Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx$
$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = x^2; \; v = \cos 2x \end{aligned}\right] \\\\ & = \left[x^2 \int \cos 2x \; dx\right]_{0}^{\frac{\pi}{2}} - \int \limits_{0}^{\frac{\pi}{2}} \left[\int \cos 2x \; dx \times \dfrac{d}{dx} \left(x^2\right)\right] dx \\\\ & = \dfrac{1}{2} \left[x^2 \; \sin 2x\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin 2x}{2} \times 2 x \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi^2}{4} \; \sin \left(\pi\right) - 0\right] - \int \limits_{0}^{\frac{\pi}{2}} x \; \sin \left(2x\right) \; dx \\\\ & \hspace{10em} \left[\text{Here } u = x; \; v = \sin \left(2x\right)\right] \\\\ & = 0 - \left[x \int \sin \left(2x\right) \; dx\right]_{0}^{\pi / 2} + \int \limits_{0}^{\frac{\pi}{2}} \left[\sin \left(2x\right) \; dx \times \dfrac{d}{dx} \left(x\right)\right] \; dx \\\\ & = \dfrac{1}{2} \left[x \; \cos \left(2x\right)\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos \left(2x\right)}{2} \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi}{2} \; \cos \left(0\right) - 0\right] - \dfrac{1}{4} \left[\sin \left(2x\right)\right]_{0}^{\pi / 2} \\\\ & = \dfrac{- \pi}{4} - \dfrac{1}{4} \left[\sin \left(\pi\right) - \sin \left(0\right)\right] \\\\ & = \dfrac{- \pi}{4} \end{aligned}$