Real integral by contour integration

Let I = ∫₀^2π cos 2θ/(5 + 4 sin θ) dθ. Put z = e^(iθ) on the unit circle |z| = 1:

cos 2θ = (z² + z⁻²)/2,   sin θ = (z − 1/z)/(2i),   dθ = dz/(iz)
5 + 4 sin θ = 5 − 2iz + 2i/z = (−2i z² + 5z + 2i)/z = −i (2z² + 5iz − 2)/z

So

I = ∮_C g(z) dz,   g(z) = (z⁴ + 1) / [2 z² (2z² + 5iz − 2)]

Factor: 2z² + 5iz − 2 = 2(z + 2i)(z + i/2). Poles: z = 0 (double), z = −i/2 (simple), z = −2i (simple). Inside |z| = 1: z = 0 and z = −i/2.

Res at z = −i/2:  (z⁴ + 1) / [2z² · 2(z + 2i)] at z = −i/2
                = (17/16) / [(−1/2)(3i)] = 17i/24

Res at z = 0 (double pole): d/dz [ (z⁴ + 1) / (2(2z² + 5iz − 2)) ] at 0
                = [0 − 1·2·(5i)] / [4·(−2)²] = −5i/8

Sum of residues = 17i/24 − 15i/24 = 2i/24 = i/12.

I = 2πi × (i/12) = −π/6

Answer: I = −π/6 (numerical integration gives −0.5236, which agrees).