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Câu hỏi: Xét tích phân $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.\cos 2xdx}$, nếu đặt $t=\sin x$ thì $I$ bằng
A. $\int\limits_{0}^{1}{\left( 1-2{{t}^{2}} \right)dt}.$
B. $2\int\limits_{0}^{1}{\left( 1-{{t}^{2}} \right)dt}.$
C. $-2\int\limits_{0}^{1}{\left( 1-{{t}^{2}} \right)dt}.$
D. $\int\limits_{0}^{1}{\left( 2{{t}^{2}}-1 \right)dt}$
A. $\int\limits_{0}^{1}{\left( 1-2{{t}^{2}} \right)dt}.$
B. $2\int\limits_{0}^{1}{\left( 1-{{t}^{2}} \right)dt}.$
C. $-2\int\limits_{0}^{1}{\left( 1-{{t}^{2}} \right)dt}.$
D. $\int\limits_{0}^{1}{\left( 2{{t}^{2}}-1 \right)dt}$
Ta có $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.\cos 2xdx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.\left( 1-2{{\sin }^{2}}x \right)dx}$
Đặt $t=\sin x\Rightarrow dt=\cos xdx$
Với $x=0\Rightarrow t=0;x=\dfrac{\pi }{2}\Rightarrow t=1$
Vậy $I=\int\limits_{0}^{1}{\left( 1-2{{t}^{2}} \right)dt}.$
Đặt $t=\sin x\Rightarrow dt=\cos xdx$
Với $x=0\Rightarrow t=0;x=\dfrac{\pi }{2}\Rightarrow t=1$
Vậy $I=\int\limits_{0}^{1}{\left( 1-2{{t}^{2}} \right)dt}.$
Đáp án A.