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Câu hỏi: Một nguyên hàm của hàm số $f\left( x \right)=\sin \left( 2x-\dfrac{\pi }{3} \right)$ là
A. $F\left( x \right)=\dfrac{1}{2}\cos \left( 2x-\dfrac{\pi }{3} \right).$
B. $F\left( x \right)=\cos \left( 2x-\dfrac{\pi }{3} \right).$
C. $F\left( x \right)=-\dfrac{1}{2}\cos \left( 2x-\dfrac{\pi }{3} \right).$
D. $F\left( x \right)=-\cos \left( 2x-\dfrac{\pi }{3} \right).$
A. $F\left( x \right)=\dfrac{1}{2}\cos \left( 2x-\dfrac{\pi }{3} \right).$
B. $F\left( x \right)=\cos \left( 2x-\dfrac{\pi }{3} \right).$
C. $F\left( x \right)=-\dfrac{1}{2}\cos \left( 2x-\dfrac{\pi }{3} \right).$
D. $F\left( x \right)=-\cos \left( 2x-\dfrac{\pi }{3} \right).$
Ta có $\int{f\left( x \right)dx=\int{\sin \left( 2x-\dfrac{\pi }{3} \right)}}dx=-\dfrac{1}{2}\cos \left( 2x-\dfrac{\pi }{3} \right)+C$. Chọn $C=0$.
Đáp án C.