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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How do I integrate something like this:
$\displaystyle \int e^{-2t}cos(t) dt$
I have no idea how to even approach this.
See here for very similar... http://www.mathhelpforum.com/math-he...tml#post658188
Have a look at this,Quote:
Originally Posted by Lancet
... or, better still, this... integrate e^(-2t) cos(t) - Wolfram|Alpha
Aside from the missing sign, that tells me *what* it is, but it doesn't tell me how to get there.Quote:
Originally Posted by Plato
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'.Quote:
Originally Posted by Lancet
__________________________________________________ __________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
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!
You need to use Integration by Parts twice. Call this integral $\displaystyle \displaystyle I$. ThenQuote:
Originally Posted by Lancet
$\displaystyle \displaystyle \begin{align*} I &= \int{e^{-2t}\cos{t}\,dt} \\ I&= e^{-2t}\sin{t} - \int{-2e^{-2t}\sin{t}\,dt} \\ I&= e^{-2t}\sin{t} - \left(2e^{-2t}\cos{t} - \int{-4e^{-2t}\cos{t}\,dt}\right) \\ I&= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} - 4\int{e^{-2t}\cos{t}\,dt} \\ I &= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} - 4I \\ 5I &= e^{-2t}\sin{t} - 2e^{-2t}\cos{t} \\ I &= \frac{1}{5}e^{-2t}\sin{t} - \frac{2}{5}e^{-2t}\cos{t} \end{align*}$
and don't forget the integration constant...
$\displaystyle \begin{aligned}& \int e^{{-2t}}\cos{t}\;{dt} = \Re \bigg(\int e^{-2t}e^{it}\;{dt}\bigg) = \Re \bigg(\int e^{(i-2)t}\;{dt}\bigg)\\& = \Re \bigg(\frac{e^{(i-2)t}}{i-2}\bigg)+\mathcal{C} = \frac{1}{5}\bigg(e^{-2t}\sin{t}-2e^{-2t}\cos{t}\bigg)+\mathcal{C}.\end{aligned}$Quote:
Originally Posted by Lancet
