I need some help with this integral.

$\displaystyle \int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta $

I've tried using integration by parts:

$\displaystyle u=\sin ^{2}2\theta $

$\displaystyle \frac{dv}{d\theta }=\cos 2\theta $

And got the integral as:

$\displaystyle \frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right)$.

I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks

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