原式=∫(-1,1)√(1-x^2)dx=2∫(0,1)√(1-x^2)dx【令x=sint,dx=costdt,x=0,t=0；x=1,t=pi/2】=2∫(0,pi/2)(cost)^2dt=2∫(0,pi/2)[1/2+1/2cos2t]dt=∫(0,pi/2)dt+(1/2)∫(0,pi/2)cos2t)d(2t)=pi/2+(1/2)sin2t|(0,pi/2)=pi/2原式=∫(-1,1)√(1-x^2)dx=2∫(0,1)√(1-x^2)dx【令x=sint,dx=costdt,x=0,t=0；x=1,t=pi/2】=2∫(0,pi/2)(cost)^2dt=2∫(0,pi/2)[1/2+1/2cos2t]dt=∫(0,pi/2)dt+(1/2)∫(0,pi/2)cos2t)d(2t)=pi/2+(1/2)sin2t|(0,pi/2)=pi/2