Answer
$y = {e^{ - \sin x}}\int {x{e^{\sin x}}} dx + C{e^{ - \sin x}}$
Work Step by Step
$$\eqalign{
& y' + y\cos x = x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + y\cos x = x \cr
& {\text{The differential equation is in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \cos x{\text{ and }}Q\left( x \right) = x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\cos xdx} }} = {e^{\sin x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{\sin x}}\frac{{dy}}{{dx}} + {e^{\sin x}}y\cos x = x{e^{\sin x}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{\sin x}}y} \right] = x{e^{\sin x}} \cr
& d\left[ {{e^{\sin x}}y} \right] = \left( {x{e^{\sin x}}} \right)dx \cr
& {\text{Integrate both sides}} \cr
& {e^{\sin x}}y + k = \int {x{e^{\sin x}}} dx \cr
& {\text{The integral }}\int {x{e^{\sin x}}} dx{\text{ can't be solved by the methods}} \cr
& {\text{known up to now}}{\text{, so solving for }}y \cr
& {e^{\sin x}}y = \int {x{e^{\sin x}}} dx - k \cr
& y = {e^{ - \sin x}}\int {x{e^{\sin x}}} dx - k{e^{ - \sin x}} \cr
& {\text{Let }}C = - k \cr
& y = {e^{ - \sin x}}\int {x{e^{\sin x}}} dx + C{e^{ - \sin x}} \cr} $$
