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up vote 1 down vote favorite If $alpha, beta, gamma$ are real numbers such that $alpha^2> beta^2+gamma^2$ show that $$int_0^{2pi}frac{dtheta}{alpha+beta cos theta +gamma sin theta} = frac{2 pi}{sqrt{alpha^2-beta^2-gamma^2}}$$ My attempt: change the variable $z = e^{itheta}$ , i.e. $cos theta = frac{z+z^{-1}}{2}, sin theta = -ifrac{z-z^{-1}}{2}$ in the given integral to obtain $$int_0^{2pi}frac{dtheta}{alpha+beta cos theta +gamma sin theta} = 2int_{C} frac{-iz^{-1} times z, dz}{2alpha z+beta(z^2+1)-igamma(z^2-1)} = 2int_{C} frac{-i, dz}{z^2(beta-igamma)+2alpha z+beta+i gamma} $$ where $C$ is unit circle. Further, from above integral we get $$= frac{-2i}{beta-igamma}int_{C} frac{dz}{bigg(z+frac{alpha-sqrt{alpha^2-beta^2-gamma^2)}}{beta-igamma}bigg) bigg(z+frac{alpha+sqrt{alpha^2-beta^2-gamma^2)}}{beta-igamma}bigg)} $$ ...