第1个回答 2008-10-09
设x=sint,-π/2<t<π/2,则√(1-x²)=cost,dx=cost dt,则有
∫√(1-x²)dx=∫cos²t dt=∫[(1+cos2t)/2]dt
=1/2(∫dt+∫cos2t dt)=1/2∫dt+1/4∫cos2t d(2t)
=t/2+(sin2t)/4+C=t/2+sintcost/2+C
由于x=sint, -π/2<t<π/2,所以t=arcsinx,
Cost=√(1-sin²t)= √(1-x²)
故所求积分为∫√(1-x²)dx=arcsinx/2+x√(1-x²)/2+C本回答被提问者采纳