Cho \(f\left( x \right)=\frac{x}{{{\cos }^{2}}x}\)...

A

- Hướng dẫn giải

Phương pháp giải:

Sử dụng phương pháp tích phân từng phần tính \(F\left( x \right)\).

Giải chi tiết:

Theo bài ra ta có: \(F\left( x \right) = \int {xf\left( x \right)dx} \).

Đặt \(\left\{ \begin{array}{l}u = x\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = f\left( x \right)\end{array} \right.\).

Khi đó ta có:

\(\begin{array}{*{20}{l}}\begin{array}{l}F\left( x \right) = x.f\left( x \right) - \int {f\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - \int\limits_{}^{} {\frac{x}{{{{\cos }^2}x}}dx}  + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - \int\limits_{}^{} {xd\left( {\tan x} \right)}  + C\end{array}\\\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x + \int\limits_{}^{} {\tan dx}  + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x + \int\limits_{}^{} {\frac{{\sin x}}{{\cos x}}dx}  + C\end{array}\\\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \int\limits_{}^{} {\frac{{d\left( {\cos x} \right)}}{{\cos x}}}  + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \ln \left| {\cos x} \right| + C\end{array}\\{F\left( 0 \right) = C = 0 \Rightarrow F\left( x \right) = \frac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \ln \left| {\cos x} \right|}\\{F\left( x \right) = \int\limits_{}^{} {xf'\left( x \right)dx}  = \int\limits_{}^{} {xd\left( {f\left( x \right)} \right)}  = xf\left( x \right) - \int\limits_{}^{} {f\left( x \right)dx}  + C}\\\begin{array}{l}\tan a = 3 \Rightarrow \frac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1 = 10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \cos a = \frac{1}{{\sqrt {10} }}\left( {a \in \left( { - \frac{\pi }{2};\frac{\pi }{2}} \right)} \right)\end{array}\end{array}\)

\(\begin{array}{l} \Rightarrow F\left( a \right) = 10{a^2} - 3a - \ln \frac{1}{{\sqrt {10} }}\\ \Rightarrow F\left( a \right) - 10{a^2} + 3a =  - \ln \frac{1}{{\sqrt {10} }} =  - \frac{1}{2}\ln \frac{1}{{10}} = \frac{1}{2}\ln 10\end{array}\)

Chọn A.