$$
\begin{aligned}
\int x \sin x \sec ^{3} x d x &=\int x \sin x \frac{1}{\cos ^{3} x} d x \\
&=\int x \tan x \cdot \sec ^{2} x d x \\
\text { Put } \tan x=t & \Rightarrow \sec ^{2} x d x=d t \\
\text { and } \quad x &=\tan ^{-1} t
\end{aligned}
$$
Then, it reduces to
$$
\begin{array}{l}
\int \tan ^{-1} t \cdot t d t=\frac{t^{2}}{2} \tan ^{-1} t-\int \frac{t^{2}}{2\left(1+t^{2}\right)} d t \\
=\frac{x \tan ^{2} x}{2}-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t+c \\
=\frac{x\left(\sec ^{2} x-1\right)}{2}-\frac{1}{2} \tan x+\frac{1}{2} x+c \\
=\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c
\end{array}
$$