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Having trouble with an integral

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫√(x^2 - t^2)dt/(xt) x > t > 0
    2. Relevant equations



    3. The attempt at a solution

    So I noticed that the integrand had the form a^2 - b^2x^2, and I can apply trig substitution, so I did this:

    t = xsin(θ), dt = xcos(θ), and therefore, x^2 - t^2 = x^2 - x^2sin(θ)^2.
    The last formula can be rearranged into x^2 cos(θ)^2 (From the identity 1 - sin(θ)^2)
    After simplification, I obtain the integral
    ∫cos(θ)^2 dθ/sinθ

    From here, I don't know where to go. After rearranging multiple times, there is no integral and I keep getting (cosθ) - ∫cscθdθ.

    Thank you for any clarity you can provide for me.
     
  2. jcsd
  3. Feb 7, 2014 #2
    ∫cscx dx = -ln( csc x + cot x)
     
    Last edited: Feb 7, 2014
  4. Feb 7, 2014 #3

    Dick

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    I think you should get (cosθ) + ∫cscθdθ. And ∫cscθdθ can be done. You can look it up if nothing else.
     
  5. Feb 7, 2014 #4
    That's actually where i'm stuck, because after the cosecant integral, this is for a multiple choice question and there is no choice for the - ln(csc x + cot x), so I was just wondering if there was any other way to do this.
     
  6. Feb 7, 2014 #5

    Dick

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    What are your choices? You aren't done when you express it in terms of ##\theta##. You probably want to express it in terms of t.
     
    Last edited: Feb 7, 2014
  7. Feb 7, 2014 #6
    Yup, but none of the choices have the term "ln" in it, and there's also a "none of the above" option, but i'm always nervous about picking that type of option.

    I have:
    (x^2 - t^2)^(3/2) / (3t) + C
    t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C
    - ((t-x)(t+x))^(3/2) / (3t) + C
    and (x^2 - t^2) + t tan(t/ sqrt(x^2-t^2))^-1 / t + C
     
  8. Feb 8, 2014 #7

    haruspex

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    I interpreted that as ##\int_0^x\frac{\sqrt{x^2-t^2}}{xt}dt##, but seeing the choice of answers it must mean ##\int^t\frac{\sqrt{x^2-z^2}}{xz}dz##
    But that's a different x. You mean - ln(csc θ + cot θ) where t = x sin θ.
     
  9. Feb 8, 2014 #8
    Yeah I just have a habit of using x as my variable, but where did the z come from? I also don't know why they provided the x > t > 0, I assumed that they provided that information so that I would know the denominator wouldn't be 0, but does it have another meaning?
     
  10. Feb 8, 2014 #9

    Dick

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    I don't think x > t > 0 is supposed to be a range of integration. The answers are indefinite integrals. It's just to insure the square root is defined and there no division by zero.

    You could try differentiating those to see if they give you anything like you want. But I think you are pretty safe with the "none of the above".
     
  11. Feb 8, 2014 #10

    haruspex

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    Isn't that what I said? It looked like a range of integration until I saw the choice of answers.
    ##\int^t f(t).dt## is a "pun". The t in the upper bound is different from the other two t's, which are 'dummy variables'. In the present problem it seemed to me that could be a source of confusion, so I tried to make it clearer by using a different dummy variable.
     
  12. Feb 8, 2014 #11

    Dick

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    I guess I'm not sure what you are trying to say. Given there is a +C in the answers it's just a plain old indefinite integral without limits. ##\int \frac{\sqrt{x^2-t^2}}{xt}dt##, isn't it? I've never seen the notation with only an upper limit.
     
  13. Feb 8, 2014 #12

    haruspex

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    All of the proposed answers have t in them. I don't know how to get that in there without putting a t in a bound. When someone writes ∫cos(x).dx = sin(x), as they do, what is really meant is ∫x cos(t).dt = sin(x). It's still an indefinite integral because there is no lower bound specified.
    Re the offered answers, I suspect t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C is to be read as ##t (\sqrt{x^2 - t^2} + t \arctan(\frac t{\sqrt{x^2 - t^2}}) + C##, and similarly the fourth choice. (The parentheses don't match up, so I'm still not sure how to interpret it.) I mention this because arctan and ln are connected: arctan(x) = ln((1-x)/(1+x))/(2i), or something like that.
     
  14. Feb 8, 2014 #13

    Dick

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    Well, I would say ##\int x \cos(t) dt = x sin(t) +C##, and that has a ##t## in it. Sure, there's a relation between arctan and log, but I don't see any i's around.
     
  15. Feb 8, 2014 #14

    haruspex

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    That only works if you interpret ##\int x \cos(t) dt## as meaning ##\int^t x \cos(t) dt##, and that in turn is exactly the same as ##\int^t x \cos(u) du## (since the t in the bound does not mean the same as the t in the integrand), which is shorthand for ##\int^{u=t} x \cos(u) du##.
    Agreed - just noting that the absence of a ln function in the offered solutions is not quite conclusive.
     
  16. Feb 8, 2014 #15

    Dick

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    Yeah, but they don't look right. I'd check by differentiating, but as you've noted some of them may not even be transcribed right. Hardly seems worth it. On the other hand notation like ##\int x \cos(t) dt = x sin(t) +C## is what I've seen in integral tables all my life with no need for special comment or special notation.
     
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