设t=-x、I=∫(-π/2,π/2)sin²tdt/(1+e^t)。∴I=∫(-π/2,π/2)sin²xdx/[1+e^(-x)]=∫(-π/2,π/2)(e^x)sin²xdx/(1+e^x)。与未换元前的I相加，∴2I=∫(-π/2,π/2)sin²xdx=∫(0,π/2)2sin²xdx=∫(0,π/2)(1-cos2x)dx=π/2。∴原式=3I=3π/4。供参考。
