## Calculus 10th Edition

$\frac{1}{6}\tan^32x + \frac{1}{2}\tan2x +C$
Find the indefinite integral $\int sec^42x dx$ Split into two factors of $sec^22x$ $\int sec^22x sec^22x dx$, Convert secants to tangents $\int (tan^22x +1) sec^22x dx$, Divide into two integrations $\int (tan^22x sec^22x dx + \int sec^22x dx$ Use u-substitution for both integrals For the first, let $u=tan2x$, and $du=2sec^22x$ For the second, let $u=2x$, and $du=2dx$ $\frac{1}{2}\int u^2 du + \frac{1}{2}\int \sec^2u du$, Integrate $\frac{1}{6}u^3 + \frac{1}{2}\tan u +C$ $\frac{1}{6}\tan^32x + \frac{1}{2}\tan2x +C$, Resubstitute