Compute the following integral.
![\[ \int \frac{dx}{1+a \cos x} \qquad \text{for} \quad a > 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0fe8f9a9bb5398b96629a9111162a3f0_l3.png "Rendered by QuickLaTeX.com")
Using the previous exercise we know that with the substitution 
we obtain
![\[ \int \frac{dx}{1 + a \cos x} = \frac{2}{1+a} \int \frac{du}{\left( \frac{1-a}{1+a} \right) u^2 +1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7600b366b1decf5ecb4ea5d62b634cf9_l3.png "Rendered by QuickLaTeX.com")
This time, since 
we make the substitution
![\[ t = u\sqrt{\frac{a-1}{a+1}} \qquad dt = \sqrt{\frac{a-1}{a+1}} \, du. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-43acd1c1e4b72c6c6093916efac62772_l3.png "Rendered by QuickLaTeX.com")
This gives us, for ,
![\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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d2c5db27f8816112c353b7c8995fbcb1_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b7aafadea2b476516ae4bf7b3e336bbb_l3.png "Rendered by QuickLaTeX.com")
i guess denominator in log is wrong it should be -1-acosx and the sign is to be positive at the end where it was changed to negative for perfect answer
There are 2 typos. The $s$ in the third row should be $u$, and there should be no minus sign at the binning of the last 2 rows (i.e. |x| = |-x|)