Determine the value of the integral $I=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx$ Determine the value of the integral $$I(a)=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx, \: |a|\leq 1$$ 
My try:
$\to I'(a)=\int_{0}^{1}\frac{-2ax^2}{\left(1-a^2x^2\right)\sqrt{1-x^2}}dx$
Set $x=\cos t\to dx=-\sin tdt$
Hence $I'(a)=\int_{0}^{\frac{\pi}{2}}\frac{-2a\cos^2t}{1-a^2\cos^2t}dt=\pi\left(\frac{1}{a}-\frac{1}{\sqrt{1-a^2}}\right)\to I(a)=\pi\left(\ln|a|-\arcsin a\right)+C$
Question: Find C?
 A: $$\begin{align}I(a) &= -\sum_{k=1}^{\infty} \frac{a^{2 k}}{k} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}}\\ &= -\frac{\pi}{2}\sum_{k=1}^{\infty} \frac1{k} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} \end{align} $$
$$I'(a) = -\frac{\pi}{2} \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k-1} $$
$$a I'(a) = -\pi \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} = -\pi \left (\frac1{\sqrt{1-a^2}} -1\right )$$
$$\implies \begin{align}I(a) &= -\pi \int da \left (\frac1{a\sqrt{1-a^2}}-\frac1{a} \right )\\ &= \pi \log{\left (1+\sqrt{1-a^2}\right )} +C \end{align}$$
$$I(0)=0 \implies C=-\pi \log{2}$$
$$\therefore I(a) = \pi \log{\left (\frac{1+\sqrt{1-a^2}}{2} \right )}$$
To derive the result for the integral in the first line, see here.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\,{\rm I}\pars{a}
     =\int_{0}^{1}{\ln\pars{1 - a^{2}x^{2}} \over \root{1 - x^{2}}}\,\dd x:
     \ {\large ?}.\qquad\verts{a} \leq 1}$.

\begin{align}\color{#66f}{\large\,{\rm I}\pars{a}}
&=-a^{2}\int_{0}^{1}{x^{2} \over \root{1 - x^{2}}}
\int_{0}^{1}{\dd t \over 1 - a^{2}x^{2}t}\,\dd x
\\[5mm]&=-a^{2}\int_{0}^{1}
\int_{0}^{1}{x^{2} \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}}\,\dd x\,\dd t
\\[5mm]&=\int_{0}^{1}
\int_{0}^{1}
{\pars{1 - a^{2}t\,x^{2}} - 1 \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}}
\,\dd x\,{\dd t \over t}
\\[5mm]&=\int_{0}^{1}\ \overbrace{\int_{0}^{1}\bracks{
{1 \over \root{1 - x^{2}}} - {1 \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}}}
\,\dd x}
^{\ds{\dsc{x}\ \equiv\ \dsc{\cos\pars{\theta}}}}\
\,{\dd t \over t}
\\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2}
-\int_{0}^{\pi/2}{\dd\theta \over 1 - a^{2}t\,\cos^{2}\pars{\theta}}}
\,{\dd t \over t}
\\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2}
-\int_{0}^{\pi/2}
{\sec^{2}\pars{\theta}\,\dd\theta \over \sec^{2}\pars{\theta} - a^{2}t}}
\,{\dd t \over t}
\\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2}\ -\
\overbrace{\int_{0}^{\pi/2}
{\sec^{2}\pars{\theta}\,\dd\theta \over \tan^{2}\pars{\theta} + 1 - a^{2}t}}^{\ds{\dsc{\tan\pars{\theta}}\ \equiv\ \dsc{\root{1 - a^{2}t}\ \xi}}}}\
\,{\dd t \over t}
\\[5mm]&=\int_{0}^{1}\pars{{\pi \over 2}
-{1 \over \root{1 - a^{2}t}}\ \overbrace{\int_{0}^{\infty}{\,\dd\xi \over \xi^{2} + 1}}^{\dsc{\pi \over 2}}}
\,{\dd t \over t}
={\pi \over 2}\ \overbrace{
\int_{0}^{1}\pars{1 - {1 \over \root{1 - a^{2}t}}}\,{\dd t \over t}}
^{\ds{\dsc{t}\ \equiv \dsc{1 - y^{2} \over a^{2}}}}
\\[5mm]&={\pi \over 2}\int_{1}^{\root{1 - a^{2}}}\pars{1 - {1 \over y}}
\,{-2y\,\dd y/a^{2} \over \pars{1 - y^{2}}/a^{2}}
=\pi\int^{\root{1 - a^{2}}}_{1}{\dd y \over 1 + y}
\\[5mm]&=\color{#66f}{\large\pi\ln\pars{1 + \root{1 - a^{2}} \over 2}}
\end{align}
