Question:

Let $I = \int_{0}^{\pi/2} \log(\sin x) dx$ ...(i)<br/>then $I = \int_{0}^{\pi/2} \log \sin \left( \frac{\pi}{2} - x \right) dx = \int_{0}^{\pi/2} \log(\cos x) dx$ ...(ii)<br/>adding (i) and (ii), we get<br/>$2I = \int_{0}^{\pi/2} \log \sin x dx + \int_{0}^{\pi/2} \log \cos x dx = \int_{0}^{\pi/2} (\log \sin x + \log \cos x) dx$<br/>$\Rightarrow 2I = \int_{0}^{\pi/2} \log(\sin x \cos x) dx = \int_{0}^{\pi/2} \log \left( \frac{2 \sin x \cos x}{2} \right) dx$