<s>[INST] How to solve $\int\cot^5x\sin^2x\ dx$? I'm not quite sure how to approach this without it getting extremely messy... and even then, I don't know if it will come out right.
The best I can think of is to use IBP, but neither of those functions are easy to integrate.
I did integrate $\sin^2x$ to get $$\int\sin^2x\ dx=\frac{2x+\sin{2x}}{4}+C$$
but I dare not try to go further as it looks like hell to apply IBP after this point.
I did consider converting $\cot^5x$ to $\displaystyle\frac{\cos^5x}{\sin^5x}$, but then I wasn't any better off...
$$\cot^5x\sin^2x=\frac{\cos^5x}{\sin^5x}\sin^2x=\cos^5x\csc^3x=\cos^2x\cot^3x$$
 [/INST] We have:
\begin{align}
I = \int \cot^5 x \sin^2 x \,dx &= \int \frac{\cos^5 x}{\sin^5 x} \sin^2 x \,dx \\
&= \int \frac{\cos^5 x}{\sin^3 x} \,dx \\
&= \int \frac{\left(1 - \sin^2 x\right)^2}{\sin^3 x} \cos x \,dx
\end{align}
Now put $u = \sin x$ to get:
\begin{align}
I &= \int \frac{\left(1 - u^2\right)^2}{u^3} \,du \\
&= \int \left(u^{-3} -2 u^{-1} + u\right) \,du \\
&= -\frac{u^{-2}}{2} -2 \log\left|u\right| + \frac{u^2}{2} + C
\end{align}
Put $u = \sin x$ to get the final result:
$$
I = -\frac{\sin^{-2} x}{2} -2 \log\left|\sin x\right| + \frac{\sin^2 x}{2} + C
$$
</s>