Explanation

(b)
Let $I = \int_{}^{}{\frac{\cos{4x + 1}}{\cot{x - \tan x}}dx}$
$= \int_{}^{}{\frac{2\cos^{2}{2x}}{\frac{\cos^{2}{x - \sin^{2}x}}{\sin{x\cos x}}}dx}$
$= \int_{}^{}{\frac{{2\ \cos^{2}}{2x}}{\cos{2x}}.\sin x\cos xdx}$
$= \int_{}^{}{\cos{2x}}.\sin{2x}\ dx$
$= \frac{1}{2}\ \int_{}^{}{2\cos{2x}}\sin{2x}\ dx$
$= \frac{1}{2}\ \int_{}^{}{\sin{4x}}\ dx = \frac{1}{2}\ \frac{( - \cos{4x)}}{4} + c$
$\Rightarrow I = \frac{- \cos{4x}}{8} + c = k\cos{4x + c}$
$\therefore\ k = - \frac{1}{8}$