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Câu hỏi: Giá trị của tích phân $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}x}dx}$ là
A. $I=\dfrac{\pi \sqrt{3}}{6}+\ln 2$
B. $I=\dfrac{2\pi \sqrt{3}}{6}+\ln 2$
C. $I=\dfrac{\pi \sqrt{3}}{6}-\ln 2$
D. $I=\dfrac{\pi \sqrt{3}}{2}+\ln 2$
A. $I=\dfrac{\pi \sqrt{3}}{6}+\ln 2$
B. $I=\dfrac{2\pi \sqrt{3}}{6}+\ln 2$
C. $I=\dfrac{\pi \sqrt{3}}{6}-\ln 2$
D. $I=\dfrac{\pi \sqrt{3}}{2}+\ln 2$
Ta có $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}x}dx}$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=\dfrac{1}{{{\sin }^{2}}x}dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
$I=-x\cot x\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\cot xdx}=\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx}=\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{d\left( \sin x \right)}{\sin x}}=\dfrac{\pi \sqrt{3}}{6}+\ln \left| \sin x \right|\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} \right. \right.$
$=\dfrac{\pi \sqrt{3}}{6}-\ln \dfrac{1}{2}=\dfrac{\pi \sqrt{3}}{6}+\ln 2$. Vậy $I=\dfrac{\pi \sqrt{3}}{6}+\ln 2$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=\dfrac{1}{{{\sin }^{2}}x}dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
$I=-x\cot x\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\cot xdx}=\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx}=\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{d\left( \sin x \right)}{\sin x}}=\dfrac{\pi \sqrt{3}}{6}+\ln \left| \sin x \right|\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} \right. \right.$
$=\dfrac{\pi \sqrt{3}}{6}-\ln \dfrac{1}{2}=\dfrac{\pi \sqrt{3}}{6}+\ln 2$. Vậy $I=\dfrac{\pi \sqrt{3}}{6}+\ln 2$
Đáp án A.