Prove $\int_{0}^{\pi/2} \ln \left(x^{2} + (\ln\cos x)^2 \right) \, dx=\pi\ln\ln2 $

Here is a real-analytic method.

We have

$$ \int_{0}^{\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, {\rm d}x=\pi\ln(\ln2) \tag1 $$

Proof. Let $s$ be a real number such that $-1<s<1$. One may use the following theorem (proved here) $$ \int_{0}^{\Large \frac{\pi}{2}} \frac{\cos \left( s \arctan \left(\frac{x}{-\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}} \mathrm{d}x = \frac{\pi}{2}\frac{1}{\ln^{s}\!2}. \tag2 $$

We are then allowed to differentiate both sides of $(2)$ $$ \begin{align} \partial_s \left. \left( \frac{\cos \left( s \arctan \left(\frac{x}{-\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}}\right) \right|_{s=0} &=-\frac 12 \ln \left(x^{2} + \ln^2\cos x\right) \\\\ \partial_s \left. \left( \frac{\pi}{2}\frac{1}{\ln^{s}\!2}\right) \right|_{s=0} &=-\frac{\pi}{2}\ln(\ln2) \end{align} $$ which gives the result $(1)$.

As noted in the comments, the integral is: $$2\Re\int_0^{\pi/2} \ln\ln\left(\frac{1+e^{2ix}}{2}\right)\,dx=2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx$$ Consider $$f(x)=\ln(\ln(1+x)-\ln2)$$ Around $x=0$, the taylor expansion can be written as: $$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+....$$ Replace $x$ with $e^{2ix}$. Notice that integrating the powers of $e^{2ix}$ would result in either zero or a purely imaginary number and since the derivatives of $f(x)$ at $0$ are real, we need to consider only the constant term i.e $f(0)$. Since $f(0)=\ln(-\ln 2)=\ln\ln 2+i\pi$, hence, $$2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx=2\int_0^{\pi/2} \ln\ln 2\,dx=\boxed{\pi\ln\ln 2}$$

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A solution using complex analysis is given here by sos440.