Google
Câu hỏi: Tích phân $\int\limits_{0}^{\dfrac{\pi }{12}}{\sin 3xdx}$ bằng
A. $\dfrac{2+\sqrt{2}}{6}.$
B. $\dfrac{2-\sqrt{2}}{6}.$
C. $\dfrac{2+\sqrt{2}}{2}.$
D. $\dfrac{2-\sqrt{2}}{2}.$
A. $\dfrac{2+\sqrt{2}}{6}.$
B. $\dfrac{2-\sqrt{2}}{6}.$
C. $\dfrac{2+\sqrt{2}}{2}.$
D. $\dfrac{2-\sqrt{2}}{2}.$
Ta có $\int\limits_{0}^{\dfrac{\pi }{12}}{\sin 3\text{xdx}}=\left. -\dfrac{\cos 3x}{3} \right|_{0}^{\dfrac{\pi }{12}}=\dfrac{2-\sqrt{2}}{6}$.
Đáp án B.