Tricky Integrals [Part 1]

Note that ( $\large 1 + \tan (\frac{\pi}{4} - \theta)$ ) * ( $\large 1 + \tan \theta$ ) = 2.

We then write 2I = $\large \int _{ 0 }^{ { \pi }/{ 4 } }{ \ln { 2} }~ {dx}$ . Simplify.

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$\begin{aligned} &\text{I}= \int_{0}^{\frac{\pi}{4}}\ln\bigg(1+\tan x\bigg)dx &&(1)\\ &\text{I}= \int_{0}^{\frac{\pi}{4}}\ln\bigg(1+\tan\big(\frac{\pi}{4} -x\big)\bigg)dx =\int_{0}^{\frac{\pi}{4}}\ln\bigg(1+\frac{1-\tan x}{1+\tan x}\bigg)dx\\ &\text{I}=\int_{0}^{\frac{\pi}{4}}\ln\bigg(\frac{2}{1+\tan x}\bigg)dx && (2)\\ &\text{Adding (1) and(2)} \\ &\text{2I}= \int_{0}^{\frac{\pi}{4}}\ln\bigg(\big(1+\tan x\big)\times\big(\frac{2}{1+\tan x}\big)\bigg)dx= \int_{0}^{\frac{\pi}{4}}\ln(2)=\frac{\pi}{4}\ln(2)\\ &\text{I}=\frac{\pi}{8}\ln(2)\\ \end{aligned}$