Explanation

(b)
$\int_{}^{}{{\cos^{4}x}{dx = \frac{1}{4}\left( 2\cos^{2}x \right)^{2}}}dx$
$= \frac{1}{4}\int_{}^{}\left( 1 + \cos{2x} \right)^{2}dx$
$= \frac{1}{4}\int_{}^{}{(1 + 2\cos{2x + \cos^{2}{2x)dx}}}$
$= \frac{1}{8}\int_{}^{}{\{ 2 + 4\cos{2x + (1 + \cos{4x)\} dx}}}$
$= \frac{1}{8}\left\{ 3x + 2\sin{2x + \frac{1}{4}\sin{4x}} \right\} + D$
$= \frac{3}{8}x + \frac{1}{4}\sin{2x + \frac{1}{32}\sin{4x + D}}$
On comparing, we get
$A = \frac{3}{8},\ B = \frac{1}{4},\ C = \frac{1}{32}$