I am stuck with this integral, that I must solve with a complex one. First I thought it would get easier, as a polynomial in the denominator and I would get the old and good residue times $2\pi i$ but it didn't get and I refuse to invert the obtained series. Is there any other way.
What I've done so far. $$ \int_{0}^{2\pi}\frac{1}{\left( 3+2sin(t)\right )^2}dt$$ $$\oint \frac{1}{\left( 9+12\frac{z-z^{-1}}{2i}+4(\frac{z-z^{-1}}{2i})^2 \right )}\frac{dz}{iz} $$ $$ \oint \frac{1}{\left(7iz+6iz^2-6+iz^3+iz^{-1} \right )}dz$$
Should I invert this series, or try to find the roots modifying the denominator, I think I am missing something at plain sight.
$\dfrac {1}{(3 + 2\frac {e^{it} - e^{-it}}{2i})^2}\ dt\\ z = e^{it}\\ dz = i e^{it} dt\\ -iz^{-1} dz = dt$
$\oint_{|z| = 1} \frac {-iz^{-1}}{(3 -iz + iz^{-1})^2} \ dz$
multiply numerator and denominator by $(iz)^2$
$\oint \frac {iz}{(3iz + z^2 - 1)^2} \ dz\\ \oint \frac {iz}{(z-a)^2(z-b)^2} \ dz$
There is one pole inside the contour and it is of order 2.
$\oint \frac {f(z)}{(z-a)^2} = 2\pi i f'(a)$
$f(z) = \frac {iz}{(z -b)^2}\\ f'(z) = \frac {-i(z + b)}{(z-b)^3}\\ f'(a) = \frac {-i(a+b)}{(a-b)^3}$
$a = -\frac {3 - \sqrt 5}{2}i\\ b = -\frac {3 + \sqrt 5}{2}i\\ a+b = 3i\\ a-b = -\sqrt 5i\\ 2\pi i f'(a) = \frac {6\pi}{5\sqrt5}$