Contour integral (3)

$\begin{aligned} I & = \int_0^{2\pi} \frac {\sin^2 \theta - 2\cos \theta}{2+\cos \theta} d \theta \\ & = \int_0^\pi \frac {\sin^2 \theta - 2\cos \theta}{2+\cos \theta} d \theta + \int_{\color{#3D99F6}\pi}^{\color{#3D99F6}2\pi} \frac {\sin^2 {\color{#3D99F6}\theta} - 2\cos {\color{#3D99F6}\theta}}{2+\cos {\color{#3D99F6}\theta}} d {\color{#3D99F6}\theta} \\ & = \int_0^\pi \frac {\sin^2 \theta - 2\cos \theta}{2+\cos \theta} d \theta + \int_{\color{#D61F06}0}^{\color{#D61F06}\pi} \frac {\sin^2 {\color{#D61F06}(\theta+\pi)} - 2\cos {\color{#D61F06}(\theta+\pi)}}{2+\cos {\color{#D61F06}(\theta+\pi)}} d {\color{#D61F06}(\theta+\pi)} \\ & = \int_0^\pi \frac {\sin^2 \theta - 2\cos \theta}{2+\cos \theta} d \theta + \int_0^\pi \frac {\sin^2 \theta + 2\cos \theta}{2-\cos \theta} d \theta \end{aligned}$

$\begin{aligned} \ \ & = \int_0^\pi \frac {4\sin^2 \theta + 4\cos^2 \theta}{(2+\cos \theta)(2-\cos \theta)} d \theta \\ & = \int_0^\pi \frac {4 \ d \theta}{4-\cos^2 \theta} & \small \color{#3D99F6} \text{Multiply up and down by }\sec^2 \theta \\ & = 4 \int_0^\pi \frac {\sec^2 \theta \ d \theta }{4\sec^2 \theta -1} \\ & = 4\int_{\color{#3D99F6}0}^{\color{#3D99F6}\pi} \frac {\sec^2 \theta \ d \theta }{4\tan^2 \theta +3} & \small \color{#3D99F6} \text{Since the integral is symmetrical at }\theta = \frac \pi 2 \\ & = {\color{#D61F06}8} \int_{\color{#D61F06}0} ^{\color{#D61F06} \frac \pi 2} \frac {\sec^2 \theta \ d \theta }{4\tan^2 \theta +3} & \small \color{#3D99F6} \text{Let }t = \tan \theta \implies dt = \sec^2 \theta \ d \theta \\ & = 8 \int_0^\infty \frac {dt}{4t^2+3} \\ & = \frac 83 \int_0^\infty \frac {dt}{\left(\frac {2t}{\sqrt 3}\right)^2+1} & \small \color{#3D99F6} \text{Let }u = \frac {2t}{\sqrt 3} \implies du = \frac 2{\sqrt 3}dt \\ & = \frac 4{\sqrt 3} \int_0^\infty \frac {du}{u^2+1} \\ & = \frac 4{\sqrt 3} \bigg[\tan^{-1} u \bigg]_0^\infty \\ & = \frac {2\pi}{\sqrt 3} \approx \boxed{3.628} \end{aligned}$