how to compute this definite integral how to compute $\displaystyle I=\int\limits_{0}^{\pi/2}\frac{x}{\tan x}\,dx$
i made $f(x)=\frac{x}{\tan x}$ and then i see that
$$\begin{align}
\lim_{x\to0}f(x)&=\lim_{x\to0}\frac{x}{\tan x}\\
&=\lim_{x\to0}\frac{x\cos x}{\sin x}\\
&=\lim_{x\to0}\frac{\cos x}{\frac{\sin x}{x}}\\
&=\frac{\lim\limits_{x\to0}\cos x}{\lim\limits_{x\to0}\frac{\sin x}{x}}=1\\
\lim_{x\to\pi/2}f(x)&=\lim_{x\to\pi/2}\frac{x}{\tan x}=0
\end{align}$$
we have that $f(x)$ is decreasing into $(0,\pi/2)$ and $f(x)\in(0,1)$ for $x\in(0,\pi/2)$, then i think that
$$0<\displaystyle \int\limits_{0}^{\pi/2}\frac{xdx}{\tan x}<\frac{\pi}{2}$$
i think the primitive can not be expressed in terms of elementary functions, then i think that other method like residue theorem or power series would be used, but i dont know how to use any of these, how i compute this integral?
 A: Integration by parts gives
$$\int_0^{\pi/2}\frac{x}{\tan x}dx=\left. x\log (\sin x)\right|_0^{\pi/2}-\int_0^{\pi/2}\log(\sin x)dx=\frac{\pi}{2}\log 2$$
A: Define $I(b)$ as: 
$I(b)=\int_0^{\pi/2}{\frac{\tan^{-1}(b\tan(x))}{\tan(x)}dx}$ 
It is not hard to show that $f(x,b) = \frac{\tan^{-1}(b\tan(x))}{\tan(x)}$ and its derivative $f'(x)$ are both continuous in $x$ and $b$. Then 
$
\begin{align*}
I'(b) &= \frac{d}{dx}\int_0^{\pi/2}{\frac{\tan^{-1}(b\tan(x))}{\tan(x)}dx} \\
      &= \int_0^{\pi/2}{\frac{\partial}{\partial x}\frac{\tan^{-1}(b\tan(x))}{\tan(x)}dx} \\
      &= \int_0^{\pi/2}{\frac{dx}{(b\tan(x))^2+1}} \\
      &= \frac{\pi}{2(b+1)}
\end{align*}
$ 
Integrating with respect to $b$ leads to: $I(b) = \frac{\pi}{2}\log(b+1)$ 
So the original integral is:
$\int_0^{\pi/2}{\frac{xdx}{tan(x)}} = \frac{\pi}{2}\log(2) $

Ref. "INTEGRATION: THE FEYNMAN WAY" from one of the MIT open course web sites
A: Hint: Letting $x=\tan t$ and integrating by parts with regard to $~\dfrac{\arctan t}{t^2+1}~dt=d\bigg(\dfrac{\arctan^2t}2\bigg)$, then doing the Cauchy product of $\dfrac{\arctan t}t$ with itself after expanding $\arctan t$ into its Taylor series and dividing each term by t, we switch the order of summation and integration only to obtain an infinite series which is immediately recognized as the Cauchy product of the Mercator series for $\ln2$ and the Leibniz series for $\dfrac\pi4$.
