Prove that the integral of $\sin^2(x)/(5+3\cos(x))$ from $0$ to $2\pi$ is $2\pi/9$ I'm not really unsure of how to approach this problem. I was thinking of reparametrizing the sin and the cos to its exponential form but I realize that it becomes even messier and leads sort of nowhere.  There are no singularities for this function f(x) I believe, so there's not really a way to use the Residue theorem either.  Can anyone give me some help on this?

Suppose we seek to evaluate
$$\int_0^{2\pi} \frac{\sin^2 x}{5+3\cos x} dx.$$
Put  $z   =  \exp(ix)$  so  that   $dz  =  i\exp(ix)  \; dx$  and  hence
$\frac{dz}{iz} = dx$ to obtain
$$\int_{|z|=1}
\frac{(z-1/z)^2/4/(-1)}{5+3/2(z+1/z)} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{(z-1/z)^2}{20+6(z+1/z)} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{z^2-2+1/z^2}{20+6z+6/z} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{z^4-2z^2+1}{20z^2+6z^3+6z} \frac{dz}{iz}
\\ = -\frac{1}{i} \int_{|z|=1}
\frac{1}{z^2} \frac{z^4-2z^2+1}{6z^2+20z+6} \; dz.$$
There  are three  poles, one  at $z=-3$,  another one  at  $z=-1/3$ and
another one at $z=0.$ Only the latter two contribute.

The pole at $z=-1/3$ is simple and hence the residue is
$$\left.\frac{1}{z^2}
\frac{z^4-2z^2+1}{12z+20}\right|_{z=-1/3}
= \frac{4}{9}.$$
The remaining contribution is from
$$\int_{|z|=1}
\frac{1}{z^2} \frac{1}{6z^2+20z+6} \; dz
=\frac{1}{6}
\int_{|z|=1}
\frac{1}{z^2} \frac{1}{z^2+10/3z+1} \; dz
\\ =\frac{1}{6}
\int_{|z|=1}
\frac{1}{z^2} \sum_{q\ge 0} (-1)^q z^q (z+10/3)^q \; dz.$$
The only contribution in the series is from $q=1$ and it is
$$-\frac{1}{6} 10/3 = -\frac{5}{9}.$$
Collecting everything we get
$$-\frac{1}{i} \times 2\pi i\times
\left(\frac{4}{9}-\frac{5}{9}\right)
= \frac{2\pi}{9}.$$
