Question

高数定积分？

为什么第五题第三步直接变成了二分之π变成了π，积分上下限变成了0-π/2，主要是积分上下限不知道怎么变的

Answer 1

见下图：

Answer 2

(5)
let
u=π-x
du=-dx
x=0, u=π
x=π, u=0
I
=∫(0->π) x.√[(sinx)^2-(sinx)^4] dx
=∫(π->0) (π-u).√[(sinu)^2-(sinu)^4] (-du)
=∫(0->π) (π-u).√[(sinu)^2-(sinu)^4] du
=∫(0->π) (π-x).√[(sinx)^2-(sinx)^4] dx
2I
=∫(0->π) π.√[(sinx)^2-(sinx)^4] dx
I =(π/2).∫(0->π) √[(sinx)^2-(sinx)^4] dx =π.∫(0->π/2) √[(sinx)^2-(sinx)^4] dx
let
t= π-x
dt= -dx
x=π, t=0
x=π/2, t= π/2
∫(0->π) √[(sinx)^2-(sinx)^4] dx
=∫(0->π/2) √[(sinx)^2-(sinx)^4] dx +∫(π/2->π) √[(sinx)^2-(sinx)^4] dx
=∫(0->π/2) √[(sinx)^2-(sinx)^4] dx +∫(π/2->0) √[(sint)^2-(sinx)^4] ( -dt)
=∫(0->π/2) √[(sinx)^2-(sinx)^4] dx +∫(0->π/2) √[(sint)^2-(sinx)^4] dt
=∫(0->π/2) √[(sinx)^2-(sinx)^4] dx +∫(0->π/2) √[(sinx)^2-(sinx)^4] dx
=2∫(0->π/2) √[(sinx)^2-(sinx)^4] dx