# College Math Teaching

## December 13, 2012

### Domains and Anti Derivatives (Indefinite Integration)

Grading student exams sometimes inspires me to revisit elementary topics. For example, I recently spoke about some unusual (but mostly correct) integration techniques used by students on a final exam.

I’ll recap (and adjust the example slightly): on a recent exam, a student encountered $\int \frac{2}{1-x^2} dx$. I had expected the student to use the usual partial fractions expansion to obtain $\int \frac{1}{1+x} dx + \int \frac{1}{1-x} dx = ln|1+x| - ln|1-x| + C$ which is valid when $x \ne \pm 1$. I admit to being a bad professor and not being picky about domains.

But one student noticed the $1 - x^2$ in the denominator of the fraction and so used the trig substitution $x = sin(\theta), dx = cos(\theta) d\theta$ which leads to the following integral: $\int \frac{2}{cos(\theta)} d\theta = 2ln|sec(\theta) + tan(\theta)| + C$ which leads to $2ln|\frac{1}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}| + C = 2ln|\frac{1+x}{\sqrt{1-x^2}}| = ln|1+x| - ln|1-x| + C$ for $x \in (-1,1)$. Note that, strictly speaking, the “final answer” is really defined for all $x \ne \pm 1$ though the equalities do not hold outside of the domain for $x$ used in the original trig substitution.

And yes, I was a bad professor; I gave full credit to this answer even though we “lost domain” during the string of equalities.

But that got me to wondering: is there a trig substitution that works for $|x| > 1$? Answer: of course:

$\int \frac{2}{1-x^2} dx = -\int \frac{2}{x^2 -1} dx$. Now use $x = sec(\theta), dx = sec(\theta) tan(\theta) d\theta$ which leads to $-2\int csc(\theta) d\theta = 2ln|csc(\theta) + cot(\theta)| + C = 2ln|\frac{x}{\sqrt{x^2-1}} + \frac{1}{\sqrt{x^2 -1}}| + C$ which leads us to our ultimate solution for $|x| > 1$

So, if one REALLY wanted to use trig substituions for this problem, one could and do it in a way to cover the entire domain.

But…as our existence and uniqueness theorems imply, once we get a candidate for an anti-derivative that “works” or the domain, it really doesn’t matter if we did “illegal” steps to get it; we need only show that it is an anti derivative and is valid for the entire domain for the integrand.

Now if one wants a more detailed discussion on domain issues for anti-derivatives, I can recommend the article The Importance of Being Continuous by D. J. JEFFREY which appeared in Mathematics Magazine, Vol 67, pp 294 – 300. (reprint can be found here, scroll down a bit; this mathematician has written quite a bit!). Note: I can recommend this little paper as it talks about the domains of the anti derivatives themselves and not just the domains assumed in doing the calculations along the way or the domains of validity of the substitutions. Note: integral tables and computer algebra systems don’t always give the anti derivative with the “largest” possible domain. One has to watch for that.