A mathematical description of motion motivated the creation of Calculus. Problem of Motion:

1. 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.

2. 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

3. James Gregory Pierre de Fermat

4. Joseph Fourier Cauchy

5. Bernhard Riemann Henri Lebesgue

6. 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.

7. Mathematical description (Model): 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.

8. 2. Reduction formulae are useful to compute the following:

10. 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

11. In [1 + (n – 1)] = - sinn-1 x cos x + ( n – 1) In-2 Therefore In =  sinn x dx = • is the required reduction formula. • Illustration (i): To find  sin4 x dx.

12. I4 =  sin4 x dx = We need to apply the result (1) again by taking n = 2 That is, I4 = I0 =  sin0 x dx =  1 dx = x Thus I4 =  sin4 x dx =

13. Illustration (ii): To find  sin5 x dx Solution: I5 =  sin5 x dx = But I1 =  sin1 x dx = - cos x.

14. Corollary : To evaluate From (1) , In = But cos /2 = 0 = sin 0. Thus In =

15. Now, In-2 = In = Continuing the process we get: I1 if n is odd. In = I0 if n is even. But I1 = - [cos x]0/2 = - (0 – 1) = 1 and I0 =

16. .1 if n is odd. = if n is even.

17. Exercise : Prove the following:

18. Evaluation of Integrals: where n is a positive integer. • We put x = sin  • Note that when x = 0,  = 0 and when x = 1,  = /2. • we get

19. We put x = tan  Note that when x = 0,  = 0 and when x , /2

21. Reduction formula for Im,n =  sinm x cosn x dx: Write Im,n =  (sinm-1 x) (sin x cosn x)dx Then Im,n = -  (m – 1) sinm-2 x cos x  sinm-2 x cosn x (1 – sin2 x) dx

22. Evaluation of Thus we get

23. Changing m to m – 2 successively, we have …… Finally I3,n = if m is odd I2,n = if m even

24. Im,n =  sinm x cosn x dx

25. Case (i): When m is odd (and n is even or odd), Case (ii): When m is even and n is odd, Case (iii): When m and n are both even,

26. Illustrations:

27. Exercise : Prove the following:

28. Evaluation of Integrals : Put x = tan , Sol: These values can be computed.

29. Put x = tan , dx = sec2  d

30. Put x = a sin2 Then dx = 2a sin  cos  d  ;  varies from 0 to /2. = a sin  cos .

31. Example : If n is a positive integer, show that Solution: First we note that Now we put a – x = a cos  . Then x = a (1 – cos ) = 2a sin2 (/2); when x = 0,  = 0 and when x = 2a,  =  .

33. 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 This is the reduction formula .

34. On changing n to n – 2 successively, The last expression is I1 if n is odd and I0 if n is even .

35. = [ log sec x]0/4 = - … …..I where I = I1 if n is odd, I = I0 if n is even andI appears with appropriate sign