Evaluate: ∫(3-2x)√(2+x-x²)dx

∫(3-2x)√(2+x-x²)dx = ∫(3-2x)√(3²/2-(x-1/2)²)dx

Let x-1/2 = (3/√2)sinθ ⇒ dx = (3/√2)cosθdθ

Then,
∫(3-2x)√(9/2-(x-1/2)²)dx = ∫(3-2(3/√2sinθ+1/2))√(9/2-9/2sin²θ)(3/√2)cosθdθ
= ∫(3-3√2sinθ-1)√(9/2(1-sin²θ))(3/√2)cosθdθ
= ∫(2-3√2sinθ)(3/√2)cosθ(3/√2)cosθdθ
= (9/2)∫(2-3√2sinθ)cos²θdθ
= (9/2)∫(2-3√2sinθ)(1+cos2θ)/2dθ
= (9/4)∫(2+2cos2θ-3√2sinθ-3√2sinθcos2θ)dθ
= (9/4)[2θ+sin2θ+3√2cosθ+(3√2/2)cos²θ]+C
= (9/4)[2θ+sin2θ+3√2cosθ+(3√2/4)(1+cos2θ)]+C
= (9/4)[2θ+sin2θ+3√2cosθ+(3√2/4)+(3√2/4)cos2θ]+C

Substituting back,
x-1/2 = (3/√2)sinθ ⇒ sinθ = √2(x-1/2)/3
cosθ = √(1-sin²θ) = √(1-2(x-1/2)²/9) = √(9-2(x²-x+1/4))/3 = √(8+2x-2x²)/3
θ = sin⁻¹(√2(x-1/2)/3)

Therefore,
∫(3-2x)√(2+x-x²)dx = (9/4)[2sin⁻¹(√2(x-1/2)/3)+2sinθcosθ+3√2√(1-2(x-1/2)²/9)+(3√2/4)(1+2cos²θ-1)]+C
=(9/4)[2sin⁻¹(√2(x-1/2)/3)+2√2(x-1/2)/3√(1-2(x-1/2)²/9)+3√2√(1-2(x-1/2)²/9)+(3√2/2)(1-2(x-1/2)²/9)]+C