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How to solve int. (sqrt(x^2-1))dx

  1. Dec 15, 2005 #1
    how to solve the integral: int. (sqrt(x^2-1))dx
     
  2. jcsd
  3. Dec 15, 2005 #2
  4. Dec 16, 2005 #3

    Curious3141

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    A plain trig substitution makes the problem fairly hard. Try instead a hyperbolic trig sub. Hint x = cosh(u)
     
  5. Dec 16, 2005 #4
    I think trig substituion works fine. I'll start you out. Like Yann suggested, draw a right triangle with the horizontal leg being 1, the other leg [itex]\sqrt{x^2-1}[/itex] and the hypotenuse x. Then make a substitution for x, which is [tex]x=sec(\theta)[/tex]. You must replace x and dx for this integral to make a correct substitution. So now we get that [tex]dx=sec(\theta)tan(\theta)d\theta[/tex]. Plug in your values, play around with trig identities and you should be fine!
     
  6. Dec 16, 2005 #5
    integral(x^2-1)^(1/2)dx

    just rewrite it like that and I think you will se how to solve it. Remember to divide with the inner derivate of the function. Unless Im missing something in the formulation of the question no variable substitution is needed.
     
  7. Dec 16, 2005 #6
    Azael, this cannot be done by u-substitution because the derivative has a 2x involved. Trig substitution, hyperbolic substitution, or some other form must be used.
     
  8. Dec 16, 2005 #7
    ohh yes your right. I se what a stupid error in thinking I did. I must blame it on the late night lol :). Nevermind my post!!!
     
  9. Dec 17, 2005 #8

    Curious3141

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    I'm sorry, but I would like a little clarification on this. I get [tex]\int{\tan^2\theta \sec\theta}d\theta[/tex] How do you integrate that "easily" ?
     
  10. Dec 17, 2005 #9
    Oops, that isn't done easily. My fault. You did the correct substitution, but for this integral there is no trig inverse in the answer, so it isn't necessary to use. I did the integral on Mathematica, and now I'm thinking by parts is the best way, with u=[itex]\sqrt{x^2-1}[/itex] and dv=1. If anyone else has a more elegant way, please speak up.
     
  11. Dec 17, 2005 #10
    The easiest way imo(now when I am fully awake lol) to solve F(x)=int. (sqrt(x^2+a))dx is to rewrite it as
    F(x)=int.(1*sqrt(x^2+a))dx

    Now partialy integrate over that and you get F(x)=[x*sqrt(x^2+a)]-int.((x^2+a-a)dx/sqrt(x^2+a))

    With some meddling with the integral you can split it into 2 integrals. One is identical to the begining integral(in other words its the same as F(x)) the other is quite easily solvable by a variable substitution. That way one doesnt have to use any trigonometric substitutions.
     
  12. Dec 17, 2005 #11

    Curious3141

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    Yeah, I was pretty sure a straight trig sub would get nowhere fast.

    The hyperbolic sub is the fastest "elegant" way.

    Substitute [tex]x = \cosh u[/tex], keeping in mind that [tex]d(\cosh u) = \sinh u du[/tex] and that [tex]\cosh^2 u - 1 = \sinh^2 u[/tex]
    [tex]\int{\sqrt{x^2 - 1} dx} = \int{\sinh^2 u} du = \frac{1}{2}\int{(\cosh 2u - 1)}du = \frac{1}{2}[\frac{1}{2}\sinh 2u - u] + const = \frac{1}{2}[\sinh u \cosh u - u] + const = \frac{1}{2}[x\sqrt{x^2 - 1} - arccosh(x)] + const [/tex]

    You can see from the fact that there is an inverse hyperbolic trig function in the result that a plain trig sub probably won't work.

    I'll leave it as an exercise to the orig. poster to differentiate that and verify that it works out (it does).

    Interestingly, there is a way to do this with normal trig, transforming into hyperbolic trig with a complex substitution.

    You can quite easily evaluate [tex]\int{\sqrt{1 - x^2}}dx[/tex] with a simple substitution like [tex]x = \cos \theta[/tex] Now observe that [tex]\int{\sqrt{x^2 - 1}}dx = (i)\int{\sqrt{1 - x^2}}dx[/tex] and use the relationship between circular and hyperbolic trig in the complex plane to work it out.

    This is the way I initially tried to solve the problem, but I later decided the straight hyperbolic sub was easiest. But this way helps you conceptualise the problem better and avoid mistakes like using wrong straight trig subs.
     
    Last edited: Dec 17, 2005
  13. Dec 17, 2005 #12
    If you have a square root and the coefficient of the x^2 term is positive then hyperbolic substitution is the best way IMO. The reason is that hyperbolic substitutions (usually) require less knowledge of integrals than when you use trigonometric substitutions. I'd much rather integrate "1" than something like tan^2(t)sec(t).
     
  14. Dec 17, 2005 #13
    I didn't look at the original integral but if after trig substitution there is a secant and a tangent squared term substitute again, secant is u du is tangent squared and it doesn't get much easier than integrating u. You could also try polar coordinates.
     
  15. Dec 17, 2005 #14

    GCT

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    the derivative of sec@ is sec@tan@.

    nice suggestions so far, yet, the integral above is definitely managable....transform to everything to secant, distribute, and you'll have two separate integrals

    [tex] \int sec \theta d \theta~and~ \int sec^3 \theta d \theta [/tex]

    euler's "trick" can be applied to the first, the second is somewhat of a table integral (in that a form exist for all powers of the secant integral).
     
  16. Dec 17, 2005 #15

    Curious3141

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    Try it, show the working. It isn't as easy as you make it out to be. The integral of secant is an unwieldy log function. Differentiating tangent gives you secant^2. The product of those two is not easy to integrate. Doing it the other way is no easier.

    I mean, I could be wrong here, but it just doesn't strike me as being an easy way to do the problem. I'd like you to clarify by showing the working with your method, please. :)
     
  17. Dec 17, 2005 #16

    Curious3141

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    Again, integrating secant requires a little insight to multiply by the correct product to get it into a manageable form, but it's not that difficult. When you want to use that result and integrate by parts to integrate secant cubed, the tediousness just adds up.

    I know how to do all this, it's just that it's fairly tedious and certainly harder than a simple hyperbolic sub. I never said the circ trig sub was impossible, just that it was hard. And I still think it is.

    In fact, there's a math page out there with exactly this method worked out. He doesn't show a lot of intermediate steps so it looks easier than it actually is. http://www.mathreference.com/ca-int,trigsub.html

    IMHO, much easier and more straightforward to use hyperbolic trig here.
     
  18. Dec 18, 2005 #17

    VietDao29

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    Hmmm, I don't think we should integrate any secants or integrate by parts.
    [tex]\int \tan ^ 2 t \sec t dt = \int \frac{\sin ^ 2 t}{\cos ^ 3 t} dt = \int \frac{\sin ^ 2 t}{\cos ^ 4 t} d(\sin t) = \int \frac{\sin ^ 2 t}{(1 - \sin ^ 2 t) ^ 2} d(\sin t) = \int \frac{\sin ^ 2 t}{(1 - \sin t) ^ 2 (1 + \sin t) ^ 2} d(\sin t)[/tex].
    Partial-fraction it gives:
    [tex]\int \left( -\frac{1}{4 (\sin t + 1)} + \frac{1}{4 (\sin t + 1) ^ 2} + \frac{1}{4 (\sin t - 1)} + \frac{1}{4 (\sin t - 1) ^ 2} \right) d(\sin t)[/tex]
    Not very tedious as you probably think... o:)
     
  19. Dec 18, 2005 #18

    Curious3141

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    I have to say that is an excellent and insightful way to do it. I didn't see that initially. Very nice. :approve:
    Of course, simplifying the result takes a bit of work, whereas the hyperbolic trig way gives you an immediately usable expression. :cool:
     
  20. Dec 18, 2005 #19

    GCT

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    Why are we going with partial functions when you can distribute the secants, do any of you realize that there's a specific table integral form to solve secant functions of any power? Look in the back of your calculus text and find the form, apply the form and there you go.
     
  21. Dec 18, 2005 #20

    Curious3141

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    But that wouldn't be "fundamental" or "from first principles" or whatever you want to call it. It's nice to know how it's being done.:smile:
     
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