Answer
$\frac{1}{6}\tan^32x + \frac{1}{2}\tan2x +C$
Work Step by Step
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
