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Thread: Inserting Limits Into Integral Result

  1. #1
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    Inserting Limits Into Integral Result

    What would this give: $\displaystyle \left[\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$.

    The solution given is $\displaystyle (1 + \cos n\pi)$.

    I don't understand what to do with the $\displaystyle n\pi$ because $\displaystyle \cos(0) = 1$ but what does $\displaystyle \cos n\pi (0)$ equal?
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  2. #2
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    What is your question?.

    $\displaystyle cos(n{\pi})=-1$ for odd values of n

    $\displaystyle cos(n{\pi})=1$ for even values of n.

    Is that what you meant?.
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  3. #3
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    Quote Originally Posted by galactus View Post
    What is your question?.

    $\displaystyle cos(n{\pi})=-1$ for odd values of n

    $\displaystyle cos(n{\pi})=1$ for even values of n.

    Is that what you meant?.
    I had integral of:

    $\displaystyle \int^0_{-1} (1+t)\cos n \pi t \ \mathrm{d}t$ Through which I used integration by parts and got a similar result to $\displaystyle \left[\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$.

    The correct is $\displaystyle \left[-\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$ but what would I get if I put the limits in?
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  4. #4
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    $\displaystyle \int_{-1}^{0} (1+t)cos(n{\pi}t)dt$

    The result should be:

    $\displaystyle \frac{1-cos(n\pi)}{(n{\pi})^{2}}$
    Last edited by galactus; Jan 10th 2009 at 01:03 PM. Reason: wrong integral
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