Cho hàm số $f \left( x \right)$ có đạo hàm trên $\left( - \dfrac{\pi}{2} ; \dfrac{\pi}{2} \right)$ thỏa mãn $f \left( 0 \right) = 1$ và mọi $x \in \left( - \dfrac{\pi}{2} ; \dfrac{\pi}{2} \right)$ $\left(\text{cos}\right)^{2} x f^{'} \left( x \right) = \text{sin} 2 x f \left( x \right) + \text{cos} x + 2$. Biết $\int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} f \left( x \right) d x = m \sqrt{n}$ với $m , n \in \left(\mathbb{N}\right)^{*} , m > 1$. Giá trị của biểu thức $T = m + n$ bằng

A.  

9 .

B.  

4 .

C.  

6 .

D.  

5 .

Đáp án đúng là: D

Từ $\left(\text{cos}\right)^{2} x f^{'} \left( x \right) = \text{sin} 2 x f \left( x \right) + \text{cos} x + 2$ tìm hàm $f \left( x \right)$ và tính nguyên hàm
Cách giải:
$\left(\text{cos}\right)^{2} x f^{'} \left( x \right) = \text{sin} 2 x f \left( x \right) + \text{cos} x + 2$
$\Leftrightarrow - \text{sin} 2 x f \left( x \right) + \left(\text{cos}\right)^{2} x f^{'} \left( x \right) = \text{cos} x + 2$
$\Leftrightarrow \left(\right. \left(cos\right)^{2} x . f \left( x \right) \left.\right)^{'} = \text{cos} x + 2$
$\Leftrightarrow \left(\text{cos}\right)^{2} x . f \left( x \right) = \int_{}^{​} \left( \text{cos} x + 2 \right) d x$
$\Leftrightarrow \left(\text{cos}\right)^{2} x . f \left( x \right) = \text{sin} x + 2 x + c$
Do $f \left( 0 \right) = 1$ nên $1 . 1 = 0 + 2 . 0 + c \Rightarrow c = 1$
$\Rightarrow f \left( x \right) = \dfrac{\text{sin} x + 2 x + 1}{\left(\text{cos}\right)^{2} x}$
$\Rightarrow \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} f \left( x \right) d x = \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} \dfrac{sin x + 2 x + 1}{\left(cos\right)^{2} x} d x = \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} \dfrac{sin x}{\left(cos\right)^{2} x} d x + \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} \dfrac{2 x + 1}{\left(cos\right)^{2} x} d x$
$= \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} \dfrac{- 1}{\left(\text{cos}\right)^{2} x} d \left( \text{cos} x \right) + \left( 2 x + 1 \right) \left( \text{tan} x \left|\right.\right)_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} - 2 \int_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} \text{tan} x d x$
$= \left( \dfrac{1}{\text{cos} x} \left|\right.\right)_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} + \left( 2 x + 1 \right) \left( \text{tan} x \left|\right.\right)_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} + \left( 2 \text{ln} \left|\right. \text{cos} x \left|\right. \left|\right.\right)_{\dfrac{- \pi}{3}}^{\dfrac{\pi}{3}} = 2 \sqrt{3}$
Suy ra $m = 2 , n = 3 \Rightarrow m + n = 5$
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