you need to do integration by parts twice, and then solve for the integral as in an algebraic equation
set u=e^x and dv=sin 4x dx du = e^x dx and v=-1/4 cos4x
the integral becomes:
-e^x cos4x/4 + 1/4Integral[e^x cos 4x dx]
now, we have to do this second integral by parts, and we set u=e^x and dv=cos 4x
du=e^x dx and v=1/4 sin 4x
this gives us:
-e^x cos4x/4+1/4[e^x sin 4x/4 - 1/4Integral[e^x sin4x dx]
at this point your instinct might be to think you are in an infinite loop of integrals, but look closely at the last integral on the right, it is a multiple of the original integral you had, so:
Integral[e^x sin 4x dx] =
-e^x cos4x/4+1/16 e^x sin 4x -1/16 Integral[e^x sin 4x dx]
collect terms:
(1+ 1/16)Integral[e^x sin 4x dx] =
-e^x cos 4x/4 + e^x sin4x/16
Integral[e^x sin 4x dx] =
1/17 e^x(sin 4x - 4 cos 4x)