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Evaluate the integral of 1 / (1 + a cos x) for 0 < a < 1

Compute the following integral.

    \[ \int \frac{dx}{1+a \cos x} \qquad \text{where} \quad 0 < a < 1. \]


We make the same substitution as in the previous exercise, letting u = \tan \frac{x}{2}, which gives us

    \begin{align*}  x &= 2 \arctan u \\ dx &= \frac{2}{1+u^2} \, du \\ \cos x &= \frac{1-u^2}{1+u^2}. \end{align*}

Therefore, we have

    \begin{align*}  \int \frac{dx}{1+a \cos x} &= \int \frac{2 \, du}{(1+u^2) \left( 1 + a \left( \frac{1-u^2}{1+u^2} \right) \right)} \\[9pt]  &= \int \frac{2 \, du}{1+u^2 + a - au^2} \\[9pt]  &= \int \frac{2 \, du}{(1-a)u^2 + (1+a)} \\[9pt]  &= \frac{2}{1+a} \int \frac{du}{\left( \frac{1-a}{1+a} \right)u^2 + 1}. \end{align*}

The, since 0 < a < 1, we know \frac{1-a}{1+a} > 0, so we may let

    \[ t = u \sqrt{\frac{1-a}{1+a}}, \qquad dt = \sqrt{\frac{1-a}{1+a}} \, du. \]

Therefore,

    \begin{align*}  \int \frac{dx}{1+a \cos x} &= \frac{2}{1+a} \int \frac{du}{\left( \frac{1-a}{1+a} \right)u^2 + 1} \\[9pt]  &= \frac{2}{\sqrt{(1+a)(1-a)}} \int \frac{dt}{t^2+1} \\[9pt]  &= \frac{2}{\sqrt{1-a^2}} \arctan t + C \\[9pt]  &= \left( \frac{2}{\sqrt{1-a^2}} \right) \arctan \left( \frac{\sqrt{1-a}u}{\sqrt{1+a}} \right) + C \\[9pt]  &= \left( \frac{2}{\sqrt{1-a^2}} \right) \arctan \left( \frac{\sqrt{1-a}}{\sqrt{1+a}} \tan \left( \frac{x}{2} \right) \right) + C. \end{align*}

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