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Math Help - Integral arcsin

  1. #1
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    Integral arcsin

    Integral arcsin-imageuploadedbytapatalk-21370464403.048676.jpgbook answer is -.134
    They use u=1-x^2

    But what is wrong with what I did?
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  2. #2
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    Re: Integral arcsin

    Is this the actual problem?

    \int_{\frac{-1}{2}}^0 {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}}
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  3. #3
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    Re: Integral arcsin

    Quote Originally Posted by Plato View Post
    Is this the actual problem?

    \int_{\frac{-1}{2}}^0 {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}}
    What is the difference? Aren't they the same.
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  4. #4
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    Re: Integral arcsin

    By taking v= x^2, you wind up with \int \frac{dv}{\sqrt{1- v}} NOT \int\frac{dv}{\sqrt{1- v^2}}dv

    (Plato wasn't saying there was a difference, he was asking you to verify what your problem is.)
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  5. #5
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    Re: Integral arcsin

    Yes he wrote it correctly. The book uses u=1-x^2

    I'm just confused because I did it a different way and got a different answer. What did I do wrong?
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    Re: Integral arcsin

    Quote Originally Posted by minneola24 View Post
    Yes he wrote it correctly. The book uses u=1-x^2

    I'm just confused because I did it a different way and got a different answer. What did I do wrong?
    You can do this without that blasted u-substitution

    What is the derivative of -\sqrt{1-x^2}~?
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  7. #7
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    Re: Integral arcsin

    I understand you can do it without the u sub I did, but if let's say this is all that came to your mind at the time what did I do wrong in my work?
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  8. #8
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    Re: Integral arcsin

    Anyone have an idea?

    Thanks
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  9. #9
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    Re: Integral arcsin

    Quote Originally Posted by minneola24 View Post
    I understand you can do it without the u sub I did, but if let's say this is all that came to your mind at the time what did I do wrong in my work?
    You need to put the whole integrand in terms of u. So (avoiding the question of the limits)
    \int \frac{x}{\sqrt{1 - x^2}}~dx = \int \frac{1/2}{\sqrt{1 - u}}~du

    You can't call u \sqrt{x} when you do the u integral.

    -Dan
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