Contour ∫dθ/(2+sinθ)

Method. Let z = e^{iθ}, so the integral becomes one around the unit circle |z| = 1:

dθ = dz/(iz), sin θ = (z² − 1)/(2iz)

Transform:

2 + sin θ = 2 + (z² − 1)/(2iz) = (z² + 4iz − 1)/(2iz)

I = ∮_C [ 2iz/(z² + 4iz − 1) ] · dz/(iz) = ∮_C 2 dz/(z² + 4iz − 1)

Poles: solve z² + 4iz − 1 = 0:

z = [ −4i ± √(−16 + 4) ] / 2 = −2i ± i√3 = i(−2 ± √3)

• z₁ = i(−2 + √3), |z₁| = 0.268 < 1 : inside C
• z₂ = −i(2 + √3), |z₂| = 3.73 > 1 : outside C

Residue at z₁ (simple pole):

Res = 2/(z₁ − z₂) = 2/(2i√3) = 1/(i√3)

Result: I = 2πi · 1/(i√3) = 2π/√3.

Answer: ∫₀^{2π} dθ/(2 + sin θ) = 2π/√3 ≈ 3.628

(Check: 2π/√(a² − b²) with a = 2, b = 1.)