Question

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

Answer 1

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

$(\cot x-\tan x=\frac{1}{\tan x}-\tan x=\frac{1-{\tan }^{2}x}{\tan x}=\frac{2(1-{\tan }^{2}x)}{2\tan x}=\frac{2}{\tan 2x}=2\cot 2x)$

$=\int \frac{1+\cos 4x}{2\cot 2x}dx$

$=\int \frac{2{\cos }^{2}2x}{2\cot 2x}dx$

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

$=\int \sin 2x\cos 2xdx$

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

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

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

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

Hence, the answer is $\frac{-\cos 4x}{8}+c.$