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Quick Problem with an integral

  1. Oct 15, 2012 #1
    I am just having problems factoring the constants out of the integral:

    ∫Fsinwt dt

    where F and w are constants.

    I can obviously take F right out of the integral but I forget how to take the w out of the sin, and I can't seem to find anywhere on the internet that will tell me how to do so.

    If anyone could quickly help me that would be great. Thanks.

    EDIT: If anyone could explain to me how to enter an integral like this into wolfram alpha so I can have steps that would be just as good.
  2. jcsd
  3. Oct 15, 2012 #2
    What is the final porpose of doing that?
    You could use the half-angle formula:


    for particular 'w', but I don't think that there is a general (resonable) formula for any w. The only thing that comes to my mind is using de Moivre's theorm:

    (cosx+isinx)^w=(coswx+isinwx)=(cosx+isinx)^w, and now you can multiply out the bracket and take the imaginary part what wil leave u with:

    sinwx=(powers of sinx)
  4. Oct 15, 2012 #3
    You don't need to factor out the w. You really just have [tex]F\intsinwtdt[\tex]. You just need to integrate sinwt. So you'll get -coswt, and accounting for the w and F gives (-Fcoswt)/w. For wolfram alpha just type in integrate and then the function
  5. Oct 15, 2012 #4
    Yeah, but I think that he has some hidden purpose in doing that trick :P It's pretty simple to integrate, so propably he needs it in that form.
  6. Oct 15, 2012 #5


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    The way I find easiest is 'everywhere I see a t, replace it with t/w'. Of course, that includes in the bounds of the integral.
  7. Oct 15, 2012 #6


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    u = wt, so dt = du/w. The integral becomes (F/w)∫sinudu = -(F/w)coswt + C

    Not acquainted with Wolfram alpha.
  8. Oct 15, 2012 #7


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    In other words, even if the constants are abstractly depicted through letters, you must still be able to apply the elementary methods like part integration or substitution.
  9. Oct 15, 2012 #8
    Well, I still think that the problem was not to simply evaluate the integral, which is easy, but to evaluate it by a given method (why else struggle with dragging that constant out of the sin?).
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