Ta có: \(\displaystyle \tan x = 0 \Leftrightarrow x = 0\) do \(\displaystyle x \in \left[ { - \frac{\pi }{4};\frac{\pi }{4}} \right]\).
Khi đó \(\displaystyle S = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left| {\tan x} \right|dx} \) \(\displaystyle = \int\limits_{ - \frac{\pi }{4}}^0 {\left| {\tan x} \right|dx} + \int\limits_0^{\frac{\pi }{4}} {\left| {\tan x} \right|dx} \) \(\displaystyle = - \int\limits_{ - \frac{\pi }{4}}^0 {\tan xdx} + \int\limits_0^{\frac{\pi }{4}} {\tan xdx} \)
\(\displaystyle = - \int\limits_{ - \frac{\pi }{4}}^0 {\frac{{\sin x}}{{\cos x}}dx} + \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x}}{{\cos x}}dx} \) \(\displaystyle = \int\limits_{ - \frac{\pi }{4}}^0 {\frac{{d\left( {\cos x} \right)}}{{\cos x}}} - \int\limits_0^{\frac{\pi }{4}} {\frac{{d\left( {\cos x} \right)}}{{\cos x}}dx} \) \(\displaystyle = \left. {\ln \left| {\cos x} \right|} \right|_{ - \frac{\pi }{4}}^0 - \left. {\ln \left| {\cos x} \right|} \right|_0^{\frac{\pi }{4}}\)
\(\displaystyle = \ln 1 - \ln \frac{{\sqrt 2 }}{2} - \ln \frac{{\sqrt 2 }}{2} + \ln 1\) \(\displaystyle = - 2\ln \frac{{\sqrt 2 }}{2} = \ln 2\)
Chọn C.
