$$\begin {aligned} \int _{0}^{\frac {\pi }2}\ln \left (2\sin \frac {x}2\right )\ln \left (2\cos \frac {x}2\right )dx &=\frac {1}4\int _{-\pi }^{\pi }\ln \left |2\sin \frac {x}2\right |\ln \left |2\cos \frac {x}2\right |dx\\ &=\frac {1}4\int _{-\pi }^{\pi }\sum _{n=1}^\infty \sum _{m=1}^\infty \frac {(-1)^n\cos (nx)\cos (mx)}{nm}dx\\ &=\frac {1}4\sum _{n=1}^{\infty }\sum _{m=1}^\infty \frac {(-1)^n}{nm}\int _{-\pi }^{\pi }\cos (nx)\cos (mx)dx\\ &=\frac {1}4\sum _{n=1}^\infty \frac {(-1)^{n}}{n^{2}}\pi \\ &=\frac {\pi }4\cdot \frac {-\pi ^{2}}{12}\\ &=\textcolor {blue}{-\frac {\pi ^{3}}{48}}. \end {aligned}$$

$$\begin {aligned} \frac {\pi ^{3}}{96}&=-\int _{0}^{\frac {\pi }4}\ln (2\sin x)\ln (2\cos x)dx\\ &=- \int _{0}^{\frac {\pi }4}\left (\ln ^{2}2+\ln (2)(\ln \sin x+\ln \cos x)+\ln (\sin x)\ln (\cos x)\right )dx\\ \int _{0}^{\frac {\pi }4}\ln (\sin x)\ln (\cos x)dx &=-\frac {\pi ^{3}}{96}-\frac {\pi \ln ^{2}2}4-\ln (2)\int _{0}^{\frac {\pi }4}(\ln \sin x+\ln \cos x)dx\\ &=-\frac {\pi ^{3}}{96}-\frac {\pi \ln ^{2}2}4-\ln (2)\int _{0}^{\frac {\pi }2}\ln \sin xdx\\ &=-\frac {\pi ^{3}}{96}-\frac {\pi \ln ^{2}2}4+\frac {\pi \ln ^{2}2}2\\ &=\textcolor {blue}{-\frac {\pi ^{3}}{96}+\frac {\pi \ln ^{2}2}4}. \end {aligned}$$