Answer
$$\frac{1}{4}\tan^3x-\frac{1}{2}\tan^2x+\ln \sec x+c$$
Work Step by Step
Given
$$\int \tan^5xdx $$
Since
$$1+\tan^2x=\sec^2x $$
Then
\begin{align*}
\int \tan^5xdx &=\int \tan^3x \tan^2xdx \\
&=\int \tan^3x (\sec^2x-1)dx \\
&=\int( \tan^3x \sec^2x-\tan x(\sec^2x-1))dx \\
&=\int( \tan^3x \sec^2x-\tan x \sec^2x +\tan x )dx \\
&=\frac{1}{4}\tan^3x-\frac{1}{2}\tan^2x+\ln \sec x+c
\end{align*}