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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn đồng thời các điều kiện $f'\left( x \right)={{\cos }^{2}}x.\sin x$ và $f\left( 0 \right)=1$. Tìm $f\left( x \right)$.
A. $f\left( x \right)=\dfrac{{{\cos }^{3}}x}{3}+\dfrac{11}{3}$.
B. $f\left( x \right)={{\cos }^{3}}x+4$.
C. $f\left( x \right)=-\dfrac{{{\cos }^{3}}x}{3}+\dfrac{13}{3}$.
D. $f\left( x \right)=-{{\cos }^{3}}x+5$.
A. $f\left( x \right)=\dfrac{{{\cos }^{3}}x}{3}+\dfrac{11}{3}$.
B. $f\left( x \right)={{\cos }^{3}}x+4$.
C. $f\left( x \right)=-\dfrac{{{\cos }^{3}}x}{3}+\dfrac{13}{3}$.
D. $f\left( x \right)=-{{\cos }^{3}}x+5$.
Ta có $\int{{{\cos }^{2}}x\sin x\text{d}x}=-\int{{{\cos }^{2}}x\text{d}\left( \cos x \right)}=-\dfrac{{{\cos }^{3}}x}{3}+C.$
$f\left( 0 \right)=4$ $\Leftrightarrow -\dfrac{1}{3}+C=4$ $\Leftrightarrow C=\dfrac{13}{3}$.
Vậy $f\left( x \right)=-\dfrac{{{\cos }^{3}}x}{3}+\dfrac{13}{3}$.
$f\left( 0 \right)=4$ $\Leftrightarrow -\dfrac{1}{3}+C=4$ $\Leftrightarrow C=\dfrac{13}{3}$.
Vậy $f\left( x \right)=-\dfrac{{{\cos }^{3}}x}{3}+\dfrac{13}{3}$.
Đáp án C.