Find $\displaystyle \int_{0}^{\pi /2}{\frac{\tan ^{\alpha }x-1}{\ln (\tan x)}\,dx},\,0<\alpha <1.$
$\displaystyle \int_{0}^{\pi /2}{\frac{\tan^{n}x-1}{\ln (\tan x)}\,dx}$
$\displaystyle \frac{\tan ^{n}x-1}{\ln (\tan x)} = \int_0^n \tan^y x~dy$
It becomes,
$\displaystyle \int_{0}^{\pi /2} \int_0^n \tan^y x~dy ~dx$
$\displaystyle \int_0^n \int_{0}^{\pi /2} \tan^y x~dx ~dy$
I'm not in a condition to go further. It's been too late I better go sleep some ..