Contour ∫cos3θ/(5-4cosθ) dθ

Method. Let z = e^{iθ} so that θ from 0 to 2π traces the unit circle |z| = 1. Then

cos θ = (z + 1/z)/2, dθ = dz/(iz), cos 3θ = Re(e^{3iθ}) = Re(z³).

Solution

Let I = ∫₀^{2π} cos 3θ/(5 − 4 cos θ) dθ. Write I = Re J, where

J = ∫₀^{2π} e^{3iθ}/(5 − 4 cos θ) dθ (the imaginary part is sin 3θ/(5 − 4cos θ), which integrates to 0 by symmetry, but we only need the real part).

Denominator: 5 − 4 cos θ = 5 − 2(z + 1/z) = −(2z² − 5z + 2)/z = −(2z − 1)(z − 2)/z.

J = ∮_C z³ · [−z/((2z − 1)(z − 2))] · dz/(iz)
  = ∮_C [ i z³ / ((2z − 1)(z − 2)) ] dz

(because −1/i = i.)

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

Residue at z = 1/2:

Res = i (1/2)³ / [2 (1/2 − 2)] = i(1/8)/(−3) = −i/24

By the residue theorem:

J = 2πi · (−i/24) = 2π/24 = π/12

This is real, so I = Re J = π/12.

Answer: ∫₀^{2π} cos 3θ/(5 − 4 cos θ) dθ = π/12 ≈ 0.2618

(Numerical integration confirms 0.26180.)