integration - $int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$???

I saw in a paper by @Jack D'aurizio the following integral

$$I=\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$$
Below is my attempt.

$$I=4\int_0^{\pi/2}\log^2(\cos x)\mathrm{d}x$$
Then we define
$$F(a)=\int_0^{\pi/2}\log^2(a\cos x)\mathrm{d}x$$
So we have
$$F'(a)=\frac2a\int_0^{\pi/2}\log(a\cos x)\mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.

Answer

Let $$I(a)=\int_0^{\frac {\pi}{2}} (\cos^2 x)^a dx$$

Hence we need $I''(0)$.

Now recalling the definition of Beta function we get $$I(a)=\frac 12 B\left(a+\frac 12 ,\frac 12\right)=\frac {\sqrt {\pi}}{2}\frac {\Gamma\left(a+\frac 12\right)}{\Gamma(a+1)}$$

Hence we have $$I''(a) =\frac {\sqrt {\pi}}{2}\frac {\Gamma\left(a+\frac 12\right)}{\Gamma(a+1)}\left(\left[\psi^{(0)}\left(a+\frac 12 \right)-\psi^{(0)}(a+1)\right]^2 +\psi^{(1)}\left(a+\frac 12 \right)-\psi^{(1)}(a+1)\right)$$

Substituting $a=0$ in above formula yields the answer.