Question

请问这个积分怎么做

Answer 1

let
u= π/2 -x
du = -dx
x=0, u=π/2
x=π/2 , u=0
∫(0->π/2) dx/[ 1+ (cotx)^3 ]
=∫(0->π/2) (sinx)^3/[ (sinx)^3+ (cosx)^3 ] dx

=∫(0->π/2) { 1 - (cosx)^3/[ (sinx)^3+ (cosx)^3 ] } dx
=π/2 - ∫(0->π/2) (cosx)^3/[ (sinx)^3+ (cosx)^3 ] dx
=π/2 -∫(0->π/2) dx/[ 1+ (tanx)^3 ]
=π/2 -∫(π/2->0) -du/[ 1+ (cotu)^3 ]
=π/2 -∫(0->π/2) dx/[ 1+ (cotx)^3 ]
2∫(0->π/2) dx/[ 1+ (cotx)^3 ] =π/2
∫(0->π/2) dx/[ 1+ (cotx)^3 ] =π/4