If ∫ cos 4 x + 1 cot x - tan x = k cos 4 x + c , then " k " is equal to

$\int \frac{\cos 4x+1}{\cot x-\tan x}$

$=\int \frac{\cos 4x+1}{{\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{\cos x\sin x}}}$

Use:

$\left[ \cos 2x=2{\cos }^{2}x-1=1-2{\sin }^{2}x\right]$

$=\int \frac{2{\cos }^{2}2x}{\cos 2x}\times \sin x\cos xdx$

$=\frac{1}{2}\int 2\cos 2x\sin 2xdx$

$=\frac{1}{2}\int \sin 4xdx$

$=\frac{1}{2}\times \frac{-\cos 4x}{4}+c$

$=\frac{-1}{8}\cos 4x+c$

$\Rightarrow k=\frac{-1}{8}$