Prove $\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$ How to prove
$$\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$$
I got this result using Fourier representation of
$$|\log(\cos(\frac{x}{2}))|$$
and Parseval's identity. But I am wondering if there is a direct way to calculate this integral.
 A: With $t= \frac x2$
\begin{align}
I& = \int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx}\\
&= 4\int_0^{\frac\pi2}\ln^2 (\cos t) dt= 2\int_0^{\frac\pi2}(\ln^2 (\cos t)+ \ln^2 (\sin t)  )dt \\
&=\int_0^{\frac\pi2}\left( \ln^2 (\sin t\cos t) + \ln^2 \frac{\sin t}{\cos t} \right)dt
=J+K\tag1
\end{align}
where
\begin{align}
J &= \int_0^{\frac\pi2}\ln^2 (\sin t\cos t) dt 
\overset{2t\to t}= \frac12 \int_0^{\pi}\ln^2 (\frac12\sin t)dt= \int_0^{\frac\pi2}\ln^2 (\frac12\sin t)dt\\
&= \int_0^{\frac\pi2}\left( \ln^2 2 -2\ln2 \ln\sin t +\ln^2(\sin t ) \right)dt\\
&= \frac\pi2\ln^22 -2\ln 2(-\frac\pi2\ln2) +\int_0^{\frac\pi2}\ln^2(\sin t )dt
= \frac{3\pi}2\ln^22 +\frac14I\\
K&=\int_0^{\frac\pi2} \ln^2 (\frac{\sin t}{\cos t}) dt\overset{u=\tan t}= \int^\infty_0 \frac{\ln^2 u} {1+u^2} du=\frac{\pi^3}8
\end{align}
Plug $J$ and $K$ into (1) to obtain
$$I = 2\pi\ln^22 +\frac{\pi^3}6$$
A: Let us write $$I=\int_{-\pi}^{\pi} \ln^2(\cos(x/2)) dx=4\int_{0}^{\pi/2} \ln^2 \cos y dy$$ Let $\cos y=t$, then
$$I=4\int_{0}^{1} \frac{ln^2 t}{\sqrt{1-t^2}} dt =4\int_{0}^{1} \sum_{k=0}^{\infty} \ln^2 t ~ C_k ~(-1)^k~t^{2k} dt, C_k={-1/2,\choose k}$$ Let $t=e^{-u}$, then
$$I=4\int_{0}^{1}\sum_{k=0}^{\infty} (-1)^k C_k u^2 e^{-(2k+1)u} =8\sum_{k=0}^{\infty} \frac{(-1)^k ~ C_k}{(2k+1)^3}$$
I may get back.
A: In my favorite CAS is has two identic forms integrate in the sense of an unbounded general integral. I can only do screenshots because the MathML is ill-posed. This can be looked up in books with integreal relationsships and representations like Abramovitch, Stegun.
First a longer one:

Now are either integration done automatically or looked up in formula collections.
Evaluating this at the limits of the given integral is:
+∞

doubly singular. The signs do not change for -/+ . So the limits have to be taken. These are not too easy and need complex theory and residue methodology.
The problem gets simpler with the substitution t=x/2 and the knowledge that the function is even.

For t=0 and     t=/2 the term is positive ∞.
This is the graph of the function of the question:

With the substitution     z=Cos[t] from @z-ahmed we get something even easier:

The process path can be gone in this direction and reverse. Mind the minus sign.
Our limit for     z->0 is now 0.
The limit for     z->1 is     1/24  (^2 + 3 Log5^2)
We remember our factor 4.
So the result is
.
This can be simplified with 4 Log[2]^2 - Log[4]^2==0.
