Solution:

Assume $I=\int \tan ^{5} x d x$
We can write the above equation as
$\Rightarrow I=\int \tan ^{2} x \tan ^{3} x d x$
By using the standard formula
$\Rightarrow I=\int\left(\sec ^{2} x-1\right) \tan ^{3} x d x$
On splitting the above equation we obtain
$\begin{array}{l}
\Rightarrow I=\int \tan ^{3} x \sec ^{2} x d x-\int \tan ^{3} x d x \\
\Rightarrow I=\int \tan ^{3} x \sec ^{2} x d x-\int\left(\sec ^{2} x-1\right) \tan x d x
\end{array}$
$\Rightarrow \mathrm{I}=\int \tan ^{3} \mathrm{x} \sec ^{2} \mathrm{x} \mathrm{dx}-\int\left(\sec ^{2} \mathrm{x} \tan \mathrm{x}\right) \mathrm{dx}+\int \tan \mathrm{x} d \mathrm{x}$
Assume $\tan \mathrm{x}=\mathrm{t}$, therefore
$\begin{array}{l}
\Rightarrow \sec ^{2} x d x=d t \\
\Rightarrow I=\int t^{3} d t-\int t d t+\int \tan x d x \\
\Rightarrow I=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x|+c
\end{array}$
$\Rightarrow \mathrm{I}=\frac{\tan ^{4} \mathrm{x}}{4}-\frac{\tan ^{2} \mathrm{x}}{2}+\log |\sec \mathrm{x}|+\mathrm{c}$
Hence, $\int \tan ^{5} x d x=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x|+c$