Re: Indefinite Integration

BrandonFromFlorida <BrandonShw@xxxxxxx> writes:

On Dec 11, 11:50 pm, Robert Israel <isr...@xxxxxxxxxxx> wrote:
On Dec 11, 4:46 am, BrandonFromFlorida <Brandon...@xxxxxxx> wrote:





On Dec 11, 1:54 am, Robert Israel

<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
BrandonFromFlorida <Brandon...@xxxxxxx> writes:
Let's say I want to integrate a function I(x) by substituting x =
f(u). Now, this isindefiniteintegration. I am only seeking the
anti-
derivative and never intend to plug in numbers. If I choose a
function
f(u) which has a range which doesn't match the values of x for
which
the integrand is defined, might I get the wrong anti-derivative? An
example might be trying to substitute x = sin(u) for an integrand
which is defined for all values of x. I'm not sure whether the
range
of the function substituted has to match the allowed values of x in
the integrand to be sure of getting the right answer.

Your answer will not be defined except for x in the range of f.
But where your answer is defined, you will get a valid
antiderivative.

For example, look at what happens if you substitute x = sin(u) into
I(x) = int x dx. You get dx = cos(u) du so I(x) = int sin(u) cos(u)
du
= 1/2 int sin(2u) du = -1/4 cos(2u) + C. To express this in terms of
= x,
you must solve x = sin(u) for u. Thus you would get
I(x) = -1/4 cos(2 arcsin(x)) + C. If you only allow real variables
arcsin(x) is not defined for x outside the interval [-1,1]. However,
the complex-variables version of this is OK for all x, and in fact
-1/4 cos(2 arcsin(x)) simplifies to x^2/2 - 1/4, a formula that is
clearly valid for all x.
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Thank you, Dr. Israel, this is a very enlightening answer, but let me
get to the essence of my question. If you were teaching a freshman
calculus course, and only trying to introduce them to basic techniques
ofintegrationwithout going much into the subtleties (I'm not a
teacher), would you give your class an example solution of an integral
using a substitution in which the range of the function substituted
was not identical to the domain of the integrand, without mentioning
that there was an issue with ranges and domains? If you were writing
a basic book on calculus, would you give an example of such a
substitution? There are numerous opportunitues to do this unwittingly
when demonstrating trigonometric or hyperbolic substitutions.

Been there, didn't do that...
Typically in the calculus courses I teach, we mainly do simple
"direct" substitutions of the form u = g(x) where the integral is of
the form int h(g(x)) g'(x) dx. We only do a few particular "inverse"
substitutions x = f(u), e.g. x = a sin(u) is done only
in integrals involving sqrt(a^2 - x^2), which at this level is only
defined for -a <= x <= a, and therefore the issue does not usually
arise. I don't recall ever using this substitution in a calculus
class in a situation where the domain and range issue arose. Nor did
it arise in the one calculus book I wrote ("Calculus the Maple Way").
I'm not saying that it couldn't happen, but I wouldn't go out of my
way to confront the issue in a calculus class. This is not to imply
any criticism of an instructor who does confront it.

Robert Israel isr...@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada- Hide
quoted text -

- Show quoted text -

Thank you again, Dr. Israel. It just seems to me that after you show
how the substitution x = a sin(u) facilitates an integral involving
sqrt(a^2 - x^2) it would be natural for the student to assume that it
would work the same way for integrals involving (a^2 - x^2) without
the square root sign, such as 1/(a^2 - x^2). Most likely you would
solve that with partial fractions, but, still, it would be natural for
the student to wonder why it wasn't solved or whether it could be
solved with the sine substitution. When an integration technique
works beautifully for a certain integrand, at some point the student
will likely wonder whether it works as well for integrands with a
similar appearance.

Yes, that would be natural, and if it did arise I'd have to point out
that there's a problem with domains and ranges, but if (after simplification)
you get a formula that makes sense for |x| > a, it will probably work as
an antiderivative; and the way to check whether it does is to differentiate.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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