Answer
$\frac{\sec^52t}{10} - \frac{\sec^32t}{6} +C$
Work Step by Step
Find the indefinite integral
$\int tan^32t sec^32tdt$
$\int tan^2(2t)sec^2(2t) sec(2t)tan(2t)dt$, Break the integral up
$\int (sec^22t-1) sec^2(2t) sec(2t) tan(2t) dt$
$\int(sec^42t -sec^22t) sec(2t) tan(2t)dt$
Let $u=sec2t$, and $du=2sec(2t)tan(2t)dt$
$\frac{1}{2}\int u^4du - \frac{1}{2}\int u^2du$, U-substitution
$\frac{u^5}{10} - \frac{u^3}{6}+C$
$\frac{\sec^52t}{10} - \frac{\sec^32t}{6} +C$
