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Solve dx/y = dy/x

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi I am trying to solve an indefinite integral of the form 1/ sqrt[ (x^2 +1)] dx..




    2. Relevant equations



    3. The attempt at a solution

    Different ways of solving it are posted on the link below. I would like to know the following result :

    dx/y = dy/x = d (x+y)/ (x+y) ???

    http://math.stackexchange.com/questions/610733/how-to-integrate-int-dx-over-sqrt1-x2?lq=1

    Thanks.
     
  2. jcsd
  3. Aug 17, 2014 #2

    Zondrina

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    If ##1 = 1^2##, then denominator has the form ##\sqrt{x^2 + a^2}##.

    Make the substitution ##x = a \tan(\theta)## and use the identity ##\sec^2(x) - \tan^2(x) = 1##.
     
  4. Aug 17, 2014 #3
    You have misunderstood my post. I want to know why [itex] dx/y = dy/x = \int d (x+y) / (x+y) = ln (x+y) + C [/itex]

    Thanks.
     
  5. Aug 17, 2014 #4

    HallsofIvy

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    Probably the reason he misunderstood is that the equation you have is NOT correct. In fact it doesn't even makes sense- the first two parts are differentials, the last two are not. You need to separate it into two different equations. These are two completely different problems.

    If dx/y= dy/x then, multiplying both sides by xy, xdx= ydy. Integrate both sides to find y as a function of x.

    The other parts says that [itex]\int d(x+y)/(x+ y)= ln(x+ y)+ C[/itex]. To see that that is true, let u= x+ y and use the fact that [itex]\int du/u= ln(u)+ C[/itex]

    But the two have nothing to do with each other.
     
  6. Aug 17, 2014 #5
    This is why I am stumped.
    I am referring to the bottom most solution posted on the following link by juantheron (user): http://math.stackexchange.com/questions/610733/how-to-integrate-int-dx-over-sqrt1-x2?lq=1
     
  7. Aug 17, 2014 #6

    Zondrina

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    Observing the integral:

    ##\int \frac{d(x+y)}{(x+ y)} = \int \frac{1}{(x+ y)} d(x+y) = ln(x+ y)+ C##

    Where ##(x + y)## is being treated as the independent. Back subbing for ##y = \sqrt{1 + x^2}## will give you the answer.

    Does this alleviate the problem or is the concern something else entirely?
     
  8. Aug 18, 2014 #7
    Hey there,

    This can be done using the technique of trigonometric integration. Please try it out and let me know if you have anymore difficulty.
    Thanks
     
  9. Aug 18, 2014 #8

    vanhees71

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    Another (more convenient?) way is to substitute [itex]x=\sinh u[/itex]. As turns out, this is in fact a basic integral, involving hyperbolic functions.
     
  10. Aug 18, 2014 #9
    Guys thanks for the responses.

    My issue is still not resolved. Let me rephrase the question.

    I want to know how one can show that [itex] \int dx/y = \int d(x+y) / (x+y) [/itex]
     
  11. Aug 18, 2014 #10

    HallsofIvy

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    You can't. Unless there is some additional relation between x and y you haven't told us, such as "x= y", that equation is not true. What is true is that [itex]\int dy/y= \int d(x+ y)/(x+ y)[/itex]. That equation is true but is confusing notation because we are using y to represent two different "dummy" variables. I would prefer to write it in the equivalent form [itex]\int du/u= \int d(x+ y)/(x+ y)[/itex] which is clearly true with u= x+ y.
     
  12. Aug 18, 2014 #11
    I was under the impression that maybe there is a way of showing the above relation, but now it all makes sense.

    Thanks for the clarification.
     
  13. Aug 18, 2014 #12

    Curious3141

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    It IS correct.

    It is a simple consequence of the fact that ##\frac ab = \frac cd \implies \frac ab = \frac cd = \frac{a+c}{b+d}## for nonzero ##a,b,c,d##. This is trivial to prove.

    Juantheron (on MathSE) is merely replacing some of the variables with infinitesimals to come up with a rather interesting way to evaluate the integral. He starts by defining ##y^2 = x^2 + 1##, which then leads to ##ydy = xdx##. That allows him to go to ##\frac{dx}{y} = \frac{dy}{x}= \frac{dx + dy}{x + y} = \frac{d(x+y)}{x+y}##, and he uses that to evaluate the original integral.

    Some may balk at this "fast and loose" use of the Leibniz notation, but it is sound.
     
    Last edited: Aug 18, 2014
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