Evaluate the anti derivative of e^2x * cos 3x.

We have to find Int [e^2x * cos 3x
dx]

Here the best way to solve would be to use integration
by parts.

Int [u dv] = u*v – Int [v
du]

take u = e^2x, du = 2*e^2x
dx

dv = cos 3x dx, v = (1/3)* sin
3x

Int [e^2x * cos 3x
dx]

=> [e^2x*sin 3x]/3 – (2/3)*Int [e^2x * sin 3x
dx]

We have again landed up with an integral like the
original with Int [e^2x * sin 3x dx].

So using integration
by parts again, this time we take U = e^2x, dU = 2*e^2x
dx

dV = sin 3x dx, V = (-1/3) cos
3x

Int [e^2x * sin 3x dx] = (- e^2x * cos 3x)/3 + (2/3)*Int
[e^2x * cos 3x dx]

So we
have

Int [e^2x * cos 3x dx] = [e^2x*sin 3x]/3 – (2/3) [(-
e^2x * cos 3x)/3 + (2/3)*Int [e^2x * cos 3x dx]]

=>
Int [e^2x * cos 3x dx] = [e^2x*sin 3x]/3 + (2/3)*e^2x * cos 3x)/3 - (2/3)*(2/3)*Int
[e^2x * cos 3x dx]]

=> Int [e^2x * cos 3x dx] +
(2/3)*(2/3)*Int [e^2x * cos 3x dx] = [e^2x*sin 3x]/3 + (2/3)*(e^2x * cos
3x)/3

=> (13/9)* Int [e^2x * cos 3x dx] = [e^2x*sin
3x]/3 + (2/3)*(e^2x * cos 3x)/3

=> Int [e^2x * cos
3x dx] =3*[e^2x*sin 3x]/13 + 2*(e^2x * cos 3x)/13

=>
Int [e^2x * cos 3x dx] = [3*(e^2x*sin 3x) + 2*(e^2x * cos
3x)]/13

The required result
is

[3*(e^2x*sin 3x) + 2*e^2x*cos 3x]/13 +
C