How do I integrate something like this:

$\displaystyle \int e^{-2t}cos(t) dt$

I have no idea how to even approach this.

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- Jun 25th 2011, 11:15 AMLancetHow do I integrate this?
How do I integrate something like this:

$\displaystyle \int e^{-2t}cos(t) dt$

I have no idea how to even approach this. - Jun 25th 2011, 11:20 AMtom@ballooncalculusRe: How do I integrate this?
See here for very similar... http://www.mathhelpforum.com/math-he...tml#post658188

- Jun 25th 2011, 11:21 AMPlatoRe: How do I integrate this?
- Jun 25th 2011, 11:28 AMtom@ballooncalculusRe: How do I integrate this?
... or, better still, this... integrate e^(-2t) cos(t) - Wolfram|Alpha

- Jun 25th 2011, 11:29 AMLancetRe: How do I integrate this?
- Jun 25th 2011, 11:32 AMtom@ballooncalculusRe: How do I integrate this?
Then look at my first link, but I'll draw similar for exactly the present problem. Give me two ticks.

Just in case a picture helps...

http://www.ballooncalculus.org/draw/...twice/four.png

Larger

... where (key in spoiler) ...

__Spoiler__:

PS: There's a button on the Wolfram page that say's 'show steps'.

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jun 25th 2011, 11:38 AMLancetRe: How do I integrate this?
Oh! I see! That's one of those tricks where you do integration by parts until you get a duplicate term, toss that on the other side to combine with the original integration and then divide out the coefficient.

I didn't see that at first. Thanks guys! - Jun 25th 2011, 11:49 AMProve ItRe: How do I integrate this?
You need to use Integration by Parts twice. Call this integral $\displaystyle \displaystyle I$. Then

$\displaystyle \displaystyle \begin{align*} I &= \int{e^{-2t}\cos{t}\,dt} \\ I&= e^{-2t}\sin{t} - \int{-2e^{-2t}\sin{t}\,dt} \\ I&= e^{-2t}\sin{t} - \left(2e^{-2t}\cos{t} - \int{-4e^{-2t}\cos{t}\,dt}\right) \\ I&= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} - 4\int{e^{-2t}\cos{t}\,dt} \\ I &= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} - 4I \\ 5I &= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} \\ I &= \frac{1}{5}e^{-2t}\sin{t} - \frac{2}{5}e^{-2t}\cos{t} \end{align*}$

and don't forget the integration constant... - Jun 25th 2011, 12:07 PMTheCoffeeMachineRe: How do I integrate this?
$\displaystyle \begin{aligned}& \int e^{{-2t}}\cos{t}\;{dt} = \Re \bigg(\int e^{-2t}e^{it}\;{dt}\bigg) = \Re \bigg(\int e^{(i-2)t}\;{dt}\bigg)\\& = \Re \bigg(\frac{e^{(i-2)t}}{i-2}\bigg)+\mathcal{C} = \frac{1}{5}\bigg(e^{-2t}\sin{t}-2e^{-2t}\cos{t}\bigg)+\mathcal{C}.\end{aligned}$