Question

If $\int \frac{\cos 4x+1}{\cot x-\tan x}dx=$ ?

Answer 1

$I=\int \frac{2{\cos }^{2}2x-1+1}{\left(\cos 2x\right)}\times \sin x.\cos x.dx[\because \cot x-\tan x=\frac{\cos 2x}{\sin x\cos x}]$

$=\int \cos 2x\times \sin x\times \cos x.dx$

$=\frac{1}{2}\int 2.\cos 2x.\sin 2x.dx=\frac{1}{2}\int \sin 4x.dx$

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

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