Evaluate the following integrals:

Let $I=\int \tan x \sec ^{4} x d x$

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

$\Rightarrow I=\int \tan x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$

$\Rightarrow I=\int\left(\tan x+\tan ^{3} x\right) \sec ^{2} x d x$

Let $\tan x=t$, then

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

$\Rightarrow I=\int\left(t+t^{3}\right) d t$

$\Rightarrow I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+c$

$\Rightarrow I=\frac{\tan ^{2} x}{2}+\frac{\tan ^{4} x}{4}+c$

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