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Extremely difficult double integral.

  1. Nov 26, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    [itex]\int_{0}^{8} \int_{y^{1/3}}^{2} \frac{1}{x^4+1} dxdy[/itex]

    2. Relevant equations

    Completing the square.

    3. The attempt at a solution

    This integral is disgusting. It literally took me 4 sheets of paper to do the partial fraction decomposition and then integrate the inner integral giving me an expression involving ln(compicated y expression) + ln(a number) + 1/2arctan(more garbage) + 1/2arctan(even more garbage).

    Now I have an even more complicated integral to do and I have no clue how to approach it as I've never integrated arctan and integrating ln is nuts.

    Any pointers on this one?
     
  2. jcsd
  3. Nov 26, 2012 #2
  4. Nov 26, 2012 #3

    Zondrina

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    Yeah I know I could easily look it up, but I don't think that's going to help.
     
  5. Nov 26, 2012 #4
    The way I see it, your 1/2 arctan(garbage) expressions have two quite similar sorts of garbage, (x√2-1) and (x√2+1). How about using this?

    arctan(a) + arctan(b) = arctan( (a+b)/(1-ab) )

    That should lead you to arctan( (something with x) / (1 - something with x²) ). Partial fractions and then integrating two expressions like arctan(something) should be possible.

    The ln expression I get looks like this:

    ln ( (something with x² + something with x) / (something with x² - something with x) )

    Try multiplying top and bottom of that fraction by the numerator to simplify the denominator. Maybe that'll help.
     
  6. Nov 26, 2012 #5

    gabbagabbahey

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    A good general rule of thumb is to think about what you are doing before you try to do it.

    Bearing that in mind, what are you doing by computing that integral? Answer: integrating the function [itex]f(x)=\frac{1}{x^4+1}[/itex] over some region in the [itex]xy[/itex]-plane. Is the given form of the integral the only way to accomplish that? Is it the best way? What happens if you break the region down into vertical strips instead of horizontal strips and integrate over [itex]y[/itex] first?
     
  7. Nov 26, 2012 #6
    Unless I'm mistaken, something very nice happens then, and the double integral suddenly becomes

    (some number) times ln (some other number)

    Very clever. That's a strategy well worth remembering.
     
  8. Nov 26, 2012 #7

    Zondrina

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    Yes i completely forgot about changing my limits! Thank you very much, I solved it :).
     
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