Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 52

$$\frac{2}{15}$$

\begin{align*} \int_{0}^{ \pi/2}\sin^2x\cos^3 xdx&= \int_{0}^{ \pi/2}\sin^2x\cos^2 x\cos xdx \\ &= \int_{0}^{ \pi/2}\sin^2x (1-\sin^2x) \cos xdx \\ &=\int_{0}^{ \pi/2} (\sin^2x-\sin^4x) \cos xdx \\ &= \left( \frac{1}{3}\sin^3x -\frac{1}{5}\sin^5x \right) \bigg|_{0}^{\pi/2}\\ &=\frac{2}{15} \end{align*}