Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$ 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.
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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}

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

A: Use the expansion for $|z| < 1$
$$\log{\left ( \frac{1+z}{1-z}\right )} = 2 \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$
Then the integral is equal to
$$2 \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \int_0^{\pi/2} dx \, \cos^{2 k}{x}$$
It is straightforward to show that
$$\int_0^{\pi/2} dx \, \cos^{2 k}{x} = \frac{1}{2^{2 k}} \binom{2 k}{k} \frac{\pi}{2}$$
Thus the integral $I(a)$ is
$$I(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \frac{1}{2^{2 k}} \binom{2 k}{k}$$
We may evaluate this sum by considering
$$I'(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k}}{2^{2 k}} \binom{2 k}{k} = \pi \left (1-a^2\right)^{-1/2}$$
Integrating with respect to $a$ and noting that $I(0)=0$, we find that 
$$I(a) = \pi \arcsin{a}$$
A: $$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &=  \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$
Let $ \displaystyle u = \tan \frac{x}{2}$.
$$\begin{align} &= \int_{b}^{a} \int_{0}^{1} \frac{1}{1+ t \left(\frac{1-u^{2}}{1+u^{2}} \right)} \frac{2}{1+u^{2}} \ du \ dt \\ &= 2 \int_{b}^{a} \int_{0}^{1} \frac{1}{1+t} \frac{1}{1+ \frac{1-t}{1+t} u^{2}}  du \ dt \end{align}$$
Let $\displaystyle w = \sqrt{\frac{1-t}{1+t}} u $.
$$ \begin{align} &= 2 \int_{b}^{a} \int_{0}^\sqrt{\frac{1-t}{1+t}} \frac{1}{\sqrt{1-t^{2}}} \frac{1}{1+w^{2}} \ dw \ dt \\ &= 2 \int_{b}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}}\ dt \\ &= \int_{b}^{a} \frac{\arccos t}{\sqrt{1-t^{2}}} \ dt \\ &= \frac{1}{2} \Big(\arccos^{2} (b)- \arccos^{2} (a)\Big) \end{align}$$
Then 
$$ \begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1-a \cos x}\right)\frac{1}{\cos x}dx &= \frac{1}{2} \Big(\arccos^{2} (-a)- \arccos^{2} (a)\Big) \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} - \arcsin (-a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} + \arcsin (a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big(2 \pi \arcsin a\Big) = \pi \arcsin a \end{align}$$
