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 $\sin(x)\tan(x/2)$.
 A: 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 $u=1-t$.
A: This integral becomes deceptively easy when you simplify the integrand and use Feynman's Trick to differentiate with respect to its parameter $a$. First off, use the double angle identity for $\sin x$$$\sin x=2\sin\frac x2\cos\frac x2$$and call our integral $I$. What's left becomes$$\begin{align*}I(a) & =\int\limits_0^{\pi/2}dx\,\cos\frac x2\sqrt{2\sin^2\frac x2}\log\left(\frac 1a\tan x\tan\frac x2\right)\\ & =\sqrt2\int\limits_0^{\pi/2}dx\,\sin\frac x2\cos\frac x2\log\left(\frac 1a\tan x\tan\frac x2\right)\\ & =\frac 1{\sqrt2}\int\limits_0^{\pi/2}dx\,\sin x\log\left(\frac 1a\tan x\tan\frac x2\right)\end{align*}$$
Now differentiate with respect to $a$. The natural log then becomes $-1/a$ and the resulting integral becomes trivial
$$I'(a)=-\frac 1{a\sqrt2}\int\limits_0^{\pi/2}dx\,\sin x=-\frac 1{a\sqrt2}$$
Integrate with respect to $a$ and we get
$$I(a)=-\frac {\log a}{\sqrt2}+C$$
When $a=0$, then $I(a)=0$ and the right-hand side also equals zero. Therefore, $C=0$ and the identity has been proven$$\int\limits_0^{\pi/2}dx\,\cos\frac x2\log\left(\frac 1a\tan x\tan\frac x2\right)\sqrt{\sin x\tan\frac x2}\color{blue}{=-\frac {\log a}{\sqrt2}}$$
A: Using 
$$\tan(x) \, \tan(x/2) = \frac{2 \, \sin^{2}(x/2)}{\cos(x)}$$
and 
$$\sin(x) \, \tan(x/2) = 2 \, \sin^{2}(x/2)$$ 
then
\begin{align}
I &= \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 \\
&= 2\sqrt{2} \, \int_{0}^{\pi/2} \sin(x/2) \, \cos(x/2) \, \ln\left(\frac{2}{\alpha} \, \frac{\sin^{2}(x/2)}{\cos(x)} \right) \, dx 
\end{align}
Let $u = \sin(x/2)$ to obtain
\begin{align}
I &= 4 \sqrt{2} \, \int_{0}^{1/\sqrt{2}} u \, \ln u \, du - 2\sqrt{2} \, \ln\left(\frac{\alpha}{2}\right) \, \int_{0}^{1/\sqrt{2}} u \, du - \sqrt{2} \,\int_{0}^{\pi/2} \sin(x) \, \ln(\cos(x)) \, dx \\
&= - \frac{1 + \ln 2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \, \ln\left(\frac{\alpha}{2}\right) + \frac{1}{\sqrt{2}} \\
&= - \frac{\ln \alpha}{\sqrt{2}}.
\end{align}
A: Try to apply $\frac{d}{d \alpha}$ on both sides observing that $\sin x = 2 \sin \left(\frac x2\right) \cos \left(\frac x 2\right)$, the integral is then solvable.
A: Let $u=\frac{1}{2}x \implies \int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sqrt{2\sin(u)\cos(u)\tan(u)}du$
$$\int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sqrt{2\sin^2(u)}du$$
$$\sqrt{2}\int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sin(u)du$$
$$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]du$$
$$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{1}{\alpha}\frac{2\sin(u)\cos(u)}{cos(2u)}tan(u)]du$$
$$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{2}{\alpha}\frac{\sin^2(u)}{\cos(2u)}]du$$
$$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + \int_0^{\pi/4} \sin(u)\cos(u)ln[\frac{\sin^2(u)}{\cos(2u)}]du)$$
$$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + 
\frac{1}{2} \int_0^{\pi/4} \sin(2u)ln[\frac{\sin^2(u)}{\cos(2u)}]du)$$
$$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + 
\frac{-1}{4} \int_1^{0} ln[\frac{1-\frac{1}{2}(1+a)}{a}]da)$$
Now is almost immediate.
