不定积分∫㏑﹙sin x﹚dx怎么求啊~有谁知道啊。谢谢啦。速度求

A = ∫(0→π/2) ln(sinx) dx
令x = π/2 - u，dx = - du
A = ∫(π/2→0) ln(sin(π/2 - u)) (- du)
A = ∫(0→π/2) ln(cosx) dx
2A = ∫(0→π/2) ln(sinx) dx + ∫(0→π/2) ln(cosx) dx
2A = ∫(0→π/2) ln(sinxcosx) dx
2A = ∫(0→π/2) ln(1/2 • sin2x) dx
2A = ∫(0→π/2) ln(1/2) dx + ∫(0→π/2) ln(sin2x) dx
2A = (- π/2)ln(2) + B
令z = 2x，dx = 1/2 dz
B = (1/2)∫(0→π) ln(sinz) dz
= (1/2)∫(0→π/2) ln(sinx) dx + (1/2)∫(π/2→π) ln(sinx) dx
= (1/2)A + (1/2)C
令x = π - y，dx = - dy
C = ∫(π/2→0) ln(sin(π - y)) (- dy)
= ∫(0→π/2) ln(sinx) dx
= A
于是
B = (1/2)(A + C) = A
2A = (- π/2)ln(2) + A
A = (- π/2)ln(2)
=> ∫(0→π/2) ln(sinx) dx = (- π/2)ln(2)