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Math Help - Help Getting Started on: 9dx/Sqrt(81-x^4)

  1. #1
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    Help Getting Started on: 9dx/Sqrt(81-x^4)

    Not sure how to get this one going exactly, we just started getting into Integrals that really require manipulation before solving them and I believe this is definitely one of those! Thanks in advance.

    \int{9dx/\sqrt{81-x^4}}
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  2. #2
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    This seems like a rather messy problem, are you sure you have written it correctly?

    Anyway, you could rewrite the nominator as:

    \frac{9}{\sqrt{3-x} \sqrt{x+3} \sqrt{x^2+9}}

    But I akm not sure how much of a helpt this would be.
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  3. #3
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    Yeah I just went ahead and double checked, the final answer is fairly complicated, it includes a tan function as well.
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  4. #4
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    Right I'll post working as I go along... Haven't checked this yet so just making it up as I go...

    Let x = 3\sin(u).

    then dx = 3\cos(u)du.

    So out integral becomes...

    \int \frac{9 \cdot 3\cos(u)}{\sqrt{81 - 81\sin^4(u)}}du

    = \int \frac{27\cos(u)}{\sqrt{81(1 - \sin^4(u))}}du

    = \int \frac{27\cos(u)}{9\sqrt{cos^4(u)}}du

    = \int \frac{27\cos(u)}{9\cos^2(u)}du
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  5. #5
    Super Member Deadstar's Avatar
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    = \int \frac{3}{\cos(u)}du

    = 3\int \frac{1}{\cos(u)}du

    = 3\int \sec(u) du.

     = 3\ln|\sec(u) + \tan(u)|. (proof http://math2.org/math/integrals/more/sec.htm)
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  6. #6
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    Thanks a bunch, could you explain the first part to me? When we do trig subs our professor always makes us use a triangle like noobs so I'm a little confused.

    (Anyone else could chime in as well!)

    Thanks again!!!
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