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.

For trigonometric monomials, the standard method is linearisation when both exponents are even, substitution when an exponent is odd ($u=\sin x$ or $u=\cos x$, depending on which exponent is odd).
You can linearise either with trigonometry formulæ or via the complex exponential. I'll show both:


*

*with trigonometry: use repeatedly the linearisation formulæ
$$\cos^2x=\tfrac12(1+\cos 2x),\qquad\sin^2x=\tfrac12(1-\cos 2x).$$
So
\begin{align}
\cos^2x\sin^4x&=\tfrac12(1+\cos 2x)\tfrac14(1-\cos 2x)^2=\tfrac 18(1+\cos 2x)(1-2\cos 2x+\cos^22x)\\
&=\tfrac 18(1+\cos 2x)\bigl(\tfrac32-2\cos 2x+\tfrac12\cos4x)\\
&=\tfrac1{16}(3-4\cos 2x+\cos 4x+3\cos 2x-4\cos^2 2x+\underbrace{\cos2x\cos 4x}_{\frac12(\cos 6x+\cos 2x)}) \\ 
&=\tfrac1{32}(2-\cos 2x-2\cos 4x+\cos 6x).
\end{align}

*with the complex exponential: to have lighter computations, set $u=\mathrm e^{ix}$. We'll use the definitipn of sine and cosine in terms of $u$:
$$\sin x=\tfrac1{2i}(u-\bar u),\qquad \cos x =\tfrac 12(u+\bar u),$$
so that, taking into accoount that $u\bar u=1$,
\begin{align}
\cos^2x\sin^4x &=\tfrac14(u+\bar u)^2\cdot\tfrac1{16}(u-\bar u)^4=\tfrac1{64}(u^2+\bar u^2+2)(u^4-4^2+6-4\bar u^2+\bar u^4) \\
&=
\end{align}
