Use complex numbers.
Write cos(2x) = Re(e^(i 2 x))
So,
int e^(-x) cos(2x) dx = Re int e^(-x)e^(i 2 x) dx = Re int e^((-1+2i)x) dx
Now it's easy to integrate because you know that if a ne 0, int e^(ax) dx = 1/a e^(ax) + cst. So :
int e^(-x)cos(2x) dx = Re(1/(-1+2i) e^((-1+2i)x)) + cst
After that, you have to simplify 1/(-1+2i) e^((-1+2i)x) :
e^(-x)/(-1+2i) = e^(-x)(cos(2x) + i sin(2x))/(-1+2i)
Multiply by (-1-2i) everywhere and use the fact that (a-bi)(a+bi) = a^2+b^2 :
(cos(2x) + i sin(2x))/(-1+2i) = ((cos(2x) +i sin(2x))(-1-2i))/(1+4)
Keep only the real part :
Re((cos(2x) + i sin(2x))/(-1+2i)) = (-cos(2x) + 2 sin(2x))/5
Finally,
int e^(-x)cos(2x) dx = e^(-x)/5 (-cos(2x) + 2 sin(2x)) + cst
