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Indefinite integration substitutions: 'any' permissible?

  1. Jul 28, 2012 #1
    Hello!

    Permissible is probably the wrong word, but here's what I am having difficulty with:

    With substitutions in integration to make the integral easier to solve, there doesn't seem to be a restriction on the substitution one can use with indefinite integration. If one function, or part of a function, is replaced with a different function of a different variable,
    e.g. [tex]\sqrt(3-4/5x)[/tex], where sin(y) = 4/5x, as a random example, and they have different ranges for values, this doesn't seem to matter:
    You can replace one function with another, solve the integral, and convert back. There is one anti derivative for the function, so if you convert back, whereas it might have worked for a limited range of values for a definite integral with the substitution, in the indefinite form, once converted back, can work for all values.

    Does that make any sense?
    It seems to me any substitution can be made, the integral solved, and then convert back to the original variable. This can't be true, because when reading about subsitutions to solve integrals the range of values is important to consider.

    I hope this makes sense,
    Any help appreciated.
     
  2. jcsd
  3. Jul 28, 2012 #2

    Simon Bridge

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    The solution you obtain for an integral has to work for the entire range specified to be considered a general solution. This means the substitutions will have to be valid for the entire range that the integral is likely to be performed over. For an indefinite integral that would mean everywhere.

    It is, however, valid to present a partial solution that works only for a specific range. You just have to say so.

    Apart from that you can make any substitution you like.
     
  4. Aug 1, 2012 #3
    Thanks for the response Simon. Does that mean that it is possible to have a number of different functions that are solutions to an integral depending on the substitution made, and which will be valid for a certain range depending upon the range dictated by the particular substitution?
    I can't give a specific example, because I have never seen this explained - but, say I used a trig substitution, which may limit the range for the original integral, and in another case I used some other substitution that also limited the range, but a different range. Could I get two different functions, with two different ranges by using these two different substitutions?

    Thanks again!
     
  5. Aug 1, 2012 #4

    SammyS

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    The anti-derivative of a function is unique up to the added constant.

    Looking at the example you propose in your original post:
    Using standard techniques [itex]\displaystyle \int\, \sqrt{3-(4/5)x}\,dx =-\ \frac{5}{6} \left(3-\frac{4 x}{5}\right)^{3/2}+C\,,[/itex] the domain of the integrand and the result being x ≤ 15/4 .

    Using the substitution [itex]\displaystyle \sin(y)=-(4/5)x[/itex] gives the following. (I realize this is a bit different than the substitution you proposed.)

    [itex]\displaystyle x=-\,\frac{5}{4}\sin(y)\quad \to \quad
    dx=-\,\frac{5}{4}\cos(y)\,dy[/itex]

    The integral becomes, [itex]\displaystyle -\frac{5}{4} \int\,\sqrt{3+\sin(y)}\,\cos(y)\,dy\ .[/itex]

    Integrating gives, [itex]\displaystyle -\frac{5}{6} \left(3+\sin(y) \right)^{3/2}+C\ .[/itex]

    Substituting -(4/5)x for sin(y) gives the result, [itex]\displaystyle -\frac{5}{6} \left(3-\frac{4}{5}x \right)^{3/2}+C\ .[/itex]

    This looks like the same result we got initially, but sin(y) can only take on values between -1 and 1, inclusive. That corresponds to a domain of -5/4 ≤ x ≤ 5/4 .
    So, what can we conclude?

    Since, the anti-derivative of a function is unique up to the added constant, I suppose it's OK to use a somewhat restricted substitution such as [itex]\displaystyle \sin(y)=-(4/5)x\,.[/itex] That is to say it's OK for finding the anti-derivative.

    However, there is a potential problem if you are evaluating a definite integral.

    With the substitution, [itex]\displaystyle \sin(y)=-(4/5)x\,,[/itex]
    [itex]
    \displaystyle \int_{a}^{b}\, \sqrt{3-(4/5)x}\,dx =-\frac{5}{4} \int_{y(a)}^{y(b)}\,\sqrt{3+\sin(y)}\,\cos(y)\,dy
    [/itex]
    [itex]
    \displaystyle =-\frac{5}{4} \int_{-\arcsin(4a/5)}^{-\arcsin(4b/5)}\,\sqrt{3+\sin(y)}\,\cos(y)\,dy
    [/itex]​
    This only makes sense if a & b are in the closed interval, [-5/4 , 5/4]

     
  6. Aug 1, 2012 #5

    Simon Bridge

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    Thanks SammyS;
    ... the short form: you can get different results from the definite integral for different substitutions but they are valid only within their restrictions. The "correct" substitution will be determined by the purpose of doing the integration in the first place and you need to be careful to keep track of the consequences.
     
  7. Aug 4, 2012 #6
    Thanks Sammy and Simon, your responses address both of my concerns/questions. Many thanks to both of you.
     
  8. Aug 4, 2012 #7

    Simon Bridge

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    No worries - you sound like you are getting used to using math as a language :)
     
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