∫_0^(π)/4log_e(1+tan x) dx is equal to

The correct answer is Option (1) → \(\displaystyle\frac{\pi}{8}\log_e 2\)

Let

    \( \displaystyle I = \int_{0}^{\pi/4}\log_e(1+\tan x)\,dx \qquad (1) \)

By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\), replace x by \(\frac{\pi}{4}-x\). Since

    \( \displaystyle 1+\tan\left(\frac{\pi}{4}-x\right) = 1+\frac{1-\tan x}{1+\tan x} = \frac{2}{1+\tan x} \)

we get

    \( \displaystyle I = \int_{0}^{\pi/4}\left[\log_e 2 - \log_e(1+\tan x)\right]dx \qquad (2) \)

Adding (1) and (2):

    \( \displaystyle 2I = \int_{0}^{\pi/4}\log_e 2\,dx = \frac{\pi}{4}\log_e 2 \)

    \( \displaystyle I = \frac{\pi}{8}\log_e 2 \)