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Math Help - general integration formula

  1. #1
    Member disclaimer's Avatar
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    general integration formula

    Hi all,

    Is there any general formula for calculating integrals of the type \int{\frac{dx}{x\sqrt{x^2-a^2}}} other than \frac{1}{a}arc\sec{\frac{|x|}{a}}+C (guess the function is not even included in LaTeX)? I was trying to figure it out and got something like -\frac{1}{a}\arcsin{\frac{a}{|x|}}+C but I'm not sure if it's (always) correct and where absolute values should be put.

    Thank you.

    EDIT:
    Guess I got it: -\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C. Can someone confirm if it's correct? Thanks.
    Last edited by disclaimer; April 19th 2008 at 07:33 AM.
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  2. #2
    Eater of Worlds
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    I get the arcsec.

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

    Let x=asec(u), \;\ dx=asec(u)tan(u)du

    \frac{asec(u)tan(u)}{asec(u)\sqrt{a^{2}(sec^{2}(u)-1)}}du

    This all whittles down to:

    \int\frac{1}{a}du=\frac{u}{a}+C

    But, u=sec^{-1}(\frac{x}{a})

    And we get \frac{sec^{-1}(\frac{x}{a})}{a}+C
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  3. #3
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    Thanks. It's just that using arc\sec or \sec is very uncommon in my neck of the woods. So I wanted to get something connected with \sin. Here's my reasoning:

    \int{\frac{dx}{x\sqrt{x^2-a^2}}}
    Let \frac{1}{x}=t
    So x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}
    That way we get \int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt which yields -\int{\frac{dt}{\sqrt{1-a^2t^2}}}
    Now we can write it as -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}} and use the formula \int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C
    That said, -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C
    And, finally coming back to our substitution, we get:
    \boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}
    What do you think about it? I'm not sure about the absolute values, to my mind it should be <br />
-\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C<br />
.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by disclaimer View Post
    Thanks. It's just that using arc\sec or \sec is very uncommon in my neck of the woods. So I wanted to get something connected with \sin. Here's my reasoning:

    \int{\frac{dx}{x\sqrt{x^2-a^2}}}
    Let \frac{1}{x}=t
    So x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}
    That way we get \int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt which yields -\int{\frac{dt}{\sqrt{1-a^2t^2}}}
    Now we can write it as -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}} and use the formula \int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C
    That said, -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C
    And, finally coming back to our substitution, we get:
    \boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}
    What do you think about it? I'm not sure about the absolute values, to my mind it should be <br />
-\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C<br />
.
    I think you mean to say arcos\bigg(\frac{a}{x}\bigg) using the identity taht arcos(u(x))=arcsec\bigg(\frac{1}{u(x)}\bigg)...but I think the |a| is incorrect the only absolute values I have seen in relation to this is the derivative of arsec(x) or arccsc(x)
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  5. #5
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    That looks better. Since \left(\arcsin{x}\right)'=\frac{1}{\sqrt{1-x^2}} and \left(\arccos{x}\right)'=-\frac{1}{\sqrt{1-x^2}}, the integral \int{\frac{dx}{x\sqrt{x^2-a^2}}} would be \frac{1}{a}\arccos{\frac{a}{x}}+C regardless of the sign of a. Thanks.
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  6. #6
    Math Engineering Student
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    Substitution z^2=x^2-a^2 kills the problem.
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    Substitution z^2=x^2-a^2 kills the problem.
    Your substitution leads to \frac{1}{a}\arctan{\frac{\sqrt{x^2-a^2}}{a}}+C, am I right?
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  8. #8
    Math Engineering Student
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    That's what you need to find by yourself. Differentiate your result, does yield the integrand?
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