Đáp án:
Giải thích các bước giải:
\(\frac{1}{2}\int {(\cos 2x + \cos 6x)} dx = \frac{1}{2}\int {\cos 2xdx + } \frac{1}{2}\int {\cos 6xdx} \)
Đặt :
\(\begin{array}{l}
t = 2x \to \frac{{dt}}{2} = dx\\
u = 6x \to \frac{{du}}{6} = dx
\end{array}\)
\(\begin{array}{l}
\to \frac{1}{2}\int {\frac{1}{2}} .\cos tdt + \frac{1}{2}\int {\frac{1}{6}} \cos udu = \frac{1}{4}\sin t + \frac{1}{{12}}\sin u + C\\
= \frac{1}{4}\sin 2x + \frac{1}{{12}}\sin 6x + C
\end{array}\)
