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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?

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    $\begingroup$ 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. $\endgroup$
    – Blue
    Nov 9 '20 at 16:22
  • $\begingroup$ On second thought what @Blue said would be easier $\endgroup$
    – Math Lover
    Nov 9 '20 at 16:25
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With $c:=\cos x$, the integral becomes polynomial,

$$I=-\int (1-c^2)(2c^2-1)^2dc.$$

Then

$$\int(4c^6-8c^4+5c^2-1)\,dc$$ is immediate.

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