Evaluate the following integrals:

Let I $=\int \tan ^{5} x d x$

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

$\Rightarrow I=\int\left(\sec ^{2} x-1\right) \tan ^{3} x d x$

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

$\Rightarrow I=\int \tan ^{3} x \sec ^{2} x d x-\int\left(\sec ^{2} x-1\right) \tan x d x$

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

Let $\tan x=t$, then

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

$\Rightarrow I=\int t^{3} d t-\int t d t+\int \tan x d x$

$\Rightarrow I=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x|+c$

$\Rightarrow I=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x|+c$

Therefore, $\int \tan ^{5} x d x=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x|+c$