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Help me integrate please

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data

    So while doing arc length integrals I can't figure out how to solve them. For instance:
    f(x) = 3x^2
    f'(x) = 6x
    So.. the length, L = ∫ √(1+36x^2) dx

    How do you solve this indefinite integral?

    Notice: I know you can solve this easily with bounds by approximation but how do you solve it indefinitely

    Thank you


    2. Relevant equations


    3. The attempt at a solution
     
    Last edited: Sep 21, 2016
  2. jcsd
  3. Sep 21, 2016 #2

    berkeman

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    Can you show how you would solve it with bounds?
     
  4. Sep 21, 2016 #3

    Ray Vickson

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    If you say you can solve it easily with bounds, why can't you solve it without bounds? Usually we get the "bounds" result by substituting the bounds into the "no-bounds" result!

    Anyway, arc-length problems should have bounds, since you usually want the arc length between two points ##(x_1,y_1)## and ##(x_2,y_2)## on the curve.

    Finally: you have the wrong integral. Go back and check the arc-length formulas in your textbook or course notes (or in on-line sources).
     
  5. Sep 21, 2016 #4
    I would just type it in to the good ol calculator or wolfram. And you're right, I just didn't square it. But I don't know how to solve it without approximation
     
  6. Sep 21, 2016 #5

    Ray Vickson

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    If you rely on tools like Wolfram to do elementary integrations for you, then you cannot possibly learn the subject. That sounds to me like a recipe for failure.
     
  7. Sep 21, 2016 #6

    Mark44

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    Two words that I hope will ring a bell: trig substitution
     
  8. Sep 22, 2016 #7
    Well thank you for responding, but I am a junior in high school and am currently taking Ap Calculus (CALC 1) and have not yet covered integrals/antiderivatives. However, I love math (the reason I'm a year ahead everyone else in math) and I just don't know how to take this integral. I do know how to take others though and also the applications.

    Thank you sir.
     
  9. Sep 22, 2016 #8

    epenguin

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    First thing to do is to get the right function to integrate, see #3.

    If that too is a mystery how to integrate, draw a picture of what it is on a circle, you're doing an exercise like this I think you must know the equation of a circle, and the integral will be almost obvious.
    And if you prefer to do it by substitution, or to do that as a check and exercise which is not a bad idea, you will see why anyone would hit on that particular substitution, instead of it being in the being told and 'I would never have thought of that' category.
    Then it's not finished!
    Do not leave an integral like that without differentiating it as a check. This will save you making a lot of errors, and should bring you some enlightenment in this case.
    Then it's still not finished! Do not miss this occasion to check up how to do the integral of the function that you mistakenly gave, and which is closely related to the correct one. Because doing all this will be giving you an insight and empowerment over this little area of some integrals quite important in geometry and Physics (and even chemistry some of them).
    Edit: on second thoughts having read now where you are, that last bit might involve too many things you have not done yet, but you are quite close to, so have a look and if it's too much dear it in mind for the future.

    Good luck. Don't omit to come back when you have the answer or are stuck again.
     
    Last edited: Sep 22, 2016
  10. Sep 23, 2016 #9

    Ray Vickson

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    The suggested trig substitution will not work well; you need hyperbolic functions rather than trigonometric functions for this problem.
     
  11. Sep 23, 2016 #10

    Mark44

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    How so?
    The integral in post #1 was ##\int \sqrt{1 + 6x}dx##, but was later edited by the OP to ##\int \sqrt{1 + 36x^2}dx##. After a suitable trig substitution, you end up with ##\int \sec^3(\theta) d\theta##, which can be integrated using int. by parts or by looking it up in a table of integrals.
     
  12. Sep 23, 2016 #11

    Ray Vickson

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    I was thinking of the substitution ##x = (1/6)\sinh(y)##.
     
  13. Sep 25, 2016 #12

    epenguin

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    I come back to clear up confusion I may have caused the OP. I have been a bit thrown off by the comments that he had the wrong integral - it is in fact OK isn't it? Anyway I had been led to consider the integral ∫√(1 - x2) dx. This too was not self evident to me at first glance.

    And then I was exercised by Ray's comment
    . I wonder rather if that is not something that in the next decades will go the way of not being allowed to use calculators. My experience at least was that you can get pretty skilled at this at school, and within a year you'll have totally forgotten it.

    In any case it ought to be made as easy as possible. So taking a look at the integral I quote above I realised that you can make it quite obvious by a diagram. And I have noticed this before with some other integrals that turn out to be 'trigonometrical'. But I was not told this at school, it is not in the books I have nor any I have seen.

    Asked the above integral you can justify the book method by saying 'look for some function that squared and subtracted from one is a perfect square so then you're not bugged by the square root'. The majority of us however would not get that without being told, or at least nudged. But, yes, trigonometrical functions, sine or cosine have that property. Still it depends rather on hindsight or familiarity. I think a graph that makes it obvious is better.

    Then for the integral ∫√(1 + x2) dx the hyperbolic functions have the property required in the previous paragraph when we say 'add to' instead of 'subtract from'. Diagrammatically I am not seeing it yet, due to the hyperbola being a lot less familiar than the circle. Am working on that and other approaches.

    I was however assuming the OP was not familiar with the hyperbolic functions. Actually this second integral can be done by a trigonometrical substitution:
    x = tan u. You do not need explicitly hyperbolic functions, there are other forms. It is not exactly a doddle however.
     
    Last edited: Sep 25, 2016
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