Answer
$$\frac{{1 - e\cos \left( 1 \right) + e\sin \left( 1 \right)}}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{e^x}\sin x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& dv = \sin x,{\text{ }}v = - \cos x \cr
& {\text{ Integration by Parts Formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{e^x}\sin x} dx = - {e^x}\cos x - \int {{e^x}\left( { - \cos x} \right)} dx \cr
& \int {{e^x}\sin x} dx = - {e^x}\cos x + \int {{e^x}\cos x} dx \cr
& \cr
& {\text{Integrate by parts again for }}\int {{e^x}\cos x} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& dv = \cos x,{\text{ }}v = \sin x \cr
& \int {{e^x}\sin x} dx = - {e^x}\cos x + \left( {{e^x}\sin x - \int {{e^x}\sin x} dx} \right) \cr
& \int {{e^x}\sin x} dx = - {e^x}\cos x + {e^x}\sin x - \int {{e^x}\sin x} dx \cr
& {\text{Add }}\int {{e^x}\sin x} dx{\text{ to both sides}} \cr
& 2\int {{e^x}\sin x} dx = - {e^x}\cos x + {e^x}\sin x \cr
& {\text{Divide both sides by }}2 \cr
& \int {{e^x}\sin x} dx = - \frac{1}{2}{e^x}\cos x + \frac{1}{2}{e^x}\sin x + C \cr
& \cr
& {\text{Therefore,}} \cr
& \int_0^1 {{e^x}\sin x} dx = \left[ { - \frac{1}{2}{e^x}\cos x + \frac{1}{2}{e^x}\sin x} \right]_0^1 \cr
& = \left[ { - \frac{1}{2}{e^1}\cos \left( 1 \right) + \frac{1}{2}{e^1}\sin \left( 1 \right)} \right] - \left[ { - \frac{1}{2}{e^0}\cos \left( 0 \right) + \frac{1}{2}{e^1}\sin \left( 0 \right)} \right] \cr
& = - \frac{e}{2}\cos \left( 1 \right) + \frac{e}{2}\sin \left( 1 \right) + \frac{1}{2} \cr
& = \frac{{1 - e\cos \left( 1 \right) + e\sin \left( 1 \right)}}{2} \cr} $$
