求助帮我看看这道题

let
u=-x
du=-dx
x=-π/2, u=π/2
x=π/2, u=-π/2
I
=∫(-π/2->π/2) 3e^x. (sinx)^2/(1+e^x) dx
=∫(π/2->-π/2) { 3e^(-u). (sinu)^2/[1+e^(-u)] } (-du)
=∫(-π/2->π/2) 3e^(-u). (sinu)^2/[1+e^(-u)] du
=∫(-π/2->π/2) 3e^(-x). (sinx)^2/[1+e^(-x)] dx
=∫(-π/2->π/2) { (3/e^x). (sinx)^2/(1+ 1/e^x) dx
=∫(-π/2->π/2) 3(sinx)^2/(1+ e^x) dx
2I
=∫(-π/2->π/2) 3e^x. (sinx)^2/(1+e^x) dx +∫(-π/2->π/2) 3(sinx)^2/(1+ e^x) dx
=∫(-π/2->π/2) 3(1+e^x). (sinx)^2/(1+e^x) dx
=∫(-π/2->π/2) 3(sinx)^2 dx
=6∫(0->π/2) (sinx)^2 dx
=3∫(0->π/2) (1-cos2x) dx
=3[x- (1/2)sin2x]|(0->π/2)
=3π/2
I=3π/4