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. (>‿◠)✌

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
$$
\int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=2\sum_{k=1}^\infty\! \frac{(-1)^{k}}{k}\!\!\int_0^{\pi/2}\frac{\sin^2(kx)}{\sin u}\cos u \:\mathrm{d}u=-\sum_{k=1}^\infty\frac{\frac11+ ...+\frac1{2k+1}}{2k(2k-1)} =-\frac{\pi^2}{12}.
$$
