Answer
$$\frac{1}{3}\tan ^{3} x+\frac{1}{5}\tan ^{5} x+c$$
Work Step by Step
\begin{align*}
\int \tan ^{2} x \sec ^{4} x d x&=\int \tan ^{2} x (1+\tan ^{2} x)\sec ^{2} x d x\\
&= \int \tan ^{2} x \sec ^{2} x d x+\int \tan ^{4} x \sec ^{2} x d x\\
&=\frac{1}{3}\tan ^{3} x+\frac{1}{5}\tan ^{5} x+c
\end{align*}
