Tính tích phân sau: sin 3x . cos x dx

* Hướng dẫn giải

Chọn C

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 3x \cos xdx \\ =\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\sin 4x+\sin 2x\right]dx \\ =\frac{1}{2}\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 4xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2xdx\right] \\ =\frac{1}{2}\left[-\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x\right]\overline{)\begin{array}{c}\frac{\mathrm{\pi }}{2}\\ 0\end{array}} \\ =\frac{1}{2}\left[\left(-\frac{1}{4}\cos 2\mathrm{\pi }-\frac{1}{2}\mathrm{cos\pi }\right)-\left(-\frac{1}{4}\cos 0-\frac{1}{2}\cos 0\right)\right] \\ =\frac{1}{2}\left(-\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}\right)=\frac{1}{2} \end{gathered}$$