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Double integral of cos(x/y)

  1. Apr 19, 2012 #1
    Suppose we have:
    ∫∫ cos(x/y) dydx

    where the first integral is of x and is 0→1, while the second is of y and is x→1. Could someone tell me how to get the first integration (i.e. of cos(x/y) w.r.t. y) done??
  2. jcsd
  3. Apr 19, 2012 #2
    If you wish to calculate the integral

    [tex]\int \cos(1/y)dy[/tex]

    then it can't be done. This integral can not be calculated using elementary functions.

    However, you could use Fubini to switch the two integrals and integrate to x first. Maybe that gives something nice.
  4. Apr 19, 2012 #3


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    [tex]\int_{x= 0}^1\int_{y=x}^1 cos(x/y) dy dx[/tex]
    is the same as
    [tex]\int_{y=0}^1\int_{x= 0}^y cos(x/y) dx dy[/tex]
    and the second is much easier to do.
    Last edited by a moderator: Apr 20, 2012
  5. Apr 19, 2012 #4

    [tex]\int_{0}^1\int_{0}^y \cos\left(\dfrac xy\right)\ \mathrm{d}x\ \mathrm{d}y[/tex]

    Sure, some of what I just did was a matter of style, but you did run into an error parsing your LaTeX, using cos(x/y} instead of cos(x/y). I suggest you preview all your posts, especially those with LaTeX, before submitting them. Also, use \cos, not cos. (We don't really need those x=0 and y=0 because we already know what we're integrating with respect to from those d's.)
  6. Apr 20, 2012 #5
    This form avoids confusion, especially with people new to integration. I remember my Calc II professor saying once that he changed his teaching method to including the variables of integration because students constantly mixed up variables on exams of his. I'm sure others have had the same problem. Sure, it's obvious that you integrate from the inside out so you don't really need the variables, but it's just like subtracting an integer from two sides of an equation instead of doing the subtraction in your head from step to step or using u-substitution when only constants are involved, etc.
  7. Apr 20, 2012 #6
    Very true, and they do help to avoid confusion a lot. My point was we don't need those. But we're starting to get a bit off the original question.
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