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Homework Help: Integral: sqrt(a^2+x^2)

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    integral: sqrt(x^2+1)

    2. Relevant equations

    3. The attempt at a solution

    I tried using x=tan[y], dx=sec^2[y]dy


    and then i was stuck.
  2. jcsd
  3. Jan 14, 2009 #2


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    [tex]\int sec^3 y dy[/tex]

    Write it as

    [tex]\int (secy) (sec^2y dy)[/tex]

    and then use integration by parts.
  4. Jan 14, 2009 #3


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    I would write sec(y)^3 as dy/cos(y)^3=cos(y)dy/cos(y)^4=cos(y)*dy/(1-sin(y)^2)^2. Putting u=sin(y) that becomes du/(1-u^2)^2. Now use partial fractions.
  5. Jan 15, 2009 #4

    Gib Z

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    subwaybusker, Do you know about hyperbolic trig functions? I would let [itex]x=\sinh u[/itex].
  6. Jan 15, 2009 #5
    Well, my way of approaching this integral would be,
    [tex]\int \sqrt{x^2+1} . 1 [/tex]

    So, now the above integral can be solved by using integral by parts, try doing it!

    P.S: Here,[tex]u=\sqrt{x^2+1} [/tex] and v=1.
  7. Jan 15, 2009 #6


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    Ok, so after an integration by parts, now you have to integrate x^2/sqrt(x^2+1). How is that a simplification? I think you are doing the parts integration wrong. You need to have u=sqrt(x^2+1) and v=x.
  8. Jan 15, 2009 #7
    Nope, once you have,
    [tex]\int \sqrt{x^2+1}. 1~ =~ x\sqrt{x^2+1}~-~\int \frac{x^2}{\sqrt{x^2+1}}[/tex]

    Add and subtract 1 in the numerator, that'll help you, I'd solve it further if you want!
    Last edited: Jan 16, 2009
  9. Jan 16, 2009 #8

    Gib Z

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    Well, putting back in our differentails, which is a bad idea to forget, and evaluting the trivial integral, our integral I through your method is given by

    [tex] I = \frac{(x^2+1)^{\frac{3}{2}}}{3} - I + \int \frac{1}{\sqrt{x^2+1}}} dx [/tex]

    So Now you have to integrate that last integral, which takes the same substitution as suggested earlier. This method is not the shortest way.
  10. Jan 16, 2009 #9
    Well, Gibz thanks for making me realise my mistake! There shouldnt have been any integral before [tex] x.\sqrt{x^2+1} [/tex], my bad :sad:

    Thats already integrated, going according to the rule it'd be:
    [tex]\int u.v=~u\int v -\int \frac{\delta u}{\delta x} \int v [/tex]

    so, considering,[tex]
    u=\sqrt{x^2+1} [/tex] and v=1,

    it'd be
    [tex] x.\sqrt{x^2+1} -\int\frac{1}{2\sqrt(x^2+1)}.2x.x[/tex]
    where, 2 gets cancelled in the numerator and the denominator, and the sum continues..
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