Cho $f\left( x \right) = \dfrac{x}{{{{\cos }^2}x}}$ trên $\left( { - \dfrac{\pi }{2};\dfrac{\pi }{2}} \right)$ và $F\left( x \right)$ là một nguyên hàm của hàm số $xf'\left( x \right)$ thỏa mãn $F\left( 0 \right) = 0$. Biết $a \in \left( { - \dfrac{\pi }{2};\dfrac{\pi }{2}} \right)$ thỏa mãn $\tan a = 3$. Tính $F\left( a \right) - 10{a^2} + 3a$.

$F\left( x \right) = \int {xf'\left( x \right)dx} $$ = \int {xd\left( {f\left( x \right)} \right)}  = xf\left( x \right) - \int {f\left( x \right)dx}  + C$

$ = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - \int\limits_{}^{} {\dfrac{x}{{{{\cos }^2}x}}dx}  + C$$ = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - \int\limits_{}^{} {xd\left( {\tan x} \right)}  + C$

$F\left( x \right) = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x + \int\limits_{}^{} {\tan xdx}  + C$$ = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x + \int\limits_{}^{} {\dfrac{{\sin x}}{{\cos x}}dx}  + C$

$F\left( x \right) = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \int\limits_{}^{} {\dfrac{{d\left( {\cos x} \right)}}{{\cos x}}}  + C$$ = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \ln \left| {\cos x} \right| + C$

$F\left( 0 \right) = C = 0$$ \Rightarrow F\left( x \right) = \dfrac{{{x^2}}}{{{{\cos }^2}x}} - x\tan x - \ln \left| {\cos x} \right|$

$\tan a = 3 \Rightarrow \dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1 = 10$$ \Leftrightarrow \cos a = \dfrac{1}{{\sqrt {10} }}\left( {a \in \left( { - \dfrac{\pi }{2};\dfrac{\pi }{2}} \right)} \right)$

$ \Rightarrow F\left( a \right) = 10{a^2} - 3a - \ln \dfrac{1}{{\sqrt {10} }}$ $ \Rightarrow F\left( a \right) - 10{a^2} + 3a =  - \ln \dfrac{1}{{\sqrt {10} }}$ $ =  - \dfrac{1}{2}\ln \dfrac{1}{{10}} = \dfrac{1}{2}\ln 10$