The integration by parts states that #int\ u\ dv=uv-int\ v\ du#.
In this problem, call #u=sin(x)# and #dv=e^x\ dx# (since in the above formula #v# is integrated and #e^x\ dx# is easier to integrate).
Just plug into the formula: #int\ sin(x)e^x\ dx=sin(x)e^x-int\ e^x\ cos(x)\ dx#.
Now, we can repeat the integration-by-parts process to the integral on the right-hand side (this time using #u=cos(x)# and #dv=e^x\ dx#): #int\ sin(x)e^x\ dx=sin(x)e^x-int\ e^x\ cos(x)\ dx=sin(x)e^x-cos(x)e^x-int\ e^xsin(x)\ dx#.
Clean up: #int\ sin(x)e^x\ dx=sin(x)e^x-cos(x)e^x-int\ e^xsin(x)\ dx#. Note that both sides contain #int\ sin(x)e^x\ dx#, albeit opposite in sign.
Thus, add #int\ sin(x)e^x\ dx# to both sides to get #2int\ sin(x)e^x\ dx=sin(x)e^x-cos(x)e^x#.
Finally, divide both sides by #2# to get the final answer (don't forget to add the integration constant #C#): #int\ sin(x)e^x\ dx=1/2e^x(sin(x)-cos(x))+C#.
