The integrating factor of differential equation $[y(1-x\, tan x)+x^2cos x)dx-xdy =0 $ is :

$\frac{1}{x\, cos \, x}$

The correct answer is Option (3) → $\frac{1}{x\cos x}$

$y(1-x\tan x)+x^2\cos x)dx=xdy$

so $\frac{dy}{dx}=y(\frac{1}{x}-\tan x)+x\cos x$

so $\frac{dy}{dx}+(\tan x-\frac{1}{x})=x\cos x$

I.F. $e^{∫\tan x-\frac{1}{x}dx}$

$=e^{\log\sec x-\log x}$

$=\frac{1}{x\cos x}$