∫√(a²-x²)dx=a∫√[1-(x/a)²]dx
令x/a=sint,则x=asint,dx=acostdt
故原式=a²∫[√(1-sin²t)]costdt=a²∫cos²tdt=(a²/2)∫(1+cos2t)dt=(a²/2)[∫dt+(1/2)∫cos2td(2t)
=(a²/2)[t+(1/2)sin2t]+C=(a²/2)(t+sintcost)+C=(a²/2)[arcsin(x/a)+(x/a)√[1-(x/a)²]+C
=(a²/2)arcsin(x/a)+(x/2)√(a²-x²)+C
只有这么令,才能脱掉根号!