Answer
$$\frac{1}{7}\tan^7x+ \frac{1}{9}\tan^9x+c$$
Work Step by Step
\begin{align*}
\int \tan^6x\sec^4 xdx&= \int \tan^6x\sec^2 x\sec^2 x dx\\
&= \int \tan^6x(1+\tan^2x)\sec^2 x dx\\
&= \int (\tan^6x+\tan^8x)\sec^2 x dx\\
&=\frac{1}{7}\tan^7x+ \frac{1}{9}\tan^9x+c
\end{align*}
