<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] The poles of the function are $z_1 = 2\sqrt{2}-3$ and $z_2 = -3-2\sqrt{2}$. Only $z_1$ lies inside the unit circle.
You must find the residue of this pole. It's easy to see that its a pole of order $2$, so, the residue has a value of $\displaystyle \frac{-3i}{16\sqrt{2}}$
Hence, the value of your integral is $\displaystyle \pi i \left(\frac{-3i}{16\sqrt{2}}\right) = \frac{3\pi}{16\sqrt{2}}$
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