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...

(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.
