Answer
$y = \ln \left( {\frac{1}{{C - {e^{\sin x}}}}} \right)$
Work Step by Step
$$\eqalign{
& \left( {\sec x} \right)\frac{{dy}}{{dx}} = {e^{y + \sin x}} \cr
& {\text{First use the exponential property }}{e^{a + b}} = {e^a}{e^b}{\text{ to }}{e^{y + \sin x}} \cr
& \left( {\sec x} \right)\frac{{dy}}{{dx}} = {e^y}{e^{\sin x}} \cr
& {\text{Now}}{\text{, separate the variables}} \cr
& \frac{{dy}}{{{e^y}}} = \frac{{{e^{\sin x}}}}{{\sec x}}dx \cr
& {\text{Rewrite}} \cr
& {e^{ - y}}dy = {e^{\sin x}}\cos xdx \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {{e^{ - y}}} dy = \int {{e^{\sin x}}\cos } xdx \cr
& - {e^{ - y}} = {e^{\sin x}} + C_1 \cr
& {e^{ - y}} = C - {e^{\sin x}} \cr
& {\text{Solve for }}y,{\text{ take the natural logarithm on both sides}} \cr
& \ln {e^{ - y}} = \ln \left( {C - {e^{\sin x}}} \right) \cr
& - y = \ln \left( {C - {e^{\sin x}}} \right) \cr
& y = - \ln \left( {C - {e^{\sin x}}} \right) \cr
& or \cr
& y = \ln \left( {\frac{1}{{C - {e^{\sin x}}}}} \right) \cr} $$
