Bài tập 3.58 trang 184 SBT Toán 12

Bài tập 3.58 trang 184 SBT Toán 12

Ta có: 

\(\tan x = 0 \Leftrightarrow x = 0\) do \(x \in \left[ { - \frac{\pi }{4};\frac{\pi }{4}} \right]\)

Khi đó 

\(\begin{array}{l}
S = \int \limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} \left| {\tan x} \right|dx\\
 = \int \limits_{ - \frac{\pi }{4}}^0 \left| {\tan x} \right|dx + \int \limits_0^{\frac{\pi }{4}} \left| {\tan x} \right|dx\\
 =  - \int \limits_{ - \frac{\pi }{4}}^0 \tan xdx + \int \limits_0^{\frac{\pi }{4}} \tan xdx\\
 =  - \int \limits_{ - \frac{\pi }{4}}^0 \frac{{\sin x}}{{\cos x}}dx + \int \limits_0^{\frac{\pi }{4}} \frac{{\sin x}}{{\cos x}}dx\\
 = \int \limits_{ - \frac{\pi }{4}}^0 \frac{{d\left( {\cos x} \right)}}{{\cos x}} - \int \limits_0^{\frac{\pi }{4}} \frac{{d\left( {\cos x} \right)}}{{\cos x}}dx\\
 = \ln \left. {\left| {\cos x} \right|} \right|_{ - \frac{\pi }{4}}^0 - \ln \left. {\left| {\cos x} \right|} \right|_0^{\frac{\pi }{4}}\\
 = \ln 1 - \ln \frac{{\sqrt 2 }}{2} - \ln \frac{{\sqrt 2 }}{2} + \ln 1\\
 =  - 2\ln \frac{{\sqrt 2 }}{2} = \ln 2
\end{array}\)

Chọn C.

-- Mod Toán 12