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