Evaluate the following integrals:

Let $I=\int \tan ^{3} x \sec ^{2} x d x$

Let $\tan x=t$, then

$\Rightarrow \sec ^{2} x d x=d t$

$\Rightarrow I=\int t^{3} d t$

$\Rightarrow I=\frac{t^{4}}{4}+c$

$\Rightarrow I=\frac{\tan ^{4} x}{4}+c$

Therefore, $\int \tan ^{3} x \sec ^{2} x d x=\frac{\tan ^{4} x}{4}+c$