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A difficult integral

  1. Mar 23, 2006 #1
    I have been stumped trying to find the following integral:

    [tex] \int \sqrt{1+x^2} dx [/tex]

    I put it into my TI-89 calculator and it gave me and answer (that checks), but I cannot figure out how to do it by hand (is it possible?).

    I tried using substitution and got:

    [tex] \int \frac{\sqrt{u}}{\sqrt{u-1}} du [/tex]

    If you then use integration by parts, it just flips the fraction. Any suggestions?

    Thanks!
    Tom
     
    Last edited: Mar 23, 2006
  2. jcsd
  3. Mar 23, 2006 #2

    Hootenanny

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    Does writing it as

    [tex]\int (1+x^{2})^{\frac{1}{2}}[/tex]

    Help?
     
  4. Mar 23, 2006 #3

    robphy

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    What substitution did you try?
     
  5. Mar 23, 2006 #4

    Hootenanny

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    I think he used [itex] u = 1+x^2[/itex] becasue I've just tried it and arrived at the same result. :smile:
     
  6. Mar 23, 2006 #5
    You can also try starting with integration by parts, and that gives you:

    [tex] x \sqrt{1+x^2} - \int \frac{x^2}{\sqrt{1+x^2}} dx [/tex]

    but that doesn't help
     
  7. Mar 23, 2006 #6

    Hootenanny

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    For this type of problem you can use the inverse of the chain rule to give;

    [tex]\int (ax + b)^n \;\; dx = \frac{1}{a(n+1)}\cdot (ax+b)^{n+1} +c[/tex]

    Hope this helps
    -Hoot:smile:
     
  8. Mar 23, 2006 #7

    TD

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    Have you seen the hyperbolic functions? You can use cosh²a-sinh²a = 1 to substitute x = sinh(a).
    If you haven't seen those, using 1+tan²a = sec²a is an alternative to use x = tan(a).
     
  9. Mar 23, 2006 #8

    nrqed

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    But he (she) has a x^2 so this does not help... (or I might be missing something). A trig substitution seems the way to go to me....

    Pat
     
  10. Mar 23, 2006 #9

    Hootenanny

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    Okay I could also write it like this;

    [tex]\int [f(x)]^n \;\; dx = \frac{1}{f'(x)\cdot (n+1)} \cdot [f(x)]^{n+1}[/tex]

    I trig ident is not required :smile:. I could show the full derivation if the OP likes.
     
    Last edited: Mar 23, 2006
  11. Mar 23, 2006 #10

    nrqed

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    I am sorry, but this is incorrect! (*unless f'(x) is a constant!!!!)

    Pat
     
  12. Mar 23, 2006 #11

    Hootenanny

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    Indeed you are correct. I;ve obviuosly not been listening in class:bugeye:
     
  13. Mar 23, 2006 #12

    nrqed

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    To Kb1jij:

    TD is right.
    That's indeed the way to go. Using [itex] x= tan (\theta) [/itex], one gets [itex] {\sqrt {1 + x^2}} = sec (\theta) [/itex].

    Then find dx in terms of [itex] d \theta [/itex]. The resulting integral is easy to do. Then you need to reexpress [itex] \theta [/itex] in terms of x.

    Pat
     
  14. Mar 23, 2006 #13
    So the result is [tex] \int sec^3 (\theta) d \theta [/tex], right?
    I don't think that this is really "easy to do", but I'll let you know if I get stuck.

    Thanks!
    Tom
     
  15. Mar 23, 2006 #14

    nrqed

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    You are right, it does not look pretty. Sorry.

    Then, you must use hyperbolic trig functions. Set x = sinh(u). You will get the integral of cosh(u)^2. Using the hyperbolic trig identity for cosh^2, you will get a very easy integral to do.

    Pat
     
  16. Mar 23, 2006 #15
    Ok, I found the integral, and I am left with this:
    [tex]\frac{\theta + cosh \theta sinh \theta}{2}[/tex]
    which can then be turned into (arcsinh x + x cosh(arcsinh x))/2. Can I simplify that further? I'm not too familiar with the hyperbolic trig functions, but if those were regular trig functions that second term would become
    [tex]x\sqrt{1-x^2}[/tex]. Can I do the same here?
     
  17. Mar 23, 2006 #16

    Hurkyl

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    [tex]\int \sec^3 \theta \, d\theta[/tex]
    is one of the standard (circular) trigonometric integrals. Your textbook probably explicitly states an algorithm for simplifying integrals of the type

    [tex]\int \sec^a \theta \tan^b \theta \, d\theta[/tex]

    and the one of interest is certainly in any integral table.

    I think the recommended step is to rewrite the integrand using [itex]\sec^2 \theta = 1 + \tan^2 \theta[/itex].


    As for the hyperbolic substitution...
    You can use the same idea that was used to derive that result for the circular trig functions. However, the result will probably not be identical.
     
  18. Mar 23, 2006 #17

    nrqed

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    Do not turn it back in terms of x yet! You must still integrate! *After* you have integrated you will go back to x. It's easy to integrate because cosh is the derivative of sinh!!! (or vice versa). That's the beauty of this trick...
     
  19. Mar 23, 2006 #18

    nrqed

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    I don't think that this trick works if the difference of exponents a-b is 3. It works only for specific differences of exponents (I am too busy right now to work it out)... But I may be missing something in which case I would be interested in seeing how this works out...

    Pat
     
  20. Mar 23, 2006 #19

    Hurkyl

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    Oh right, I made a mistake. (That's what I get for doing it in my head) I needed to also use integration by parts to evaluate [itex]\int \sec^3 \theta \, d\theta[/itex].
     
  21. Mar 23, 2006 #20

    Curious3141

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    If I'm not mistaken, we had a thread very similar to this one a while back. I came to the conclusion that hyperbolic trig was the most direct and elegant way. Let me search...
     
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