Question

Integration by Parts Evaluate e-2x cos(2x) dx.​

Answer 1

Let

`I = \displaystyle \int e^(-2x) \cos(2x) \, dx[/]tex</p><p>Integrate by parts:</p><p>[tex]\displaystyle \int u \, dv = uv - \int v \, du`

with

`u = e^(-2x) \implies du = -2 e^(-2x) \, dx \\\\ dv = \cos(2x) \, dx \implies v = \frac12 \sin(2x)`

Then

`\displaystyle I = \frac12 e^(-2x) \sin(2x) + \int e^(-2x) \sin(2x) \, dx + C`

Integrate by parts again, this time with

`u = e^(-2x) \implies du = -2 e^(-2x) \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\frac12 \cos(2x)`

so that

`\displaystyle I = \frac12 e^(-2x) \sin(2x) - \frac12 e^(-2x) \cos(2x) - \int e^(-2x) \cos(2x) \, dx + C\\\\ \implies I = (\sin(2x)-\cos(2x))/(2e^(2x)) - I + C \\\\ \implies 2I = (\sin(2x) - \cos(2x))/(2e^(2x)) + C \\\\ \implies I = \boxed{(\sin(2x) - \cos(2x))/(4e^(2x)) + C}`