<s>[INST] 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. (>‿◠)✌
 [/INST] 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.
</s>