x=2sina
√(4-x²)=2cosa
dx=2cosada
x=0,a=0
x=1,a=π/6
所以原式=∫(0,π/6)4cos²ada
=∫(0,π/6)4(1+cos2a)/2da
=∫(0,π/6)(1+cos2a)d(2a)
=2a+sin2a(0,π/6)
=π/3+√3/2