Answer
$$\frac{1}{5}\sec^5x-\frac{2}{3}\sec^3x+\sec x+C$$
Work Step by Step
Given $$\int \tan ^{5} x \sec x d x$$
Then
\begin{align*}
\int \tan ^{5} x \sec x d x&=\int \tan ^{4} x \sec x\tan x d x\\
&= \int (\sec^2 x-1) ^{2} x \sec x\tan x d x,\ \ \ \ \text{let } u = \sec x,\ \ d\sec x\tan xdx\\
&= \int (u^2-1)^2du\\
&= \int (u^4-2u^2+1)du\\
&= \frac{1}{5}u^5-\frac{2}{3}u^3+u+C\\
&= \frac{1}{5}\sec^5x-\frac{2}{3}\sec^3x+\sec x+C
\end{align*}
