<s>[INST] How to solve $\int \csc^{4} \frac{3}{2} x \,dx$ using integration by parts How to solve $\int \csc^{4} \frac{3}{2} x \,dx$
This is what I have so far:
$$\int \csc^{4} \frac{3}{2} x \,dx=\frac{3}{2} \int \csc^{2} x \csc^{2}x\,dx.$$
Let $u=\csc^{2}$ and $dv=\csc^{2} \,dx$. Then we have $du=-2\csc^{2}x\cot x\,dx$ and $v=-\cot x$.
\begin{align}
\frac{3}{2} \int \csc^{2} x \csc^{2} x \,dx&=\frac{3}{2}\left(-\csc^{2}x\cot x-\int-\cot x(-2\csc^{2}x\cot x\,dx)\right)\\ 
&= \frac{3}{2}\left(-\csc^{2}x\cot x-2\int \csc^{2}x\cot^{2}x\,dx\right)
\end{align}
At this point, I am uncertain about how to continue. Do I find the integral in the right hand side again? Or perhaps I missed something? Thank you.
 [/INST] Let $t=\frac32x$ and integrate by parts as follows
\begin{align}
 \int \csc^{4} \frac{3}{2} x \,dx
&= \frac23\int \csc^{4} t \,dt=- \frac29\int \sec^{2} t \>d(\cot^3 t)\\
&= -\frac29\csc^2 t\cot t+\frac49 \int\csc^2 t dt\\
 &= -\frac29\csc^2 t\cot t-\frac49 \cot t +C
\end{align}
</s>