Show that $\displaystyle \int_0^{\pi/2}\ln(\tan x)\,dx=0$
well, $\displaystyle \int \ln (\tan x)~dx = \int \ln \left( \frac {\sin x }{\cos x} \right)~dx = \int \ln (\sin x ) ~dx - \int \ln (\cos x)~dx$
by symmetry, the area "under" the curve $\displaystyle \ln (\sin x)$ is the same as the area "under" the curve of $\displaystyle \ln (\cos x)$ between the limits $\displaystyle 0$ and $\displaystyle \frac {\pi}2$. thus,
$\displaystyle \int_0^{\frac {\pi}2} \ln (\sin x)~dx = \int_0^{\frac {\pi}2} \ln (\cos x )~dx$
$\displaystyle \Rightarrow \int_0^{\frac {\pi}2} \ln (\sin x)~dx - \int_0^{\frac {\pi}2} \ln (\cos x)~dx = 0 = \int_0^{\frac {\pi}2} \ln (\tan x )~dx$
Or we can say
$\displaystyle u=\int_0^{\pi /2} {\ln (\sin x)\,dx} .$
So $\displaystyle u=\int_0^{\pi /2} {\ln (\cos x)\,dx} .$
(This last it's easy to prove with a simple substitution.)
Finally
$\displaystyle \int_0^{\pi /2} {\ln (\sin x)\,dx} - \int_0^{\pi /2} {\ln (\cos x)\,dx} =\int_0^{\pi /2} {\ln (\tan x)\,dx} = 0\,\blacksquare$
I agree. I like this integral
Define $\displaystyle \varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} .$
Set $\displaystyle u=\frac\pi2-x,$
$\displaystyle \varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} = \int_0^{\pi /2} {\ln (\cos x)\,dx} .$
Now adding this integrals,
$\displaystyle 2\varphi = \int_0^{\pi /2} {\ln (\sin 2x)\,dx} - \frac{\pi }{2}\ln 2.$
For the last integral, make a substitution according to $\displaystyle \alpha=2x,$
$\displaystyle 2\varphi = \frac{1}{2}\int_0^\pi {\ln (\sin \alpha )\,d\alpha } - \frac{\pi }{2}\ln 2.$
Since $\displaystyle \int_0^\pi {\ln (\sin \alpha )\,d\alpha } = \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } + \int_{\pi /2}^\pi {\ln (\sin \alpha )\,d\alpha } ,$
let's make a final substitution defined by $\displaystyle \beta=\pi-\alpha$ for the second integral, so
$\displaystyle \int_0^\pi {\ln (\sin \alpha )\,d\alpha } = 2\int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = 2\varphi .$
Therefore
$\displaystyle 2\varphi = \varphi - \frac{\pi }{2}\ln 2\,\therefore \,\varphi = - \frac{\pi }{2}\ln 2\,\blacksquare$