∫(0~1) 2x√(1 - x²)arcsinx dx
令x = siny,dx = cosy dy,√(1 - x²) = √(1 - sin²y) = cosy
x∈[0,1] → y∈[0,π/2]
= ∫(0~π/2) 2ysinycosy • cosy dy
= -2∫(0~π/2) ycos²y dcosy
= (-2/3)∫(0~π/2) y dcos³y
= (-2/3)[ycos³y] + (2/3)∫(0~π/2) cos³y dy
= (2/3)∫(0~π/2)∫ (1 - sin²y) dsiny
= (2/3)[siny - 1/3 • sin³y] |(0~π/2)
= (2/3)(1 - 1/3)
= 4/9