第1个回答 2020-09-26
let
y=π-x
dy=-dx
x=0, y=π
x=π, y=0
∫(0->π) x.(sinx)^2 dx
=∫(π->0) (π-y).(siny)^2 (-dy)
=∫(0->π) (π-y).(siny)^2 dy
=∫(0->π) (π-x).(sinx)^2 dx
2∫(0->π) x.(sinx)^2 dx =π∫(0->π) (sinx)^2 dx
∫(0->π) x.(sinx)^2 dx =(π/2)∫(0->π) (sinx)^2 dx