Answer
$$\int\sec^4\theta d\theta=\frac{\tan^3\theta}{3}+\tan\theta+C$$
Work Step by Step
$$A=\int\sec^4\theta d\theta$$ $$A=\int\sec^2\theta(\sec^2\theta d\theta)$$ $$A=\int(\tan^2\theta+1)d(\tan \theta)$$
We set $u=\tan\theta$, which means $$A=\int(u^2+1)du$$ $$A=\frac{u^3}{3}+u+C$$ $$A=\frac{\tan^3\theta}{3}+\tan\theta+C$$
