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Math Help - How do I integrate this?

  1. #1
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    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.
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  2. #2
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    Re: How do I integrate this?

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  3. #3
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    Re: How do I integrate this?

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

    \int e^{-2t}cos(t) dt
    Have a look at this,
    Last edited by Plato; June 25th 2011 at 11:35 AM.
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    Re: How do I integrate this?

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  5. #5
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    Re: How do I integrate this?

    Quote Originally Posted by Plato View Post
    Aside from the missing sign, that tells me *what* it is, but it doesn't tell me how to get there.
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  6. #6
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    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...



    Larger

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

    Spoiler:

    ... is the product rule, straight continuous lines differentiating downwards with respect to x. And...



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



    PS:
    Quote Originally Posted by Lancet View Post
    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'.

    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; June 25th 2011 at 11:51 AM. Reason: pic
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    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!
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  8. #8
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    Re: How do I integrate this?

    Quote Originally Posted by Lancet View Post
    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...
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    Re: How do I integrate this?

    Quote Originally Posted by Lancet View Post
    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}
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