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Câu hỏi: Cho tích phân $I=\int\limits_{0}^{\pi }{{{x}^{2}}\cos xdx}$ và đặt $u={{x}^{2}},dv=\cos xdx$. Mệnh đề nào sau đây là mệnh đề đúng?
A. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-\int\limits_{0}^{\pi }{x\sin xdx}.$
B. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}+\int\limits_{0}^{\pi }{x\sin xdx}.$
C. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}+2\int\limits_{0}^{\pi }{x\sin xdx}.$
D. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-2\int\limits_{0}^{\pi }{x\sin xdx}.$
A. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-\int\limits_{0}^{\pi }{x\sin xdx}.$
B. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}+\int\limits_{0}^{\pi }{x\sin xdx}.$
C. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}+2\int\limits_{0}^{\pi }{x\sin xdx}.$
D. $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-2\int\limits_{0}^{\pi }{x\sin xdx}.$
Đặt$\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv=\cos xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=\sin x \\
\end{aligned} \right. $. Từ đó, $ I={{x}^{2}}\left. \sin x \right|_{0}^{\pi }-2\int\limits_{0}^{\pi }{x\sin xdx}.$
Nhưng ${{x}^{2}}\left. \sin x \right|_{0}^{\pi }={{x}^{2}}\left. \sin x \right|_{\pi }^{0}=0$, nên $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-2\int\limits_{0}^{\pi }{x\sin xdx}.$
& u={{x}^{2}} \\
& dv=\cos xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=\sin x \\
\end{aligned} \right. $. Từ đó, $ I={{x}^{2}}\left. \sin x \right|_{0}^{\pi }-2\int\limits_{0}^{\pi }{x\sin xdx}.$
Nhưng ${{x}^{2}}\left. \sin x \right|_{0}^{\pi }={{x}^{2}}\left. \sin x \right|_{\pi }^{0}=0$, nên $I={{x}^{2}}\left. \sin x \right|_{\pi }^{0}-2\int\limits_{0}^{\pi }{x\sin xdx}.$
Đáp án D.