<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] Expanding on @Dr.SonnhardGraubner's answer, since $\sin^2x=\frac{1-\cos 2x}{2}$ we have $$\sin^4x=\frac{1-2\cos 2x+\cos^22x}{4}=\frac{3-4\cos 2x+\cos4x}{8}$$ while $$\sin^6x=\frac{1-3\cos 2x+3\cos^22x-\cos^32x}{8}=\frac{10-15\cos 2x+6\cos4x-\cos6x}{32}.$$Thus $$\int(\sin^4x-\sin^6x)dx\\=\int\frac{2-\cos2x-2\cos4x+\cos6x}{32}dx=\frac{12x-3\sin2x-3\sin4x+\sin6x}{192}+C.$$
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