Answer
$$\frac{2}{15}$$
Work Step by Step
Given $$\int_{0}^{\pi/2}\sin^3\theta\cos^2\theta d \theta$$
Since
\begin{align*}
\int_{0}^{\pi/2}\sin^3\theta\cos^2\theta d \theta&=\int_{0}^{\pi/2}\sin^2\theta\cos^2\theta \sin\theta d \theta\\
&=\int_{0}^{\pi/2}(1-\cos^2\theta)\cos^2\theta \sin\theta d \theta\\
&=\int_{0}^{\pi/2}\cos^2\theta \sin\theta d \theta-\int_{0}^{\pi/2}\cos^4\theta \sin\theta d \theta \\
&=-\frac{1}{3}\cos^3\theta + \frac{1}{5}\cos^5 \theta \bigg|_{0}^{\pi/2}\\
&=\frac{1}{3} - \frac{1}{5}\\
&=\frac{2}{15}
\end{align*}
