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

Compute the following integral.

    \[ \int \frac{dx}{1+a \cos x} \qquad \text{for} \quad a > 1. \]


Using the previous exercise we know that with the substitution u = \tan \frac{x}{2} we obtain

    \[ \int \frac{dx}{1 + a \cos x} = \frac{2}{1+a} \int \frac{du}{\left( \frac{1-a}{1+a} \right) u^2 +1}. \]

This time, since a > 1 we make the substitution

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

This gives us, for a > 1,

    \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}{1+a} \sqrt{\frac{a+1}{a-1}} \int \frac{dt}{t^2-1}. \end{align*}

Using partial fractions we can rewrite the integrand in this to obtain,

    \begin{align*} \frac{-2}{1+a} \sqrt{\frac{a+1}{a-1}} \int \frac{dt}{t^2-1} &= \frac{-2}{\sqrt{a^2-1}} \int \left( \frac{1/2}{t-1} - \frac{1/2}{t+1} \right) \, dt \\[9pt]  &= \frac{-1}{\sqrt{a^2-1}} \left( \log |t-1| - \log |t+1| \right) + C \\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{s+1}{s-1} \right| + C \\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{\sqrt{\frac{a-1}{a+1}}u + 1}{\sqrt{\frac{a-1}{a+1}} u - 1} \right| + C \\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{\sqrt{a-1} u + \sqrt{a+1}}{\sqrt{a-1} u - \sqrt{a+1}} \right| + C\\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{ \left( \sqrt{a-1} u + \sqrt{a+1} \right)^2}{(a-1)u^2 - (a+1)} \right| + C \\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{au^2-u^2 + 2u \sqrt{a^2-1} + a + 1}{au^2 - u^2 - a - 1} \right| + C\\[9pt]  &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{a (u^2+1) - (u^2-1) + 2u\sqrt{a^2-1}}{a(u^2-1) - (u^2+1)} \right| + C \\[9pt]  &= \frac{-1}{\sqrt{a^2-1}} \log \left| \frac{a + \frac{1-u^2}{1+u^2} + \frac{2u}{1+u^2} \sqrt{a^2-1}}{a \frac{1-u^2}{1+u^2} + \frac{1+u^2}{1+u^2}} \right| + C\\[9pt]  &= \frac{-1}{\sqrt{a^2-1}} \log \left| \frac{a + \cos x + \sqrt{a^2-1} \sin x}{1+ a \cos x} \right| + C. \end{align*}

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