<s>[INST] Help evaluating $ \int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $ $\int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $ 
so I approached  
$\frac{1}{2}\int_{0}^{2π} \frac{1}{({3+cosθ})^2}dt $ 
$z=e^{iθ}$,
$\\dθ=\frac{1}{iz}dz\\\cosθ = \frac{e^{iθ}+e^{-iθ}}{2} =\frac{z+z^{-1}}{2}$ 
$\frac{2}{i}\oint\frac{z}{(z^2+6z+1)^2}dz$
then I cannot solve the problem....
 [/INST] A geometric approach: since $\rho(\theta)=\frac{1}{3+\cos\theta}$ is the polar equation of an ellipse with major axis $2a=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ and semi-latus rectum $\frac{b^2}{a}=\frac{1}{3}$, we have:
$$ \int_{0}^{\pi}\rho(\theta)^2\,d\theta = \frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2\,d\theta = \pi a b = \pi\cdot\frac{3}{8}\cdot\frac{1}{2\sqrt{2}}=\color{red}{\frac{3\pi}{16\sqrt{2}}} $$
since $\int_{0}^{2\pi}\frac{\rho(\theta)^2}{2}\,d\theta$ is the area enclosed by such ellipse.
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