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Math Help - Trig Substitution Verification

  1. #1
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    Trig Substitution Verification

    I have just worked the following integral and I believe my text has a wrong answer. If anyone would take the time to verify, I would appreciate it.

    Initial Problem: \int\sqrt{4 + 9x^2} dx

    My solution: \frac{1}{8}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C

    Book solution: \frac{1}{6}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C

    As you can see, the solutions differ only by the factor's 1/8 and 1/6.
    I would appreciate someone verifying for me.
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  2. #2
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    please find derivative, to prove, what is correct
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  3. #3
    Super Member Deadstar's Avatar
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    If you gave us working it would be easier but Maple gives...

    \frac{1}{2}x\sqrt{4+9x^2} + \frac{2}{3} \textrm{arcsinh}  \left(\frac{3x}{2} \right)

    which would seem the book is right...
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  4. #4
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    Thanks. That's what I needed to know. My work was rather lengthy and on 2 pages, so I didn't include it. Thanks again.
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  5. #5
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    Back to the drawing board!
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  6. #6
    Super Member Deadstar's Avatar
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    Quote Originally Posted by kaiser0792 View Post
    Back to the drawing board!
    Just go over your working again, it's not a massive error so you likely just wrote down the wrong number or something... If you can't find the error post up what substitution you used and if I stop raging at my uni work I might have a look at it.
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  7. #7
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    Z= \int\sqrt{4+9x^2}dx= x\sqrt{4+9x^2}- \int{x}d(\sqrt{4+9x^2})=
    =(1)- \int\frac{9x^2dx}{\sqrt{4+9x^2}}=(1)- \int\frac{9x^2+4-4}{\sqrt{4+9x^2}}dx=
    =(1)-Z+4 \int\frac{dx}{\sqrt{4+9x^2}}
    then we get
    2Z= x\sqrt{4+9x^2}+4 \int\frac{dx}{\sqrt{4+9x^2}}
    the last int is standart.
    Last edited by zzzoak; March 31st 2010 at 02:53 PM.
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