求反常积分∫ln(sinx)dx,上限是π⼀2,下限是0，希望给出过程，多谢

∫ln(sinx)dx上限是π/2,下限是0，将x都改成π/2-x;得
∫ln(sinx)dx上限是π/2,下限是0 = -∫ln(cosx)dx上限是0,下限是π/2
= ∫ln(cosx)dx上限是π/2,下限是0;(*)
同理（* ）式中再将x都变成x-π/2,得：
∫ln(sinx)dx上限是π/2,下限是0 =∫ln(sinx)dx上限是π,下限是 π/2；
于是∫ln(sinx)dx上限是π/2,下限是0 =二分之一倍的 ∫ln(sinx)∫ln(sinx)dx上限是π/2,下限是0

∫ln(sinx)dx上限是π/2,下限是0 + ∫ln(cosx)dx上限是π/2,下限是0 +∫ln(2)dx上限是π/2,下限是0
=∫ln(2*sinx*cosx)dx上限是π/2,下限是0
=∫ln(sin2x)dx上限是π/2,下限是0
=(1/2)*∫ln(sinx)dx上限是π,下限是0
=∫ln(sinx)dx上限是π/2,下限是0

所以可得：
∫ln(sinx)dx上限是π/2,下限是0=- ∫ln(2)dx上限是π/2,下限是0
=-π*ln(2)/2

I=∫(0,π/2) ln(sinx)dx =∫(0,π/4) ln(sinx)dx+∫(π/4,π/2) ln(sinx)dx =∫(0,π/4) ln(sinx)dx+∫(π/4,0) ln(sin(π/2-x))d(π/2-x)=∫(0,π/4) ln(sinx)dx+∫(0,π/4) ln(cosx)dx=∫(0,π/4) ln(1/2sin2x)dx=∫(0,π/4) ln(sin2x)dx+πln(1/2)/4=1/2∫(0,π/2) ln(sinx)dx-πln2/4=I/2-πln2/4即I=I/2-πln2/4，所以I=-πln2/2