Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx$
$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 - \cos 2x}{2}\right) \; dx \\\\ & = \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} dx - \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} \cos 2x \; dx \\\\ & = \dfrac{1}{2} \left[x\right]_{0}^{\pi / 2} - \dfrac{1}{2} \left[\dfrac{\sin 2x}{2}\right]_{0}^{\pi / 2} \\\\ & = \dfrac{1}{2} \left(\dfrac{\pi}{2} - 0\right) - \dfrac{1}{4} \left(\sin \pi - \sin 0\right) \\\\ & = \dfrac{\pi}{4} - \dfrac{1}{4} \left(0 - 0\right) \\\\ & = \dfrac{\pi}{4} \end{aligned}$