Question

How do you integrate `int e^(5x)cos3x`?

Answer 1

You can integrate by parts twice and algebraically solve for the integral to get `int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C`

Explanation:

Let `I=int\ e^{5x}cos 3x\ dx`. Now let `u=e^{5x}` and `dv=cos 3x\ dx` so that `du=5e^{5x}\ dx` and `v=1/3 sin 3x`.

Then `I=1/3 e^{5x}sin 3x-5/3int\ e^{5x}sin 3x\ dx`.

For this next integral, let `u=e^{5x}` and `dv=sin 3x\ dx` so that `du=5e^{5x}` and `v=-1/3 cos 3x`. It follows that

`I=1/3 e^{5x}sin 3x-5/3(-1/3 e^{5x}cos 3x+5/3 int e^{5x}cos 3x\ dx)`

`=1/3 e^{5x}sin 3x+5/9e^{5x}cos 3x-25/9 I`.

Therefore, `34/9 I=1/3 e^{5x}sin 3x+5/9 e^{5x}cos 3x`.

Multiplying both sides by `9/34`, rearranging, and tacking on a `+C` at the end gives

`I=int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C`.