Answer
$$y = 1 + C{e^{ - \sin x}}$$
Work Step by Step
$$\eqalign{
& y\cos x - \cos x + \frac{{dy}}{{dx}} = 0 \cr
& \frac{{dy}}{{dx}} + y\cos x = \cos x \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \cos x{\text{ }}Q\left( x \right) = \cos x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\cos xdx} }} = {e^{\sin x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{\sin x}}\frac{{dy}}{{dx}} + y{e^{\sin x}}\cos x = {e^{\sin x}}\cos x \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {y{e^{\sin x}}} \right] = {e^{\sin x}}\cos x \cr
& d\left[ {y{e^{\sin x}}} \right] = \left( {{e^{\sin x}}\cos x} \right)dx \cr
& {\text{Integrate both sides}} \cr
& y{e^{\sin x}} = \int {\left( {{e^{\sin x}}\cos x} \right)} dx \cr
& y{e^{\sin x}} = {e^{\sin x}} + C \cr
& {\text{Solve for }}y \cr
& y = 1 + \frac{C}{{{e^{\sin x}}}} \cr
& y = 1 + C{e^{ - \sin x}} \cr} $$
