Answer
$$ \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \sqrt{\theta^{2}-1}+C$$
Work Step by Step
Given $$\int \theta \sec ^{-1} \theta d \theta$$
Let
\begin{align*}
u&= \sec ^{-1} \theta\ \ \ \ \ \ dv= \theta \, d\theta\\
du&= \frac{1}{|\theta| \sqrt{\theta^{2}-1}} d \theta\ \ \ \ \ \ v=\frac{1}{2}\theta^2
\end{align*}
Then
\begin{align*}
\int x \csc x \cot x \, d x&= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \int \frac{\theta^2}{|\theta| \sqrt{\theta^{2}-1}} d \theta\\
&= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \int \frac{\theta }{ \sqrt{\theta^{2}-1}} d \theta\\
&= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \sqrt{\theta^{2}-1}+C
\end{align*}
