Find the Definite Integral Of A Logarithmic Function Involving Trigonometry

\begin{aligned}
I&=\int^{\frac{\pi}{4}}_{0}{\log(1+\tan x)} dx\\
\end{aligned}

\begin{aligned}
\int^{b}_{a}{f(x)} dx=\int^{b}_{a}{f(a+b-x)} dx
\end{aligned}

\begin{aligned}
I&=\int^{\frac{\pi}{4}}_{0}{\log(1+\tan x)} dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log\bigg(1+\tan \bigg(\frac{\pi}{4}-x\bigg)} \bigg) dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log \bigg(1+\frac{\tan (\frac{\pi}{4})-\tan x}{1+{\tan (\frac{\pi}{4})\tan x}}\bigg) } dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log\bigg(1+ \frac{1-\tan x}{1+{\tan x}}} \bigg) dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log\bigg( \frac{1+{\tan x}+1-\tan x}{1+{\tan x}}\bigg)} dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log\bigg( \frac{2}{1+{\tan x}}\bigg)} dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{(\log 2 - \log(1+\tan x))} dx\\
\\
&=\int^{\frac{\pi}{4}}_{0}{\log 2} \ dx-\int^{\frac{\pi}{4}}_{0}{\log(1+\tan x)} dx\\
\\
\Rightarrow I&=\int^{\frac{\pi}{4}}_{0}{\log 2} \ dx-I\\
\\
\Rightarrow 2I&=\int^{\frac{\pi}{4}}_{0}{\log 2} \ dx\\
\\
&=\log 2 \times \bigg[x\bigg]_{0}^{\frac{\pi}{4}}\\
\\
I&=\frac{\pi \log 2}{8}\\
\end{aligned}