I was wondering about how I would go on evaluating
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln{(\tan{(x)}})}{\tan{(x)}}dx$$
I used a calculator to find it evaluates to $\frac{\pi^2}{48}$, but I'm unable to prove this result. I've tried basic integration technique and Feynman's rule of integration to no avail. Thoughts?
Substitute $t=\frac1{\tan^2 x}$ $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln{(\tan{x}})}{\tan{x}}dx=-\frac14 \int_0^1 \frac{\ln t}{1+t}dt\overset{ibp}=\frac14 \int_0^1 \frac{\ln (1+t)}tdt $$ where $\int_0^1 \frac{\ln (1+t)}tdt =\frac{\pi^2}{12}$.
Letting $t=\cot x$ yields $$ \begin{aligned} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln (\tan x)}{\tan x} d x & =-\int_0^1 \frac{t \ln t}{1+t^2} d t \\ & =-\sum_{n=0}^{\infty}(-1)^n \int_0^1 t^{2 n+1} \ln t d t \\ & =-\left.\sum_{n=0}^{\infty}(-1)^n\left[\frac{d}{d a} \int_0^1 t^a d t\right]\right|_{a=2 n+1} \\ & =-\sum_{n=0}^{\infty}(-1)^n \cdot\left.\left[-\frac{1}{(a+1)^2}\right]\right|_{a=2 n +1} \\ & =\frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \\ & =-\frac{1}{4} \cdot \frac{\pi^2}{12} \\ & =-\frac{\pi}{48} \end{aligned} $$
