Using the identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$ $$\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx$$ $$= \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2}$$ $$= \frac{1}{2}\left[\left(\frac{\pi}{2} - \frac{\sin \pi}{2}\right) - \left(0 - \frac{\sin 0}{2}\right)\right]$$ $$= \frac{1}{2}\left[\frac{\pi}{2} - 0 - 0\right] = \frac{\pi}{4}$$