Prove :
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) $