Answer to $\int_0^{{\pi}/{4}} (\pi x-4x^2)\ln(1+\tan x)\,\mathrm dx$

Question

In this post, I asked $\displaystyle\int_0^{{\pi}/{2}} (\pi x-4x^2)\ln(1+\tan x)\mathrm dx$. I received a large amount of positive response accompanied with quite a interesting answers involving Clausen Functions $\operatorname{Cl}_2(z)$,etc. However, I went back to my book. And to my surprise, I found that the answer was simply $\pi^3\frac{\ln(2)}{192}$. After a little research on the internet, I found that this was the answer to

$$\int_0^{{\pi}/{4}} (\pi x-4x^2)\ln(1+\tan x)\,\mathrm dx$$

Sometimes mistakes lead to discoveries! (In this case incredible solutions!) Make sure you check out the amazing answers giving by the community. I am including the answer to the question below. Thanks! :D

Answer 1

Using the property $\displaystyle\int_0^af(x)\,\mathrm dx=\int_0^af(a-x)\,\mathrm dx$

$ \begin{align*} I &= \displaystyle\int_0^{{\pi}/{4}} [\pi(\frac{\pi}{4} -x) - 4(\frac{\pi}{4} - x)^2]\ln(1+\tan (\frac{\pi}{4}-x))\mathrm dx\\ &=\displaystyle\int_0^{{\pi}/{4}} [\frac{\pi^2}{4} -\pi x - 4(\frac{\pi^2}{16} + - \frac{\pi x}{2}+ x^2)]\ln(1+ \frac{1-\tan(x)}{1+\tan(x)})\mathrm dx\\ &=\displaystyle\int_0^{{\pi}/{4}} (\pi x-4x^2)[\ln(2) - \ln(1+\tan(x))]\mathrm dx\\ & = \displaystyle\int_0^{{\pi}/{4}}(\pi x-4x^2)\ln(2) \mathrm dx -\displaystyle\int_0^{{\pi}/{4}} (\pi x-4x^2)\ln(1+\tan x)\mathrm dx \end{align*}$

$2I = \displaystyle\int_0^{{\pi}/{4}}(\pi x-4x^2)\ln(2)\mathrm dx$

$\begin{align*} I &= \dfrac{\ln(2)}{2}[\dfrac{\pi x^2}{2} - \dfrac{4x^3}{3}]_0^{\frac{\pi}{4}}\\ &= \pi^3\frac{\ln(2)}{192} \end{align*}$