1. ## Prove

Prove :

$\120dpi \int_{0}^{\frac{\pi }{2}}{\color{red} \ln } \left ( \tan x \right+\cot x )dx={\color{red}\pi }\ln 2$

2. Try this.. tan(x)=sin(x)/cos(x) and cot(x) = cos(x)/sin(x) add them both you get c^2+s^2/c*s = 1/c*s

then it would become integral between 0 and pi/2 --> ln (1/cos(x)*sin*(x)) dx ...

now you know sin(2x)=2sin(x)cos(x)

see what you get form here...

3. I think of another method , old method is to transform it to this famous integral :$\displaystyle \int_0^{\pi/2} \ln[\sin(x)]~dx = - \frac{\pi}{2} \ln(2)$ .

Let

$\displaystyle I(a) = \int_0^{\pi/2} \ln[ \tan(x) + a \cot(x) ] ~dx$

Then $\displaystyle I'(a) = \int_0^{\pi/2} \frac{ \cot(x) }{ \tan(x) + a \cot(x) }~dx$

$\displaystyle = \int_0^{\pi/2} \frac{dx }{ \tan^2(x) + a }$

Sub. $\displaystyle t = \tan(x)$ we have

$\displaystyle I'(a) = \int_0^{\infty} \frac{dt}{ (1+t^2)(t^2 + a ) }$

$\displaystyle = \frac{1}{a-1 } \int_0^{\infty} \left[ \frac{1}{t^2 + 1 } - \frac{1}{t^2 + a} \right] ~dt$

$\displaystyle = \frac{1}{a - 1 } [ \frac{\pi}{2} - \frac{1}{\sqrt{a}} ~ \frac{\pi}{2} ]$

$\displaystyle = \frac{\pi}{2} \cdot \frac{\sqrt{a} - 1 }{\sqrt{a} (a-1) }$

$\displaystyle = \frac{\pi}{2} \cdot \frac{1 }{\sqrt{a} (\sqrt{a}+1) }$

Therefore , $\displaystyle I(1) - I(0) = \frac{\pi}{2} ~ \int_0^1 \frac{da }{\sqrt{a} (\sqrt{a}+1) }$

Sub. $\displaystyle a = u^2$

$\displaystyle = \frac{\pi}{2} \cdot 2\ln(2) = \pi \ln(2)$

Finally note that $\displaystyle \int_0^{\pi/2} \ln[\tan(x)] ~dx = 0$ so

$\displaystyle \int_0^{\pi/2} \ln[ \tan(x) + \cot(x) ] ~dx = \pi \ln(2)$