Prove that $\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}$ 
Prove that
  \begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}
\end{equation}


I tried to use by parts method and ended with
\begin{equation}
\int \ln^2(\cos x)\,dx=x\ln^2(\cos x)+2\int x\ln(\cos x)\tan x\,dx
\end{equation}
The latter integral seems hard to evaluate. Could anyone here please help me to prove it preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
Addendum:
I also found this nice closed-form
\begin{equation}
-\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3)
\end{equation}
I hope someone here also help me to prove it. (>‿◠)✌

Let's get all powers of $\ln(\cos(x))$ at once, using an exponential generating function:
$$ G(z) = \int_0^{\pi/2} \sum_{j=0}^\infty \dfrac{\ln^j(\cos(x)) z^j}{j!}\; dx = 
\int_0^{\pi/2} \exp(z \ln(\cos(x)))\; dx$$
Change variables: $\cos(x) = t^{1/2}$, and using the Beta function:
$$\eqalign{\dfrac{1}{2} &\int_0^1 \dfrac{t^{(z-1)/2}}{\sqrt{1-t}}\; dt =
\dfrac{1}{2} B\left(\dfrac{1}{2}, \dfrac{z+1}{2}\right) = \dfrac{\Gamma(1/2)\Gamma((z+1)/2)}{2\; \Gamma(1+z/2)}\cr
&= {\frac {\pi }{2}}-{\frac {\pi \,\ln  \left( 2 \right) }{2}}z+ \left( 
{\frac {{\pi }^{3}}{48}}+{\frac {\pi \, \left( \ln  \left( 2 \right) 
 \right) ^{2}}{4}} \right) {z}^{2}-{\frac {\pi \, \left( 4\, \left( 
\ln  \left( 2 \right)  \right) ^{3}+{\pi }^{2}\ln  \left( 2 \right) +6
\,\zeta  \left( 3 \right)  \right) }{48}}{z}^{3}+\ldots 
} $$
