Tính tích phân sau I = (2x+3).sin 4x dx

* Hướng dẫn giải

Chọn C

Đặt $\left\{\begin{array}{l}u=2x+3\\ dv=\sin 4x.dx\end{array}\Rightarrow \left\{\begin{array}{l}du=2.dx\\ v=-\frac{1}{4}\cos 4x.dx\end{array}\right.\right.$

$$\begin{gathered} \Rightarrow I=\left(-\frac{1}{4}(2x+3)\cos 4x\right){|}_0^{\frac{\mathrm{\pi }}{4}}+\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{4}}\cos 4xdx \\ =\left(-\frac{1}{4}(2x+3)\cos 4x+\frac{1}{2}.\frac{1}{4}.\sin 4x\right){|}_0^{\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{\pi }}{8}+\frac{3}{2} \end{gathered}$$