How to show that this integral is correct? How can one show that
$$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$
assume $\alpha\ge1$.
I can't see how to simplify $\tan(x)\tan(x/2)$ and $\sin(x)\tan(x/2)$.

Use properties of trig functions to get
\begin{equation*}
\sin(x)\tan\left(\frac{x}{2}\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=2\sin^2\left(\frac{x}{2}\right).
\end{equation*}
Now we can simplify the integral to
\begin{equation*}
\int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)\left(\ln \tan(x)\tan\left(\frac{x}{2}\right)-\ln\alpha\right) dx.
\end{equation*}
It is easy to verify that
\begin{equation*}
\int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)(-\ln\alpha)dx=-\frac{\ln \alpha}{\sqrt{2}}.
\end{equation*}
which is your answer. Now we only need to show
\begin{equation}\label{x}
\int_0^{\pi/2} \sin(x) \ln \tan(x)\tan\left(\frac{x}{2}\right) dx=0.
\end{equation}
Let $t=\cos(x)$, we can draw a right triangle with sides $1,t,\sqrt{1-t^2}$ and draw a bisector of the angle $x$. Then
\begin{equation*}
\tan(x)=\frac{\sqrt{1-t^2}}{t},\quad\tan\left(\frac{x}{2}\right)=\frac{\sqrt{1-t^2}}{1+t}
\end{equation*}
and the integral becomes 
\begin{equation*}
\int_0^1 \ln \left(\frac{1-t}{t}\right)dt=\int_0^1 \ln(1-t)dt-\int_0^1 \ln tdt=0
\end{equation*}
by substitution $u=1-t$.
