Question

How do you integrate `[(e^(2x))sinx]dx`?

Answer 1

Integrate by parts twice using `u = e^(2x)` both times.

Explanation:

After the second integration by parts, you'll have

`int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx`

Note that the last integral is the same as the one we want. Call it `I` for now.

`I = -e^(2x)cosx + 2e^(2x)sinx - 4 I`

So `I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C`

You may rewrite / simplify / factor as you see fit.