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Prove that $$ \int_0^\pi \frac{dx}{\alpha - \cos(x)}\ = \frac{\pi}{\sqrt{\alpha^2 - 1}}$$ for $\alpha > 1$

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There is a general approach to this kind of problem, the Weierstrass substitution $t=\tan(x/2)$. That changes the integral of a rational function of $\sin x$ and/or $\cos x$ into the integral of a rational function of $t$.

We have $\cos x=\dfrac{1-t^2}{1+t^2}$ and $dx=\dfrac{2\,dt}{1+t^2}$. When we substitute, we get, after a little simplification, $$\int_0^\infty \frac{2\,dt}{(\alpha+1)t^2+\alpha-1}.$$ Now there is the natural substitution $t=\sqrt{\frac{\alpha-1}{\alpha+1}}\,u$, and we end up integrating $\dfrac{1}{1+u^2}$.

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Hint: Use $\cos x=2\cos^2(\frac{x}{2})-1$, multiply the denominator and numerator by $\sec^2(\frac{x}{2})$ and use the substitution $\tan(\frac{x}{2})=z$.

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Put $z = e^{ix}$ so that $dz = i e^{ix} dx$ and $dx = 1/z/i \,dz$. The integral becomes $$\frac{1}{2} \int_{|z|=1} \frac{2}{2\alpha - z - 1/z} \frac{1}{iz} dz $$ Simplifying this yields $$\frac{1}{2i} \int_{|z|=1} \frac{2}{2\alpha z - z^2 - 1} dz$$ The poles are at $$z = \alpha \pm \sqrt{\alpha^2-1},$$ Of these, only the smaller one is inside the unit circle and the residue is $$ \text{Res}_{z=\alpha - \sqrt{\alpha^2-1}} \frac{2}{2\alpha z - z^2 - 1} = \frac{1}{\sqrt{\alpha^2-1}}.$$ Hence by the Cauchy Residue Theorem the integral is $$ 2\pi i \frac{1}{2i} \frac{1}{\sqrt{\alpha^2-1}} = \frac{\pi}{\sqrt{\alpha^2-1}}.$$ Here we have used the fact that $$\text{Res}_{z=\rho_0} \frac{1}{(z-\rho_0)(z-\rho_1)} = \lim_{z\rightarrow \rho_0} \frac{1}{z-\rho_1} = \frac{1}{\rho_0-\rho_1}.$$

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