Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên...

Phương pháp:
- Tính $\left( f\left( x \right)\tan x \right)'$
- Sử dụng định lí Newton-Leibniz: $\int\limits_{a}^{b}{f'\left( x \right)}dx=f\left( b \right)-f\left( a \right)$
Cách giải:
Ta có:
$\left( f\left( x \right)\tan x \right)'=f'\left( x \right)\tan x+f\left( x \right).\dfrac{1}{{{\cos }^{2}}x}$
$=$ $f'\left( x \right)\tan x+f\left( x \right)\left( {{\tan }^{2}}x+1 \right)$
$=f'\left( x \right)\tan x+f\left( x \right){{\tan }^{2}}x+f\left( x \right)$
$f'\left( x \right)\tan x+f\left( x \right){{\tan }^{2}}x=\left( f'\left( x \right)\tan x \right)'-f\left( x \right)$
$\Rightarrow I=\int\limits_{0}^{3}{\left[ f\left( x \right){{\tan }^{2}}x+f'\left( x \right)\tan x \right]}dx$
$\Leftrightarrow I=\int\limits_{0}^{3}{\left[ \left( f'\left( x \right)\tan x \right)'-f\left( x \right) \right]}dx$
$\Leftrightarrow I=\int\limits_{0}^{3}{\left[ \left( f'\left( x \right)\tan x \right)' \right]}dx\int\limits_{0}^{3}{f\left( x \right)}dx$
$\Leftrightarrow I=f\left( 3 \right)\tan 3-f\left( 0 \right)\tan 0-10$
$\begin{aligned}
& \Leftrightarrow I=\cot 3.\tan 3-0-10 \\
& \Leftrightarrow I=1-10 \\
& \Leftrightarrow I=-9 \\
\end{aligned}$