Question

The integral of (e^-x) cos(4x) dx.

Answer 1

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)