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Integration by recognition

  1. Sep 2, 2006 #1
    I need to find the integral of the following using integration by recognition:

    Now I have the following equation (after using recognition) which I am 100% sure is correct:

    I was wondering can I multiply both sides by the following

    , is it mathematically correct? Because it would make integrating the L.H.S soo much easier and then I can just dived the R.H.S across.

    If not what do I do?
  2. jcsd
  3. Sep 2, 2006 #2
    You can multiply by it, but a product of integrals is not the integral of the product so I don't think it would help you much. The easiest way to do that integral is probably to make a substitution x=tan(u), but it's not the easiest integral, letting x=sinh(u) might be easier, but I'm not used to making hyperbolic function substitutions so I canb't judge that one.
  4. Sep 2, 2006 #3
    I did it with a much simpler example and realised that you cannot do what I was trying to do.

    I am not too sure on x=sinh(u) and how to substitute it (I am a bit lost) so you are saying replace all the x's with sinh(u) but then I end up with terms such as sqroot(sinh(u)^2+1)

    here is the full question if anyone wants:
    Last edited: Sep 2, 2006
  5. Sep 2, 2006 #4
    Ummm... Why are you integrating when you're given f and trying to find f'? Hmm.. I don't want to work out this problem but I have a feeling it's going to work out rather nicely. Is there a word missing before hence?
  6. Sep 2, 2006 #5
    If you are doing things by recognition, then go to a standard table of integrals & set the correct variables to suit the table form.

    Your form is there - I've just checked... :-)

    Gives you a rather long, nasty-looking result...

  7. Sep 2, 2006 #6
    Try to find the derivative first, then the integral should be relatively easy.
  8. Sep 2, 2006 #7
    I have nearly tried everything that I have learnt, integration by parts etc.
    Now this seems to get me somewhere, but I know there is something wrong with it (i checked some values on my calc). Does anyone know?

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  9. Sep 2, 2006 #8


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    When you do substitution, you need to change dx to du. (of course, you need to include dx in the first place)
  10. Sep 2, 2006 #9
    You didn't somehow change dx into du, so you didn't substitute correctly. Also, where did this new problem come from? Did you figure out the last one?
  11. Sep 2, 2006 #10
    Its the same problem. :smile:
    I differentiated this equation f(x):

    Then used integration by recognition to get (the sinh's ended up cancelling):

    Now if I can integrate the first term in the above equation I can find the integral of (x^2 + 1)^.5

    if I want to change dx into du I end up with 1/2x du in my equation with u's what happens with the 1/2x?
  12. Sep 2, 2006 #11
    OK I finally worked this out, and this is a very simple problem. Differentiate f(x), simplify it as much as possible and the integral you want should be obvious.
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