Show that $(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}$ How do we show that $(1)$

$$(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}?\tag1$$

$n\ge0$ and G is the Catalan's constant
$u=\tan{x}$ then $\mathrm \cos^2{x}du=\mathrm dx$ [0,1]
$$-\int_{0}^{1}(u)^{2n}\ln(u)\cos^2{x}\mathrm du\tag2$$
$$-\int_{0}^{1}u^{2n}\ln(u){\mathrm du\over 1+u^2}\tag3$$
$$-\int_{0}^{1}\ln(u){\mathrm du\over 1+u^2}=G\tag4$$
 A: $n\geq 1$,
$\displaystyle J_n=(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx$
Perform the change of variable $y=\tan x$,
$\begin{align} J_n&=(-1)^{n+1} \int_0^1 \frac{x^{2n}\ln x}{1+x^2}\,dx\\
&=-\int_0^1 \frac{(-x^2)^{n}\ln x}{1+x^2}\,dx\\
&=\int_0^1 \frac{\Big(1-(-x^2)^{n}\Big)\ln x}{1+x^2}\,dx-\int_0^1 \frac{\ln x}{1+x^2}\,dx\\
&=\int_0^1  \left(\sum_{k=0}^{n-1}(-x^2)^k\ln x\right)\,dx+\text{G}\\
&=\sum_{k=0}^{n-1}\left((-1)^k\int_0^1 x^{2k}\ln x\,dx\right)+G\\
&=\sum_{k=0}^{n-1} \frac{(-1)^{k+1}}{(2k+1)^2}+\text{G}\\
&=\sum_{k=1}^{n} \frac{(-1)^{k}}{(2k-1)^2}+\text{G}\\
\end{align}$
NB:
For $s\geq 0$,
$\displaystyle \int_0^1 x^s \ln x\,dx=-\frac{1}{(s+1)^2}$
A: This is a partial solution. I can take the integral as far as expressing it in terms of polygamma functions of order one (the trigamma function) but are currently unable to simplify it further into the form the OP gives in terms of Catalan's constant $G$ and a finite sum term.
Let
$$I_n = \int^{\pi/4}_0 \tan^{2n} x \ln (\cot x) \, dx,$$
and setting $u = \tan x$ the integral can be rewritten as
$$I_n = -\int^1_0 \frac{u^{2n}}{1 + u^2} \ln u \, du.$$
Writing the term $1/(1+u^2)$ that appears in the integrand as a geometric series, namely
$$\frac{1}{1+u^2} = \sum^\infty_{k = 0} (−1)^k u^{2k},\quad |u|<1,$$
after interchanging the summation with the integration (by Tonelli's theorem) we have
$$I_n = -\sum^\infty_{k = 0} (-1)^k \int^1_0 u^{2k + 2n} \ln u \, du.$$
After integrating by parts, we find
$$I_n = \sum^\infty_{k = 0} \frac{(-1)^k}{(2k + 2n + 1)^2}.$$
To evaluate the sum we write
\begin{align*}
I_n &= \sum^\infty_{k = 0} \frac{(-1)^k}{(2k + 2n + 1)^2} = \frac{1}{4} \sum^\infty_{k = 0} \frac{(-1)^k}{(k + n + \frac{1}{2})}\\
&= \frac{1}{4} \sum^\infty_{k = 0, k \in \text{even}} \frac{1}{(k + n + \frac{1}{2})^2} - \frac{1}{4} \sum^\infty_{k = 0, k \in \text{odd}} \frac{1}{(k + n + \frac{1}{2})^2}.
\end{align*}
Reindexing, in the first sum let $k \mapsto 2k$ while in the second sum let $k \mapsto 2k + 1$. Thus 
\begin{align*}
I_n &= \frac{1}{4} \sum^\infty_{k = 0} \frac{1}{(2k + n + \frac{1}{2})^2} - \frac{1}{4} \sum^\infty_{k = 0} \frac{1}{(2k + n + \frac{3}{2})^2}\\
&= \frac{1}{16} \sum^\infty_{k = 0} \frac{1}{(k + \frac{n}{2} + \frac{1}{4})^2} - \frac{1}{16} \sum^\infty_{k = 0} \frac{1}{(k + \frac{n}{2} + \frac{3}{4})^2}\\
&= \frac{1}{16} \left [\zeta \left (2, \frac{2n + 1}{4} \right ) - \zeta \left (2, \frac{2n + 3}{4} \right ) \right ],
\end{align*}
where $\zeta (s,z)$ is the Hurwitz zeta function.
Now the polygamma function $\psi^{(m)}(z)$ is related to the Hurwitz zeta function by
$$\zeta (1 + m, z) = \frac{(-1)^{m + 1}}{m!} \psi^{(m)}(z),$$
so in terms of polygamma (trigamma) functions our integral can be expressed as
$$I_n = \frac{1}{16} \left [\psi^{(1)} \left (\frac{2n + 1}{4} \right ) - \psi^{(1)} \left (\frac{2n + 3}{4} \right ) \right ].$$
It now remains to prove that
$$\psi^{(1)} \left (\frac{2n + 1}{4} \right ) - \psi^{(1)} \left (\frac{2n + 3}{4} \right ) = (-1)^n 16 \left [G + \sum^n_{k = 1} \frac{(-1)^k}{(2k - 1)^2} \right ].$$
