Answer
$$
\int \sin^2\theta \cos \theta d\theta
=\frac{1}{3}\sin^3\theta +c
$$
Work Step by Step
Since $ u=\sin\theta $, then $ du=\cos \theta d\theta $ and hence, $$
\int \sin^2\theta \cos \theta d\theta=\int u^2 d u= \frac{1}{3}u^3 +c\\
=\frac{1}{3}\sin^3\theta +c.
$$
