Answer
$\frac{\pi}{4}$
Work Step by Step
Recall the identity
$\cos 2x=1-2\sin^{2} x$ which gives $\sin^{2}x=\frac{1-\cos 2x}{2}$.
Therefore, $\int \sin^{2}x\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 2x\,dx$
$= \frac{x}{2}-\frac{1}{4}\sin 2x+C$
Then $\int_{0}^{\pi/2} \sin^{2}x\,dx=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi/2}_{0}=(\frac{\pi}{4}-0)-0=\frac{\pi}{4}$
