Question

How do you find the integral of `int [sin^2(pi x) cos^5(pi x)]dx`?

Answer 1

This belongs in your mathematical toolbox: `int sin^m u cos^n u du` with at least one of `m, n` odd.

Explanation:

`int sin^m u cos^n u du` with at least one of `m, n` odd.
Integrate by substitution. Do this by pulling off one from the odd power, then convert the remaining even power to the other function. Integrate the resulting polynomial in `sinu` or `cosu` term by term.

`I = int [sin^2(pi x) cos^5(pi x)]dx = int [sin^2(pi x) cos^4(pi x)]cos(pix)dx`

`cos^4(pi x) = (cos^2 pix)^2 = (1-sin^2 pix)^2 = 1-2sin^2 pix+sin^4 pix`

`I = int [sin^2(pi x) (1-2sin^2 pix+sin^4 pix)]cos(pix)dx`

`= int [sin^2pix -2sin^4 pix+sin^6 pix]cos(pix)dx`

`= 1/pi int (u^2-2u^4+u^6)du`

`= 1/pi [(sin^3 pix)/3-(2sin^5 pix)/5 + (sin^7 pix)/7] +C`