Need help evaluating $ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} $. From Churchill and Brown's Complex Analysis I am trying to conclude:
$$ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} = \frac{2\pi}{3}$$
I have tried substituting in:
$$\sin{\theta} = \frac{z + z^{-1}}{2i}$$
$$d\theta = \frac{dz}{zi}$$
To get (note $C$ is $|z| = 1$):
$$ \int_{C} \frac{\frac{dz}{zi}}{5 + 4\frac{z + z^{-1}}{2i}} $$
$$ \int_{C} \frac{dz}{5 - 2i(z + z^{-1})zi} $$
$$ \int_{C} \frac{dz}{5 + 2(z + z^{-1})z} $$
$$ \int_{C} \frac{dz}{5 + 2(z^2 + 1)} $$
$$ \int_{C} \frac{dz}{7 + 2z^2} $$
$$ \frac12 \int_{C} \frac{dz}{z^2 + \frac72}$$
But then I concluded the singularities of $\pm i\frac72$ were outside of the circle so the integral became $0$ instead of $\frac{2\pi}{3}$. Where did I go wrong?

Note that 
$$\left(5+4\left(\frac{z-z^{-1}}{2i}\right)\right)iz=2z^2+i5z-2$$
