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Trying an alternative to Trig Substitutions

  1. May 13, 2012 #1
    I wanted to try an alternative method to the proverbial technique used in trig substitution. Is this method a dead-end or is there hope for it?

    [tex] \int \frac{\sqrt{x^2-3}}{x} dx[/tex]
    Using trig substitution
    [tex]c^2=a^2+b^2[/tex]
    [tex]a = \sqrt{c^2-b^2}[/tex]
    [tex]∴ c = x, b = \sqrt{3}[/tex]
    Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..
    [tex] \int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx[/tex]

    And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.
     
  2. jcsd
  3. May 13, 2012 #2

    sharks

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    You could try the substitution: [tex]x=\sqrt 3 \cosh \theta[/tex]
     
  4. May 13, 2012 #3

    SammyS

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    Your substitution appears to be [itex]\displaystyle \cos(\theta)=\frac{\sqrt{x^2-3}}{x}\ .[/itex]

    In that case, how is dx related to dθ ?
     
  5. May 13, 2012 #4
    I just did, it starts becoming ugly. And your still plagued by the dx so it hasn't made a difference.

    Edit: I think I over-complicated things, I'm reviewing my method.

    Edit: Either way, I feel a bit uncomfortable with the approach because I don't know how

    [tex] x = \sqrt{3} cosh \theta [/tex]
     
    Last edited: May 13, 2012
  6. May 13, 2012 #5
    Well, it doesn't seem that I can know until I can explicitly define the function in terms of x.
     
  7. May 13, 2012 #6

    sharks

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    It works out by the substitution that i suggested. What is the final answer given in your book/notes?
     
  8. May 13, 2012 #7
    [tex] \sqrt{3} \int (sec^2 \theta - 1) d\theta [/tex]
    [tex] \sqrt{3} (tan \theta - \theta) [/tex]
     
  9. May 13, 2012 #8

    dextercioby

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    The substitution reccomended in post #2 should solve the problem.
     
  10. May 13, 2012 #9
    I can use any of the proverbial substitutions given in the book but that is besides the point. I was trying to do it in a different way and just need to know if my method is a dead end or not.

    And at any rate, I'm not too familiar with the hyperbolic functions and how he derived that substitution equivalent.
     
  11. May 13, 2012 #10

    HallsofIvy

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    Why are the substitutions in your book "proverbial"? I don't recally any proverbs related to Calculus!

    In any case, your method is not a "dead end" but it is exactly the trig substitution you apparently are trying to avoid. Taking x to be the hypotenuse of your right triangle and the "near side" to be [itex]\sqrt{x^3- 3}[/itex], the "opposite side" has length 3 so [itex]sin(\theta)= 3/x[/itex], [itex]x= 3/sin(\theta)= 3csc(\theta)[/itex]. From that, [itex]dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta[/itex].
     
  12. May 13, 2012 #11
    Proverbial as in widely known and common.

    Hmm, I see now. ^.^ Thanks.
     
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