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