Answer
$$y = - 2\ln \left( {\frac{1}{2}\cos t + C} \right)$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = {e^{y/2}}\sin t \cr
& \frac{{dy}}{{dt}} = {e^{y/2}}\sin t \cr
& {\text{separate}} \cr
& \frac{{dy}}{{{e^{y/2}}}} = \sin tdt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{dy}}{{{e^{y/2}}}}} = \int {\sin t} dt \cr
& \int {{e^{ - y/2}}dy} = \int {\sin t} dt \cr
& - 2{e^{ - y/2}} = - \cos t + C \cr
& {\text{solve for }}y \cr
& 2{e^{ - y/2}} = \cos t + C \cr
& {e^{ - y/2}} = \frac{1}{2}\cos t + C \cr
& \ln {e^{ - y/2}} = \ln \left( {\frac{1}{2}\cos t + C} \right) \cr
& - \frac{y}{2} = \ln \left( {\frac{1}{2}\cos t + C} \right) \cr
& y = - 2\ln \left( {\frac{1}{2}\cos t + C} \right) \cr} $$
