Integralshard
0:00.0

Consider the integral J=0π/2ln(sinx)dxJ = \int_0^{\pi/2} \ln(\sin x) dx. Given that J=π2ln2J = -\frac{\pi}{2} \ln 2, evaluate K=0π/2ln(sinx+cosx)dxK = \int_0^{\pi/2} \ln(\sin x + \cos x) dx.