Answer
$\displaystyle \frac{2}{15}$
Work Step by Step
Leave one sine and use $\sin^{2}t+\cos^{2}t=1.$
$\displaystyle \int_{0}^{\pi/2}\sin^{3}\theta\cos^{2}\theta d\theta=\int_{0}^{\pi/2}\left(1-\cos^{2}\theta\right)\cos^{2}\theta\sin\theta d\theta=$
$\left[\begin{array}{ll}
u=\cos\theta & \\
du=-\sin\theta d\theta & \\
\theta=0\Rightarrow u=1, & \theta=\pi/2\Rightarrow u=0
\end{array}\right]$
$=\displaystyle \int_{1}^{0}\left(1-u^{2}\right)u^{2}(-du)\quad $
$=\displaystyle \int_{0}^{1}\left(u^{2}-u^{4}\right)du$
$=\displaystyle \left[\frac{1}{3}u^{3}-\frac{1}{5}u^{5}\right]_{0}^{1}$
$=\displaystyle \left(\frac{1}{3}-\frac{1}{5}\right)-0$
$=\displaystyle \frac{2}{15}$
