Question

Integration of $\int \frac{1+\cos 4x}{\cot x-\tan x}dx$ yields $-\frac{1}{8}\cos kx$.
Find k

Answer 1

$I=\int \frac{2{\cos }^{2}2x}{{\cos }^{2}x-{\sin }^{2}x}\sin x\cos xdx$
$=\int 2\cos 2x\cdot \frac{1}{2}\sin 2xdx$
$=\frac{1}{2}\int \sin 4xdx=-\frac{1}{8}\cos 4x$