Contour ∫dθ/(5-4sinθ)

Method (unit circle substitution). Put z = e^{iθ}. As θ goes from 0 to 2π, z traverses the unit circle |z| = 1 once. Then

dθ = dz/(iz), sin θ = (z − 1/z)/(2i)

Transform the integrand:

5 − 4 sin θ = 5 − (2/i)(z − 1/z) = 5 + 2i(z − 1/z) = (2i z² + 5z − 2i)/z

I = ∮ [ z/(2i z² + 5z − 2i) ] · dz/(iz) = ∮ dz / ( i (2i z² + 5z − 2i) ) = ∮ dz / ( −2z² + 5iz + 2 )

Factor: −2z² + 5iz + 2 = −2 (z² − (5i/2) z − 1) = −2 (z − 2i)(z − i/2).

So I = ∮_C (−1/2) · 1 / ((z − 2i)(z − i/2)) dz, C: |z| = 1.

Poles: z = 2i (|z| = 2, outside C) and z = i/2 (|z| = 1/2, inside C).

Residue at z = i/2:

Res = (−1/2) · 1/(i/2 − 2i) = (−1/2) · 1/(−3i/2) = 1/(3i)

Result: I = 2πi · 1/(3i) = 2π/3.

Answer: ∫₀^{2π} dθ/(5 − 4 sin θ) = 2π/3

(Check with the standard formula ∫₀^{2π} dθ/(a + b sin θ) = 2π/√(a² − b²) = 2π/√(25 − 16) = 2π/3.)