Answer
$$y = C{e^{\sin x}} - 1$$
Work Step by Step
$$\eqalign{
& \left( {y + 1} \right)\cos xdx - dy = 0 \cr
& \left( {y + 1} \right)\cos xdx = dy \cr
& {\text{Separating the variables}} \cr
& \cos xdx = \frac{1}{{y + 1}}dy \cr
& \frac{1}{{y + 1}}dy = \cos xdx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{{y + 1}}} dy = \int {\cos x} dx \cr
& \ln \left| {y + 1} \right| = \sin x + {C_1} \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| {y + 1} \right|}} = {e^{\sin x}}{e^{{C_1}}} \cr
& y + 1 = {e^{\sin x}}C \cr
& y = C{e^{\sin x}} - 1 \cr} $$
