Verified answer

Let I= INTEGRAL[ e^(2x) cos(2x) dx]

u=e^(2x); dv = cos(2x)

du= 2e^(2x) dx; v= (1/2)sin(2x)

(1/2)e^(2x)sin(2x) - INTEGRAL [e^(2x) sin(2x) dx]

u=e^(2x); dv =sin(2x) dx

du= 2e^(2x) dx; v= (-1/2)cos(2x)

(1/2)e^(2x)sin(2x) + (1/2)e^(2x)cos (2x) - I (from top line)

Now since I= line above, we get

2I= (1/2)e^(2x)sin(2x) + (1/2)e^(2x)cos(2x)

I= (1/4)e^(2x) [sin(2x)+cos(2x)] + C