Question

三角函数微分方程

Answer 1

令t=π-x 原式=-∫(π->0)(π-t)sint/(1+(cost)^2)dt=-1∫(0->pai)tsint/(1+(cost)^2)dt+∫(0->pai)πsint/(1+(cost)^2)dt 所以原式=0.5*∫(0->pai)πsint/(1+(cost)^2)dt=π^2/8(你自己在计算一遍)
实际上这个公式∫(0->pai)xf(sinx)dx=pai/2 ∫(0->pai)f(sinx)dx(运用t=π-x变换证明即可)

Answer 2

∫(0->π)xsinx/(1+(cosx)^2) dx
let
y= π-x
dy = -dx
x=0 ,y=π
x=π, y=0
∫(0->π)xsinx/(1+(cosx)^2) dx
=∫(π->0)(π-y)siny/(1+(cosy)^2) (-dy)
=∫(0->π)(π-y)siny/(1+(cosy)^2) dy
=∫(0->π)(π-x)sinx/(1+(cosx)^2) dx
2∫(0->π)xsinx/(1+(cosx)^2) dx = ∫(0->π)πsinx/(1+(cosx)^2) dx
∫(0->π)xsinx/(1+(cosx)^2) dx
= (π/2) ∫(0->π)sinx/(1+(cosx)^2) dx

let z= cosx
dz = -sinx dx
x=0 , z=1
x=π , z=-1
∫(0->π)sinx/(1+(cosx)^2) dx
=∫(-1->1)dz/(1+z^2)
= π/2

∫(0->π)xsinx/(1+(cosx)^2) dx
= (π/2) ∫(0->π)sinx/(1+(cosx)^2) dx
= (π/2) (π/2)
=π^2/4