# 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...

$$\cot^5x\sin^2x=\frac{\cos^5x}{\sin^5x}\sin^2x=\cos^5x\csc^3x=\cos^2x\cot^3x$$

Now put $u = \sin x$ to get:
Put $u = \sin x$ to get the final result:
$$I = -\frac{\sin^{-2} x}{2} -2 \log\left|\sin x\right| + \frac{\sin^2 x}{2} + C$$