Answer
$$\int \sin^2\theta\cos\theta d\theta=\frac{\sin^3\theta}{3}+C$$
Work Step by Step
$$A=\int \sin^2\theta\cos\theta d\theta$$
Let $u=\sin\theta$
Then $du=(\sin\theta)'d\theta=\cos\theta d\theta$.
Substitute into $A$, we have $$A=\int u^2du$$ $$A=\frac{u^3}{3}+C$$ $$A=\frac{\sin^3\theta}{3}+C$$