Explanation

(b)
Given, $\int_{}^{}{\frac{\cos{4x + 1}}{\cos{x - \tan x}}\ dx = A\cos{4x + B}}$
Let $I = \int_{}^{}{\frac{\cos{4x + 1}}{\cos{x - \tan x}}dx}$
$= \int_{}^{}{\frac{2\cos^{2}{2x}}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}dx = \int_{}^{}{\frac{2\cos^{2}x}{\frac{\cos{2x}}{\sin{x\cos x}}}dx}}$
$= \int_{}^{}{\sin{2x\cos{2x\ dx}}}$
$= \frac{1}{2}\int_{}^{}{2\sin{2x\cos{2x\ dx}}}$
$= \frac{1}{2}\int_{}^{}{\sin{4x\ dx = \frac{1}{2}\ \left\lbrack \ - \frac{\cos{4x}}{4}\ \right\rbrack + B}}$
$= - \frac{1}{8}\cos{4x + B}$
But $I = A\cos{4x + B}$ (given)
$\therefore A = - \frac{1}{8}$