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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{4} \right)=-2$ và $f'\left( x \right)=\dfrac{4\cos 2x}{{{\sin }^{2}}2x}\forall x\in \left( 0;\dfrac{\pi }{2} \right).$ Khi đó $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{f\left( x \right)dx}$ bằng
A. $\dfrac{\pi }{3}-\ln 2$
B. $-\ln 3$
C. $\dfrac{\pi }{6}$
D. $\dfrac{\pi }{2}-\ln 3$
A. $\dfrac{\pi }{3}-\ln 2$
B. $-\ln 3$
C. $\dfrac{\pi }{6}$
D. $\dfrac{\pi }{2}-\ln 3$
Cách giải:
Ta có $f\left( x \right)=\int\limits_{{}}^{{}}{f'\left( x \right)dx}=\int\limits_{{}}^{{}}{\dfrac{4\cos 2x}{{{\sin }^{2}}2x}dx}=\int\limits_{{}}^{{}}{\dfrac{2d\left( \sin 2x \right)}{{{\sin }^{2}}2x}}=\dfrac{-2}{\sin 2x}+C$
Mà $f\left( \dfrac{\pi }{4} \right)=-2\Rightarrow C=0\Rightarrow f\left( x \right)=-\dfrac{2}{\sin 2x}$
Nên $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{f\left( x \right)dx}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{-2}{\sin 2x}dx}=-\ln 3$
Ta có $f\left( x \right)=\int\limits_{{}}^{{}}{f'\left( x \right)dx}=\int\limits_{{}}^{{}}{\dfrac{4\cos 2x}{{{\sin }^{2}}2x}dx}=\int\limits_{{}}^{{}}{\dfrac{2d\left( \sin 2x \right)}{{{\sin }^{2}}2x}}=\dfrac{-2}{\sin 2x}+C$
Mà $f\left( \dfrac{\pi }{4} \right)=-2\Rightarrow C=0\Rightarrow f\left( x \right)=-\dfrac{2}{\sin 2x}$
Nên $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{f\left( x \right)dx}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{-2}{\sin 2x}dx}=-\ln 3$
Đáp án B.