Evaluate: $\;$ $\displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$
Let $\; I = \displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$
$\left[\text{Note: Integration by parts: } \displaystyle \int u \cdot v \; dx = u \displaystyle \int v \; dx - \displaystyle \int \left\{\dfrac{du}{dx} \times \displaystyle \int v \; dx \right\} dx \right]$
$\left[ \text{Here: } u = \log \left(\sin x\right), \; v = \cos 2x \right]$
$\begin{aligned}
\therefore \; I & = \log \left(\sin x\right) \int \cos 2x \; dx - \int \left\{\dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \times \int \cos 2x \; dx \right\} dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1}{\sin x} \times \cos x \times \dfrac{\sin 2x}{2} \; dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{\cos x \times 2 \times \sin x \times \cos x}{2 \sin x} \; dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \cos^2 x \; dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1 + \cos 2x}{2} \; dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{1}{2} \int dx - \dfrac{1}{2} \int \cos 2x \; dx \\\\
& = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{x}{2} - \dfrac{\sin 2x}{4} + c
\end{aligned}$