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Tricky Antiderivative

  1. Mar 19, 2013 #1
    1. The problem, the whole problem, and nothing but the problem

    $$ \int \sqrt{\frac{1+x^2}{1-x^2}} dx $$

    2. Relevant equations

    ?

    3. The attempt at a solution

    I tried a partial fraction decomp, but this obviously didn't get me very far because this isn't a rational function. Is there some kind of ingenious substitution applicable here?
     
  2. jcsd
  3. Mar 19, 2013 #2

    Dick

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    I don't see any way to get an antiderivative. And Wolfram Alpha says it's an elliptic integral. I don't disagree. If it's a definite integral there might be a trick. Is that the whole problem?
     
  4. Mar 19, 2013 #3
    ... I made it up

    I've changed the problem to the following:

    $$ \int \sqrt{\frac{1+x}{1-x}} dx $$

    According to wolfram, this does have a (nasty) elementary antiderivative. Any thoughts?
     
  5. Mar 19, 2013 #4

    Dick

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    You are making up HARD PROBLEMS.
     
  6. Mar 19, 2013 #5
    Sorry for adding new detail to a previous post by the way, I just remembered that that's a big no-no
     
  7. Mar 19, 2013 #6
    Making your own integrals isn't a good idea because you tend not to know if you can solve them. For example: [itex]\int e^{x^{2}} dx[/itex]
     
  8. Mar 20, 2013 #7
    The new integral you made up is solvable however. Do a u substitution.
     
  9. Mar 20, 2013 #8

    Curious3141

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    Substitute something that would make that square root on the denominator much easier to deal with (or eliminate).
     
  10. Mar 20, 2013 #9
    How about the substitution ## x = -2u - u^2##, ##dx = -2 - 2u##

    $$ \int \sqrt{ \frac{1 - 2u-u^2}{1+2u + u^2} } (-2-2u) \, du $$

    $$ -2 \int \frac{ \sqrt{1-2u-u^2} }{1+u}(1+u) \, du $$

    $$ -2 \int \sqrt{1-2u-u^2} \, du$$

    Alright, that looks fairly simple, but I'm blanking as to how I should deal with it.
     
    Last edited: Mar 20, 2013
  11. Mar 20, 2013 #10

    Dick

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    I'd say complete the square and then use a trig substitution.
     
  12. Mar 20, 2013 #11

    Curious3141

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    Why not ##x = 1 - u^2##?
     
  13. Mar 20, 2013 #12
    Alright, I completed the square, changing ##1-2u-u^2## to ## 2-(1+u)^2##.

    $$ -2\int \sqrt{2-(1+u)^2} \, du $$

    $$1+u = \sqrt{2} \sin\theta, \quad du = \sqrt{2} \cos \theta \, d\theta $$

    $$ -2\sqrt{2} \int \cos \theta \sqrt{2-2 \sin^2\theta} \, d\theta $$

    $$ -4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta $$

    $$ v = \sin \theta, \quad dv = \cos \theta \, d\theta$$

    $$ -4 \int 1 - v^2 \, dv $$

    $$ -4( v - v^3/3) $$

    $$ 4(v^3/3 - v) $$

    Before I go back and perform the various substitutions, is that correct?
     
  14. Mar 20, 2013 #13
    That would have truly been a wise choice
     
  15. Mar 20, 2013 #14
    Isn't 1 - sin2θ just cos2θ?
     
  16. Mar 20, 2013 #15
    -.- yes

    $$ -4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta $$
    $$ -4 \int \cos^2 \theta \, d\theta $$
    $$ -4 \int \frac{1+\cos(2\theta)}{2} d\theta $$
    $$ -2\Big(\theta + \int \cos(2\theta) d\theta \Big)$$
    $$ -2(\theta + \frac{ \sin(2\theta)}{2}) $$
    $$ - \sin(2\theta) -2\theta $$
     
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