Answer
$$\int\sec^22\theta d\theta=\frac{1}{2}\tan(2\theta)+C$$
Work Step by Step
$$A=\int\sec^22\theta d\theta$$
Let $u=2\theta$
Then $du=2d\theta$. So $d\theta=\frac{1}{2}du$
Substitute into $A$, we have $$A=\int\sec^2u(\frac{1}{2})du$$ $$A=\frac{1}{2}\int \sec^2udu$$ $$A=\frac{1}{2}(\tan u)+C$$ $$A=\frac{1}{2}\tan(2\theta)+C$$