\(\int_{0}^{\frac{\pi}{2}} \log \left(\sin x\right)dx\) is equal to

\(-\frac{\pi}{2}\log 2\)

\(I=\int\limits_{0}^{\frac{\pi}{2}} \log\sin xdx=\int\limits_{0}^{\frac{\pi}{2}} \log\left(\sin\left(\frac{\pi}{2}-x\right)\right)dx=\int\limits_{0}^{\frac{\pi}{2}} \log\cos xdx\)

so $2I=\int\limits_{0}^{\frac{\pi}{2}} \log\sin x\cos xdx$

$2I=\int\limits_{0}^{\frac{\pi}{2}} \log\sin 2x-\log 2dx$

$2I=\frac{1}{2}\int\limits_{0}^{\pi}\log\sin xdx-\frac{\pi}{2}\log 2$

so $2I=\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\log\sin xdx-\frac{\pi}{2}\log 2$

$2I=I-\frac{\pi}{2}\log 2⇒I=-\frac{\pi}{2}\log 2$