We have the integral
$$\int_{0}^{\pi/4}{\tan^{2e}x-2\sin^2 x\over \sin(2x)\ln{\tan x}}\mathrm dx={1\over 2}+{1\over 2}\ln{\pi\over 2}\tag1$$
What other methods can employ to prove $(1)$?
An attempt:
Rewrite $(1)$ as
$$\int_{0}^{\pi/4}{\tan^{2e} x\over \sin(2x)\ln{\tan x}}\mathrm dx-2\int_{0}^{\pi/4}{\sin^2x\over \sin(2x)\ln{\tan x}}\mathrm dx=I_1+I_2\tag2$$
$u=\tan x$ $\implies du=(1+u^2)dx$, then $I_1$ becomes
$${1\over 2}\int_{0}^{1}{u^{-e-1}\over \ln u}\mathrm du\tag3$$
Again, $u=e^v \implies du=e^vdv$, then $(3)$ becomes
$${1\over 2}\int_{0}^{\infty}e^{-2ev}\mathrm dv={1\over 4e}\tag4$$
$$I_2=\int_{0}^{\pi/4}{\tan x\over \ln \tan x}\mathrm dx$$
$u=\tan x \implies (1+u^2)dx$, then $I_2$ becomes
$$\int_{0}^{1}{\mathrm du\over \ln u }{\cdot {1\over u^{-1}+u}}\tag5$$
Again, $u=e^v \implies du=e^vdv$, then $(5)$ becomes
$${1\over 2}\int_{0}^{\infty}{e^{-v}\over v}\cdot{\mathrm dv\over \cosh v}=\int_{0}^{\infty}{\mathrm dv\over v(1+e^{2v})}\tag6$$
Not sure how to deal with $(6)$...(diverge?)
Hint: Noting that $$ \sin^2x=\frac{\tan^2x}{1+\tan^2x},\sin(2x)=\frac{2\tan x}{1+\tan^2x} $$ after changing variable $x\to \tan x$ and $\ln x\to-x$, one has \begin{eqnarray} &&\int_{0}^{\pi/4}{\tan^{2e}x-2\sin^2 x\over \sin(2x)\ln{\tan x}}\mathrm dx\\ &=&\int_0^{\pi/4}{\tan^{2e}x(1+\tan^2x)-2\tan^2 x\over 2\tan x\ln{\tan x}}\mathrm dx\\ &=&\frac12\int_0^1{x^{2e-1}(1+x^2)-2x\over \ln{x}}\frac{1}{x^2+1}\mathrm dx\\ &=&\frac12\sum_{n=0}^\infty(-1)^n\int_0^1{x^{2e-1}(1+x^2)-2x\over \ln{x}}x^{2n}\mathrm dx\\ &=&\frac12\sum_{n=0}^\infty(-1)^n\int_0^1{(x^{2(n+e)-1}-x^{2n+1})+(x^{2(n+e)+1}-x^{2n+1})\over\ln x}\mathrm dx\\ &=&\frac12\sum_{n=0}^\infty(-1)^n\bigg[\ln\frac{n+e}{n+1}+\ln\frac{n+e+1}{n+1}\bigg]\\ &=&\frac12\bigg(1+\ln\prod\frac{(2n)^2}{(2n)^2-1}\bigg)\\ &=&\frac12(1+\ln\frac{\pi}{2}) \end{eqnarray} Here $$ \int_0^1{x^{p}-x^q\over \ln{x}}\mathrm dx=\ln\frac{p+1}{q+1} $$ and the Wallis product $$ \prod\frac{(2n)^2}{(2n)^2-1}=\frac{\pi}{2} $$ are used.