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Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=\sin 2x$ biết $F\left( \dfrac{\pi }{6} \right)=0$.
A. $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}$.
B. $F\left( x \right)=\text{co}{{\text{s}}^{\text{2}}}x-\dfrac{1}{4}$.
C. $F\left( x \right)=-\dfrac{1}{2}\text{cos2}x+\dfrac{\pi }{6}$.
D. $F\left( x \right)=-\dfrac{1}{2}\text{cos2}x$.
A. $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}$.
B. $F\left( x \right)=\text{co}{{\text{s}}^{\text{2}}}x-\dfrac{1}{4}$.
C. $F\left( x \right)=-\dfrac{1}{2}\text{cos2}x+\dfrac{\pi }{6}$.
D. $F\left( x \right)=-\dfrac{1}{2}\text{cos2}x$.
$F\left( x \right)=\int{f\left( x \right)\text{dx}}=\int{\sin 2x\text{dx}}=-\dfrac{1}{2}\text{cos}2x+C=-\dfrac{1}{2}\left( 1-2{{\sin }^{2}}x \right)+C={{\sin }^{2}}x-\dfrac{1}{2}+C$.
Suy ra: $F\left( \dfrac{\pi }{6} \right)=0$ $\Leftrightarrow C-\dfrac{1}{4}=0$ $\Leftrightarrow C=\dfrac{1}{4}$.
Vậy $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}$.
Suy ra: $F\left( \dfrac{\pi }{6} \right)=0$ $\Leftrightarrow C-\dfrac{1}{4}=0$ $\Leftrightarrow C=\dfrac{1}{4}$.
Vậy $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}$.
Đáp án A.