# Development of Integration by Substitution

## Main Question or Discussion Point

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?

This is reverse of the chain rule.
(For simplicity sake assume that $$f(x)$$ and $$g(x)$$ are differentiable everywhere).

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

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

Gib Z
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.
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We suppose that a new variable u is introduced into a function F(x) by means of the equation $x=\phi (u)$ so that F(x) becomes a function of u:
$$F(x) = F( \phi (u) ) = G(u)$$.

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

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

then on one hand the chain rule takes the form $$g(u) = f(x) \phi ' (u)$$
and on the other hand $G(u) = F(x)$ by definition, that is,
$$\int g(u) du = \int f(x) dx$$, and we obtain the integral formula equivalent to the chain rule: $$\int f( \phi (u)) \phi '(u) du = \int f(x) dx , {x=\phi (u)}$$.
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Trig substitution is merely the case where the substitution happens to be a trigonometric function.

Here is how trigonometric substitution works formally using the substitution rule.

Consider $$\int \sqrt{1-x^2} \ dx$$ (on $$(-1,1)$$.)

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

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