Answer
$$\int 2\sin^2 t\sec^4t dt=\frac{2}{3}\tan^3t+C$$
Work Step by Step
$$A=\int 2\sin^2 t\sec^4t dt$$ $$A=2\int\frac{\sin^2t}{\cos^2t}\sec^2tdt$$ $$A=2\int\tan^2t\sec^2tdt$$
We set $u=\tan t$, which means $$du=\sec^2tdt$$
Therefore, $$A=2\int u^2du=\frac{2u^3}{3}+C$$ $$A=\frac{2}{3}\tan^3t+C$$
