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Tough one

  1. Aug 21, 2006 #1

    lo2

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    I find this one tough so could anyone please help me solving this intregral:

    [tex]\int{\sqrt{x^2-1}}dx[/tex]

    I would like the procedure.
     
  2. jcsd
  3. Aug 21, 2006 #2
    A trig substitution seems to be the best method, and I guess the best substitution would be to let x = sec(y), but I think that would get a bit messy. Maybe a hyperbolic function substitution would work better is you try letting x = cosh(y).
     
  4. Aug 21, 2006 #3

    lo2

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    Yeah it is a bit tricky.
     
  5. Aug 21, 2006 #4
    Well what are you having trouble with? We could all be more help if you tell us where you are getting stuck. And I've thought it through that this shouldn't really be a very difficult problem if you use the second substitution I recommended above letting x = cosh(y).
     
  6. Aug 21, 2006 #5
    I've seen this before. Ok first keep in mind that you are integrating a semicircle...so...you can cheat and just geometrically find the area...or you can use a u-substitution.

    y=sqrt(x^2-1) can be worked backwards into

    y^2=x^2-1
    y^2-x^2=-1
    x^2-y^2=1

    In "conic Sections" you learn this as being a hyperbola. It would be a circle if it had a +. But you only deal with half of it so circle or hyperbola is same thing...the answer would be in terms of Pi but integrals are not exact either...

    you can continue from here if you choose this way.
     
    Last edited: Aug 21, 2006
  7. Aug 21, 2006 #6
    No. It is a hyperbola not a semicircle, and this is why the hyperbolic substitution will work best. Also as far as we know this is an indefinite integral so that makes it a bit harder to interpret as an area.

    I said above that we can't really treat this as an area to get an answer, but if this were a definite integral we integrate and find the area under the curve, and a hyperbola looks nothing like a circle so that doesn't work. Also what do you mean integrals aren't exact? When we write pi we mean the irrational number exactly not some approximation of it so how is that not exact?
     
  8. Aug 24, 2006 #7
    But it's a half only. It's not Positive and negative so it's actually a parabola that forms a semicircle. There is no Semihyperbola...

    It's not both Plus and Minus so it can't be a hyperbola. The equation of a hyperbola is not a function. It fails the Vertical Axis text...unless it's an oblique one like graph of 1/x but this is a straight parabola, forming a semicircle with the x-axis...
     
  9. Aug 24, 2006 #8

    HallsofIvy

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    ??? A "parabola that forms a semicircle"???

    The function [itex]y= \sqrt{1- x^2}[/itex] gives, squaring both sides, [itex]y^= 1- x^2[/itex] or [itex]x^2+ y^2= 1[/itex], the equation of a circle. Since the y in [itex]y= \sqrt{1- x^2}[/itex] must be positive, that would give a semicircle.

    The function [itex]y= \sqrt{x^2- 1}[/itex], being discussed here, gives [itex]y^2= x^2- 1[/itex] or [itex]x^2- y^2= 1[/itex], the equation of a hyperbola, center at the origin, having y= x and y= -x as asymptotes. Since y must be positive in the original equation, the graph is two halves of the two parts of a hyperbola. Not a "parabola", not a "semicircle". Robokapp, please check that you are using the right definitions before you contradict people.
     
  10. Aug 24, 2006 #9
    ummm has any 1 solved it. if not isnt it just
    1/3 * 1/(2x) (x^2 - 1)^(3/2)
    = 1/(6x)(x^2 - 1)^(3/2)

    correct me if im wrong :)
     
    Last edited: Aug 24, 2006
  11. Aug 24, 2006 #10
    [tex]
    \displaystyle \int\displaystyle \sqrt{x^{\displaystyle2}-a^{\displaystyle2}}\,dx =
    x \frac {\sqrt{x^2-a^2}} {2} - \frac {a^2}
    {2}\ln\left(x+\displaystyle \sqrt{x^{\displaystyle2}-a^{\displaystyle2}}\right)
    [/tex]

    from a table look up at http://www.sosmath.com/tables/integral/integ12/integ12.html
     
    Last edited: Aug 24, 2006
  12. Aug 24, 2006 #11

    uart

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    You're wrong. Differentiate your answer making sure you use the quotient rule and you'll soon see why.

    I think the best approach would be to rewrite the integrand as,

    [tex]\sqrt{x^2 - 1} = \frac{x^2}{\sqrt{x^2 - 1}} - \frac{1}{\sqrt{x^2 - 1}} [/tex]

    I'm pretty sure the first term is amenable to integration by parts and the second term is something like an inverse sinh or cosh. Are you sure the original integrand wasn't supposed to be sqrt(1-x^2) instead of sqrt(x^2-1)? It's just that your integrand isn't defined on the reals around the origin, not that that's necessarily a problem.
     
    Last edited: Aug 24, 2006
  13. Aug 24, 2006 #12

    uart

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    ok, working it out as outlined above I got the integral equal to,

    [tex]I = \frac{ x \sqrt{x^2 - 1} - {\rm acosh}(x)}{2}[/tex]

    Note that here "acosh" represents the inverse hyperbolic cosine.
     
    Last edited: Aug 24, 2006
  14. Aug 24, 2006 #13

    uart

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    The solution using the above decomposition and integrations by parts was kind of "curly" so I'll post the details in case anyone interested.


    I started with
    [tex]I = \int \sqrt{x^2 - 1}\, dx = \int \frac{x^2}{\sqrt{x^2 - 1}}\, dx \ - \ \int \frac{1}{\sqrt{x^2 - 1}} \, dx [/tex]

    Using the substitution, x=cosh(u), it's easy to see that the integral of the second term is acosh(x).

    So the overall integral can be expressed as,

    I = I1 - acosh(x) ,

    where I1 represents the integral of the first term (the x^2/sqrt(x^2-1) term).

    Hitting I1 with integration by parts gives

    [tex]I1 = \int x^2 (x^2-1)^{-1/2} \, dx[/tex]
    [tex]I1 = \int x \, \frac{d}{dx}\left(\sqrt{x^2-1}\right) \, dx[/tex]
    [tex]I1 = x \sqrt{x^2-1} - I [/tex]

    So when you put it all together you get

    [tex]2I = x \sqrt{x^2-1} - acosh(x) [/tex]
     
    Last edited: Aug 24, 2006
  15. Aug 24, 2006 #14

    VietDao29

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    There's another way though. In general, say you want to integrate:
    [tex]\int \sqrt{x ^ 2 + \alpha} dx[/tex], where [tex]\alpha \in \mathbb{R}[/tex]
    First, we'll try to integrate:
    [tex]\int \frac{dx}{\sqrt{x ^ 2 + \alpha}}[/tex]
    Now, we let:
    [tex]x + \sqrt{x ^ 2 + \alpha} = t[/tex]
    [tex]\Rightarrow \left( 1 + \frac{x}{\sqrt{x ^ 2 + \alpha}} \right) dx = dt[/tex]
    [tex]\Rightarrow \frac{x + \sqrt{x ^ 2 + \alpha}}{\sqrt{x ^ 2 + \alpha}} dx = dt[/tex]
    [tex]\Rightarrow \frac{t}{\sqrt{x ^ 2 + \alpha}} dx = dt[/tex]
    [tex]\Rightarrow \frac{1}{\sqrt{x ^ 2 + \alpha}} dx = \frac{dt}{t}[/tex]
    Now, integrate both sides, we have:
    [tex]\Rightarrow \int \frac{1}{\sqrt{x ^ 2 + \alpha}} dx = \int \frac{dt}{t} = \ln |t| + C = \ln |x + \sqrt{x + \alpha}| + C[/tex]
    ------------------
    Now, we'll integrate
    [tex]\int \sqrt{x ^ 2 + \alpha} dx[/tex] by parts:
    Let [tex]u = \sqrt{x ^ 2 + \alpha} , \quad \mbox{and} \quad dv = dx[/tex]
    [tex]\Rightarrow du = \frac{x dx}{\sqrt{x ^ 2 + \alpha}} , \quad \mbox{and} \quad v = x[/tex]
    [tex]= x \sqrt{x ^ 2 + \alpha} - \int \frac{x ^ 2}{\sqrt{x ^ 2 + \alpha}} dx[/tex]
    [tex]= x \sqrt{x ^ 2 + \alpha} - \int \frac{x ^ 2 + \alpha - \alpha}{\sqrt{x ^ 2 + \alpha}} dx = x \sqrt{x ^ 2 + \alpha} + \alpha \ln |x + \sqrt{x ^ 2 + \alpha}| - \int \sqrt{x ^ 2 + \alpha} dx[/tex]
    Now let [tex]I = \int \sqrt{x ^ 2 + \alpha} dx[/tex], we have:
    [tex]I = x \sqrt{x ^ 2 + \alpha} + \alpha \ln |x + \sqrt{x ^ 2 + \alpha}| - I[/tex]
    [tex]\Rightarrow I = \frac{x \sqrt{x ^ 2 + \alpha}}{2} + \frac{\alpha}{2} \ln | x + \sqrt{x ^ 2 + \alpha} | + C[/tex]
    Ok, can you get it? :)
     
  16. Aug 24, 2006 #15

    uart

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    Well if you look at what you've done it really was pretty much the same as what I did. Only the first step in your solution {to get the integral of 1/sqrt(x^2+a) } was different. But personally I prefer to use a sinh or cosh substitution (as appropriate depending on the sign of "a") to do that one. I think it's easier.

    BTW. Note that ln(x + sqrt(x^2-1)) is just a closed form expression for acosh(x). So the solutions given by vietdao, stefan and myself are all equivalent.
     
  17. Aug 24, 2006 #16

    VietDao29

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    Yes, I agree to your opinion. Using cosh, or sinh substitution is much faster. However, what if the OP hasn't learnt hyperbolic functions?
     
  18. Aug 25, 2006 #17
    HallsOfIvy all I'm saying is since it's in a y= form and it's a function it cant be a hyperbola. A hyperbola has 2 parabolas...therefore the formula includes 2 functions. I only learned 4 conic sections:

    Circle
    Parabola
    Hyperbola
    Ellipse

    Out of these 4, Parabola is only one that can be expressed as one function...meaning it passes the "Vertical line" test.

    Here in this problem we have a function. it's in a y= form and we are antiderivatizing it. So it is a continuous function.

    A hyperbola is not a function...it's 2 functions to my knowledge.

    Now the way you form a circle...or a hyperbola is by having 2 parabolas. Correct? Well here we have one...so it can be a part of a circle or of a hyperbola. It doesn't matter because the other part is not something we are looking at.

    x^2-y^2=1 is the equation of a hyperbola. Agreed 100%. But...when we put it in y= we dont consider the negative...because we are given the problem in the y= form...showing we only want positive.

    So by considering half of a hyperbola...wouldn't we consider a parabola?
    And does that parabola not look like ... a parabola that would form a circle if theere would be its negative counterpart?

    All i'm trying to do (i know i go way too far and i'm sorry for that) is try to set it up in terms of Pi. I don't know anything about hyperbolas. I dont even know if they have areas. I don't know if parabolas do either...but i know circles and semicircles do. I'm just relating it to something I can work with.

    well I guess I cant explain it. I know my reasoning makes snese to me, I know I can produce the right answer but...i dont know why. :(

    Sorry for all this mess.
     
  19. Aug 25, 2006 #18
    First off, I don't really mean to be offensive, but you really need to stop posting gibberish that you think might work as opposed to mathematically sound reasoning that makes logical sense. You have done this in numerous threads and always been corrected, if you're not sure you shouldn't just write out random math that could confuse the person trying to get help.

    Explain to me how you can make a circle out of 2 parabolas? A parabola and half of a hyperbola are definitely not the same thing, it's late and I don't want to work this out but it isn't difficult so I'll leave it to you to see why, take the hyperbola y2-x2=1 solve for y taking the positive square root and compare this to the parabola y=x2+1 and you should see the obvious differences, it might be better to take both general equations solved for y and then compare but my example should help you see why half a hyperbola isn't a parabola.

    This is rather ambiguous, if you take the integral of the expression for a parabola or hyperbola you will get the area under the curve, but you can easily adjust this to find the area bounded by a hyperbola or parabola.
     
  20. Aug 25, 2006 #19
    I am not writing gibberish on purpose...when I write my paragraphs I am sure that what I say is at least worth reading. I'm being told it's not, and I'm trying to fix that, by asking questions which create more confusion.

    Now how is a parabola a half circle part:

    -Edited-

    I worked the math...I was saying a big stupidity. I feel awful now... I was not doing an operation to both sides of the equal as I should have. I was getting one power less on y.
     
  21. Aug 25, 2006 #20
    I could have sworn there was another post here this morning in response to mine, and I would like to apologize I realize what you were trying to do. You took a shot at what you thought made sense and that is reasonable, so I guess all that I really have to say is maybe you should think through the math and try your best to be sure it's correct before you post it. But again I apologize for saying that what you wrote was gibberish, and that other post I saw this morning was absolutely correct that noones work really deserves to be called that.

    I'm glad you can see the mistake you made, and again I'm sorry for my earlier post, but I see you've learned from that as we all should do.
     
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