题目

∫（0-2）x^2/(2x-x^2）^1/2 dx
我设x-1=sint,然后∫（-π/2-π/2）(1+sint)^2 dt,然后就不会了.

提问时间：2020-11-17

答案

∫(0->2)x^2 dx/√(2x-x^2)
=-∫(0->2)(2x-x^2)/√(2x-x^2) dx + ∫(0->2) 2x/√(2x-x^2) dx
=-∫(0->2)√(2x-x^2) dx -2 ∫(0->2) d√(2x-x^2) + 2∫(0->2) dx/√(2x-x^2)
=-∫(0->2)√(2x-x^2) dx + 2∫(0->2) dx/√(2x-x^2)
consider
2x-x^2 = -(x^2-2x) = 1-(x-1)^2
let
siny = x-1
cosy dy = dx
x=0,y=-π/2
x=2,y= π/2
∫(0->2)x^2 dx/√(2x-x^2)
=-∫(0->2)√(2x-x^2) dx + 2∫(0->2) dx/√(2x-x^2)
=-∫(-π/2->π/2) (cosy)^2 dy +2∫(-π/2->π/2) dy
= -(1/2)∫(-π/2->π/2) ( 1+cos2y) dy +2π
= -(1/2) [ y + sin(2y)/2 ](-π/2->π/2) +2π
= -π/2 + 2π
=3π/2