TITLE: Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$
QUESTION [15 upvotes]: How can I evaluate the following integral?

$$
\int_0^{\pi/2}
\log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\,
\frac{1}{\cos\left(x\right)}\,{\rm d}x\,,
\qquad\left\vert\,a\,\right\vert \le 1$$

I tried differentiating under the integral with respect to the parameter $a$, and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches.

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_0^{\pi/2}\ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\,
{1 \over \cos\pars{x}}\,\dd x:\ {\large ?}\,.\qquad\qquad\verts{a}\ <\ 1}$.

The general idea is to derivate respect of $\ds{\quad a\quad}$ in order 
  " to kill " the " $\ds{\cos\pars{x}}$ term " in the denominator:

\begin{align}&\color{#c00000}{\partiald{}{a}\bracks{\int_0^{\pi/2}
\ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}}}}
\\[3mm]&=\int_0^{\pi/2}\bracks{{\cos\pars{x} \over 1 + a\cos\pars{x}}
-{-\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}}
=2\int_{0}^{\pi/2}{\dd x \over 1 - a^{2}\cos^{2}\pars{x}}
\\[3mm]&=2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \sec^{2}\pars{x} - a^{2}}
=2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \tan^{2}\pars{x} + 1 - a^{2}}
=2\int_{0}^{\infty}{\dd x \over x^{2} + 1 - a^{2}}
\\[3mm]&={2 \over \root{1 - a^{2}}}\
\overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\ =\
\color{#c00000}{\pi \over \root{1 - a^{2}}}
\end{align}

$$\color{#66f}{\large%
\int_0^{\pi/2}\ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\,
{1 \over \cos\pars{x}}\,\dd x}
=\int_{0}^{a}{\pi\,\dd t \over \root{1 - t^{2}}}
=\color{#66f}{\large\pi\ \arcsin\pars{a}}
$$