<s>[INST] Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$
I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w}
$$ and was asked to find the value $kw$. Here is my try on it:
Substituting $\theta = \dfrac{\pi}{4}+x$, we get
$$\begin{equation} \label{eq1}
\begin{split}
I &= \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{(-4x) \tan (\frac{\pi}{4}+x)}{1-\tan (\frac{\pi}{4}+x)} \, dx \\
  &= -4\displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x.\dfrac{1+ \tan x}{1- \tan x}}{1- \dfrac{1+ \tan x}{1- \tan x}} \, dx \\
&= 2 \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
\end{split}
\end{equation}$$
From here I am stuck on this, I don't know how to proceed from here to get answer in terms of $\pi \ln k-\frac{\pi^2}{w}$
Thank you for any help.
 [/INST] letting $\theta=\frac{\pi}{4}-x$ changes the integral into
$$
I=2 \int_0^{\frac{\pi}{2}} \frac{x(1-\tan x)}{\tan x} d x
$$
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{x}{\tan x} d x &=\int_0^{\frac{\pi}{2}} x d(\ln (\sin x)) \\
&= \left[x \ln (\sin x) \right] _0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \ln (\sin x) d x\\
&=\frac{\pi}{2} \ln 2
\end{aligned}
$$
where the last integral using the result: $\int_0^{\frac{\pi}{2}} \ln (\sin x) d x=-\frac{\pi}{2} \ln 2$.
Hence $$\boxed{\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta =\pi \ln 2-\frac{\pi^2}{4}}
$$
</s>