Question

高数题不会？

？？

Answer 1

let
u=-x
du =-dx
∫(-π/2->π/2) (sinx)^4/ [1+e^(-x) ] dx
=∫( π/2->-π/2) (sinu)^4/ [1+e^u ] (-du)
=∫( -π/2->π/2) (sinx)^4/ [1+e^x ] dx
//
∫(-π/2->π/2) (sinx)^4/ [1+e^(-x) ] dx
=∫(-π/2->π/2) e^x. (sinx)^4/ [1+e^x ] dx
=∫(-π/2->π/2) (sinx)^4 dx - ∫(-π/2->π/2) (sinx)^4/ [1+e^x ] dx
=∫(-π/2->π/2) (sinx)^4 dx - ∫(-π/2->π/2) (sinx)^4/ [1+e^(-x) ] dx
2∫(-π/2->π/2) (sinx)^4/ [1+e^(-x) ] dx = ∫(-π/2->π/2) (sinx)^4 dx
∫(-π/2->π/2) (sinx)^4/ [1+e^(-x) ] dx
=(1/2) ∫(-π/2->π/2) (sinx)^4 dx
=∫(0->π/2) (sinx)^4 dx
=(1/4)∫(0->π/2) (1- cos2x)^2 dx
=(1/4)∫(0->π/2) [1- 2cos2x + (cos2x)^2 ] dx
=(1/8)∫(0->π/2) (3- 4cos2x + cos4x) dx
=(1/8) [ 3x - 2sin2x +(1/4)sin4x ]|(0->π/2)
=3π/16

Answer 2

这个积分没办法算的。