Contour integral (2)

$\begin{aligned} I & = \int_0^{2\pi} \frac {\sin^2 \theta}{5-4\cos \theta} d \theta \\ & = \int_0^\pi \frac {\sin^2 \theta}{5-4\cos \theta} d \theta + \int_{\color{#3D99F6}\pi}^{\color{#3D99F6}2\pi} \frac {\sin^2 \color{#3D99F6}\theta}{5-4\cos \color{#3D99F6}\theta} d \color{#3D99F6}\theta \\ & = \int_0^\pi \frac {\sin^2 \theta}{5-4\cos \theta} d \theta + \int_{\color{#D61F06}0}^{\color{#D61F06}\pi} \frac {\sin^2 \color{#D61F06}(\theta+\pi)}{5-4\cos \color{#D61F06}(\theta+\pi)} d \color{#D61F06}(\theta+\pi) \\ & = \int_0^\pi \frac {\sin^2 \theta}{5-4\cos \theta} d \theta + \int_0^\pi \frac {\sin^2 \theta}{5+4\cos \theta} d \theta \\ & = \int_0^\pi \frac {10\sin^2 \theta}{25-16\cos^2 \theta} d \theta & \small \color{#3D99F6} \text{Divide up and down by }\cos^2 \theta \\ & = {\color{#3D99F6}10} \int_{\color{#3D99F6}0}^{\color{#3D99F6}\pi} \frac {\tan^2 \theta}{25\sec^2 \theta -16} d \theta & \small \color{#3D99F6} \text{Since the integral is symmetrical at }\theta = \frac \pi 2 \\ & = {\color{#D61F06} 20} \int_{\color{#D61F06}0}^{\color{#D61F06}\frac \pi 2} \frac {\tan^2 \theta}{25\tan^2 \theta +9} d \theta & \small \color{#3D99F6} \text{Let }t = \tan \theta \implies dt = \sec^2 \theta \ d \theta \\ & = 20 \int_0^\infty \frac {t^2}{(25t^2+9)(t^2+1)} dt & \small \color{#3D99F6} \text{By partial fractions} \\ & = \frac {20}{16} \int_0^\infty \left(\frac 1{t^2+1} - \frac 9{25t^2+9} \right) dt \\ & = \frac 54 \left[\tan^{-1} t - \frac 35 \tan^{-1} \frac 53 t \right]_0^\infty \\ & = \frac \pi 4 \approx \boxed{0.7853} \end{aligned}$

This is the solution to Contour Integral 1. I got it wrong on my first attempt, but later found a way of solving.

This problem can be solved similarly. Use the Weierstrass Substitution,i.e-replace the trigonometric denominator with a $tan\frac{\theta}{2}$ equivalent and then make the substitution $u= tan\frac{\theta}{2}$ . It is easy from there.