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Câu hỏi: Cho tích phân $I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin x{{\cos }^{2}}x}{1+\cos x}dx=a\ln 2+b\ln 3+c}$ với $a,b,c\in \mathbb{Q}$. Tính tích $P=abc$
A. $P=\dfrac{1}{8}$
B. $P=\dfrac{1}{4}$
C. $P=\dfrac{-1}{4}$
D. $P=\dfrac{-1}{8}$
A. $P=\dfrac{1}{8}$
B. $P=\dfrac{1}{4}$
C. $P=\dfrac{-1}{4}$
D. $P=\dfrac{-1}{8}$
Ta có $I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin x{{\cos }^{2}}x}{1+\cos x}dx=-\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{\cos }^{2}}x}{1+\cos x}d\left( \cos x \right)\xrightarrow{t=\cos x}-\int\limits_{1}^{\dfrac{1}{2}}{\dfrac{{{t}^{2}}dt}{1+t}=\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{{{t}^{2}}dt}{1+t}=\int\limits_{\dfrac{1}{2}}^{1}{\left( t-1+\dfrac{1}{t+1} \right)dt}}}}}$
$=\left. \left( \dfrac{{{t}^{2}}}{2}-t+\ln \left| t+1 \right| \right) \right|_{\dfrac{1}{2}}^{1}=-\dfrac{1}{8}+\ln \dfrac{4}{3}=2\ln 2-\ln 3-\dfrac{1}{8}\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-1 \\
& c=-\dfrac{1}{8} \\
\end{aligned} \right.\Rightarrow P=abc=\dfrac{1}{4}$. Chọn B
$=\left. \left( \dfrac{{{t}^{2}}}{2}-t+\ln \left| t+1 \right| \right) \right|_{\dfrac{1}{2}}^{1}=-\dfrac{1}{8}+\ln \dfrac{4}{3}=2\ln 2-\ln 3-\dfrac{1}{8}\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-1 \\
& c=-\dfrac{1}{8} \\
\end{aligned} \right.\Rightarrow P=abc=\dfrac{1}{4}$. Chọn B
Đáp án B.