[Tex]\int_{0}^{\pi/4} \frac{1 - \tan x}{1 + \tan x} dx[/Tex] is equivalent to

The integral \(\int_{0}^{\pi/4} \frac{1 - \tan x}{1 + \tan x} dx\) can be simplified using the identity \(\frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right)\). Substituting this, the integral becomes \(\int_{0}^{\pi/4} \tan\left(\frac{\pi}{4} - x\right) dx\).

Let \(u = \frac{\pi}{4} - x\), so \(du = -dx\). When \(x = 0\), \(u = \frac{\pi}{4}\), and when \(x = \frac{\pi}{4}\), \(u = 0\). The integral becomes \(-\int_{\pi/4}^{0} \tan u \, du = \int_{0}^{\pi/4} \tan u \, du\).

The integral of \(\tan u\) is \(-\ln|\cos u|\). Evaluating from 0 to \(\frac{\pi}{4}\):

\(-\ln|\cos(\pi/4)| + \ln|\cos(0)| = -\ln\left(\frac{\sqrt{2}}{2}\right) + \ln(1) = -\ln(2^{-1/2}) = \frac{1}{2}\ln 2\).

Thus, the value of the integral is \(\frac{1}{2}\ln 2\).