Answer
$\frac{1}{4}\sec^{4}\theta +C$
Work Step by Step
$\int \tan\theta \sec^{4}\theta d\theta$
$=\int \tan\theta \sec\theta \sec^{3}\theta d\theta$
Use u-substitution.
Let $u=\sec\theta$
$\frac{du}{dx}= \tan\theta \sec\theta$
$dx=\frac{1}{\tan\theta \sec\theta}du$
$=\int u^{3}du$
$=\frac{1}{4}u^{4}+C$
$=\frac{1}{4}\sec^{4}\theta +C$