Evaluate $\int_{0}^{2\pi}\frac{\cos(x)}{a+\cos(x)}dx$

Question

I need to show that $$\int_{0}^{2\pi}\frac{\cos(x)}{a+\cos(x)}dx = 2\pi \left(1-\frac{a}{\sqrt{a^2-1}}\right), a >1$$ I am unsure where to start. I tried using the methods described in https://en.wikipedia.org/wiki/Tangent_half-angle_substitution,

but not sure which substitutions to use. I would like to use residue calculus, but not sure which contour would work. Any help is greatly appreciated!

Answer 1

$$I=\int_{0}^{2\pi}\frac{\cos x}{a+\cos x}dx= 2\int_{0}^{\pi}\left(1-\frac{a}{a+\cos x}\right)dx=2\pi - 2a \int_{0}^{\pi}\frac{dx}{a+\cos x} $$

Use the half-angle substitution $t = \tan\frac x2$, $\cos x =\frac{1-t^2}{1+t^2}$ and $dx = \frac{2dt}{1+t^2}$ to evaluate the integral,

$$I=2\pi - 2a \int_{0}^{\infty}\frac{2dt}{(a+1)+(a-1)t^2}$$ $$=2\pi - \frac{4a}{\sqrt{a^2-1}} \int_{0}^{\infty} \frac{d\sqrt{\frac{a-1}{a+1}}t}{1+\left(\sqrt{\frac{a-1}{a+1}}t \right)^2}$$ $$=2\pi - \frac{4a}{\sqrt{a^2-1}}\tan^{-1} \sqrt{\frac{a-1}{a+1}}t \bigg|_0^{\infty} =2\pi \left(1-\frac{a}{\sqrt{a^2-1}}\right)$$

Answer 2

If you would like to use complex analysis.

$\cos x = \frac {e^{ix} + e^{-ix}}{2}$

$\int \frac {e^{ix} + e^{-ix}}{2a+e^{ix} + e^{-ix}} dx$

$z = e^{ix}, dx = \frac{1}{iz} \ dz$

This is going to make your contour be the unit circle.

$\oint_{|z| = 1} \frac {z + z^{-1}}{iz(2a+z + z^{-1})} dz\\ \oint_{|z| = 1} \frac {z^2 + 1}{iz(z^2 + 2az + 1)} dz$

The roots of the denominator (the poles of the function) are

$0, -a+\sqrt{a^2-1},-a-\sqrt{a^2-1}$

but $-a-\sqrt{a^2-1}$ is not inside the contour so we can ignore it. We have just the two residues to find.

$2\pi (1 + \frac {(-a+ \sqrt {a^2-1}) + (-a + \sqrt {a^2-1})^{-1}}{2\sqrt {a^2-1}})\\ 2\pi (1 + \frac {-2a}{2\sqrt {a^2-1}})$