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Math Help - basic integral refresher with a square root

  1. #1
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    basic integral refresher with a square root

    So I have a square root integral as follows...

    x*sqrt((x^2) + 3)dx

    Do I take the u-sub of this thing?
    like...
    u = x^2
    du = 2xdx
    xdx = du/2

    and what do I do from there?

    I know that it will be sqrt(u + 3)du/2

    but that doesnt help me much...
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Try substituting x^2 + 3 instead
    u = x^2 + 3
    du = 2x dx
    du/2 = x dx

    so the resulting integral will be sqrt(u)/2
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  3. #3
    Member mathemagister's Avatar
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    Quote Originally Posted by RET80 View Post
    So I have a square root integral as follows...

    x*sqrt((x^2) + 3)dx

    Do I take the u-sub of this thing?
    like...
    u = x^2
    du = 2xdx
    xdx = du/2

    and what do I do from there?

    I know that it will be sqrt(u + 3)du/2

    but that doesnt help me much...
    Well, you could stll do it from there, but try using u=x^2+3.

     du = 2x \; dx

     \frac{1}{2}\; du = x \; dx

    You should find the rest follows through.
    Hope that helps

    Mathemagister
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  4. #4
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    Well I should say

    The equation actually is...
    in terms of rdrd(theta)

    and I seperated theta and r out into seperate integrals

    I have r*sqrt((r^2) - (a^2))dr

    the equation was originally derived from the equation of a sphere...
    z^2 = x^2 + y^2 - a^2

    and a is some arbitrary radius.
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  5. #5
    MHF Contributor

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    Yes, and both MacstersUndead and Mathmagister have told you what to do: let u= r^2- a^2.
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