<s>[INST] Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ 
Prove that
  $$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$

I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} \ln(\sin{x})dx=\int_{0}^{\frac{\pi}2} \ln(\cos{x})dx=-\frac{\pi}2\ln2$$
but it didnt work. Any help? 
 [/INST] Here is a different solution using complex method. We can notice that
\begin{align*}
I = &\int_0^{\frac \pi 4} \ln\left(1+\tan x + \sqrt{2\tan x} \right) dx + \int_0^{\frac\pi 4} \ln(\cos x) dx \\
= &I_1 + I_2.
\end{align*}
Evaluation of $I_1$ : Let $\displaystyle\omega = e^{\frac {\pi i} 4} =\frac{1+i}{\sqrt 2}$ and make substitution $\tan x= y^2 $ to find that
\begin{align*}
I_1 =& \int_0^1 \log\left((1+\omega y)(1+\bar \omega y)\right)\frac{2y }{1+y^4}dy  \\
=& 4 \Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy +\frac {\pi\ln 2}2.
\end{align*} Let $\omega y = z$. Then using $\omega^2 = i, \omega^4 = -1$, 
\begin{align*}
\Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy =& \Re \int_0^\omega \frac{ \frac{z}{\omega} \log\left(\frac{1+z}{2}\right)}{1+ (\frac z \omega)^4} \frac{dz}\omega \\
=& \Im \int_0^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz \\
=& \frac 1 2 \Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz.
\end{align*} Note that $\displaystyle f(z) =  \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} $ is analytic in $|z|<1$, and continuous in $|z|\le 1$ except at points $\displaystyle -1, \pm i$. So by choosing a contour $z = e^{i\theta}, -\frac \pi 4\le \theta \le \frac \pi 4$ and using $\log(1+e^{i\theta}) = \log\left(2\cos\left(\frac \theta 2\right)\right) + \frac {i\theta}{2}$, etc, we get
\begin{align*}
\Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz =&\Im \int_{-\frac \pi 4}^{\frac \pi 4} \frac{ie^{i2\theta}\left(\log(\cos (\theta/2)) + i\theta/2\right)}{1-e^{i4\theta}} d\theta\\
=&-\Im  \int_{-\frac \pi 4}^{\frac \pi 4} \frac{\log(\cos (\theta/2)) + i\theta/2}{2\sin(2\theta)} d\theta\\
=& -\frac 1 4\int_{-\frac \pi 4}^{\frac \pi 4} \frac{\theta}{\sin(2\theta)} d\theta\\
=& -\frac 1 4 \int_{0}^{\frac \pi 4} \frac{\theta}{\sin\theta \cos \theta} d\theta\\
=:& -\frac 1 4 J
\end{align*} (In fact $J = G$, the Catalan's number.) This gives
$$
I_1 = -\frac 1 2 J +\frac {\pi \ln 2}2.
$$
Evaluation of $I_2$ : By integrating by parts, we have
\begin{align*}
I_2 =& x\ln(\cos x)|^{\frac \pi 4}_0 +\int_0^{\frac \pi 4} \frac {x \sin x}{\cos x} dx \\
=&-\frac{\pi \ln 2}{8} +K.
\end{align*} We observe that
\begin{align*}
K-J = & -\int_0^{\frac \pi 4} \frac{x\cos x}{\sin x} dx\\
=& -x\log(\sin x)|^{\frac \pi 4}_0 + \int_0^{\frac \pi 4} \log(\sin x) dx\\
=& \frac{\pi \ln 2}{8} + \int_{\frac \pi 4}^{\frac\pi 2} \log(\cos x) dx \tag{$\frac \pi 2 - x\mapsto x$}\\
=&-\frac{3\pi \ln 2}{8} -I_2
\end{align*} thus $\displaystyle K = -\frac{3\pi \ln 2}{8} -I_2 +J$ and 
$$
I_2 = -\frac{\pi \ln 2}{2} -I_2 +J \Longrightarrow I_2 = \frac 1 2 J -\frac{\pi \ln 2}{4}.
$$ Therefore, we have
$$
I = I_1 + I_2 = \left(-\frac 1 2 J +\frac{\pi\ln 2}2\right)+\left(\frac 1 2 J -\frac{\pi \ln 2}{4}\right) = \frac {\pi \ln 2}4.
$$
</s>