Evaluate ∫0π/2ln(sinx)dx\int_{0}^{\pi/2} \ln(\sin x) dx∫0π/2ln(sinx)dx.
−π2ln2-\frac{\pi}{2} \ln 2−2πln2
π2ln2\frac{\pi}{2} \ln 22πln2
−πln2-\pi \ln 2−πln2
ln2\ln 2ln2