Answer
$$\int\sec^2\theta\tan^3\theta d\theta=\frac{\tan^4\theta}{4}+C$$
Work Step by Step
$$A=\int\sec^2\theta\tan^3\theta d\theta$$
Let $u=\tan\theta$.
Then $du=\sec^2\theta d\theta$.
Substitute into $A$, we have $$A=\int u^3du$$ $$A=\frac{u^4}{4}+C$$ $$A=\frac{\tan^4\theta}{4}+C$$
