原式=∫[-1,1]x²sinxdx+∫[-1,1]√(1-x²)dx=0+2∫[0,1]√(1-x²)dx（令x=sint）=2∫[0,π/2]cos²t dt =∫[0,π/2]（1+cos2t）dt=(t+½ sin2t)|[0,π/2]=π/2
