Answer
$$ - \frac{1}{2}{e^{ - y}}\cos y - \frac{1}{2}{e^{ - y}}\cos y + C $$
Work Step by Step
$$\eqalign{
& \int {{e^{ - y}}\cos y} dy \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {e^{ - y}},\,\,\,\,du = - {e^{ - y}}dy,\,\,\,\,\,dv = \cos ydy,\,\,\,\,v = \sin y \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y - \int {\left( {\sin y} \right)\left( { - {e^{ - y}}} \right)dy} \cr
& \int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y + \int {{e^{ - y}}\sin ydy} \cr
& \cr
& {\text{Integration by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = {e^{ - y}},\,\,\,\,du = - {e^{ - y}}dy,\,\,\,\,\,dv = \sin ydy,\,\,\,\,v = - \cos y \cr
& \int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y + \left( { - {e^{ - y}}\cos y - \int {\left( { - \cos y} \right)\left( { - {e^{ - y}}} \right)} dy} \right) \cr
& \int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y - {e^{ - y}}\cos y - \int {{e^{ - y}}\cos y} dy \cr
& \cr
& {\text{The unknown integral now appears on both sides of the equation}}{\text{. }} \cr
& {\text{Adding the integral to both sides and adding the constant }}C{\text{ we obtain}} \cr
& \int {{e^{ - y}}\cos y} dy + \int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y - {e^{ - y}}\cos y + C \cr
& 2\int {{e^{ - y}}\cos y} dy = - {e^{ - y}}\cos y - {e^{ - y}}\cos y + C \cr
& {\text{Divide both sides by 2}} \cr
& \int {{e^{ - y}}\cos y} dy = - \frac{1}{2}{e^{ - y}}\cos y - \frac{1}{2}{e^{ - y}}\cos y + C \cr} $$
