Show that $\int_0^{\frac{\pi}{2}}\frac{ x}{ \tan x}dx=\frac{\pi}{2} \ln 2$

For $t>0$, define 
\[f(t)=\int_0^{\frac{\pi}{2}}\frac{\tan^{-1}(t \tan x)}{\tan x}dx,\]
and not that 
\[f(1)=\int_0^{\frac{\pi}{2}}\frac{x}{\tan x}dx.\]
Then 
\[f'(t)=\int_0^{\frac{\pi}{2}}\frac{\tan x}{1+(t \tan x)^2}\cdot \frac{1}{\tan x}dx=\int_0^{\frac{\pi}{2}}\frac{1}{1+(t \tan x)^2}dx\]
\[=\int_0^{\frac{\pi}{2}}\frac{\sec^2 x}{1+(t \tan x)^2}\cdot \frac{1}{1+\tan^2 x}dx\]
\[=\int_0^{\infty}\frac{1}{1+t^2 u^2}\cdot\frac{1}{1+u^2}du            (u=\tan x) \]
\[=\frac{1}{t^2-1}\int_0^{\infty} \frac{t^2}{1+t^2 u^2}-\frac{1}{1+u^2}du\]
\[=\frac{1}{t^2-1} \big( t^2\cdot \frac{\tan^{-1}(tu)}{t}-\tan^{-1}u\big)|_0^{\infty}\]
\[=\frac{1}{t^2-1} \big( t^2\cdot \frac{\frac{\pi}{2}}{t}-\frac{\pi}{2}\big)-0=\frac{1}{t^2-1} (t-1)\frac{\pi}{2}\]
\[=\frac{\pi}{2}\frac{1}{t+1}.\]
Thus 
\[f'(t)=\frac{\pi}{2}\frac{1}{t+1}   \Rightarrow  f(t)=\frac{\pi}{2}\ln (t+1)+C.\]
It is easy to see that $f(0)=0=C$. Hence
\[f(t)=\frac{\pi}{2}\ln (t+1).\]
Therefore 
\[\int_0^{\frac{\pi}{2}}\frac{x}{\tan x}dx=f(1)=\frac{\pi}{2}\ln (2).\]