Question

How do you find the integral of `int (sin(pix))^2*(cos(pix))^5 dx`?

Answer 1

Add the form `int sin^m u cos^n u du` with at least one of `m, n` odd to your mathematical cookbook.

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^2pixcos^5 pix dx = int sin^2pixcos^4 pix (cos pix )dx`

`= int sin^2pix (underbrace(cos^2 pix)_"Replace")^2 (cos pix )dx`

`= int sin^2pix underbrace((1-sin^2 pix)^2)_"Expand" (cos pix )dx`

`= int sin^2pix (1-2sin^2 pix +sin^4 pix)(cos pix )dx`

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

Let `u = sinx` so `du = cosx dx` and the integral becomes

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

Finishing is left to the student.