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