Originally Posted by

**TheEmptySet** I assume you mean

$\displaystyle I_n=\int_{0}^{\pi}\sin^{2n}\theta d\theta$

There are two ways to do this.

First method

Write the integral as for $\displaystyle n \ge 1$

$\displaystyle I_n=\int_{0}^{\pi}\sin\theta \sin^{2n-1}\theta d\theta$

Use integration by parts.

$\displaystyle dv=\sin(\theta)d\theta \implies v=-\cos(\theta)$ and $\displaystyle u=\sin^{2n-1}(\theta) \implies du=(2n-1)\sin^{2n-2}(\theta)\cos(\theta)d\theta$

This gives

$\displaystyle I_n=-\cos(\theta)\sin^{2n-1}(\theta)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(\theta)\cos^2(\theta)d\theta$

Use the pythagorean Identity from here and you are done

The 2nd is to use complex variables. If you want to see it this way I can show you that as well.