Evaluate $\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \left(2x\right) \cdot \log \left(\sin x\right) \; dx$
$\begin{aligned} \text{Let } I & = \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \left(2x\right) \cdot \log \left(\sin 2x\right) \; 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 = \log \left(\sin x\right); \; v = \cos \left(2x\right) \end{aligned}\right] \\\\ & = \left[\log \left(\sin x\right) \int \cos \left(2x\right) \; dx\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left\{\int \cos \left(2x\right) \times \dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \right\} \; dx \\\\ & = \left[\dfrac{1}{2} \sin \left(2x\right) \; \log \left(\sin x\right)\right]_{\pi / 4}^{\pi / 2} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\sin \left(2x\right)}{2} \times \dfrac{1}{\sin x} \times \cos x \; dx \\\\ & = \dfrac{1}{2} \left[\sin \left(\pi\right) \; \log \left|\sin \left(\dfrac{\pi}{2}\right)\right| - \sin \left(\dfrac{\pi}{2}\right) \; \log \left|\sin \left(\dfrac{\pi}{4}\right)\right|\right] \\ & \hspace{10em} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{2 \sin x \cos x \times \cos x}{2 \sin x} \; dx \\\\ & = \dfrac{1}{2} \left[0 - 1 \times \log \left|\dfrac{1}{\sqrt{2}}\right|\right] - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 x \; dx \\\\ & = \dfrac{1}{2} \log \left|\sqrt{2}\right| - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2x}{2}\right) \; dx \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} dx - \dfrac{1}{2} \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2x \; dx \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \left[x\right]_{\pi / 4}^{\pi / 2} - \dfrac{1}{2} \left[\dfrac{\sin 2x}{2}\right]_{\pi / 4}^{\pi / 2} \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \left[\dfrac{\pi}{2} - \dfrac{\pi}{4}\right] - \dfrac{1}{4} \left[\sin \left(\pi\right) - \sin \left(\dfrac{\pi}{2}\right)\right] \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{\pi}{8} - \dfrac{1}{4} \left[0 - 1\right] \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{\pi}{8} + \dfrac{1}{4} \end{aligned}$