∫(上限2,下限0)x根号下(2x-x^2)dx求大神解答. 弄不懂了.希望详细一点谢谢了

∫(0,2) x√(2x - x²) dx
= ∫(0,2) x√[- (x² - 2x + 1) + 1] dx
= ∫(0,2) x√[1 - (x - 1)²] dx
令x - 1 = sinθ,dx = cosθ dθ
x = 0 --> θ = - π/2
x = 2 --> θ = π/2
= ∫(- π/2,π/2) (1 + sinθ)|cosθ| * cosθ dθ
= ∫(- π/2,π/2) (1 + sinθ)cos²θ dθ
= ∫(- π/2,π/2) cos²θ dθ + ∫(- π/2,π/2) sinθcos²θ dθ
= 2∫(0,π/2) (1 + cos2θ)/2 dθ + ∫(- π/2,π/2) cos²θ d(- cosθ)
= [θ + (1/2)sin2θ] |(0,π/2) - (1/3)[cos³θ] |(- π/2,π/2)
= π/2

2x-x^2
= 1-(x^2-2x+1)
= 1- (x-1)^2
let
x-1 = sina
dx= cosa da
x=0, a= -π/2
x=2, a=π/2
∫(0->2) x√(2x-x^2) dx
=∫(-π/2->π/2) (sina+1)(cosa )cosa da
= -[(cosa)^3/3]...

原式=(-1/2)∫ [0,2] √(2x-x^2)d(2x-x^2)+∫ [0,2]√[-(1-2x+x^2)+1]dx
=(-1/2) (2x-x^2)^(3/2)*2/3[0,2]+∫ [0,2]√[1-(1-x)^2]dx
=(-1/2)(0-0)+∫ [0,2]√[1-(x-1)^2]dx
设u=x-1, du=dx,
原式=∫[-1,1]√(1-u^2)...