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Câu hỏi: Cho $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=3$, tính $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 3\cos xf\left( \sin x \right)-2 \right]\text{d}x}$
A. $I=9-\pi $.
B. $I=3-2\pi $.
C. $I=9-2\pi $.
D. $I=3+2\pi $.
A. $I=9-\pi $.
B. $I=3-2\pi $.
C. $I=9-2\pi $.
D. $I=3+2\pi $.
Ta có $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 3\cos xf\left( \sin x \right)-2 \right]\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{3\cos xf\left( \sin x \right)\text{d}x}-\int\limits_{0}^{\dfrac{\pi }{2}}{2\text{d}x}={{I}_{1}}-{{I}_{2}}$.
Đặt $t=\sin x\Rightarrow \text{d}t=\cos x\text{d}x,x=0\Rightarrow t=0;x=\dfrac{\pi }{2}\Rightarrow t=1$ suy ra ${{\text{I}}_{1}}=3\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=9$ ;
${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{2\text{d}x}=\left. 2x \right|_{0}^{\dfrac{\pi }{2}}=\pi $. Vậy $I=9-\pi $.
Đặt $t=\sin x\Rightarrow \text{d}t=\cos x\text{d}x,x=0\Rightarrow t=0;x=\dfrac{\pi }{2}\Rightarrow t=1$ suy ra ${{\text{I}}_{1}}=3\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=9$ ;
${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{2\text{d}x}=\left. 2x \right|_{0}^{\dfrac{\pi }{2}}=\pi $. Vậy $I=9-\pi $.
Đáp án A.
