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.

Use that $$\cos^2(x)=1-\sin^2(x)$$ and your integral will be
$$\int\sin^4(x)-\sin^6(x)dx$$
For your work:
$$\int \sin(x)^4dx=-1/4\, \left(  \left( \sin \left( x \right)  \right) ^{3}+3/2\,\sin
 \left( x \right)  \right) \cos \left( x \right) +3/8\,x
+C$$
and
$$\int\sin(x)^6dx=-1/6\, \left(  \left( \sin \left( x \right)  \right) ^{5}+5/4\,
 \left( \sin \left( x \right)  \right) ^{3}+{\frac {15\,\sin \left( x
 \right) }{8}} \right) \cos \left( x \right) +{\frac {5\,x}{16}}
+C$$
