Contour integral (1)

By Contour Integration

$\begin{aligned} I & = \int_0^{\color{#3D99F6}\pi} \frac {d\theta}{6+4\cos \theta} & \small \color{#3D99F6} \text{Since the integral is even} \\ & = \frac 1{\color{#3D99F6}2} \int_0^{\color{#3D99F6}2\pi} \frac {d\theta}{6+4\cos \theta} & \small \color{#3D99F6} \text{Using unit circle contour} \\ & = \frac 14 \int_0^{2\pi} \frac {d\theta}{3+2\cos \theta} & \small \color{#3D99F6} \text{Let }z=e^{i\theta} \implies dz = ie^{i\theta} \ d \theta \\ & = \frac 14 \oint \frac {dz}{iz\left(3+z+\frac 1z\right)} \\ & = \frac 1{4i} \oint \frac {dz}{z^2+3z+1} \\ & = \frac 1{4i} \oint \frac {dz}{\left(z+\frac {3-\sqrt 5}2\right)\left(z+\frac {3+\sqrt 5}2\right)} & \small \color{#3D99F6} \text{Only the pole } z- \frac {-3+\sqrt 5}2 \text{ is in the circle.} \\ & = \frac {2\pi i}{4i} \cdot \frac 1{\frac {-3+\sqrt 5}2+\frac {3+\sqrt 5}2} & \small \color{#3D99F6} \text{By residue theorem} \\ & = \frac \pi{2\sqrt 5} \approx \boxed{0.702} \end{aligned}$

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Alternative solution

$\begin{aligned} I & = \int_0^\pi \frac {d\theta}{6+4\cos \theta} & \small \color{#3D99F6} \text{Using tangent half-angle substitution and} \\ & = \int_0^\infty \frac {2 \ dt}{\left(6+4\left(\frac {1-t^2}{1+t^2} \right) \right)(t^2+1)} & \small \color{#3D99F6} \text{let }t = \tan \frac \theta 2 \implies dt = \frac 12 \sec^2 \frac \theta 2 d \theta \\ & = \int_0^\infty \frac {dt}{5+t^2} \\ & = \frac 15 \int_0^\infty \frac {dt}{1+\left(\frac t{\sqrt 5}\right)^2} & \small \color{#3D99F6} \text{Let }u = \frac t{\sqrt 5} \implies \sqrt 5 du = dt \\ & = \frac 1{\sqrt 5} \int_0^\infty \frac {du}{1+u^2} \\ & = \frac 1{\sqrt 5} \bigg[\tan^{-1} u \bigg]_0^\infty \\ & = \frac \pi {2\sqrt 5} \approx \boxed{0.702} \end{aligned}$