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!TT

Question

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

Answer 1

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