Let z = tan(x/2) and dx = 2dz/( 1+z? )，sinx = 2z/( 1 + z？ )
∫ 1/(2 + sinx) dx
= ∫ [2/( 1 + z？ )]/[2 + (2z)/( 1 + z？ )] dz
= ∫ 1/[( 1 + z？ )+ z] dz
= ∫ 1/[(z + 1/2)？ + 3/4] dz
=(2/√3)arctan[(z+ 1/2)2/√3]+C
=(2/√3)arctan[(2 tan(x/2)+ 1)/√3]+C
Between partitions, pay attention to if? (x) = 1/(2 + sinx)，？ (0) = ? (π) = ? (2π)
f(x)=(2/√3)arctan[(2 tan(x/2)+ 1)/√3]
∫(0→2π) 1/(2 + sinx) dx
=∫(0→π) 1/(2+sinx)dx+∫(π→2π) 1/(2+sinx)dx
= [F(π)-F(0)]-[F(2π)-F(π)], x = π is a discontinuous point, divided into left and right limits.
= [lim(x→0) F(x) - lim(x→π？ )F(x)] - [lim(x→2π) F(x) - lim(x→π？ )F(x)]
= [π/(3√3) - (- π/√3)] - [π/(3√3) - π/√3]
= 2π/√3