Answer
$\frac{{{e^x}}}{{{\pi ^2} + 1}}\left( {\sin \pi x - \pi \cos \pi x} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{e^x}\sin \pi x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& dv = \sin \pi xdx,{\text{ }}v = - \frac{1}{\pi }\cos \pi x \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{e^x}\sin \pi x} dx = \left( {{e^x}} \right)\left( { - \frac{1}{\pi }\cos \pi x} \right) - \int {\left( { - \frac{1}{\pi }\cos \pi x} \right)\left( {{e^x}} \right)dx} \cr
& \int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{1}{\pi }\int {{e^x}\cos \pi xdx} \cr
& \cr
& {\text{Integrate by parts }}\int {{e^x}\cos \pi xdx} \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& dv = \cos \pi xdx,{\text{ }}v = \frac{1}{\pi }\sin \pi x \cr
& \int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{1}{\pi }\left( {\int {{e^x}\cos \pi xdx} } \right) \cr
& \int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{1}{\pi }\left( {\frac{{{e^x}}}{\pi }\sin \pi x - \int {\frac{{{e^x}}}{\pi }\sin \pi x} dx} \right) \cr
& \int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{{{e^x}}}{{{\pi ^2}}}\sin \pi x - \frac{1}{{{\pi ^2}}}\int {{e^x}\sin \pi x} dx \cr
& \cr
& {\text{Add }}\frac{1}{{{\pi ^2}}}\int {{e^x}\sin \pi x} dx{\text{ to both sides}} \cr
& \int {{e^x}\sin \pi x} dx + \frac{1}{{{\pi ^2}}}\int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{{{e^x}}}{{{\pi ^2}}}\sin \pi x \cr
& \frac{{{\pi ^2} + 1}}{{{\pi ^2}}}\int {{e^x}\sin \pi x} dx = - \frac{{{e^x}}}{\pi }\cos \pi x + \frac{{{e^x}}}{{{\pi ^2}}}\sin \pi x \cr
& \cr
& {\text{Multiply both sides by }}\frac{{{\pi ^2}}}{{{\pi ^2} + 1}} \cr
& \int {{e^x}\sin x} dx = \frac{{{\pi ^2}}}{{{\pi ^2} + 1}}\left( { - \frac{{{e^x}}}{\pi }\cos \pi x} \right) + \frac{{{\pi ^2}}}{{{\pi ^2} + 1}}\left( {\frac{{{e^x}}}{{{\pi ^2}}}\sin \pi x} \right) + C \cr
& \int {{e^x}\sin x} dx = \frac{\pi }{{{\pi ^2} + 1}}\left( { - {e^x}\cos \pi x} \right) + \frac{1}{{{\pi ^2} + 1}}\left( {{e^x}\sin \pi x} \right) + C \cr
& \int {{e^x}\sin x} dx = \frac{{{e^x}}}{{{\pi ^2} + 1}}\left( { - \pi \cos \pi x + \sin \pi x} \right) + C \cr
& \int {{e^x}\sin x} dx = \frac{{{e^x}}}{{{\pi ^2} + 1}}\left( {\sin \pi x - \pi \cos \pi x} \right) + C \cr} $$
