∫[-1,1] (2+sinx)/√(4-x^2)dx
=∫[-1,1] 2/√(4-x^2)dx+∫[-1,1] sinx/√(4-x^2)dx
后一项被积函数是奇函数,积分限关于原点对称,所以积分值是0
=∫[-1,1] 2/√(4-x^2)dx
=2arcsin(x/2)[-1,1]
=2π/3