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Câu hỏi: Cho $\int\limits_{{}}^{{}}{f\left( \dfrac{x}{2} \right)\text{d}x}={{x}^{3}}+2x+C$. Một nguyên hàm của hàm số $f(\cos x)$ là
A. $3\cos 2x-8x$
B. $8\sin x+3x$
C. $3\sin 2x-8x$
D. $3\sin 2x+8x$.
A. $3\cos 2x-8x$
B. $8\sin x+3x$
C. $3\sin 2x-8x$
D. $3\sin 2x+8x$.
Ta có
$\int{f\left( \dfrac{x}{2} \right)} d\dfrac{x}{2}= \dfrac{1}{2} \left( {{x}^{3}}+2x \right)+C\Rightarrow f\left( t \right)={{\left( \dfrac{1}{2} \left( 8{{t}^{3}}+4t \right)+C \right)}^{,}}\ =12{{t}^{2}}+2$ $\Rightarrow f\left( \cos x \right)=12{{\cos }^{2}}x+2$ Do đó: $\int\limits_{{}}^{{}}{f\left( \cos x \right)\text{d}x}=\int{\left( 12{{\cos }^{2}}x+2 \right)} dx=\int{\left( \dfrac{12(1+\cos 2x)}{2}+2 \right) dx}$
$=\int{\left( 8+6\cos 2x \right)}dx=3\sin 2x+8x+C$.
$\int{f\left( \dfrac{x}{2} \right)} d\dfrac{x}{2}= \dfrac{1}{2} \left( {{x}^{3}}+2x \right)+C\Rightarrow f\left( t \right)={{\left( \dfrac{1}{2} \left( 8{{t}^{3}}+4t \right)+C \right)}^{,}}\ =12{{t}^{2}}+2$ $\Rightarrow f\left( \cos x \right)=12{{\cos }^{2}}x+2$ Do đó: $\int\limits_{{}}^{{}}{f\left( \cos x \right)\text{d}x}=\int{\left( 12{{\cos }^{2}}x+2 \right)} dx=\int{\left( \dfrac{12(1+\cos 2x)}{2}+2 \right) dx}$
$=\int{\left( 8+6\cos 2x \right)}dx=3\sin 2x+8x+C$.
Đáp án D.
