<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] 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}
</s>