I´ll start by integrating e2x and leaving cos(x) as it is and then derive it leaving the integrated part as it is.
e2x2⋅cos(x)−∫e2x2⋅(−sin(x))dx=
=e2x2⋅cos(x)+12∫e2x⋅(sin(x))dx=
by parts again:
=e2x2⋅cos(x)+12[e2x2⋅(sin(x))−∫e2x2⋅cos(x)dx]=
=e2x2⋅cos(x)+e2x4⋅(sin(x))−14∫e2x⋅cos(x)dx
So your integral is:
∫e2xcos(x)dx=e2x2⋅cos(x)+e2x4⋅(sin(x))−14∫e2x⋅cos(x)dx
Now I can take to the left the integral: −14∫e2x⋅cos(x)dx
Giving:
∫e2xcos(x)dx+14∫e2x⋅cos(x)dx=e2x2⋅cos(x)+e2x4⋅(sin(x))
54∫e2x⋅cos(x)dx=e2x2⋅cos(x)+e2x4⋅(sin(x))
and finally:
∫e2x⋅cos(x)dx=45[e2x2⋅cos(x)+e2x4⋅(sin(x))]+c