1. ## Integral

Show that $\int_0^{\pi/2}\ln(\tan x)\,dx=0$

2. Originally Posted by liyi
Show that $\int_0^{\pi/2}\ln(\tan x)\,dx=0$
well, $\int \ln (\tan x)~dx = \int \ln \left( \frac {\sin x }{\cos x} \right)~dx = \int \ln (\sin x ) ~dx - \int \ln (\cos x)~dx$

by symmetry, the area "under" the curve $\ln (\sin x)$ is the same as the area "under" the curve of $\ln (\cos x)$ between the limits $0$ and $\frac {\pi}2$. thus,

$\int_0^{\frac {\pi}2} \ln (\sin x)~dx = \int_0^{\frac {\pi}2} \ln (\cos x )~dx$

$\Rightarrow \int_0^{\frac {\pi}2} \ln (\sin x)~dx - \int_0^{\frac {\pi}2} \ln (\cos x)~dx = 0 = \int_0^{\frac {\pi}2} \ln (\tan x )~dx$

3. Or we can say

$u=\int_0^{\pi /2} {\ln (\sin x)\,dx} .$

So $u=\int_0^{\pi /2} {\ln (\cos x)\,dx} .$

(This last it's easy to prove with a simple substitution.)

Finally

$\int_0^{\pi /2} {\ln (\sin x)\,dx} - \int_0^{\pi /2} {\ln (\cos x)\,dx} =\int_0^{\pi /2} {\ln (\tan x)\,dx} = 0\,\blacksquare$

4. A more interesting problem is so actually find,
$\int_0^{\pi/2} \ln (\sin x) dx$.

5. Originally Posted by ThePerfectHacker
$\int_0^{\pi/2} \ln (\sin x) dx$.
I agree. I like this integral

Define $\varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} .$

Set $u=\frac\pi2-x,$

$\varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} = \int_0^{\pi /2} {\ln (\cos x)\,dx} .$

$2\varphi = \int_0^{\pi /2} {\ln (\sin 2x)\,dx} - \frac{\pi }{2}\ln 2.$

For the last integral, make a substitution according to $\alpha=2x,$

$2\varphi = \frac{1}{2}\int_0^\pi {\ln (\sin \alpha )\,d\alpha } - \frac{\pi }{2}\ln 2.$

Since $\int_0^\pi {\ln (\sin \alpha )\,d\alpha } = \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } + \int_{\pi /2}^\pi {\ln (\sin \alpha )\,d\alpha } ,$

let's make a final substitution defined by $\beta=\pi-\alpha$ for the second integral, so

$\int_0^\pi {\ln (\sin \alpha )\,d\alpha } = 2\int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = 2\varphi .$

Therefore

$2\varphi = \varphi - \frac{\pi }{2}\ln 2\,\therefore \,\varphi = - \frac{\pi }{2}\ln 2\,\blacksquare$