积分问题52
令x=sint,0≤x≤1,→0≤t≤π/2, ∫√(1-x^2)dx=∫√(1-sin^t)d(sint)=∫cos^tdt=∫[(1+cos2t)/2]dt=t/2+(1/4)∫cos2td(2t)=t/2+sin2t+C ∴原式=[π/4+sinπ]-[0+sin0]=π/4