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How to evaluate an indefinite integral

  1. Dec 25, 2007 #1
    how to evaluate the indefinite integral [tex]\int \frac{1}{\sqrt{x^2-1}} dx[/tex]
    Last edited: Dec 25, 2007
  2. jcsd
  3. Dec 25, 2007 #2


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    Do you have any thoughts of your own?
  4. Dec 25, 2007 #3


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    Is this a test for you or for us?
  5. Dec 25, 2007 #4
    Actually i have solved this problem by using the substitution x=sec u. i'm just thinking if there is any other way to solve this problem?

    and another question is how can I delete my post if I find it not worth discussion any more? (I'm new to this forum)

    Thanks a lot.
    Last edited: Dec 25, 2007
  6. Dec 26, 2007 #5


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  7. Dec 26, 2007 #6
    cant solve it using x=sec(theta)as it becomes

    [tex]\int \frac {sec^2 \theta}{tan \theta}d \theta [/tex]

    and further simplifyin i get

    [tex]2\int sin^{-1}2\theta[/tex]
  8. Dec 29, 2007 #7
    so, you are getting:

    2\int \frac{1}{sin 2\theta} d \theta = 2\int cosec 2\theta


    2\int cosec 2\theta d \theta = 2 \times \frac{1}{2} \times log|tan(\theta)| = log|tan(\theta)| + c
  9. Dec 30, 2007 #8
    yeah thats right. i wouldve attepmted the trig integral but have no idea how to even start them other.

    my first guess would be the integral of the cosec expansion, from the maclaurin series
  10. Dec 30, 2007 #9


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    the way this old game is played is to use trig substitutions and try to show the integral agrees with some elementary combination of trig and exponential functions. but that is just a classical game, with few important consequences.

    I think a more intelligent and practical real life solution is yours, using a power series to approximate the value, and similarly that of hundreds of others which do not yield to the tricks of the classical game.
  11. Dec 30, 2007 #10
    Here is a small and effective proof for this:


    \int cosec(\theta) d\theta = \int\frac{cosec(\theta)(cosec(\theta) - cot(\theta))}{cosec(\theta) - cot(\theta)} d\theta


    If you differentiate the denominator in this question, you get the numerator. So, this results to:

    log|cosec(\theta) - cot(\theta)|

    Having done this, you can easily show using trignometry that:

    cosec(\theta) - cot(\theta) = tan(\frac{\theta}{2})

    I guess this is what mathwonk referred to as the 'classical' game.. well.. this is a way.. which works for some cases.. in others the power series is the way to go :D
  12. Jan 3, 2008 #11

    Gib Z

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    There was no need for ANY of that :( Even if you don't choose the hyperbolic substitution, you should have stopped at [tex]\int \frac {sec^2 \theta}{tan \theta}d \theta [/tex]. Tell me you see the easier way to do that!!
  13. Jan 3, 2008 #12
    Lol. Gib Z is totally right.
    The same answer is obtained with a simple u sub of the tan function, resulting in the same answer.

    However I still like rohanprabhu way and the power series way also.
    But its a good thing to learn how to get the same answer multiple ways. This of course depends on your patience.
  14. Jan 3, 2008 #13
    OMFG.. lol :D

    I just didn't look at that.. I just looked at [itex]\int \frac{1}{sin(\theta)}[/itex] and continued with it.. I didn't pay attention to [itex]\int \frac {sec^2\theta}{tan \theta}[/itex].. which ofcourse as u said.. is of the form [itex]\int \frac {f(x)}{f'(x)}dx[/itex]...

    Last edited: Jan 3, 2008
  15. Jan 4, 2008 #14
    hi, lazypast, actually the original problem does not become [tex]\int\frac{sec^{2}\theta}{tan\theta}d\theta[/tex]by using x=sec(theta), it becomes [tex]\int sec\theta d\theta[/tex]

    however anyway, lazypast's little mistake has led to an inspiring discussion on how to deal with integration problems. and I really appreciate the power series approximation way to do it.
  16. Jan 9, 2008 #15
    i didnt actually notice [itex]\int \frac {f(x)}{f'(x)}dx[/itex] could be applied to the sec^2/tan till you just said it.

    and since i looked in more detail to it, the result is in terms of theta. and the original expression was in terms of x.

    as we said x=sec(theta), replacing the answer to in terms of x instead of theta becomes

    ln(tan(sec^-1(x)) +c

    an edit- just seen [itex]tan(sec^{-1}(x)) = sinx[/itex] ??
    Last edited: Jan 9, 2008
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