$ \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 $
Put $ \tan x=t \Rightarrow \sec ^{2} x \,d x=d t $
and $\,\,\,\,x=\tan ^{-1} t$
Then, it reduces to
$\int \tan ^{-1} t \cdot t \,d t=\frac{t^{2}}{2} \tan ^{-1} t-\int \frac{t^{2}}{2\left(1+t^{2}\right)} d t$
$=\frac{x \tan ^{2} x}{2}-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t+c$
$=\frac{x\left(\sec ^{2} x-1\right)}{2}-\frac{1}{2} \tan x+\frac{1}{2} x+c$
$=\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c$