詳解：反覆使用平方化倍角

\(\int {{{\sin }^2}x} {\cos ^2}xdx = \int {\frac{{(1 - \cos 2x)}}{2}\frac{{(1 + \cos 2x)}}{2}} dx = \frac{1}{4}\int {(1 - \cos 2x)(1 + \cos 2x)} dx\)

\( = \frac{1}{4}\int {(1 - {{\cos }^2}2x)} dx = \frac{1}{4}\int {(1 - \frac{{1 + \cos 4x}}{2})} dx = \frac{1}{8}\int {(2 - (1 + \cos 4x)} )dx\)

\( = \frac{1}{8}\int {(1 - \cos 4x)} dx = \frac{1}{8}\left( {x - \frac{1}{4}\sin 4x} \right) + C = \frac{x}{8} - \frac{1}{{32}}\sin 4x + C\)