If \(I=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\), then value of \(I\) is

\( \text { Let } I= \int_0^{\frac{\pi}{4}} \log (1+\tan \theta) \mathrm{d} \theta \)
\( = \int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] \mathrm{d} \theta \)
\( \ldots\left[\because \int_0^a \mathrm{f}(x) \mathrm{d} x=\int_0^2 \mathrm{f}(\mathrm{a}-x) \mathrm{d} x\right] \)
\( = \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) \mathrm{d} \theta \)
\( =\int_0^{\frac{\pi}{4}} \log 2 \mathrm{~d} \theta-\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) \mathrm{d} \theta \)
\( \therefore \quad 2 \mathrm{I}=\int_0^{\frac{\pi}{4}} \log 2 \mathrm{~d} \theta \Rightarrow \mathrm{I}=\frac{\log 2}{2}[\theta]_0^{\pi / 4}=\frac{\pi}{8} \log 2\)