Cho hàm số $\Large f(x)=\dfrac{x}{\cos^2x},$ với $\Large x \in \bigg(\

Cho hàm số $\Large f(x)=\dfrac{x}{\cos^2x},$ với $\Large x \in \bigg(\dfrac{-\pi}{2}; \dfrac{\pi}{2}\bigg).$ Gọi $\Large F(x)$ là một nguyên hàm của $\Large x.{f}'(x)$ thỏa mãn $\Large F(0)=0.$ Biết $\Large \tan a=7$ với $\Large a \in \bigg(\dfrac{-\pi}{2}; \dfrac{\pi}{2}\bigg).$ Biểu thức $\Large F(a)-50a^2+7a$ có giá trị là

• A.

  A. $\Large \mathrm{ln}50.$

• B.

  B. $\Large -\dfrac{1}{4}\mathrm{ln}50.$

• C.

  C. $\Large \dfrac{1}{2}\mathrm{ln}50.$

• D.

  D. $\Large -\dfrac{1}{2}\mathrm{ln}50.$

Chọn C

Đặt $\Large \left\{\begin{align} & u=x \Rightarrow \mathrm{d}u=\mathrm{d}x \\ & \mathrm{d}v={f}'(x)dx \Rightarrow v=f(x) \end{align}\right.$

Ta có $\Large \int x{f}'(x)\mathrm{d}x$ $\Large =xf(x)-\int f(x)\mathrm{d}x$ $\Large =\dfrac{x^2}{\cos^2x}-\int f(x)\mathrm{d}x$ $\Large =\dfrac{x^2}{\cos^2x}-\int \dfrac{x}{\cos^2x}\mathrm{d}x.$

Đặt $\Large \left\{\begin{align} & u_1=x \Rightarrow \mathrm{d}u_1=\mathrm{d}x \\ & \mathrm{d}v_1=\dfrac{1}{\cos^2x}\mathrm{d}x \Rightarrow v_1=\tan x \end{align}\right..$

Do đó:

$\Large \int \dfrac{x}{\cos^2x}\mathrm{d}x$ $\Large =x\tan x-\int \tan x \mathrm{d}x$ $\Large =x\tan x-\int \dfrac{\sin x}{\cos x}\mathrm{d}x$ $\Large =x\tan x+\mathrm{ln}(\cos x)+C.$

Suy ra $\Large F(x)=\dfrac{x^2}{\cos^2x}-x\tan x-\mathrm{ln}(\cos x)-C.$

Từ $\Large F(0)=0 \Rightarrow C=0.$

Do đó $\Large F(x)=\dfrac{x^2}{\cos^2x}-x\tan x-\mathrm{ln}(\cos x),$ $\Large \forall x \in \bigg(-\dfrac{\pi}{2}; \dfrac{\pi}{2}\bigg)$ 

$\Large \Rightarrow F(a)=\dfrac{a^2}{\cos^2a}-a.\tan a-\mathrm{ln}(\cos a).$

Ta có $\Large \tan a =7$ $\Large \Rightarrow 1+\tan^2a=50$ $\Large \Rightarrow \dfrac{1}{\cos^2a}=50$ $\Large \Rightarrow \cos a=\dfrac{\sqrt{2}}{10}.$

$\Large \Rightarrow F(a)=50a^2-7a-\mathrm{ln}\dfrac{\sqrt{2}}{10}$ $\Large \Rightarrow F(a)-50a^2+7a=-\mathrm{ln}\dfrac{\sqrt{2}}{10}=\mathrm{ln}(5\sqrt{2})=\dfrac{1}{2}\mathrm{ln}50.$