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Development of Integration by Substitution

  1. Oct 24, 2007 #1
    How was Integration by Substitution and Trig Substitution developed? My calc book doesn't have much info, just a short (not really complete) proof. Could someone explain and/or lead me in the right direction?
  2. jcsd
  3. Oct 24, 2007 #2
    This is reverse of the chain rule.
    (For simplicity sake assume that [tex]f(x)[/tex] and [tex]g(x)[/tex] are differentiable everywhere).

    Given [tex]f(x)[/tex] let [tex]F(x)[/tex] be any anti-derivative (which we know exists by the fundamental theorem of calculus). Now consider [tex]F(g(x))[/tex]. What is the derivative of that? Using the chain rule we fine [tex]F'(g(x))g'(x)[/tex] but [tex]F'(x) = f(x)[/tex] because it is an anti-derivative. So [tex][F(g(x))]' = f(g(x))g'(x)[/tex]. This means [tex]F(g(x))[/tex] is an anti-derivative of [tex]f(g(x))g'(x)[/tex]. So [tex]\int f(g(x))g'(x) dx = F(g(x))+C[/tex].

    Here is an example. Say we want to find [tex]\int (x+1)^5[/tex]. We can let [tex]f(x) = x^5[/tex] and [tex]g(x)=x+1[/tex]. Now confirm that [tex](x+1)^5 = f(g(x))g'(x)[/tex]. So the anti-derivative is [tex]F(g(x))+C[/tex] but [tex]F(x) = (1/6)x^6[/tex]. So [tex]F(g(x))+C = (1/6)(x+1)^6 + C[/tex]. That is is.

    Of course, I did the most basic example. But that is how I do my integrations. I never use the calculus book approach because it does not provide any insight like you said.
  4. Oct 25, 2007 #3

    Gib Z

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    Homework Helper

    Yes, as Kummer said, it is Integral calculus's counterpart of The Chain Rule from Differential Calculus. Courant's treatment of Calculus, Volume 1, has a nice proof wish I will post because I'm nice =] Just note from this point on, the following is not my own work.

    We suppose that a new variable u is introduced into a function F(x) by means of the equation [itex]x=\phi (u) [/itex] so that F(x) becomes a function of u:
    [tex]F(x) = F( \phi (u) ) = G(u)[/tex].

    By the chain rule of differential calculus:
    [tex]\frac{dG}{du} = \frac{dF}{dx} \phi ' (u)[/tex].

    If we now write
    [tex]F'(x) = f(x) \mbox{and} G'(u) = g(u)[/tex], or the equivalent expressions [tex] F(x) = \int f(x) dx \mbox{and} G(u) = \int g(u) du[/tex]

    then on one hand the chain rule takes the form [tex]g(u) = f(x) \phi ' (u)[/tex]
    and on the other hand [itex]G(u) = F(x)[/itex] by definition, that is,
    [tex]\int g(u) du = \int f(x) dx[/tex], and we obtain the integral formula equivalent to the chain rule: [tex] \int f( \phi (u)) \phi '(u) du = \int f(x) dx , {x=\phi (u)}[/tex].

    Trig substitution is merely the case where the substitution happens to be a trigonometric function.
  5. Oct 26, 2007 #4
    Here is how trigonometric substitution works formally using the substitution rule.

    Consider [tex]\int \sqrt{1-x^2} \ dx[/tex] (on [tex](-1,1)[/tex].)

    We can write, [tex]\int \frac{1-x^2}{\sqrt{1-x^2}} \ dx[/tex].
    Let [tex]g(x) = \sin^{-1} x[/tex] then [tex]g'(x) = \left( \sqrt{1-x^2} \right)^{-1}[/tex]. And [tex]f(x) = \cos^2 x[/tex] then [tex]f(g(x)) = 1-x^2[/tex]. This means,
    [tex]\int f(g(x))g'(x) dx = F(g(x))+C[/tex].
    Where [tex]F(x)[/tex] is an anti-derivative of [tex]\cos^2 x [/tex]. Which is [tex]F(x) = \frac{1}{2}x + \frac{1}{2}\sin x\cos x[/tex] now evaluate this at [tex]g(x) = \sin^{-1} x[/tex] which gives [tex]\frac{1}{2}\sin^{-1} x + \frac{1}{2} x \sqrt{1-x^2}[/tex].

    I might have a mistake because I was not careful. But you get the idea.
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