<s>[INST] Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$ 
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$

Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
 [/INST] Here is to integrate economically,
$$\cos^2x\sin^4 x = \frac 18 \sin^2 2x (1-\cos 2x)= \frac {1}{16}-\frac {1}{16}\cos 4x-\frac 18 \sin^22x\cos 2x$$
Thus,
$$\int \cos^2x\sin^4x\mathrm{d}x = \frac {x}{16}-\frac {1}{64}\sin4x-\frac{1}{48}\sin^32x +C$$
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