$\displaystyle\int_0^{\pi/2} \log\sin x\,dx$ equals

1. By $\mathbf{P_4}$, $I=\int_0^{\pi/2}\log\cos x\,dx$ too. 2. Adding: $2I=\int_0^{\pi/2}\log(\sin x\cos x)\,dx=\int_0^{\pi/2}\log\tfrac{\sin 2x}{2}\,dx$. 3. This gives $2I=\int_0^{\pi/2}\log\sin 2x\,dx-\tfrac{\pi}{2}\log 2$; the first integral reduces to $I$. 4. So $2I=I-\tfrac{\pi}{2}\log 2$, giving $I=-\dfrac{\pi}{2}\log 2$. _Source: syllabus Mathematics Ch 7 "Integrals", p.346_