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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}$ và ${f}'\left( x \right)=\dfrac{2}{{{\sin }^{2}}x}+1,\ \forall x\in \left( 0;\pi \right)$. Khi đó $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{{{\pi }^{2}}}{9}-2\ln 2.$
B. $\dfrac{{{\pi }^{2}}}{9}-2\ln \dfrac{1}{2}.$
C. $\dfrac{5{{\pi }^{2}}}{36}-2\ln 2.$
D. $-\dfrac{{{\pi }^{2}}}{9}+\ln \dfrac{1}{2}.$
A. $\dfrac{{{\pi }^{2}}}{9}-2\ln 2.$
B. $\dfrac{{{\pi }^{2}}}{9}-2\ln \dfrac{1}{2}.$
C. $\dfrac{5{{\pi }^{2}}}{36}-2\ln 2.$
D. $-\dfrac{{{\pi }^{2}}}{9}+\ln \dfrac{1}{2}.$
Ta có: ${f}'\left( x \right)=\dfrac{2}{{{\sin }^{2}}x}+1,\ \forall x\in \left( 0;\pi \right)\Rightarrow f\left( x \right)=-2\cot x+x+C,\ \forall x\in \left( 0;\pi \right)$
Mà $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}\Leftrightarrow 0+\dfrac{\pi }{2}+C=\dfrac{\pi }{2}\Rightarrow C=0\Rightarrow f\left( x \right)=-2\cot x+x.$
Xét $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( -2\cot x+x \right)\text{d}x}=\left. -2\ln \left| \sin x \right|+\dfrac{1}{2}{{x}^{2}} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+2\ln \dfrac{1}{2}-\dfrac{{{\pi }^{2}}}{72}=\dfrac{{{\pi }^{2}}}{9}-2\ln 2.$
Mà $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}\Leftrightarrow 0+\dfrac{\pi }{2}+C=\dfrac{\pi }{2}\Rightarrow C=0\Rightarrow f\left( x \right)=-2\cot x+x.$
Xét $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\left( -2\cot x+x \right)\text{d}x}=\left. -2\ln \left| \sin x \right|+\dfrac{1}{2}{{x}^{2}} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+2\ln \dfrac{1}{2}-\dfrac{{{\pi }^{2}}}{72}=\dfrac{{{\pi }^{2}}}{9}-2\ln 2.$
Đáp án A.