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Changing limits of integration - definite integral (without u sub)?

  1. Sep 13, 2014 #1
    Hello
    Can someone please tell me how is: [itex]\int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = 2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]
    where,
    m and R are positive real numbers

    This is how I'm trying to solve it...

    [itex]\int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = \int_{-R}^0 \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

    [itex]\Rightarrow \int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = - \int_{0}^{-R} \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

    Can't solve it any further :(
     
  2. jcsd
  3. Sep 13, 2014 #2

    LCKurtz

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    It's true for any even function ##f(x)##. That means ##f(-x) = f(x)## so the graph is symmetric about the y axis. In your argument try ##x=-u## in your first integral at the end.
     
  4. Sep 13, 2014 #3
    so I put ##x=-u## in one of the integrals .. this is what I get...

    [itex]\int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = \int_{0}^{R} \frac{\cos mu}{u^2 + 1}\,du + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

    but now the two integrals on RHS have different variables .. How do I add them so the resulting expression becomes
    [itex]2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]
     
    Last edited: Sep 13, 2014
  5. Sep 13, 2014 #4

    LCKurtz

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    Show us what happens when you let ##x=-u## in your integral$$
    \int_{-R}^0 \frac{\cos (mx)}{x^2+1}~dx$$
     
  6. Sep 13, 2014 #5
    I just changed my previous post .. can you please check that...
     
  7. Sep 13, 2014 #6

    LCKurtz

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    The ##x## and ##u## variables are dummy variables. You put the ##0## and ##R## limits in either and get the same answer from both.
     
  8. Sep 13, 2014 #7
    Oh .. That clears everything up .. Thanks a lot mate...
     
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