Proving $\int_{0}^{\pi/2}\left[2\cos\left({x\over 2}\right)-x\sin\left({x\over 2}\right)\right]\ln\left[2\cos^2\left({x\over 2}\right)\right] dx...$ Proposed:

$$\int_{0}^{\pi/2}\left[2\cos\left({x\over 2}\right)-x\sin\left({x\over 2}\right)\right]\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx=\color{green}{(4-\pi)\sqrt{2}}\tag1$$

My try: 
Rewrite as
$$\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx-
\int_{0}^{\pi/2}x\sin\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx=I_1-I_2\tag2$$
$u=2\cos(x/2)$, then $I_1$ becomes
$$I_1=\int_{\sqrt{2}}^{2}{u\over \sqrt{1-u^2}}\ln\left({{1\over 2}u^2}\right)\mathrm du\tag3$$
$$I_1=-\ln(2)\int_{\sqrt{2}}^{2}{u\over \sqrt{1-u^2}}\mathrm du-\int_{\sqrt{2}}^{2}{u\ln(u)\over \sqrt{1-u^2}}\mathrm du\tag4$$
I am sure $(4)$ is from standard integral table but $I_2$ seems difficult to transform 
How may one prove $(1)$?
 A: Hint. One may write
$$
\begin{align}
&\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx
\\\\&=\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\left(1-\sin^2\left({x\over 2}\right)\right)\right]\mathrm dx
\\\\&=4\int_{0}^{\sqrt{2}/2}\ln\left[2\left(1-u^2\right)\right]\mathrm du
\\\\&=4\ln2\int_{0}^{\sqrt{2}/2}\mathrm du+\int_{0}^{\sqrt{2}/2}\ln\left(1+u\right)\:\mathrm du+4\int_{0}^{\sqrt{2}/2}\ln\left(1-u\right)\:\mathrm du
\\\\&= 8 \ln \left(1+\sqrt{2}\right)-4 \sqrt{2}.
\end{align}
$$
Similarly, integrating by parts,
$$
\begin{align}
&\int_{0}^{\pi/2}x\sin\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx
\\\\&=\left[x \cdot \left(4 \cos \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\right)\right]_{0}^{\pi/2}-\int_{0}^{\pi/2}\left[4 \cos \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\right]\mathrm dx
\\\\&=\sqrt{2} \:\pi-\int_{0}^{\pi/2}4 \cos \left(\frac{x}{2}\right)\mathrm dx
+2\int_{0}^{\pi/2}\cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\mathrm dx
\\\\&=\sqrt{2} \:\pi-8\sqrt{2}+8 \ln\left(1+\sqrt{2}\right).
\end{align}
$$ Considering the preceding results leads to the announced equality.
A: Just simply use integration by parts. Noting
$$ (2x\cos(\frac x2))'=2\cos(\frac x2)-x\sin(\frac x2)$$
one has
\begin{eqnarray}
&&\int_{0}^{\pi/2}\left[2\cos\left({x\over 2}\right)-x\sin\left({x\over 2}\right)\right]\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx\\
&=&\int_{0}^{\pi/2}\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm d\left[2x\cos(\frac x2)\right]\\
&=&\ln\left[2\cos^2\left({x\over 2}\right)\right]\left[2x\cos(\frac x2)\right]\bigg|_0^{\frac\pi2}-\int_{0}^{\pi/2}\left[2x\cos(\frac x2)\right]\mathrm d\ln\left[2\cos^2\left({x\over 2}\right)\right]\\
&=&2\int_0^{\frac{\pi}{2}}x\sin\frac{x}{2}\mathrm  dx\\
&=&\color{green}{(4-\pi)\sqrt{2}}.
\end{eqnarray}
