Evaluate: $\int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx$

$\frac{\pi}{8} \log 2$

The correct answer is Option (4) → $\frac{\pi}{8} \log 2$

Let $I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx$

$I = \int_{0}^{\frac{\pi}{4}} \log\left( 1 + \tan\left( \frac{\pi}{4} - x \right) \right) dx$

$ = \int_{0}^{\frac{\pi}{4}} \log\left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) dx$

$= \int_{0}^{\frac{\pi}{4}} \log\left( \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right) dx$

$ = \int_{0}^{\frac{\pi}{4}} \log\left( \frac{2}{1 + \tan x} \right) dx$

$I= \int_{0}^{\frac{\pi}{4}} \log 2 dx - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx$

$I = \log 2 \int_{0}^{\frac{\pi}{4}}dx - I$

$2I = \log 2 [x]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} \log 2$

$∴I = \frac{\pi}{8} \log 2$