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

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### 1 Answer

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**