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First year calculus proving two formulas are equivalent with natural logs

  1. Dec 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the two formulas are equivalent
    integral [sec x dx] = ln|sec x + tan x| + C
    integral [sec x dx] = -ln|sec x - tan x| + C

    2. Relevant equations

    Pythagorean ID's?

    Log rule of addition

    3. The attempt at a solution
    Well, I realized the formulas can only be equivalent if

    -ln|sec x - tan x| + C = ln|sec x + tan x| + C

    I put the constants on one side, ln on the other:

    ln |sec x + tan x| + ln| sec x - tan x| = C

    Then,

    ln |(sec x + tan x) * (sec x - tan x)| = C

    ln |(sec x)^2 - (tan x)^2| = C

    Using pythagorean ID's,

    ln |1| = C

    But what now, 0 = C doesn't show the two identities are equivalent... am I also supposed to assume in

    -ln|sec x - tan x| + C = ln|sec x + tan x| + C

    that the constants are equivalent so that

    ln |sec x + tan x| + ln| sec x - tan x| = 0 ???

    Because then ln |1| = 0 ??

    I thought that constants must be different and that C-C doesn't necessarily = 0 so you can't assume that..

    Did I approach this whole thing wrong? Even if ln |1| = 0, how would that prove that

    -ln|sec x - tan x| + C = ln|sec x + tan x| + C it simplifies to ln |1| = 0, but just because the simplified result is a true statement (possibly not because what if ln |1| = C does that prove the original equations are equivalent? or .....?
     
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  3. Dec 17, 2011 #2

    vela

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    You really should use different symbols for the arbitrary constants.
    What you've shown is that if [itex]\ln |\sec x + \tan x| + c_1 = -\ln |\sec x - \tan x| + c_2[/itex], then c1=c2. The problem is you don't know if your assumption is true. In fact, that assumption is what you're trying to prove in this problem.

    What you actually want to show is that [itex]\ln |\sec x + \tan x|[/itex] and [itex]-\ln |\sec x - \tan x|[/itex] differ at most by a constant.

    Try something like
    [tex]\ln |\sec x + \tan x| = \ln \left|(\sec x+\tan x)\frac{\sec x-\tan x}{\sec x-\tan x}\right| = \cdots[/tex]
     
  4. Dec 17, 2011 #3

    SammyS

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    To show that [itex]\ln |\sec(x) + \tan(x)|[/itex] and [itex]-\ln |\sec(x) - \tan(x)|[/itex] differ by at most a constant, you could subtract one from the other an show that's a constant.

    Alternatively, to show that each is an anti-derivative of sec(x), take the derivative of each.
     
  5. Dec 17, 2011 #4
    But by your method then they WOULD differ and wouldn't be equal, if they differed by a constant...? Also, technically didn't I show that C1=-C2 and C2=-C1? Because ln (1) = C1+C2, so then 0=C1+C2, and then C1=-C2 or -C1=C2

    Also, taking the derivative of each would give 0 and 0, since ln of anything is a constant, and a derivative of that is 0...?

    No, that wouldn't make any sense.....using a basic differentiation formula: if derivative of tan x = sec^2x that already makes no sense... shouldn't derivative of tan x just be 0 then? Haha now I'm even more confused than I was before. Sorry guys I'm only in high school, I need a little more explanation than that. If not I can sit on it for a while but I don't feel enlightened at all.. for now.. but I'll think about it some more

    Oh also I googled the question and someone answered it like this, which is part of what I based my answer on :

    Equivalent if and only if
    ln I sec x + tan x I = - ln I sec x - tan x I + K <-->
    ln I sec x + tan x I + ln I sec x - tan x I = K <-->
    ln[ I sec x + tan x I ·I sec x - tan x I] = K <-->
    ln[ | sec²x - tan²x|]= K <-->
    ln[| 1+tan²x-tan²x|] = K <-->
    ln1= cte But ln1=0= --> It's true -->

    They are formulas equivalent.

    I don't get the last part of his explanation though, which says ln1= cte.. what is cte?



    EDIT: I did do it with multiplying by 1, which was awesome thank you, but now my mind is confused because of all the new questions that came up.. but whatever can't know everything right?

    Also can I ask how you are able to type with that clear mathematical font? is it some program or something?
     
    Last edited: Dec 17, 2011
  6. Dec 17, 2011 #5

    SammyS

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    When you write [itex]\displaystyle \int\sec(x)\,dx=\ln\left(|\sec(x) + \tan(x)|\right) + C\,,[/itex] the constant C is arbitrary, that is to say, without any additional information, it's not possible to say what numerical value the constant has.

    So, as in this case, if you have two functions which are different, but they have the same derivative, then they can differ by at most a constant, which is the case here.
     
  7. Dec 17, 2011 #6

    vela

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    Yes, they would be different. Note that the problem didn't ask you to show that the two expressions were equal but that they were equivalent.

    I'm not sure where you got the idea that the log of anything is a constant. Eradicate that misconception from your mind.

    In the simplest form of a proof, you start with your given assumptions, follow logically correct reasoning, and end up with what you were trying to show.

    To save some typing, I'm going to let F(x)=ln |sec x + tan x|+c1 and G(x)=-ln|sec x - tan x|+c2. You want to show that F(x) and G(x) are equivalent, in other words, that F(x)-G(x) is a constant. You can start with

    F(x)-G(x) = (ln |sec x + tan x|+c1) - (-ln|sec x - tan x|+c2),

    which is obviously true, and then do some algebra on the righthand side to simplify it and eventually end up with

    F(x)-G(x) = c1-c2

    which is what you wanted to show. So you started with a true statement and showed that what you wanted to prove was a logical consequence.

    What you generally don't want to do is something like this: Start with

    F(x)-G(x) = k

    This is what you're trying to prove, so you don't really know beforehand if it's valid to say it's equal to a constant k. That's the first problem. Then you do some work and end up with

    c1-c2 = k

    and conclude, "This is true; therefore, what I started with must be true." This is the second problem. That argument isn't logically correct. It is possible to start with a false premise and end up with a true conclusion.

    One caveat is that this form will work if all of the steps are reversible. In that case, you can logically go from c1-c2 = k back to F(x)-G(x) = k. This is what they did in the solution you found via Google. Note that every step is separated by <-->. Those aren't there by accident. The proof writer is explicitly saying that the steps work both ways.
    https://www.physicsforums.com/showthread.php?t=546968
     
  8. Dec 18, 2011 #7
    Oh, I thought equivalent meant equal. my bad... This helped a lot thanks a ton. And yeah that makes sense that log of some things doesn't have to be a constant -.-.
     
    Last edited: Dec 18, 2011
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