# How do I integrate this?

• June 25th 2011, 12:15 PM
Lancet
How do I integrate this?
How do I integrate something like this:

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

I have no idea how to even approach this.
• June 25th 2011, 12:20 PM
tom@ballooncalculus
Re: How do I integrate this?
• June 25th 2011, 12:21 PM
Plato
Re: How do I integrate this?
Quote:

Originally Posted by Lancet
How do I integrate something like this:

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

Have a look at this,
• June 25th 2011, 12:28 PM
tom@ballooncalculus
Re: How do I integrate this?
• June 25th 2011, 12:29 PM
Lancet
Re: How do I integrate this?
Quote:

Originally Posted by Plato

Aside from the missing sign, that tells me *what* it is, but it doesn't tell me how to get there.
• June 25th 2011, 12:32 PM
tom@ballooncalculus
Re: 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:
http://www.ballooncalculus.org/asy/prod.png
... is the product rule, straight continuous lines differentiating downwards with respect to x. And...

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

PS:
Quote:

Originally Posted by Lancet
Aside from the missing sign, that tells me *what* it is, but it doesn't tell me how to get there.

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

__________________________________________________ __________

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Balloon Calculus; standard integrals, derivatives and methods

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• June 25th 2011, 12:38 PM
Lancet
Re: 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!
• June 25th 2011, 12:49 PM
Prove It
Re: How do I integrate this?
Quote:

Originally Posted by Lancet
How do I integrate something like this:

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

I have no idea how to even approach this.

You need to use Integration by Parts twice. Call this integral $\displaystyle I$. Then

\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...
• June 25th 2011, 01:07 PM
TheCoffeeMachine
Re: How do I integrate this?
Quote:

Originally Posted by Lancet
How do I integrate something like this:

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

I have no idea how to even approach this.

\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}