I am struggling with this integral:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$
What I tried so far:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$
$\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{\cos^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
$\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{-\sin^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
$\displaystyle =\frac{1}{2}\int _{0}^{\frac{\pi }{2}}\frac{\cos 2x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
The answer should come out to be $\dfrac{-7\pi^2}{72}$.
Any help will be appreciated.
Best Answer
Substitute $t=\tan x$\begin{align} &\int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x \\ =&\int_0^1\frac{\ln t}{(1 - t + t^{2})(1 + t^{2})}dt + \int_1^\infty \frac{\ln t}{(1 - t + t^{2})(1 + t^{2})}\overset{t\to 1/t}{dt}\\ =& \int_0^1\frac{(1-t^2)\ln t}{(1 - t + t^{2})(1 + t^{2})}dt = \int_0^1\frac{2t\ln t}{1 + t^{2}}\>\overset{ibp}{dt} -\int_0^1\frac{(2t-1)\ln t}{1 - t + t^{2}}\>\overset{ibp}{dt}\\ =& \>-\int_0^1 \frac{\ln (1+t^2)}{t}\>\overset{t^2\to t}{dt} +\int_0^1 \frac{\ln (1+t^3)}t \>\overset{t^3\to t}{dt}-\int_0^1 \frac{\ln (1+t)}tdt\\ =& \>-\frac76 \int_0^1 \frac{\ln (1+t)}tdt= -\frac76 \cdot\frac{\pi^2}{12}=-\frac{7\pi^2}{72} \end{align} where $\int_0^1 \frac{\ln (1+t)}tdt=\frac{\pi^2}{12}$