Question

一道高数积分题求助在线等

求解这个过程

Answer 1

用换元积分法可以求出结果。

追问

这道题能看下吗

Answer 2

let
u=2π-t
du = -dt
t=0 , u=2π
t=2π, u=0
∫(0->2π) (t-sint)(1-cost)^2 dt
=∫(2π->0) (2π-u + sinu ) (1-cosu)^2 (-du)
=∫(0->2π) (2π-u + sinu ) (1-cosu)^2 du
=∫(0->2π) (2π-t + sint ) (1-cost)^2 dt
2∫(0->2π) (t-sint)(1-cost)^2 dt = 2π ∫(0->2π) (1-cost)^2 dt
∫(0->2π) (t-sint)(1-cost)^2 dt
= π ∫(0->2π) (1-cost)^2 dt
=π ∫(0->2π) [1-2cost +(cost)^2 ] dt
=(π/2) ∫(0->2π) ( 3-4cost +cos2t ) dt
=(π/2) [ 3t-4sint +(1/2)sin2t]| (0->2π)
=3π^2
∫(0->2π) (1-cost)^3 dt
=∫(0->2π) (1-3cost+3(cost)^2 -(cost)^3 ] dt

= [t-3sint]|(0->2π) + 3 ∫(0->2π) (cost)^2 dt -∫(0->2π) (cost)^3 dt
=2π + (3/2)∫(0->2π) (1+cos2t) dt -∫(0->2π) (cost)^2 dsint
=2π + (3/2)[t+(1/2)sin2t]|(0->2π) -∫(0->2π) [ 1-(sint)^2] dsint
=2π + 3π -[ sint -(1/3)(sint)^3]|(0->2π)
=5π +0
=5π
ie
∫(0->2π) (t-sint)(1-cost)^2 dt +∫(0->2π) (1-cost)^3 dt
=3π^2+5π