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Math Help - integral - trig substitution?

  1. #1
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    integral - trig substitution?

    Is this a trig substitution or a u-substitution problem?
    It dont seem to work out for trig substitution..

    \int \frac {x^3}{ \sqrt {1 - 4x^2}}dx

    x = sin\theta

    dx = cos\theta d\theta

    \sqrt{1-4x^2}

    \sqrt{2^2(1 - sin^2\theta)}

    = 2cos\theta

    sub back in

    \int \frac {sin^3\theta}{2 cos\theta}

    is this correct?
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  2. #2
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    Re: integral - trig substitution?

    \int \frac {x^3}{ \sqrt {1 - 4x^2}} \, dx

    u = 1-4x^2

    du = -8x \, dx

    x^2 = \frac{1-u}{4}

    -\frac{1}{8} \int \frac {x^2 \cdot (-8x)}{ \sqrt {1 - 4x^2}} \, dx

    -\frac{1}{32} \int \frac{1-u}{\sqrt{u}} \, du

    -\frac{1}{32} \int \frac{1}{\sqrt{u}} - \sqrt{u} \, du
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  3. #3
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    Re: integral - trig substitution?

    thank you i got it now
    Last edited by icelated; November 21st 2011 at 05:46 PM.
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  4. #4
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    Re: integral - trig substitution?

    You finish with the power rule. Square roots correspond to exponents of (1/2). Exponents in the denominator correspond to negative exponents in the numerator.
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  5. #5
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    Re: integral - trig substitution?

    Quote Originally Posted by icelated View Post
    How do you finish that? because i have no idea =(
    are you saying that you are unable to find the antiderivative of

    -\frac{1}{32} \int \frac{1}{\sqrt{u}} - \sqrt{u}  \, \, du

    and back-substitute (1 - 4x^2) for u ?
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  6. #6
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    Re: integral - trig substitution?

    thank you i got it finished
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  7. #7
    Super Member TheChaz's Avatar
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    Re: integral - trig substitution?

    Quote Originally Posted by icelated View Post
    thank you i got it finished
    Good deal. I think skeeter and I were starting to get worried!
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