Answer
$$\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x+\ln |\sec x|+C $$
Work Step by Step
We evaluate the trigonometric integral as follows:
\begin{align*}
\int \tan ^{5} x d x&=\int \tan ^{4} x \tan x d x=\int\left(\sec ^{2} x-1\right)^{2} \tan x d x\\
&=\int\left(\sec ^{4} x-2 \sec ^{2} x+1\right) \tan x d x\\
&=\int \sec ^{4} x \tan x d x-2 \int \sec ^{2} x \tan x d x+\int \tan x d x\\
&=\int \sec ^{3} x \sec x \tan x d x-2 \int \sec x \sec x \tan x d x+\int \tan x d x\\
&=\frac{1}{4} \sec ^{4} x-\sec ^{2} x+\ln |\sec x|+C\\
&=\frac{1}{4}\left(\tan ^{2} x+1\right)^{2}-\left(\tan ^{2} x+1\right)+\ln |\sec x|+C\\
&=\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x+\ln |\sec x|+C
\end{align*}
Where we used the fact that:
$\sec^2 x = \tan^2 x +1$
