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Math Help - inverse trig integration

  1. #1
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    inverse trig integration

    find the integral of dx/sqrt(1-4x^2). I know I should be using sin^-1 indentity when integrating but my problem is making the the coefficient of x equal to 1. I'm doing something wrong with the algebra. Can someone please show me the working. Thank you
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  2. #2
    Super Member Random Variable's Avatar
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     \int \frac {1}{\sqrt{1-4x^{2}}} \ dx = \int \frac {1}{\sqrt{1-(2x)^{2}}} \ dx

    let u = 2x

     \frac {1}{2} \int \frac {1}{\sqrt{1-u^{2}}} \ du

     = \frac {1}{2} \arcsin(u) + C = \frac {1}{2} \arcsin(2x) + C
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  3. #3
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    Quote Originally Posted by Random Variable View Post
     \int \frac {1}{\sqrt{1-4x^{2}}} \ dx = \int \frac {1}{\sqrt{1-(2x)^{2}}} \ dx

    let u = 2x

     \frac {1}{2} \int \frac {1}{\sqrt{1-u^{2}}} \ du

     = \frac {1}{2} \arcsin(u) + C = \frac {1}{2} \arcsin(2x) + C
    Thank you, but why is a factor of 1/2 removed form the integration?
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    Thank you, but why is a factor of 1/2 removed form the integration?
    You can always move a constant outside of the integral sign. You don't have to move it outside, though.
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  5. #5
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    Quote Originally Posted by Random Variable View Post
    You can always move a constant outside of the integral sign. You don't have to move it outside, though.
    Sorry, I should have been more specific. Where has the half come from? I can't see where it was originally to then move it out the front. Its probably staring me right in the face but I just can't see where it has come from, thank you for your patients
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    Sorry, I should have been more specific. Where has the half come from? I can't see where it was originally to then move it out the front. Its probably staring me right in the face but I just can't see where it has come from, thank you for your patients
    I made the substitution  u = 2x

    so  du = 2dx  which implies  dx = \frac {du}{2}
    Last edited by Random Variable; July 10th 2009 at 07:21 PM.
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