Answer
$$\frac{8}{15}$$
Work Step by Step
\begin{align*}
\int_{0}^{ \pi/2}\sin^5xdx&= \int_{0}^{ \pi/2}\sin^4x\sin xdx \\
&= \int_{0}^{ \pi/2}(1-\cos^2x)^2\sin xdx \\
&= \int_{0}^{ \pi/2} \left(1-2\cos^2x+\cos^4x\right) \sin xdx \\
&= \left(-\cos x+\frac{2}{3}\cos^3x -\frac{1}{5}\cos^5x \right) \bigg|_{0}^{\pi/2}\\
&=\frac{8}{15}
\end{align*}
