The 2 seems out of place

$\begin{aligned} \int_0^{\Large\frac{\pi}{2}} x\ln(\sin 2x)\ dx&=\int_0^{\Large\frac{\pi}{2}} x\ln(2\sin x\cos x)\ dx\\ &=\ln 2\int_0^{\Large\frac{\pi}{2}} x\ dx+\int_0^{\Large\frac{\pi}{2}} x\ln(\sin x)\ dx+\underbrace{\int_0^{\Large\frac{\pi}{2}} x\ln(\cos x)\ dx}_{\color{#D61F06}{\Large x\,\mapsto\,x-\frac{\pi}{2}}}\\ &=\frac{\pi^2}{8}\ln2+\int_0^{\Large\frac{\pi}{2}} x\ln(\sin x)\ dx+\int_0^{\Large\frac{\pi}{2}} \left(\frac{\pi}{2}-x\right)\ln(\sin x)\ dx\\ &=\frac{\pi^2}{8}\ln2+\frac{\pi}{2}\underbrace{\int_0^{\Large\frac{\pi}{2}} \ln(\sin x)\ dx}_{\color{#D61F06}{\Large \text{use symmetry argument}}}\ \Rightarrow\ \color{#D61F06}{\text{ or ask Uncle Google 😂}}\\[12pt] &=\frac{\pi^2}{8}\ln2+\frac{\pi}{2}\left[-\frac{\pi}{2}\ln2\right]\\ &=\color{#3D99F6}{-\frac{\pi^2}{8}\ln2} \end{aligned}$

Let the integral be denoted by $I$ . $I = \displaystyle \int_{0}^{\frac{\pi}{2}} x\ln(\sin 2x) dx$

$sin2x = sin(\pi - 2x) = \sin(2(\dfrac{\pi}{2} - x))$

Hence,

$I = \displaystyle \int_{0}^{\frac{\pi}{2}} \left(\dfrac{\pi}{2} - x\right) \ln(\sin 2x) dx$

The above step was achieved by using the property:

$\displaystyle \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$

$I = \displaystyle \dfrac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \ln(\sin 2x) dx - I$

$2I = \displaystyle \dfrac{\pi}{2} \int_{0}^{\frac{\pi}{2}} ( \ln 2 + \ln (\sin x) + \ln (\cos x) ) dx$

$2I = \displaystyle \dfrac{\pi}{2} \left( \dfrac{\pi}{2} \ln 2 + \dfrac{-\pi}{2} \ln 2 - \dfrac{-\pi}{2} \right)$

$2I = \displaystyle \dfrac{\pi^2}{4} \ln 2 - \dfrac{\pi^2}{4} \ln 2 -\dfrac{\pi^2}{4} \ln 2$

$\Rightarrow 2I = \dfrac{-\pi^2 \ln 2}{4} \Rightarrow I = \dfrac{-\pi^2 \ln 2}{8}$

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Note - $\int_{0}^{\frac{\pi}{2}} \ln (\sin x) = \int_{0}^{\frac{\pi}{2}} \ln (\cos x) = \dfrac{-\pi}{2} \ln 2$

I consider this to be a problem whose derivation can be done easily.