<s>[INST] Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral. 
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
 [/INST] Here is the most straight way to solve this problem:
Notice that $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x $
So $\sin^2 x \cos 4x = \sin^2 (1-2\sin^2 2x) = \sin^2 [1-2(1-2\sin^2 x)^2] = -8\sin^6 x +8\sin^4 x -\sin^2 x$
By Reduction Formula: The integral can be easily solved.
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