Prove that $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left[2 \log \left(\sin x\right) - \log \left(\sin 2x\right)\right] \; dx = - \dfrac{\pi}{2} \log \left(2\right)$
Let $I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left[2 \log \left(\sin x\right) - \log \left(\sin 2x\right)\right] \; dx$ $\;\;\; \cdots \; (1)$
Now,
$\begin{aligned}
2 \log \left(\sin x\right) - \log \left(\sin 2x\right) & = \log \left(\sin^2 x\right) - \log \left(2 \sin x \cos x\right) \\\\
& = \log \left(\dfrac{\sin^2 x}{2 \sin x \cos x}\right) \\\\
& \hspace{3em} \left[\text{Note: } \log \left(C\right) - \log \left(D\right) = \log \left(\dfrac{C}{D}\right)\right] \\\\
& = \log \left(\dfrac{\sin x}{2 \cos x}\right) \\\\
& = \log \left(\dfrac{\tan x}{2}\right) \\\\
& = \log \left(\tan x\right) - \log \left(2\right) \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as
$I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x\right) \; dx - \int \limits_{0}^{\frac{\pi}{2}} \log \left(2\right) \; dx$ $\;\;\; \cdots \; (3)$
$\begin{aligned}
\text{Let } I_1 & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x\right) \; dx \;\;\; \cdots \; (4) \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} \log \left[\tan \left(\dfrac{\pi}{2} - x\right)\right] \; dx \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\cot x\right) \; dx \;\;\; \cdots \; (5)
\end{aligned}$
Adding equations $(4)$ and $(5)$ gives
$\begin{aligned}
2 I_1 & = \int \limits_{0}^{\frac{\pi}{2}} \left[\log \left(\tan x\right) + \log \left(\cot x\right)\right] \; dx \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x \times \cot x\right) \; dx \hspace{3em} \left[\text{Note: } \log C + \log D = \log \left(C \cdot D\right)\right] \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} \log \left(1\right) \; dx = 0
\end{aligned}$
$\implies$ $I_1 = 0$ $\;\;\; \cdots \; (6)$
$\begin{aligned}
\text{Let } I_2 & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(2\right) \; dx \\\\
& = \log \left(2\right) \times \left[x\right]_{0}^{\pi / 2}
\end{aligned}$
i.e. $I_2 = \dfrac{\pi}{2} \log \left(2\right)$ $\;\;\; \cdots \; (7)$
$\therefore$ $\;$ In view of equations $(6)$ and $(7)$, equation $(3)$ becomes
$I = 0 - \dfrac{\pi}{2} \log \left(2\right) = - \dfrac{\pi}{2} \log \left(2\right)$ $\hspace{2em}$ Hence proved