计算定积分∫(0→π)(xsinx)⼀(1+sin^2x)dx

简单计算一下即可，答案如图所示

令x = π - y、dx = - dy
x = 0 → y = π
x = π → y = 0
M = ∫[0→π] (xsinx)/(1 + sin²x) dx
= ∫[π→0] [(π - y)sin(π - y)]/[1 + sin²(π - y)] (- dy)
= ∫[0→π] [(π - x)sinx]/(1 + sin²x) dx
= π∫[0→π] sinx/(1 + sin²x) dx - M
2M = π∫[0→π] sinx/[1 + (1 - cos²x)] dx
M = (- π/2)∫[0→π] d(cosx)/(2 - cos²x)
= (π/2)[1/(2√2)]∫[0→π] [(cosx + √2) - (cosx - √2)]/[(cosx - √2)(cosx + √2)] d(cosx)
= [π/(4√2)]∫[0→π] [1/(cosx - √2) - 1/(cosx + √2)] d(cosx)
= [π/(4√2)]ln[(cosx - √2)/(cosx + √2)] |[0→π]
= [π/(4√2)]{ln[(- 1 - √2)/(- 1 + √2)] - ln[(1 - √2)/(1 + √2)]}
= [π/(2√2)]ln[(√2 + 1)/(√2 - 1)] ≈ 1.9579

给个邮箱吧，发手写版的