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Câu hỏi: Giá trị của $\int\limits_{-m}^{m}{\dfrac{{{\sin }^{3}}x-x}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}dx}$ bằng
A. 0.
B. $m{{\pi }^{2}}$.
C. $-2m\pi $.
D. $\dfrac{\pi }{m}$.
A. 0.
B. $m{{\pi }^{2}}$.
C. $-2m\pi $.
D. $\dfrac{\pi }{m}$.
Xét hàm số $f\left( x \right)=\dfrac{{{\sin }^{3}}x-x}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}$
$f\left( -x \right)=\dfrac{{{\sin }^{3}}\left( -x \right)-\left( -x \right)}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}=\dfrac{x-{{\sin }^{3}}x}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}=-f\left( x \right)$
Vậy hàm $f\left( x \right)$ là hàm lẻ $\Rightarrow \int\limits_{-m}^{m}{f\left( x \right)dx=0,\forall m\in \mathbb{R}}$.
$f\left( -x \right)=\dfrac{{{\sin }^{3}}\left( -x \right)-\left( -x \right)}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}=\dfrac{x-{{\sin }^{3}}x}{{{\cos }^{4}}x+{{\cos }^{2}}x+1}=-f\left( x \right)$
Vậy hàm $f\left( x \right)$ là hàm lẻ $\Rightarrow \int\limits_{-m}^{m}{f\left( x \right)dx=0,\forall m\in \mathbb{R}}$.
Đáp án A.