+22

(2 1)(2 1)...1= (2a) . .2.

2 ( 2)( 1)...1n

n nn n

p +

+ -+ +

2(2 1)!( 2)! ! 2

n

n

n an n

p++

=+

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

1

n n-2

tanI = I

1

n xn

-

--

This is the reduction formula .

/ 4

n0

Evaluation of I = = tann xdxp

ò/ 41

n n-2

0

tanI = I

1

n xn

p-é ùê ú -ê ú-ë û

n-2

1- I

1n=

-

On changing n to n – 2 successively,

2 n-4

1- I ;

3nI n- =- 4 n-6

1- I ,..

5nI n- =-

The last expression is I1 if n is odd and I0 if n is even .

/ 4

10

I = tann xdxp

ò

= [ log sec x]0/4 = log 2

/ 4

00

I = = 4

dxp p

ò

11nI n

=-

13n

--

15n

+-

- … …..I

where I = I1 if n is odd,

I = I0 if n is even and I appears with appropriate sign

/ 45

0I = tan xdx

p

ò

14

=1

2

- + log 2

/ 46

0I = tan xdx

p

ò15

=1

3

- +1

1

.4p

-