Answer
$\frac{1}{2}(sec(2t))+c$
Work Step by Step
$u=2t$, therefore $du=2dt$. Let's rewrite the integral! $\int (sec(2t)tan(2t))dt$
We do the substitution! $\int (sec(2t)tan(2t))dt=\int (sec(u)tan(u))\frac{du}{2}=\frac{1}{2}\int (sec(u)tan(u))du=\frac{1}{2}(secu)+c$
We re-do the substitution! $\frac{1}{2}(secu)+c=\frac{1}{2}(sec(2t))+c$