<s>[INST] Integrate $\int\sin^2(x)\cos^4(x)dx$ As in the title: how does one integrate $$ \int\sin^2(x)\cos^4(x)dx? $$ 
Any hints? Integration by parts doesn't seem to be a reasonable technique here.
 [/INST] Or you could separate it into two integrals right from the beginning:
$$
\int \sin^2 x \cos^4 x \ dx = \int \cos^4 x \ dx - \int \cos^6 x \ dx.
$$
Then, we have
$$
\cos^4 x = \frac{3 + \cos 4x + 4\cos 2x}{8}
$$
and 
$$
\cos^6 x = \frac{10 + \cos 6x + 6 \cos 4x + 15 \cos 2x}{32}.
$$
On completing the integration, the answer should be:
$$
\int \sin^2 x \cos^4 x \ dx = \frac{x}{16} + \frac{\sin 2x}{64} - \frac{\sin 4x}{64} - \frac{\sin 6x}{192}+C.
$$
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