∫（0-2）x^2/(2x-x^2）^1/2dx我设x-1=sint,然后∫（-π/2-π/2）(1+sint)^2dt,然后就不会了.
更新时间： 2025-08-22 13:59:48
1人问答
问题描述：
∫（0-2）x^2/(2x-x^2）^1/2dx
我设x-1=sint,然后∫（-π/2-π/2）(1+sint)^2dt,然后就不会了.
屈立俊回答：
∫(0->2)x^2dx/√(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
cosydy=dx
x=0,y=-π/2
x=2,y=π/2
∫(0->2)x^2dx/√(2x-x^2)
=-∫(0->2)√(2x-x^2)dx+2∫(0->2)dx/√(2x-x^2)
=-∫(-π/2->π/2)(cosy)^2dy+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
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