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Câu hỏi: Tìm họ nguyên hàm của hàm số $y=\cos \left( 3x+\dfrac{\pi }{6} \right)$.
A. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{3}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
B. $\int{f\left( x \right)\text{d}x}=-\dfrac{1}{3}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
C. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{6}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
D. $\int{f\left( x \right)\text{d}x}=\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
A. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{3}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
B. $\int{f\left( x \right)\text{d}x}=-\dfrac{1}{3}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
C. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{6}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
D. $\int{f\left( x \right)\text{d}x}=\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
Ta có: $\int{f\left( x \right)\text{d}x}=\int{\cos \left( 3x+\dfrac{\pi }{6} \right)\text{d}x}=\dfrac{1}{3}\sin \left( 3x+\dfrac{\pi }{6} \right)+C$.
Đáp án A.