打扰你一下，麻烦你帮我解答一道定积分的题，好不好。从0到π∫ x(sinx)^3/(1+(cosx)^2) dx。在线等。。

Question

如题所述

Answer 1

令x = π - u，dx = - du
J = ∫(0~π) xsin³x/(1 + cos²x) dx
= ∫(π~0) (π - u)sin³u/(1 + cos²u) (- du)
= π∫(0~π) sin³u/(1 + cos²u) du - ∫(0~π) usin³u/(1 + cos²u) du
= π∫(0~π) sin³x/(1 + cos²x) dx - J
2J = π∫(0~π) (cos²x + 1 - 2)/(1 + cos²x) dcosx
2J = π∫(0~π) [1 - 2/(1 + cos²x)] dcosx
2J = π[cosx - 2arctan(cosx)] |(0~π)
2J = π{ [- 1 - 2arctan(- 1)] - [1 - 2arctan(1)] }
2J = π{ [- 1 - 2(- π/4)] - [1 - 2(π/4)] }
2J = π[- 1 + π/2 - 1 + π/2]
2J = π[- 2 + π]
J = π²/2 - π
==> ∫(0~π) xsin³x/(1 + cos²x) dx = π²/2 - π本回答被提问者采纳