Answer
$$y = \sin x + C\cos x$$
Work Step by Step
$$\eqalign{
& y' + y\tan x = \sec x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + y\tan x = \sec x \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \tan x{\text{ and }}Q\left( x \right) = \sec x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\tan xdx} }} = {e^{\ln \left| {\sec x} \right|}} = \sec x \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& \sec x\left( {\frac{{dy}}{{dx}} + y\tan x} \right) = \sec x\left( {\sec x} \right) \cr
& \sec x\frac{{dy}}{{dx}} + y\sec x\tan x = {\sec ^2}x \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {y\sec x} \right] = {\sec ^2}x \cr
& d\left[ {y\sec x} \right] = {\sec ^2}xdx \cr
& {\text{Integrate both sides}} \cr
& y\sec x = \int {{{\sec }^2}x} dx \cr
& y\sec x = \tan x + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{\tan x}}{{\sec x}} + \frac{C}{{\sec x}} \cr
& y = \left( {\frac{{\sin x}}{{\cos x}}} \right)\left( {\frac{{\cos x}}{1}} \right) + C\cos x \cr
& y = \sin x + C\cos x \cr} $$
