I am currently trying to evaluate this integral with residues but I'm not quite there. The questions asks me to show
$$\int_0^{2\pi} \frac{\sin^2 \theta}{5 + 3\cos\theta}\ d\theta = \frac{2\pi}{9} $$
So I let $z = e^{i\theta}$ and so $\frac {dz}{iz} = d\theta$
I let $\sin^2\theta = \frac {z^2 + z^{-2} -2}{-4}$ but I can't seem to get the answer! Are there any other tips to this?
$\int_0^{2\pi} \frac{\sin^2 \theta}{5+3\cos\theta} \ d\theta$
$z = e^{i\theta}\\ dz = ie^{i\theta}\ d\theta\\ d\theta = \frac 1{i z} \ dz$
$\sin \theta = \frac {1}{2i}(z - z^{-1})\\ \sin^2 \theta = \frac {1}{4}(2-z^2 - z^{-2})\\ \cos\theta = \frac 1{2}(z + z^{-1})$
$\oint_{|z| = 1} \frac{2-z^2 - z^{-2}}{(4iz)(5 + \frac 12 (3z + 3z^{-1}))} \ dz\\ \oint_{|z| = 1} \frac{-z^4 +2z^2 - 1 }{(2i)(z^2)(3z + 1)(z+3)} \ dz\\ \frac 1{2i}\oint_{|z| = 1} -\frac 13 - \frac 1{3z^2} + \frac {10}{9z} - \frac {8}{9(z+\frac 13)} + \frac 8{9(z+3)} \ dz$
$z = 0, z =-\frac 13$ are the only poles inside the countour
$\pi (\frac {10}{9} - \frac {8}{9}) = \frac {2\pi}{9}$