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Thread: Integral of sin^3 (sqrt(x)) / (sqrt(x))

  1. Jan 12th 2009, 04:19 PM #1
    storage1k
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    Integral of sin^3 (sqrt(x)) / (sqrt(x))

    Hi...I've been having problems with this question.

    The Integral of [sin^3(sqrt(x)) ]/ (sqrt(x))

    Should I set sqrt(x) = u and solve from there or should I bring the sqrt(x) in the denominator up?...I'm not even sure if either approach is correct.
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  2. Jan 12th 2009, 04:38 PM #2
    Mush
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    Quote Originally Posted by storage1k View Post
    Hi...I've been having problems with this question.

    The Integral of [sin^3(sqrt(x)) ]/ (sqrt(x))

    Should I set sqrt(x) = u and solve from there or should I bring the sqrt(x) in the denominator up?...I'm not even sure if either approach is correct.
    u = sin(\sqrt{x})

    \frac{du}{dx} = cos(\sqrt{x}) \times \frac{1}{2}x^{-\frac{1}{2}}


    \frac{du}{dx} = \frac{cos(\sqrt{x})}{2\sqrt{x}}

    And the original integral can be written:

    \int \frac{sin(\sqrt{x})sin^2(\sqrt{x})}{\sqrt{x}}dx

    Better?

    PS: This might actually be useless!
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  3. Jan 12th 2009, 05:13 PM #3
    storage1k
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    Oh ....is it possible to incorporate the reduction formula for sin into this?

    The problem I'm having is with that rotten sqrt x in the denominator...even if I do bring it to the numerator, I've still got problems as to how to do it.
    Is it possible to set sqrt x = v.
    Then approach the problem as sin^3v / v ....I don't think it's possible...but I'm unsure.


    I'm not sure on your approach to the problem?
    Last edited by mr fantastic; Apr 7th 2009 at 10:53 PM.
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  4. Jan 12th 2009, 05:24 PM #4
    Chop Suey
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    Quote Originally Posted by storage1k View Post
    Should I set sqrt(x) = u
    Yes. It might clear things up.

    By substituting u = \sqrt{x}, the integral becomes:
    2 \int \sin^3{u}~du

    Take one factor of sin out and apply the Pythagorean identity ( \sin^2{\theta}+\cos^2{\theta} = 1):
    2 \int (1-\cos^2{u})\sin{u}~du

    Can you do it now?
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  5. Jan 12th 2009, 05:37 PM #5
    storage1k
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    hehe! Thank you very much!

    Worked it out quite nicely. Although I used
    ∫sin^n dx = -1/n cos x sin^(n-1) x + (n-1)/n ∫ sin^(n-2) xdx to go to it directly.
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  6. Jan 21st 2010, 04:48 PM #6
    Christian0412
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    Duh! lol Very helpful! Thanks!
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