How may we show that $\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)?$ Simple closed form given by this complicated integral

$$\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)\tag1$$

Making an attempt:
Splitting $(1)$ results in the integral to be diverges.
$$I(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\cdot\ln(x)\,\mathrm dx\tag2$$
$$I'(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\,\mathrm dx\tag3$$
$$I'(2)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\,\mathrm dx\tag4$$
$u={2x\over \pi}$
$$I'(2)={2\over \pi}\int_{0}^{\infty}{{\cos\left({u}\right)-\cos^2\left({u}\right)}\over u^2}\,\mathrm du=I_1-I_2\tag5$$
Recall from table of integral, I was thinking of using $(6)$ for $I_1$ but it is only valid for $0\le p\le 1$
$$\int_{0}^{\infty}{\cos x\over x^p}\,\mathrm dx={\pi\over 2\Gamma(p)\cos(p\pi/2)}\tag6$$
How may we prove (1)?
 A: I will propose an alternative (but extremely similar to tired's) approach.
By the Laplace transform, for any $k>0$ and any $\alpha\in(1,3)$ we have
$$ \int_{0}^{+\infty}\frac{1-\cos(kx)}{x^\alpha}\,dx = k^{\alpha-1}\int_{0}^{+\infty}\frac{1-\cos(x)}{x^{\alpha}}\,dx =\frac{k^{\alpha-1}}{\Gamma(\alpha)}\int_{0}^{+\infty}\frac{s^{\alpha-2}}{1+s^2}\,ds\tag{1}$$
and the last integral can be computed through the Beta function. In particular we get:
$$\int_{0}^{+\infty}\frac{1-\cos(kx)}{x^\alpha}\,dx = \frac{\pi\,k^{\alpha-1}}{2\cos\left(\frac{\pi}{2}(a-2)\right)\Gamma(\alpha)} \tag{2}$$
and by differentiating both sides with respect to $\alpha$ we simply get:
$$ g(k)=\int_{0}^{+\infty}\frac{1-\cos(kx)}{x^2}\log(x)\,dx = \frac{k\pi}{2}\left(1-\gamma-\log k\right) \tag{3}$$
so the original integral equals:
$$ \int_{0}^{+\infty}\frac{\cos\left(\frac{2x}{\pi}\right)-\cos^2\left(\frac{2x}{\pi}\right)}{x^2}\log(x)\,dx = \frac{1}{2}\,g\left(\frac{4}{\pi}\right)-g\left(\frac{2}{\pi}\right)=\color{red}{-\log 2}\tag{4} $$
as wanted.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\int_{0}^{\infty}
{\cos\pars{2x/\pi} - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
{1 - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x -
\int_{0}^{\infty}
{1 - \cos\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
{\sin^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x -
\int_{0}^{\infty}
{2\sin^{2}\pars{x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x
\\[5mm] = &\
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x \over 2}\,\dd x -
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x}\,\dd x
\\[5mm] = &\
{2 \over \pi}\int_{0}^{\infty}
{\sin^{2}\pars{x} \over x^{2}}\,\bracks{\ln\pars{1 \over 2}}\,\dd x =
\bbx{\ds{-\ln\pars{2}}}
\end{align}
A: Define 
$$
I(b)=\int_0^{\infty}\frac{\cos(b x)-\cos^2(b x)}{x^2}\log(x)
$$
so the integral in question is $I(2/\pi)$. Now
$$
-I'(b)=\int_0^{\infty}\frac{\sin(b x)(1-2\cos(b x))}{x}\log(x)\underbrace{=}_{xb\rightarrow y}\\I'(1)-\log(b)\int_0^{\infty}\frac{\sin(x)(1-2\cos(x))}{x}=I'(1)
$$
or 
$$
I(b)=-bI'(1)+c
$$
now $I'(1)$ can be calculated pretty straightfowardy from using $2\cos(x)\sin(x)=\sin(2x)$ .
$$
I'(1)=\log(2)\int_0^{\infty}\frac{\sin(x)}{x}=\log(2)\frac{\pi}{2}
$$
Afterwards the only thing left is to fix $c$ which can be done by observing that $\lim_{b\rightarrow 0}I(b)=0$ which follows from the Taylorexpansion of $\cos(bx)-\cos^2(bx)$. This means

$$
I(b)=-b\frac{\pi}{2}\log(2)
$$

from which it follows that $I(2/\pi)=-\log(2)$ as proposed
A: I will propose an answer based on integration by parts.
Put $b = \frac{2}{\pi}$. We get
\begin{gather*}
I = \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\log(x)\, dx = \left[-\dfrac{\cos(bx)-\cos^2(bx)}{x}\log(x)\right]_{0}^{\infty}+\\[2ex]b\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\log(x)\, dx + \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\, dx = 0+bI_1+I_2\tag{1}
\end{gather*}
where
\begin{gather*}
I_1 =\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\log(x)\, dx = \int_{0}^{\infty}\dfrac{\sin(2bx)}{x}\log(x)\, dx -\\[2ex]
\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log(x)\, dx =
\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log\left(\dfrac{x}{2}\right)\, dx -\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log(x)\, dx =\\[2ex] -\log(2)\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\, dx = -\log(2)\int_{0}^{\infty}\dfrac{\sin(x)}{x}\, dx = -\log(2)\dfrac{\pi}{2}= -\log(2)\dfrac{1}{b}
\end{gather*}
and
\begin{gather*}
I_2 = \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\, dx = \left[-\dfrac{\cos(bx)-\cos^2(bx)}{x}\right]_{0}^{\infty} +\\[2ex] b\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\, dx = 0.
\end{gather*}
Consequently $I = -\log(2).$
