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Integral Computation

  1. Nov 14, 2013 #1
    Hi All,

    I am having some troble with the following integral

    $$\int_{-c}^{c} \frac{x^3}{(x^2+(y-s)^2)^2}\mathrm{d}s$$

    Many thanks as usual
     
  2. jcsd
  3. Nov 14, 2013 #2

    Simon Bridge

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    What's the problem?

    Please show your best attempt, annotated so we can see your thought process.
     
  4. Nov 14, 2013 #3

    arildno

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    To give you a hint:
    For some constant "a", functions f and g, do you know of functions that satisfies the identity:
    [tex]a^{2}+f^{2}=g^{2}[/tex]
     
  5. Nov 15, 2013 #4
    Arildno,

    thanks for your hint. I suggest functions $$f = \sqrt{log x}$$ and $$g = \sqrt{log (x/e^{a^2})}$$.

    Simon,

    My best attempt so far:
    $$\int \frac{1}{(x^2 + (y-s)^2)^2} \mathrm{d}s$$, change variable y-s = z,
    $$\int \frac{-1}{(x^2 + z^2)*(x^2+z^2)} \mathrm{d}s =$$
    and now I admit buckling under the pressure of the looming partial fraction attempt, trying to get to integrate terms such as $$\int \frac{1}{A+x^2}\mathrm{d}x$$...

    Many thanks
     
  6. Nov 15, 2013 #5

    Simon Bridge

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    I take it that the x and y have nothing to do with the s then?
    They are basically just constants for this integration?

    Because: $$a^2+f^2=a^2 + \ln |x| = \ln|e^{a^2}|+\ln|x| = \ln|xe^{a^2}| \neq g^2$$
    (did you make a typo?) OK - so, having found a pair of functions that have the suggested property - try using one of them as a substitution in the integrand....

    ... so did you try your substitution?

    Note that ##x^2+z^2 = (x+z)(x-z)##
    ... which will help with the partial fractions.

    Personally I'd start with the very first line and use a substitution.

    I believe you have misunderstood the hint ...

    How about trig functions:
    ##\sin^2\theta+\cos^2\theta=1## and ##1+\tan^2\theta = \sec^2\theta##

    The idea is to use one of them to make a substitution.
    i.e. starting from ##a^2-x^2## if you substitute ##x=a\sin\theta## then: $$a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2\theta$$

    ... it's probably easier than the partial fractions.
     
    Last edited: Nov 15, 2013
  7. Nov 15, 2013 #6

    Mark44

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    My first thought was a trig substitution. You can make things simpler by bringing the x3 term outside the integral.
    $$x^3\int_{-c}^{c} \frac{ds}{(x^2+(y-s)^2)^2}$$

    Note that (y - s)2 = (s - y)2, so you can make a slight change like so:
    $$x^3\int_{-c}^{c} \frac{ds}{(x^2+(s - y)^2)^2}$$

    Then by letting z = s - y, dz = ds, you have:
    $$x^3\int_{-c}^{c} \frac{dz}{(x^2+ z^2)^2}$$

    then a trig substitution, and off you go...
     
  8. Nov 15, 2013 #7

    Simon Bridge

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    Mark44: I think you misplaced a minus sign.
    [edit] oh I see you exploited that (-x)^2=x^2

    I figured trig substitutions too ... so did arildno.

    You can do the trig sub right at the start with s=y-x<trig fn> without having to do the initial z=s-y step.
    Although it is easier to evaluate - the back-substitution gets messy - needing a table of trig identities for things like sin(arctan(x)).

    Partial fraction approach is initially more labor intensive but looks like it gets a more useable result right away.
    I guess it depends what you're used to.
     
  9. Nov 15, 2013 #8

    Mark44

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    It wasn't obvious that that was where arildno was going.
    s = y - x? Must be a typo.
    You don't need a table if you can draw a right triangle.

    Let θ = arctan(x). Label a right triangle with the side opposite θ of length x, and the adjacent side of length 1. Then the hypotenuse is √(x2 + 1). It's straightforward to get sin(θ); i.e., sin(arctan(x)).
    There's more than one way to skin a cat. However, when I see a sum of squares, I immediately begin to salivate "trig substitution."
     
  10. Nov 15, 2013 #9

    Simon Bridge

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    x multipied by an appropriate trig function.
    I didn't want to spell it out any more than I had to.

    If T(θ) is a trig function - then a substitution of form: s=y-xT(θ)
    That way (y-s)=xT(θ) and ds = -xT'(θ) ... which, admittedly gives you a minus sign to carry around.

    Good point.

    I know - me too.

    Now to hear from OP.
     
  11. Nov 16, 2013 #10

    arildno

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    Actually, I'm rather a hyperbolic guy myself, but trigs work as well.
     
  12. Nov 16, 2013 #11

    Simon Bridge

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    That's OK - you can get medicine for that :)
     
  13. Nov 16, 2013 #12
    Many thanks to all of you for help, very useful and appreciated.
    Simon Bridge, yes my solution for the question posed by Arildno was wrong, not a typo but considering g as the constant, again my error.
    Thanks again
     
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