Answer
See answer below
Work Step by Step
$[2x-(x^2+y^2)\tan x]dx+2ydy=0$ ___(1)
Multiply (1) by $I=\sec x$
$\sec x[2x-(x^2+y^2)\tan x]dx+2y\sec xdy=0$ ___(2)
Here, $M_{1}(x,y)=\sec x[2x-(x^2+y^2)\tan x]$
$N_{1}(x,y)=2y\sec x$
$(M_{1})_y=-2y \tan x \sec x$
$(N_{1})_x=2y \tan x \sec x$
$(M_{1})_y \ne (N_{1})_x$
$\Rightarrow $ (2) is not Exact
$I=\sec x$ is not an integrating factor of (1).