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Câu hỏi: Nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=\sin 2x,$ biết $F\left( \dfrac{\pi }{6} \right)=0$ là
A. $F\left( x \right)=\dfrac{-1}{2}\cos 2x+\dfrac{\pi }{6}.$
B. $F\left( x \right)={{\cos }^{2}}x-\dfrac{1}{4}.$
C. $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}.$
D. $F\left( x \right)=\dfrac{-1}{2}\cos 2x.$
A. $F\left( x \right)=\dfrac{-1}{2}\cos 2x+\dfrac{\pi }{6}.$
B. $F\left( x \right)={{\cos }^{2}}x-\dfrac{1}{4}.$
C. $F\left( x \right)={{\sin }^{2}}x-\dfrac{1}{4}.$
D. $F\left( x \right)=\dfrac{-1}{2}\cos 2x.$
Ta có: $F\left( x \right)=\int{\sin 2xdx}=-\dfrac{1}{2}\cos 2x+C; F\left( \dfrac{\pi }{6} \right)=0\Rightarrow C=\dfrac{1}{4}.$
Vậy $F\left( x \right)=-\dfrac{1}{2}\cos 2x+\dfrac{1}{4}=-\dfrac{1}{2}\left( 1-2{{\sin }^{2}}x \right)+\dfrac{1}{4}={{\sin }^{2}}x-\dfrac{1}{4}.$
Vậy $F\left( x \right)=-\dfrac{1}{2}\cos 2x+\dfrac{1}{4}=-\dfrac{1}{2}\left( 1-2{{\sin }^{2}}x \right)+\dfrac{1}{4}={{\sin }^{2}}x-\dfrac{1}{4}.$
Đáp án C.