How to evaluate the integral ⌠e^(-5t) cos(2t) dt using integration by parts

What I did:

Let u= cos(2t) so that du= -2sin(2t) dt
Let dv= e^(-5t) dt so that v= (-1/5)e^(-5t)

But as I did the problem, it seemed that I would have to go on integrating forever because the e's never stop integrating. What am I doing wrong? Was it my setup? I tried reversing the u and dv but this still doesn't work. Any help appreciated!

⌠e^(-5t) cos(2t) dt

u = cos(2t)
du = - 2 sin(2t)

dv = e^(-5t) dt
v = -(1/5) e^(-5t)

⌠e^(-5t) cos(2t) dt = - (1/5)e^(-5t)cos(2t) - ⌠(2/5)e^(-5t) sin(2t) dt

again integrate by parts

u = sin(2t)
du = 2 cos(2t) dt

dv = (2/5)e^(-5t) dt
v = -(2/25) e^(-5t)


⌠e^(-5t) cos(2t) dt = - (1/5)e^(-5t)cos(2t) - [ -(2/25)e^(-5t) sin(2t) - ⌠-(4/25)e^(-5t) cos(2t) dt ]

=> ⌠e^(-5t) cos(2t) dt = - (1/5)e^(-5t)cos(2t) + (2/25)e^(-5t) sin(2t) - 4/25⌠e^(-5t) cos(2t) dt

add 4/25⌠ e^(-5t) cos(2t) both sides

=> 29/25⌠e^(-5t) cos(2t) dt = - (1/5)e^(-5t)cos(2t) + (2/25)e^(-5t) sin(2t)

divide by 29/25

=> ⌠e^(-5t) cos(2t) dt = - (5/29)e^(-5t)cos(2t) + (2/29)e^(-5t) sin(2t)

= (1/29)e^(-5t) ( 2 sin(2t) - 5cos(2t))