定积分题,求具体解法.∫x^2/√(2x-x^2) 积分上下限为 0-2.有答案但是没具体步骤,要具体的丫!T_T

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定积分题,求具体解法.∫x^2/√(2x-x^2) 积分上下限为 0-2.有答案但是没具体步骤,要具体的丫!T_T

定积分题,求具体解法.∫x^2/√(2x-x^2) 积分上下限为 0-2.有答案但是没具体步骤,要具体的丫!T_T
定积分题,求具体解法.
∫x^2/√(2x-x^2) 积分上下限为 0-2.
有答案但是没具体步骤,要具体的丫!T_T

定积分题,求具体解法.∫x^2/√(2x-x^2) 积分上下限为 0-2.有答案但是没具体步骤,要具体的丫!T_T
答:
设x-1=sint,x=1+sint
(0→2) ∫ x^2/√(2x-x^2) dx
=(-π/2→π/2) ∫ (1+sint)^2/cost d(1+sint)
=(-π/2→π/2) ∫ 1+2sint+(sint)^2 dt
=(-π/2→π/2) ∫ 1+2sint+(1-cos2t)/2 dt
=(-π/2→π/2) 3t/2-2cost-(1/4)sin2t
=(3π/4-0-0)-(-3π/4-0-0)
=3π/2

积分上限是2,积分下限是0
原式=∫(0,2) [(x-1)^2+2(x-1)+1]/√[1-(x-1)^2] dx
=∫(2,0) [1-(x-1)^2-2(x-1)-2]/√[1-(x-1)^2] dx
=∫(2,0) √[1-(x-1)^2]d(x-1)+∫(2,0) d[1-(x-1)^2]/√[1-(x-1)^2]-∫(2,0) 2d(x-1)/√[1-(x-1)...

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积分上限是2,积分下限是0
原式=∫(0,2) [(x-1)^2+2(x-1)+1]/√[1-(x-1)^2] dx
=∫(2,0) [1-(x-1)^2-2(x-1)-2]/√[1-(x-1)^2] dx
=∫(2,0) √[1-(x-1)^2]d(x-1)+∫(2,0) d[1-(x-1)^2]/√[1-(x-1)^2]-∫(2,0) 2d(x-1)/√[1-(x-1)^2]
=1/2*arcsin(x-1)+(x-1)/2*√(2x-x^2)|(2,0)+2√(2x-x^2)|(2,0)-2arcsin(x-1)|(2,0)
=-π/4-π/4+π+π
=3π/2

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