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Integral Substitution

  1. Aug 22, 2005 #1
    Integral Substitution......

    Heya people,
    I was wondering if someone here could point me in the right direction, as the book im reading on Integration isnt very thourough, and I dont really have anyone else to ask. :confused:
    Basically, im reading up on u-substitiution regarding integration, but im not really sure of the finer points.
    The texts says that I can ONLY use substituiton if the derivative of the 'u' is present in the integral-equation.

    My Question:

    When you have an integral to evaluate, and its complicated enough to have to use u-substituiton, but there is not any [tex] g'(x) [/tex] for your [tex] g(x) [/tex] in the integral,
    can we factor in the required derivative of our 'inside' function to the equation before evaluating the integral to make things easier????? :confused:

    Ie) Does this mean, (for simplicity,)
    If I have:
    [tex]\int sin(x^4) dx[/tex]
    and I want to make my [tex]\ u =x^4[/tex]
    then, can I somehow factor in the 4x^3 to make this work???

    Cheers. :smile:
    Last edited: Aug 22, 2005
  2. jcsd
  3. Aug 22, 2005 #2


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    I don't know the term "u-substitiution" (is it just a substitution?) but it's not possible in the example you gave. That integral can't be evaluated using elementary functions, there is no why of getting the 4x³ out of nowhere.
  4. Aug 22, 2005 #3


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    That's not quite right
    The derivative doesn't need to appear in the integrand, but it helps if it does!

    Example of not needing the derivative in the integrand.

    [tex]I= \int sin^3x dx[/tex]

    [tex]\mbox{let} \ u = cosx[/tex]
    [tex]du = -sinx dx[/tex]

    [tex]I = \int sin^3x\ dx[/tex]
    [tex]I = \int sin^2x.sinx\ dx[/tex]
    [tex]I = - \int sin^2x\ du[/tex]
    [tex]I = - \int (1 - cos^2x)\ du[/tex]
    [tex]I = \int (cos^2x - 1)\ du[/tex]
    [tex]I = \int u^2 - 1\ du[/tex]
    [tex]I = u^3/3 - u + C[/tex]
    [tex]I = cos^3x/3 - cosx + C[/tex]

    In the above example, the derivative didn't appear. However, this discounts manipulating the expression in order to make it appear. e.g. [tex]sin^3x = sinx(1 - cos^2x)[/tex].
    You can also use substitution to transform an expression, then use another substitution to simplify it. E.g. t = tan(x/2) to transform a trigonometric expression into one involving powers of t. Then use another substitution, for t, to simplify the espression - if needed.
  5. Aug 22, 2005 #4
    There is a Fresnel integral, which can be defined as:


    Your integral does not have any solutions dealing with elementary functions.
  6. Aug 22, 2005 #5


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    Substitution is just viewing the chain rule for derivatives as an integration formula
    [tex]\frac{dy}{dx}=\frac{dy}{du} \ \frac{du}{dx}[/tex]
    [tex]y+C_1 =\int \frac{dy}{dx} dx=\int \frac{dy}{du} \ \frac{du}{dx} dx[/tex]
    as was mentioned it is nice if such a substitution is immediately obvious but often some manipulations can yield the desired form.
    Last edited: Aug 22, 2005
  7. Aug 23, 2005 #6


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    No, that means you can't do that integral by substitution.
    Since there was no 4x3 in the integral originally, you can't just put one in it (not without putting 4x3 in the denominator also which would make the result just as complicated).

    The fact is that most integrals of simple functions can't be done "analytically".
  8. Aug 24, 2005 #7
    Thanks everyone,
    I didnt realise the text was refering to the chain-rule for integrals.

    If i can manipulate the values of [tex]\ du [/tex] to fit into the equation, it will work most of the time,
    but what about when the value of [tex]\ du [/tex] makes the equation even more complicated?

    Thanks again for the help.
  9. Aug 25, 2005 #8


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    It is not a chin rule for integral that would imply a method for integral of the form f(g(x))
    It is using the chain rule for derivatives to help find integrals
    Substitution does not always make integrals easier
    in your example
    [tex]\int \sin(x^4)dx=\int \frac{\sin(x^4)4x^3dx}{4x^3}=\frac{1}{4}\int u^{-3/4}\sin(u)du[/tex]
    to make the integral easier the integral needs to have a certain form and the right choice of u is needed.
    Last edited: Aug 25, 2005
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