What methods can be used to solve $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $ I'm seeking methods to solve the following definite integral:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $$
 A: The method I took was:
First make the substitution $t = \tan(x)$
$$ I = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt $$
Now, let
$$ I\left(\omega\right) = \int_{0}^{\infty} \frac{\arctan(\omega t)}{t\left(1 + t^2\right)} \:dt $$
Thus, 
\begin{align}
 \frac{dI}{d\omega} &= \int_{0}^{\infty} \frac{t}{t\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \:dt \\
&= \int_{0}^{\infty} \frac{1}{\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \\
&= \frac{1}{\omega^2  - 1} \int_{0}^{\infty}\left[\frac{\omega^2}{\left(1 + \omega^2t^2\right)} - \frac{1}{\left(1 + t^2\right)}\right]dt \\
&= \frac{1}{\omega^2  - 1} \left[\omega\arctan(\omega t) - \arctan(t) \right]_{0}^{\infty} \\
&= \frac{1}{\omega^2  - 1} \left[\omega\frac{\pi}{2} - \frac{\pi}{2}\right]\\
&= \frac{1}{\omega + 1}\frac{\pi}{2}
\end{align}
Hence, 
$$ I(\omega) = \int \frac{1}{\omega + 1}\frac{\pi}{2}\:d\omega = \frac{\pi}{2}\ln|\omega + 1| + C$$
Setting $\omega = 0$ we find:
$$I(0) = C = \int_{0}^{\infty} \frac{\arctan(0 \cdot t)}{t\left(1 + t^2\right)}\:dt = 0 $$
Thus, 
$$ I(\omega) = \frac{\pi}{2}\ln|\omega + 1| $$
And finally, 
$$I(1) = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt =\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \frac{\pi}{2}\ln(2)$$
A: Another method I found was to employ integration by parts:
$$\int uv' = uv - \int vu'$$
Let:
\begin{align}
 u&= x & u'&= 1 \\
 v'&= \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)} & v' &= \ln|\sin(x)| \\
\end{align}
Thus, 
\begin{align}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)}\:dx &= \bigg[x \ln|\sin(x)| \bigg]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}}\ln|\sin(x)|\:dx \\
 &= - \int_{0}^{\frac{\pi}{2}}\ln|\sin(x)|\:dx
\end{align}
From here, one can source the methods as discussed: here
