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Câu hỏi: Cho tích phân $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( x-1 \right)\sin 2xdx}$. Đẳng thức nào sau đây là đúng?
A. $I=-\left( x-1 \right)\cos 2x-\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
B. $I=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}-\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
C. $I=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
D. $I=\left. -\left( x-1 \right)\cos 2x \right|_{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\dfrac{\pi }{4}}+\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
A. $I=-\left( x-1 \right)\cos 2x-\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
B. $I=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}-\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
C. $I=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
D. $I=\left. -\left( x-1 \right)\cos 2x \right|_{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\dfrac{\pi }{4}}+\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx.}$
Đặt: $\left\{ \begin{aligned}
& u=x-1 \\
& dv=\sin 2xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\dfrac{1}{2}\cos 2x \\
\end{aligned} \right.$
Do đó: $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( x-1 \right)\sin 2xdx}=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx}$
& u=x-1 \\
& dv=\sin 2xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\dfrac{1}{2}\cos 2x \\
\end{aligned} \right.$
Do đó: $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( x-1 \right)\sin 2xdx}=\left. -\dfrac{1}{2}\left( x-1 \right)\cos 2x \right|_{0}^{\dfrac{\pi }{4}}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2xdx}$
Đáp án C.