Đáp án:

• \( \int \cos 4x \sin 2x \, dx = \frac{1}{2} \int \left[\sin 6x - \sin 2x\right] dx \)

\( = \frac{1}{2} \left[-\frac{1}{6} cos 6x + \frac{1}{2} \cos x\right] + C 
= -\frac{1}{12} \cos 6x + \frac{1}{4} \cos 2x + C \)

• Rút ra: \( \int \cos ax \sin bx \, dx, \quad \int \cos ax \cos bx \, dx, \quad \int \sin ax \sin bx \, dx: \) Đưa về tổng