Answer
$$\frac{1}{6}\tan^6x+ \frac{1}{8}\tan^8x+c$$
Work Step by Step
\begin{align*}
\int \tan^5x\sec^4 xdx&= \int \tan^5x\sec^2 x\sec^2 x dx\\
&= \int \tan^5x(1+\tan^2x)\sec^2 x dx\\
&= \int (\tan^5x+\tan^7x)\sec^2 x dx\\
&=\frac{1}{6}\tan^6x+ \frac{1}{8}\tan^8x+c
\end{align*}
