Đáp án:

• \( \cos^2 x \sin^4 x = \frac{(1 + \cos 2x)}{2}  \frac{(3 + \cos 4x - 4\cos 2x)}{8} \)

\( = \frac{1}{16} \left[3 + \cos 4x - 4\cos 2x + 3\cos 2x + \cos 4x \cos 2x - 4\cos^2 2x \right] \)

\( = \frac{1}{16} \left[3 + \cos 4x - \cos 2x + \frac{1}{2} (\cos 6x + \cos 2x) - 4\frac{(1 + \cos 4x)}{2} \right] \)

\( = \frac{1}{32} \left[2 - 2\cos 4x - \cos 2x + \cos 6x\right] \)

\( \Rightarrow \int \cos^2 x \sin^4 x \, dx = \frac{1}{32} \left[2x - \frac{1}{2} \sin 4x - \frac{1}{2} \sin 2x + \frac{1}{6} \sin 6x\right] + C \)