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Definite Integral Help?

  1. Sep 30, 2009 #1
    Hey guys,

    I was doing some practice questions and this particular one has me stumped. The topic was on integration with inverse hyperbolic identities, and I was asked to give exact solutions for the following integral:

    [tex]\int\sqrt{4x^2 -1} dx[/tex] between [tex]\frac{1}{2}[/tex] and [tex]\frac{13}{10}[/tex]

    From looking around on the internet I have found a standard integral that I can use (http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions" [Broken]) but I would quite like to know the process of deriving this.

    I have been fairly confident with these questions but I can see no way of using the basic inverse hyperbolic identities to reach this result and express the definite integral exactly.

    Could anyone please lend a hand?

    Thanks in advance,
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 30, 2009 #2


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    Well, this is, after all, in a section about hyperbolic functions! I remember that [itex]cosh^2(x)- sinh^2(x)= 1[/itex] so that (dividing through by [itex]cosh^2(x)[/itex]) [itex]1- tanh^2(x)= sech^2(x)[/itex]. I would try the substitution [itex]2x= tanh(x)[/itex].
  4. Sep 30, 2009 #3


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    [tex]x = \frac{1}{2} \cosh y[/tex]


    [tex]dx = \frac{1}{2} \sinh y dy[/tex]

    The indefinite integral then becomes

    [tex]\frac{1}{2} \int \sqrt{\cosh^2 y - 1} \sinh y dy [/tex]

    You should then be able to find the indefinite integral of this if you use

    [tex]\cosh^2 y - \sinh^2 y = 1[/tex]

    and the definition of [tex]\sinh y[/tex]. After you do that, it's pretty easy to rewrite the answer in terms of x instead of y, and plug in the endpoints in terms of x.
  5. Oct 1, 2009 #4

    Thank you for your replies.

    Solved the problem now :)

    Last edited: Oct 1, 2009
  6. Oct 1, 2009 #5


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    Well, your integral is therefore:
    We have the following anti-derivative:
    Using this, we get:

    Since [tex]cosh^{-1}(x)=\ln(x+\sqrt{x^{2}-1})[/tex], the answer can be simplified to:
    Last edited: Oct 1, 2009
  7. Oct 1, 2009 #6


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    By the way, if you want to see WHY we have the indicated anti-derivative, then use the definition of sinh:

    [tex]\int \sinh^2 y dy = \frac{1}{4} \int (e^y - e^{-y})^2 dy[/tex]

    and work from there.
  8. Oct 1, 2009 #7
    thank you again for the speedy replies :)

    I got the same answer as you, arildno, however I did it using different method...

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