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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=-1$ và $f'\left( x \right)=\dfrac{\sin x+\sin 3x}{2{{\sin }^{4}}x.\cos x},\forall x\in \left( \dfrac{\pi }{6};\dfrac{5\pi }{6} \right)$. Khi đó $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}$ bằng
A. $2$.
B. $4$.
C. $-2$.
D. $0$.
A. $2$.
B. $4$.
C. $-2$.
D. $0$.
Ta có $f'\left( x \right)=\dfrac{\sin x+\sin 3x}{2{{\sin }^{4}}x.\cos x},\forall x\in \left( \dfrac{\pi }{6};\dfrac{5\pi }{6} \right)$ nên $f\left( x \right)$ là một nguyên hàm của $f'\left( x \right)$
$\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{\sin x+\sin 3x}{2{{\sin }^{4}}x.\cos x}\text{d}x=\int{\dfrac{2\sin 2x.\cos x}{2{{\sin }^{4}}x.\cos x}\text{d}x}}=\int{\dfrac{2\sin x.\cos x}{{{\sin }^{4}}x}\text{d}x=\int{\dfrac{2\cos x}{{{\sin }^{3}}x}\text{d}x}}$ $=\int{\dfrac{2}{{{\sin }^{3}}x}\text{d}\left( \sin x \right)=\dfrac{-1}{{{\sin }^{2}}x}+C}$
Do đó $f\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}+C$ mà $f\left( \dfrac{\pi }{2} \right)=-1\Rightarrow C=0$ khi đó $f\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}$
Vậy $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{-\dfrac{1}{{{\sin }^{2}}x}}\text{d}x=\left. \cot x \right|_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}=-2$
$\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{\sin x+\sin 3x}{2{{\sin }^{4}}x.\cos x}\text{d}x=\int{\dfrac{2\sin 2x.\cos x}{2{{\sin }^{4}}x.\cos x}\text{d}x}}=\int{\dfrac{2\sin x.\cos x}{{{\sin }^{4}}x}\text{d}x=\int{\dfrac{2\cos x}{{{\sin }^{3}}x}\text{d}x}}$ $=\int{\dfrac{2}{{{\sin }^{3}}x}\text{d}\left( \sin x \right)=\dfrac{-1}{{{\sin }^{2}}x}+C}$
Do đó $f\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}+C$ mà $f\left( \dfrac{\pi }{2} \right)=-1\Rightarrow C=0$ khi đó $f\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}$
Vậy $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{-\dfrac{1}{{{\sin }^{2}}x}}\text{d}x=\left. \cot x \right|_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}=-2$
Đáp án C.