首先是求
∫√(1-x^2)dx
令x=sint,dx=costdt
原式=∫(cost)^2dt
=(1/2)∫(1+cos2t)dt
=(1/2)t+(1/4)sin(2t)+C
=(1/2)arcsinx+(1/2)sintcost+C
=(1/2)arcsinx+(1/2)x√(1-x^2)+C
则
∫ √(2x-x²) dx
=∫ √(1-1+2x-x²) dx
=∫ √(1-(x-1)²) d(x-1)
=(1/2)arcsin(x-1)+(1/2)(x-1)√(1-(x-1)^2)+C