$\cfrac{{dy}}{{dx}} + (\sec x)y = \tan x\left( {0 \le x < \cfrac{\pi }{2}} \right)$

: The given equation is $\cfrac{{dy}}{{dx}} + \sec x \cdot y = \tan x$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q.$

Where, $P = \sec x$ and $Q = \tan x$
$I.F. = {e^{\int {\sec xdx} }} = {e^{{\rm{log}}(\sec x + \tan x)}} = \sec x + \tan x$

$\therefore$

The solution is $y \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right) = \int Q \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right)dx + C$

$y \cdot (\sec x + \tan x) = \int {(\sec x\tan x + {{\tan }^2}x)} dx + C$

$\Rightarrow y(\sec x + \tan x) = \int {\sec } x\tan xdx + \int {{{\sec }^2}} xdx - \int 1 .dx + C$

$\Rightarrow y(\sec x + \tan x) = \sec x\tan x - x + C$,

which is the required solution.