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