# Integrating $\int \sin^3{x} \cos^2{2x}\,dx$

Is there a trick to this integral? $$\int \sin^3{x} \cos^2{2x}\,dx$$

I've tried to solve this by integration by parts and by expanding $$\cos^2{2x}$$ but these seem to make it more complicated.

Is there something I'm missing like a clever substitution or using the trigonometric identities?

• One trick here is to peel off one $\sin x$, then convert everything else to an expression in $\cos x$. This sets you up for a $u$-substitution.
– Blue
Nov 9 '20 at 16:22
• On second thought what @Blue said would be easier Nov 9 '20 at 16:25

With $$c:=\cos x$$, the integral becomes polynomial,
$$I=-\int (1-c^2)(2c^2-1)^2dc.$$
$$\int(4c^6-8c^4+5c^2-1)\,dc$$ is immediate.