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Math Help - Intergration Problem (Substitution I think)

  1. #1
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    Intergration Problem (Substitution I think)

    I have this integral i need to do in the middle of a engineering question and I have looked in a few directions at it and cant seem to crack it!

    The integral is
    \int 4r/\sqrt{2-r^2}

    First I took out factor of 4.
    Then multiplyed out the square to get

    4\int(2r-r^3/2-r^2)

    And tried by parts, which didn't go to well.

    So im looking at substitution making
    u = 2r -r^3

    so du = 2-3r^2 dr

    Where do I go from here ....
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  2. #2
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    Quote Originally Posted by jezzyjez View Post
    I have this integral i need to do in the middle of a engineering question and I have looked in a few directions at it and cant seem to crack it!

    The integral is
    \int 4r/\sqrt{2-r^2}

    First I took out factor of 4.
    Then multiplyed out the square to get

    4\int(2r-r^3/2-r^2)

    And tried by parts, which didn't go to well.

    So im looking at substitution making
    u = 2r -r^3

    so du = 2-3r^2 dr

    Where do I go from here ....
    What is the derivative of -4\sqrt{2-r^2} ?

    CB
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  3. #3
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    Trig sub is always an old standby.

    \int\frac{4r}{\sqrt{2-r^{2}}}dr

    Let r=\sqrt{2}sin(t), \;\ dr=\sqrt{2}cos(t)dt

    Make the subs and it whittles clean down to

    4\sqrt{2}\int sin(t)dt

    After integrating, to get it back in terms of r, make the resub:

    t=sin^{-1}(\frac{r}{\sqrt{2}})
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  4. #4
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    Why not substitute
    v=2-r^2
    dr=-\frac{du}{2r}
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  5. #5
    Eater of Worlds
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    Go ahead. I was just offerening an alternative. there are numerous ways to tackle it and most other integrals
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