The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$ 
Prove that
  $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$
  without using series expansion.


An easy way to calculate the above integral is using series expansion. Here is an example
\begin{align}
\int_0^1\frac{\ln(1-x^2)}{x}dx&=-\int_0^1\frac{1}{x}\sum_{n=0}^\infty\frac{x^{2n}}{n} dx\\
&=-\sum_{n=0}^\infty\frac{1}{n}\int_0^1x^{2n-1}dx\\
&=-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{n^2}\\
&=-\frac{\pi^2}{12}
\end{align}
I am wondering, are there other ways to calculate the integral without using series expansion of its integrand? Any method is welcome. Thank you. (>‿◠)✌
 A: Substitute $u=x^2$. Then, 
$$\begin{align}
\int_{0}^{1}\frac{\ln{(1-x^2)}}{x}\mathrm{d}x
&=\int_{0}^{1}\frac{\ln{(1-u)}}{\sqrt{u}}\cdot\frac{\mathrm{d}u}{2\sqrt{u}}\\
&=\frac12\int_{0}^{1}\frac{\ln{(1-u)}}{u}\mathrm{d}u\\
&=-\frac12\operatorname{Li}_2{(u)}\bigg{|}_{0}^{1}\\
&=-\frac12\operatorname{Li}_2{(1)}\\
&=-\frac{\pi^2}{12}.
\end{align}$$
I fully anticipate there will be a not small number of people who call this cheating, but it's certainly cheating with style!

Edit: Here's a slightly more satisfying result. Substitute $u=x^2$ first like before, and next substitute $u=1-e^{-w}$. Then, 
$$\begin{align}
\int_{0}^{1}\frac{\ln{(1-x^2)}}{x}\mathrm{d}x
&=\int_{0}^{1}\frac{\ln{(1-u)}}{\sqrt{u}}\cdot\frac{\mathrm{d}u}{2\sqrt{u}}\\
&=\frac12\int_{0}^{1}\frac{\ln{(1-u)}}{u}\mathrm{d}u\\
&=\frac12\int_{0}^{\infty}\frac{-w}{1-e^{-w}}(e^{-w})\mathrm{d}w\\
&=-\frac12\int_{0}^{\infty}\frac{w}{e^w-1}\mathrm{d}w\\
&=-\frac12\Gamma{(2)}\zeta{(2)}\\
&=-\frac12\zeta{(2)}.
\end{align}$$
So now the question becomes do you accept that $\zeta{(2)}=\frac{\pi^2}{6}$. This still leaves a bit to be desired, but a lot more has written about $\zeta{(2)}$ than $\operatorname{Li}_2{(1)}$.
A: Using the dilogarithm $\mathrm{Li}_2\;$ and the particular values for $0,1,-1\;$you get:
$$\int_0^1\frac{\ln(1-x^2)}{x}dx= \int_0^1\frac{\ln(1-x)(1+x)}{x}dx=
\int_0^1\frac{\ln(1+x)}{x}dx + \int_0^1\frac{\ln(1-x)}{x}dx=
-\mathrm{Li}_2(-x)\Big{|}_0^1 - \mathrm{Li}_2(x)\Big{|}_0^1
=\frac{\pi^2}{12}-\frac{\pi^2}{6} = -\frac{\pi^2}{12}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 - x^{2}} \over x}\,\dd x}
=\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x
\\[3mm]&=-\,\half\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}
{1 - x^{\mu} \over 1 - x}\,\dd x
=-\,\half\lim_{\mu \to 0}\partiald{\color{#f00}{\Psi\pars{\mu + 1}}}{\mu}
=-\,\half\,\Psi'\pars{1}
\\[3mm]&=-\,\half\,\color{#f0f}{\zeta\pars{2}}
=-\,\half\,\color{maroon}{\pi^{2} \over 6}=\color{#66f}{\large -\,{\pi^{2} \over 12}}
\end{align}

See $\underline{\color{#f00}{6.3.22}}$,
  $\underline{\color{#f0f}{6.4.2}}$ and
  $\underline{\color{maroon}{23.2.24}}$.

A: Here is a route without power series expansion. Observing that, by the change of variable $x=\sin u$, we have
$$
\int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=\int_0^{\pi/2}\ln(\cos^2 u)\frac{\cos u}{\sin u}\mathrm{d}u
$$
You may use the Fourier series expansion
$$
\log(\cos u)=\sum_{k=1}^\infty(-1)^{k}\frac{1-\cos(2kx)}{k}=\sum_{k=1}^\infty(-1)^{k}\frac{\sin^2(kx)}{k}
$$
to obtain
$$
\int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=2\sum_{k=1}^\infty\! \frac{(-1)^{k}}{k}\!\!\int_0^{\pi/2}\frac{\sin^2(kx)}{\sin u}\cos u \:\mathrm{d}u=-\sum_{k=1}^\infty\frac{\frac11+ ...+\frac1{2k+1}}{2k(2k-1)} =-\frac{\pi^2}{12}.
$$
A: After substituting $y=x^2$, we obtain
$$
\int_0^1\frac{\ln(1-x^2)}{x}\ dx=\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy
$$
Using the fact that
$$
\frac{\ln(1-x)}{x}=-\int_0^1\frac{1}{1-xy}\ dy
$$
then
$$
\frac12\int_0^1\frac{\ln(1-x)}{x}\ dx=-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx.
$$
Using transformation variable by setting $(u,v)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$ so that $(x,y)=(u-v,u+v)$ and its Jacobian is equal to $2$. Therefore
$$
-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx=-\iint_A\frac{du\ dv}{1-u^2+v^2},
$$
where $A$ is the square with vertices $(0,0),\left(\frac{1}{2},-\frac{1}{2}\right), (1,0),$ and $\left(\frac{1}{2},\frac{1}{2}\right)$. Exploiting the symmetry of the square, we obtain
$$
\begin{align}
\iint_A\frac{du\ dv}{1-u^2+v^2}=\ &2\int_{u=0}^{\Large\frac12}\int_{v=0}^u\frac{dv\ du}{1-u^2+v^2}+2\int_{u=\Large\frac12}^1\int_{v=0}^{1-u}\frac{dv\ du}{1-u^2+v^2}\\
=\ &2\int_{u=0}^{\Large\frac12}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\ du\\
&+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\ du.
\end{align}
$$
Since $\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)=\arcsin u$, and if $\theta=\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2\theta=\frac{1-u}{1+u}$ and $\sec^2\theta=\frac{2}{1+u}$. It follows that $u=2\cos^2\theta-1=\cos2\theta$ and $\theta=\frac12\arccos u=\frac\pi4-\frac12\arcsin u$. Thus
$$
\begin{align}
\iint_A\frac{du\ dv}{1-u^2+v^2}
&=2\int_{u=0}^{\Large\frac12}\frac{\arcsin u}{\sqrt{1-u^2}}\ du+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\left(\frac\pi4-\frac12\arcsin u\right)\ du\\
&=\bigg[(\arcsin u)^2\bigg]_{u=0}^{\Large\frac12}+\left[\frac\pi2\arcsin u-\frac12(\arcsin u)^2\right]_{u=\Large\frac12}^1\\
&=\frac{\pi^2}{36}+\frac{\pi^2}{4}-\frac{\pi^2}{8}-\frac{\pi^2}{12}+\frac{\pi^2}{72}\\
&=\frac{\pi^2}{12}
\end{align}
$$
and the result follows.
