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Câu hỏi: Cho tích phân $\int\limits_{0}^{\pi }{{{x}^{2}}.\text{cos}x\text{d}x}$. Khẳng định nào sau đây đúng ?
A. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }-\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
B. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }+2\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
C. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }-2\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
D. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }+\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
A. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }-\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
B. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }+2\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
C. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }-2\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
D. $I=\left. {{x}^{2}}\sin x \right|_{0}^{\pi }+\int\limits_{0}^{\pi }{x\sin x}\text{d}x$.
Ta có: $\int\limits_{0}^{\pi }{{{x}^{2}}.\text{cos}x\text{d}x}=\int\limits_{0}^{\pi }{{{x}^{2}}.\text{d}\left( \sin x \right)}=\left. {{x}^{2}}.\sin x \right|_{0}^{\pi }-2\int\limits_{0}^{\pi }{x.\sin x}\text{.d}x.$
Đáp án C.