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.
 A: Just to avoid the ambiguity pointed out in the comments, I'll carry out the integral of $\csc^4(x)$. (If $\frac32$ is a constant outside the argument to $\csc$, then just multiply this result by $\frac32$. If it belongs inside the argument, just make the appropriate substitution of $x$ with $\frac{3x}2$, keeping in mind the chain rule.)
For the remaining integral, recall that
$$\cot^2(x)=\csc^2(x)-1$$
Then
$$\int\csc^2(x)\cot^2(x)\,\mathrm dx=\int\csc^4(x)\,\mathrm dx-\int\csc^2(x)\,\mathrm dx$$
which gives
$$\color{red}{\int\csc^4(x)\,\mathrm dx}=-\csc^2(x)\cot(x)-2\left(\color{red}{\int\csc^4(x)\,\mathrm dx}-\int\csc^2(x)\,\mathrm dx\right)$$
i.e. a simple linear equation whose solution is the antiderivative you seek.
A: 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}
