Evaluate ∫0π/2ln(sinx)dx=−π2ln2\int_0^{\pi/2} \ln(\sin x) dx = -\frac{\pi}{2} \ln 2∫0π/2ln(sinx)dx=−2πln2. What is ∫0π/2ln(cosx)dx\int_0^{\pi/2} \ln(\cos x) dx∫0π/2ln(cosx)dx?
−π2ln2-\frac{\pi}{2} \ln 2−2πln2
π2ln2\frac{\pi}{2} \ln 22πln2
000
πln2\pi \ln 2πln2