Consider the integral J=∫0π/2ln(sinx)dxJ = \int_0^{\pi/2} \ln(\sin x) dxJ=∫0π/2ln(sinx)dx. Given that J=−π2ln2J = -\frac{\pi}{2} \ln 2J=−2πln2, evaluate K=∫0π/2ln(sinx+cosx)dxK = \int_0^{\pi/2} \ln(\sin x + \cos x) dxK=∫0π/2ln(sinx+cosx)dx.
−π4ln2-\frac{\pi}{4} \ln 2−4πln2
−π2ln2-\frac{\pi}{2} \ln 2−2πln2
−π4ln(2)-\frac{\pi}{4} \ln(\sqrt{2})−4πln(2)
000