Cho \(I=\int\limits_{0}^{\tfrac{\pi }{3}}{\left( \sin 3x+{{\cos }^{2}}x \right)dx}\)\(=\left. \left( a\cos 3x+bx\sin +c\sin 2x \right) \right|_{0}^{\frac{\pi }{6}}\). Giá trị của \(3a+2b+4c\) là:
Ta có:
\(\begin{align} & {{I}_{1}}=\int\limits_{0}^{\tfrac{\pi }{3}}{\left( \sin 3x+{{\cos }^{2}}x \right)dx}\\&=\int\limits_{0}^{\frac{\pi }{3}}{\left( \sin 3x+\frac{1+\cos 2x}{2} \right)dx}\\&=\left. \left( -\frac{1}{3}\cos 3x+\frac{1}{2}x+\frac{1}{4}\sin 2x \right) \right|_{0}^{\frac{\pi }{3}} \\ & \Rightarrow a=-\frac{1}{3},b=\frac{1}{2},c=\frac{1}{4}\\&\Rightarrow 3a+2c+4c=1 \\ \end{align}\)
Đáp án đúng là B.