Find the Definite Integral Of A Logarithmic Function Involving Trigonometry

Find the definite integral of a logarithmic function
Find the definite integral of a logarithmic function “log (1 + tan x)” from 0 to π/4
Solution
Let
\begin{aligned} I&=\int^{\frac{\pi}{4}}_{0}{\log(1+\tan x)} dx\\ \end{aligned}
We know
\begin{aligned} \int^{b}_{a}{f(x)} dx=\int^{b}_{a}{f(a+b-x)} dx \end{aligned}
So
\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}