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