## Calculus 10th Edition

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