How to put upper limit and lower limit for this integral when there are some undefined values in it? I was trying to solve integral for $$ \int_{0}^{\pi/2} (\sin 2x)(\log(\tan x)) dx $$ and in the final result I got some values which also included log(tan(x)). Now while evaluating the integral with limits these log values of both tan(pi/2) and tan(0) is going to be undefined. And after that, I am unable to get the appropriate result. Please help how to proceed with those undefined terms. 
 A: An alternative approach. By enforcing the substitution $x\mapsto\arctan t$ the given integral equals
$$ I= \int_{0}^{+\infty}\frac{2t\log t}{(1+t^2)^2}\,dt =\frac{d}{ds}\left.\int_{0}^{+\infty}\frac{2t^s}{(1+t^2)^2}\,dt\,\right|_{s=1^+}$$
but due to the substitution $\frac{1}{1+t^2}=u$, Euler's Beta function and the reflection formula for the $\Gamma$ function we have
$$ \int_{0}^{+\infty}\frac{2t^s}{(1+t^2)^2}\,dt = \frac{\pi(1-s)}{2\cos\frac{\pi s}{2}}=\frac{\pi(1-s)}{2\sin\frac{\pi(1-s)}{2}}=\frac{1}{\text{sinc}\frac{\pi(1-s)}{2}}$$
for any $-1<s<3$. The last function is an even function of the variable $(1-s)$, hence the derivative of the RHS at $s=1$ equals zero and so it does the original integral.
A: $$\mathop {\lim }\limits_{\varepsilon  \to 0;\delta  \to \frac{\pi }{4}} \int\limits_\varepsilon ^\delta  {\sin 2x\log \tan x\;{\kern 1pt} dx + } \mathop {\lim }\limits_{\delta  \to \frac{\pi }{4};\beta  \to \frac{\pi }{2}} \int\limits_\delta ^\beta  {\sin 2x\log \tan x\;{\kern 1pt} dx}   $$
$f(x)=-f\left(\dfrac{\pi}{2}-x\right)$
$\sin \left(2 \left(\frac{\pi }{2}-x\right)\right) \log \left(\tan \left(\frac{\pi }{2}-x\right)\right)=\sin(\pi-2x)\log(\cot x)=\sin 2x(-\log\tan x)=$
$=-\sin 2x\log\tan x$
In the interval $\left[0,\dfrac{\pi}{2}\right]$ the function is symmetric wrt the point $x=\dfrac{\pi}{4}$ therefore the integral is zero.
A: Note
\begin{align}
& \int_{0}^{\pi/2} \sin 2x\ln(\tan x)dx 
\overset{t=\tan^2 x}=\int_0^\infty \frac{\ln t}{2(1+t)^2}dt
 \overset{t=\frac1t}=-\int_0^\infty \frac{\ln t}{2(1+t)^2}dt=0
\end{align}
