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∫(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.
人气:184 ℃ 时间:2020-05-08 11:04:07
解答
∫(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
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