令x=2sint, t∈[0, π/2], 则√(4-x²)=√(4-4sin²t)=2costdx=2costdt∴∫(0→1) √(4-x²)dx=∫(0→π/6) 4cos²tdt=∫(0→π/6) 2(cos2t+1) dt=sin2t+2t|(0→π/6)=√3/2+π/3
