Complex Analysis to solve this integral? $\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$ Complex Analysis time! I need some help in figuring out how to proceed to calculate this integral:

$$\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$$

I tried to apply what I have been studying in complex analysis, that is stepping into the complex plane. So
$$\sin(x) \to z - 1/z$$
$$\cos(x) \to z + 1/z$$
Obtaining
$$\int_{|z| = 1} \frac{\ln(z^2-1) - \ln(z)}{\sqrt{z^4 + 3z^2 + 1}} \frac{\text{d}z}{i}$$
I found out the poles,
$$z_k = \pm \sqrt{\frac{-3 \pm \sqrt{5}}{2}}$$
But now I am confused: how to deal with the logarithms? Also, what when I have both imaginary and real poles?
I am a rookie in complex analysis so please be patient...
 A: $$\int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{(1-t^2)(2-t^2)}} \stackrel{t\mapsto\sqrt{t}}{=} \frac{1}{4}\int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{t(1-t)(2-t)}}\stackrel{t\mapsto 1-t}{=}\frac{1}{4}\int_{0}^{1}\frac{\log(1-t)}{\sqrt{t-t^3}}\,dt$$
equals
$$ \frac{1}{2}\int_{0}^{1}\frac{\log(1-t^2)}{\sqrt{1-t^4}}\,dt. $$
By replacing $t$ with the inverse function of the primitive of $\frac{1}{\sqrt{1-t^4}}$ we have that this integral, which is also a harmonic-binomial series, is related to the Weierstrass products of the lemniscate elliptic functions.
I am sure Marco Cantarini is working on the subject at the current time, so it might be a good idea to summon him.
Indeed, after some simplification of Glaisher's result (given by the relations between the complete elliptic integral of the first kind and the $\text{AGM}$ mean) we have
$$ \frac{1}{2}\int_{0}^{1}\frac{\log(1-t^2)}{\sqrt{1-t^4}}\,dt =\color{red}{ \frac{\log(2)-\pi}{16\sqrt{2\pi}}\,\Gamma\left(\frac{1}{4}\right)^2}=-\frac{1}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\cdot\frac{H_{2n+\frac{1}{2}}}{4n+1}.$$
Similarly
$$ \int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{1-t^4}} = \color{red}{-\sqrt{\frac{\pi}{2}}\,\Gamma\left(\frac{5}{4}\right)^2}=-\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(4n+1)^2}.$$
A: @Hans-André-Marie-Stamm, I hope you don't mind that I was unable to solve this problem using Complex Analysis, but here's a method that relies on the Beta Function and some algebric work.
$$\begin{align}I&=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx;\ \cos(x)\rightarrow y\\&=\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-y^2\right)}{\sqrt{1-y^4}}dy =\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{I_1}+\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{I_2}\end{align}$$
$$\begin{align}I_1=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{y=z^{1/4}}=\frac{1}{16}\int_{0}^{1}z^{1/4-1}\frac{\log\left(1-z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{16}\lim_{t \rightarrow 1/2}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{4},t\right)=\frac{1}{16}\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi^{(0)}\left(\frac{1}{2}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
$$\begin{align}I_2=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{y=\sqrt{\frac{1-\sqrt{z}}{1+\sqrt{z}}}}=\frac{1}{32}\int_{0}^{1}z^{1/4-1}\frac{\log\left(z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{32}\lim_{t \rightarrow 1/4}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{2},t\right)=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
Gathering both results:
$$\begin{align}I&=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\\&=\frac{1}{32}\frac{\Gamma\left(1/4\right)\Gamma\left(1/2\right)}{\Gamma\left(3/4\right)}\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
This result can be simplified if one applies Gamma's Reflection Formula, and Digamma's Reflection and Multiplication Formulas, obtaining:
$$I=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx=\frac{\log(2)-\pi}{16\sqrt{2\pi}}\Gamma^2\left(\frac{1}{4}\right)$$
A: (Too long to post as a comment.)
Using the Fourier series
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$
and reducing the power of $\cos(x)$ with the identity
$$\cos^2(x) = \frac{1+\cos(2x)}2$$
we have
$$\begin{align*}
\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1+\cos^2(x)}} \, dx &= -\sqrt2\ln(2) \int_0^{\pi/2} \frac{dx}{\sqrt{3+\cos(2x)}} - \sum_{k=1}^\infty \frac{\sqrt2}k \int_0^{\pi/2} \frac{\cos(2kx)}{\sqrt{3+\cos(2x)}} \, dx \\[1ex]
&= -\frac{\ln(2)}{\sqrt2} I_0 - \frac1{\sqrt2} \sum_{k=1}^\infty \frac{I_k}k 
\end{align*}$$
where for $\alpha\in\{0,1,2,\ldots\}$,
$$I_\alpha = \int_0^\pi \frac{\cos(\alpha x)}{\sqrt{3+\cos(x)}} \, dx$$
each of which apparently (according to Mathematica) evaluate to a linear combination of $E\left(\frac12\right)$ and $K\left(\frac12\right)$ ($K$ and $E$ denote the elliptic integrals of the first and second kind, respectively). Finding a closed form for $I_\alpha$ and in turn for your integral seems unlikely, though.
A: Following D'Aurizio hint, we have to deal with $$I=\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-t^{2}\right)}{\sqrt{1-t^{4}}}dt=\frac{1}{2}\int_{0}^{K(i)}\log\left(1-\text{sinlem}\left(t\right)^{2}\right)dt$$ where $\text{sinlem}\left(t\right)$ is the Lemniscate sine function. Using the relation $\text{sinlem}\left(t\right)=\text{sn}\left(t;i\right)$ where $\text{sn}(z;k)$ is the Jacobi Elliptic sine, we get$$I=\frac{1}{2}\int_{0}^{K(i)}\log\left(1-\text{sn}\left(t;i\right)^{2}\right)dt=\int_{0}^{K(i)}\log\left(\text{cn}\left(t;i\right)\right)dt$$ and now it is enough to recall a classical result of Glaisher $$\int_{0}^{K(k)}\log\left(\text{cn}\left(t;k\right)\right)dt=-\frac{1}{4}\pi K^{\prime}\left(k\right)+\frac{1}{2}K\left(k\right)\log\left(\frac{k^{\prime}}{k}\right)$$ and so $$I=-\frac{1}{4}\pi K^{\prime}\left(i\right)+\frac{1}{2}K\left(i\right)\log\left(\frac{\sqrt{2}}{i}\right)$$ $$=\color{blue}{\frac{L}{8}\left(\log\left(2\right)-\pi\right)}=\color{red} {-0.8024956186037819...}$$
where $L/2$ is the Lemniscate constant.
