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Integration by recognition

  1. Sep 10, 2013 #1
    My book has an exercise on integration by recognition

    one of the questions is this:

    ## \displaystyle \int \dfrac{dx}{2\sqrt{x}\sqrt{1-x}} ##

    I can't see at all how this is by recognition
     
  2. jcsd
  3. Sep 10, 2013 #2

    Office_Shredder

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    What is integration by recognition supposed to be? I have never heard of the term before and googling suggests that it basically means 'do integrals that I don't want to explain how I figured out what they were'.
     
  4. Sep 10, 2013 #3
    Pretty much that, you should be able to deduce what the integral is by just looking at it. You're obviously supposed to know how to prove it, but it saves time I guess.
     
  5. Sep 11, 2013 #4

    sophiecentaur

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    Some integration can be done by following 'rules', such as the first one we learn - increase the exponent by one and divide by what you get - but it isn't always possible to do it that way round. Sometimes you just can't find an integral systematically but you need to 'know' the answer. Tables of Integrals are produced to help you ( and there's always Mathematica). Using 'substitution' is a way round it but it's a specialist skill and could take up all your time becoming an expert. You'd have none left for the Science.
     
  6. Sep 11, 2013 #5

    verty

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    What is the obvious substitution to make? This gets you most of the way, now find a better substitution that gets you all the way.
     
  7. Sep 11, 2013 #6

    sophiecentaur

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    Well, if you can rely on making the right substitution, you are a better man than I. And, in any case, that's down to recognition. It ends up as big boys' stuff, either way. AND very annoying and confusing for us lesser individuals.
     
  8. Sep 11, 2013 #7

    pasmith

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    "Integration by recognition" to my mind means recognising the integrand as the derivative of some function, so that the integral is equal to that function plus a constant.

    I think you're supposed to recognise that [itex]\displaystyle\frac{1}{2\sqrt{x}}[/itex] is the derivative of [itex]\sqrt{x}[/itex]. That's of the form [itex](x^n)' = nx^{n-1}[/itex], ie. fairly basic. You may also be supposed to recognise [itex]\displaystyle\frac{1}{\sqrt{1 - u^2}}[/itex] as being the derivative of a known function.

    Although only very few will instantly recognise [itex]\displaystyle\frac{1}{2\sqrt{x}\sqrt{1-x}}[/itex] as being the derivative of
    arcsin(x^(1/2))
    .
     
  9. Sep 12, 2013 #8

    sophiecentaur

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    Like I said, annoying and humbling. You're a useful guy to have around for emergency integrations, pasmith ;-)
     
  10. Sep 13, 2013 #9

    verty

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    I agree it's a silly question. The route through it is to say, probably we should have 1 - u^2 for some u, so let x = u^2, then it reduces to ##\int {du \over \sqrt{1 - u^2}}##, at which point we can see a better substitution for x. So it should be called, integration by knowing the form of integrals given to college students.
     
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