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    $\int_0^{2\pi}\frac{1}{a+\cos\theta}\,d\theta=\frac{2\pi}{\sqrt{a^2-1}}$ for $a>1$

    P-6SR5N

    Area
    Complex Analysis
    Topics
    Contour Integration, Integrals, Residues, Trigonometry
    Status
    draft

    Show \[ \int_0^{2\pi} {1\over a + \cos(\theta)} \dtheta = {2\pi \over \sqrt{a^2-1}}, && a> 1 .\]

    Solution

    Sketch:

    • Set \(z=e^{i\theta}\) to get \[ \frac{2}{i} \int_{|z|=1} \frac{\mathrm{d} z}{z^{2}+2 a z+1} .\]

    • Factor into two roots \(r_1, r_2\). Use that without loss of generality, \(r_1\in \DD\) and \(r_2\in \DD^c\), with neither on \(S^1\) to compute the residue \(4\pi/(r_1 -r_2)\)

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