If $\alpha, \beta, \gamma$ are real numbers such that $\alpha^2> \beta^2+\gamma^2$ show that $$\int_0^{2\pi}\frac{d\theta}{\alpha+\beta \cos \theta +\gamma \sin \theta} = \frac{2 \pi}{\sqrt{\alpha^2-\beta^2-\gamma^2}}\frac{|\alpha|}{\alpha}.$$
My attempt: change the variable $z = e^{i\theta}$, i.e. $\cos \theta = \frac{z+z^{-1}}{2}, \sin \theta = -i\frac{z-z^{-1}}{2}$ in the given integral to obtain $$\int_0^{2\pi}\frac{d\theta}{\alpha+\beta \cos \theta +\gamma \sin \theta} = 2\int_{C} \frac{-iz^{-1} \times z\, dz}{2\alpha z+\beta(z^2+1)-i\gamma(z^2-1)}\\ = 2\int_{C} \frac{-i\, dz}{z^2(\beta-i\gamma)+2\alpha z+\beta+i \gamma} $$where $C$ is unit circle. Further, from above integral we get $$= \frac{-2i}{\beta-i\gamma}\int_{C} \frac{dz}{\bigg(z+\frac{\alpha-\sqrt{\alpha^2-\beta^2-\gamma^2)}}{\beta-i\gamma}\bigg) \bigg(z+\frac{\alpha+\sqrt{\alpha^2-\beta^2-\gamma^2)}}{\beta-i\gamma}\bigg)} $$ $$= -\iota \frac{1}{\sqrt{\alpha^2-\beta^2-\gamma^2}}\bigg[ \int_{C} \frac{dz}{\bigg(z+\frac{\alpha-\sqrt{\alpha^2-\beta^2-\gamma^2)}}{\beta-i\gamma}\bigg) }- \int_{C} \frac{dz}{ \bigg(z+\frac{\alpha+\sqrt{\alpha^2-\beta^2-\gamma^2)}}{\beta-i\gamma}\bigg)} \bigg]$$ How to proceed forward? Is my attempt correct?
HINT: let the following contour integrals defined in the boundary of the unit disk in the complex plane (traveled in counter clock-wise direction):
$$\oint \frac{dz}{(z-a)(z-b)}=\oint\frac{A}{z-a}\, dz+\oint\frac{B}{z-b}\, dz$$
for some constants $A,B\in\Bbb C$. Now use the Cauchy integral formula.