Show that $n\int_0^{2\pi}\sin^n\theta d \theta = (n-1)\int_0^{2\pi}\sin^{(n-2)}\theta d\theta$

Question

Let $I_n = \int_0^{2\pi}\sin^n\theta d\theta$ show that for $n\geq2$:

$$nI_n = (n-1)I_{n-2}$$

What I've tried

I think this could be done by applying integration by parts multiple times and rearranging, but I can't seem to get the right substitution. Substituting $u=\sin\theta, v'=\sin^{n-1}\theta$ in $\int uv' = uv - \int u'v$ gives: $-\sin^{n-1}\theta\cos\theta + \int(n-1)\sin^{n-2}\cos^2\theta d\theta$ but that doesn't seem any closer to showing the recursion.

Answer 1

Note that $$\int(n-1)\sin^{n-2}\theta\cos^2\theta d\theta =$$

$$\int(n-1)\sin^{n-2}\theta(1-\sin^2 \theta) d\theta =$$ $$\int(n-1)\sin^{n-2}\theta d\theta -\int(n-1)\sin^{n}\theta d\theta $$

you can take it from here.