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My new U substitution approach? is this legal?

  1. Feb 16, 2010 #1
    Allow me to explain my new theory, The "Mancini conjecture."

    Ok....lets say I have an integral like (4-x^2)^(1/2) dx.

    and letting u = 4-x^2, we get du/dx = -2x,

    and if I took the second derivative of du/dx....i would get -2

    this would be ideal, because I would then have du'' = -2 dx, or -1/2 du'' = dx.

    So what I'm trying to ask is, if a u substitution doesnt work out clean...that is, when i take du/dx and I end up with something that doesnt help me, what exactly is stopping me from going on, that is keep taking the next derivative...just keep taking the nth derivative until eventually I end up with a constant, which I can then pull out in front of the integral. Of course, if I do this, I would have to compensate for this somehow. I am having difficulty seeing how I would compensate for this, but I have a feeling like there is an obvious compensation that could be made. Maybe take the antiderivative of the answer you get using du'' instead of du. Could I just add an another integral sign each time i differentiate u another time? that is take the antiderivative of an antiderivative as many times as i differentiate u? Maybe I would have to differentiate u again each time I take the next derivative of du/dx . Is the integral of u du the same as the integral of u' du''?

    I have this feeling like I'm on to something, because if the exponent is positive, wouldn't I eventually get a constant after n differentiations? Wouldn't this make taking integrals super easy? Just differentiate n times until you get a constant, apply n compensations to the integral for doing that, and pull the constant out front. So once I figure out how to relate the integral of u du'' to u du, I will claim my nobel prize. My intuition is "if you take the derivative of du n times, just take the nth antiderivative of your answer." I just thought of it today doing my calculus 2 homework, and I became excited. We haven't learned integration by parts yet. I should sit down with a pencil and paper and test my intuition, but i'm a B student
    and i suck at integrals, and USA mens hockey is on soon. Let me know if I'm on to something. Thanks!
  2. jcsd
  3. Feb 16, 2010 #2


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    I can't quite figure out what you are trying to do. However when changing variables for integration purposes, only the first derivative makes sense. You are replacing dx by (dx/du)du. Higher derivatives are irrelevant.
  4. Feb 16, 2010 #3
    I don't believe that would work. I mean you couldn't really take a second indefinite integral because you must add a constant when integrating.
  5. Feb 17, 2010 #4
    Try it and see... You can easily check the result of an integration problem by differentiating the answer to see if you get the original integrand.

    In this case, as noted in another reply, it doesn't work.
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