Question

Integrate :
$\int \frac{cos4x+1}{cotx-tanx}dx$

Answer 1

$I=\int \frac{\cos x+1}{\cos x-\tan x}dx=\int \frac{2{\cos }^{2}x}{(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x})}dx=\int \frac{2{\cos }^{2}x}{\frac{{\cos }^{2}x+{\sin }^{2}x}{\left(\sin x\cos x\right)}}dx=\int \frac{{\cos }^{2}2x\sin 2x}{\left({\cos }^{2}2x\right)}dx$

$I=\int \left(\sin 2x\right)dx$

$\therefore I=\int \sin 2xdx$

$I=-\frac{\cos 2x}{2}+c$

$I=\frac{(1-2{\sin }^{2}x)}{2}+c$

$I={\sin }^{2}x+k={\cos }^{2}x+p$