Answer
$$\int \sin^2\theta\cos\theta d\theta=\frac{\sin^3\theta}{3}+c$$
Work Step by Step
To solve the integral $\int \sin^2\theta\cos\theta d\theta$ we will use substitution $u=\sin\theta$ which gives us $du=\cos\theta d\theta.$ Putting this into our integral we get:
$$\int \sin^2\theta\cos\theta d\theta=\int u^2du=\frac{u^3}{3}+c$$
where $c$ is arbitrary constant.
Now we have to express solution in terms of $\theta$ by expressing $u$ in terms of $\theta$:
$$\int \sin^2\theta\cos\theta d\theta=\frac{u^3}{3}+c=\frac{\sin^3\theta}{3}+c$$
