Talk:Integration by substitution

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Integration by substitution
The objective of Integration by substitution is to substitute the integrand with $$ u $$ or $$g(x)$$

Theory We want to transform the Integral from a function of x to a function of u $$\int_{x=a}^{x=b}f(x)\, dx\,\rightarrow\,\int_{u=c}^{u=d}h(u)\, du$$ Starting with $$u\,=\,g(x)$$ and

Procedure
 * Calculate $$g'(x)\,=\,{\operatorname{d}\!u\over\operatorname{d}\!x}$$
 * Calculate $$h(u)$$ which is $$f(x)\,{\operatorname{d}\!x\over\operatorname{d}\!u}\,=\,\frac{f(x)}{g'(x)}$$ and make sure you express the result in terms of the variable u
 * Calculate $$c\,=\,g(a)$$
 * Calculate $$d\,=\,g(b)$$
 * Looks like somebody forgot to sign. Nerd271 (talk) 19:14, 4 April 2020 (UTC)

conditions on f
In most calculus books f is required to be continuous (not just integratable), mainly because their proofs are based on the fundamental theorem of calculus which requires continuity. See also: http://eom.springer.de/I/i051740.htm http://mathworld.wolfram.com/ChangeofVariablesTheorem.html

However i've seen abstract generalization (based on lesbegue integrals) that seems to skip the contintuity condition for f (i do not fully understand the terms involved): http://planetmath.org/encyclopedia/ChangeOfVariablesFormula.html

It would be nice if somebody knowledgeable could confirm/comment this and modify the article (there should be at least a note why or under which exact condition continuity can be dropped for f).
 * Another person forgot to sign. Nerd271 (talk) 19:14, 4 April 2020 (UTC)

integrability of phi
Is it necessary to suppose integrability of (phi)'? Isn't it guaranteed by its assumption of continuity?

most general set of conditions
I'm thinking that in order to be a definitive source on the subject, there should be a theorem which states, in $$\mathbf{R}^1$$, the most general set of circumstances for F under which the change of variables works in the Reimann-Stiltjes sense

$$ \int_a^b \psi(F(t)) dF(t) = \int_{F(a)}^{F(b)} \psi(\xi) d\xi $$

that is to say that in order for the left hand side to exist as a Reimann-Stiltjes integral, the $$\psi(F(t))$$ must be integrable on (a,b) and the function, F, must be of bounded variation. Nothing about differentiability, and far less, nothing even about absolute continuity.

According to an exercise in Royden, the above is true for all F continuous and increasing (no need for absolute continuity and of course the requirement that $$\psi(F(t))$$ is integrable). Thus the above change of variables works when F is the Cantor function.

The Hewitt & Stromberg (1965, Theorem 20.3) citation attempts to be a general reference--its stated for functions on Polish Spaces, which is fine but more general than required. Something gives me the suspicion that this synopsis is mis-stated (c.f. the requirement that a function must be continuous _and_ absolutely continuous is redundant) also absolute continuity is not strictly required. Izmirlig (talk)izmirlig@mail.nih.gov


 * Someone signed incorrectly. Nerd271 (talk) 19:14, 4 April 2020 (UTC)

A section for discussing the geometry of trigonometric substitutions?
So consider the arctangent integral,

\int \frac{dx}{1+x^2} $$ The "standard" substitution is $$u(x) = \tan(x)$$, which then simplifies the integral to integration of a constant function via the derivative, $$\sec^2(x)$$.

We see is is useful by direct computation, but one could also look at a right triangle, with angle $$x$$ and side lengths $$1,u(x),\sqrt{1+u(x)^2}$$.

This corresponds to the form of the integrand itself. From here, it is easy to see that $$u(x) = \tan(x) \text{ and that } u'(x) = \sec^2(x)$$, which leads to our desired solution.

I think this perspective is important in teaching the method of substitution not as a rote algebraic trick, but one motivated by geometric intuition. Legatostaccato (talk) 06:40, 11 February 2020 (UTC)


 * We do have a page for trigonometric substitutions. Nerd271 (talk) 19:14, 4 April 2020 (UTC)