College Math Teaching

December 7, 2012

“Unusual” Student Integral Tricks….that are correct!

Filed under: academia, calculus, editorial, elementary mathematics, integrals, integration by substitution, pedagogy — Tags: — collegemathteaching @ 2:52 am

To make this list, the student has to do the integral correctly, but choose a painfully inefficient way of doing it.

On today’s final exam alone (from the most innocent to the most unusual and inefficient….)

1. \int x\sqrt{x+1} dx =
Ok, there are two standard methods. The first (and easiest) is to do the change of variable u = x+1 which transforms this to \int (u-1)\sqrt{u} du which is very easy to do. The second method: parts, let u = x, dv = \sqrt{x+1} etc. It is an algebraic exercise to see that one gets the same answer either way, though the answers look different at first.

One answer that I saw: u = \sqrt{x+1}, u^2-1 = x, 2udu = dx which leads to \int 2(u^2 -1)u^2 du which of course is doable. So this isn’t that far off of the easiest path, hence this entry only gets an “honorable mention”.

2. \int \frac{1}{9-x^2} dx . of course, I thought that I was testing “partial fractions” which leads to an answer of \frac{1}{6}(ln|3+x| - ln|3-x|)+C . Fair enough. But what did one of my students do? Well, this looked like trig substitution to him so: x = 3sin(t), dx = 3cos(t) so this was transformed to \int \frac{3cos(t)}{9cos(t)}dt = \frac{1}{3}\int sec(t) dt = \frac{1}{3}ln|sec(t) + tan(t)|+C which transforms back to \frac{1}{3}ln|\frac{3}{\sqrt{9-x^2}} + \frac{x}{\sqrt{9-x^2}}| = \frac{1}{3}(ln|3+x| - \frac{1}{2}(ln|3-x|+ln|3+x|))+C which is, of course, the correct answer.

Yes, I know that there are domain issues with the trig substitution (that is, the integral exists for all values of x \ne \pm 3 but I wasn’t being that picky. Besides, this trig substitution is really setting t = arcsin{\frac{x}{3}} and we are really just choosing a convenient “branch” (meaning: viewing the domain “mod (-1,1)”) of the arcsin(x) function.

3. \int \frac{arcsin(x))^2}{\sqrt{1-x^2}} dx . Easy, you say? Why not let u = arcsin(x), du = \frac{1}{\sqrt{1-x^2}}, etc. Yes, most did it that way. But then we had a couple do the following: x = sin(t), dx = cos(t)dt, arcsin(x) = t which lead to \int t^2 dt = \frac{t^3}{3} + C which transforms to \frac{(arcsin(x))^3}{3} + C which is the correct answer. πŸ™‚

Well, I tell my classes that “this isn’t a gymnastics meet; there are no “degree of difficulty points”” but some insist on trying to entertain me anyway. πŸ™‚



  1. […] I’ve had some “fun” at my math blog though. […]

    Pingback by Finals Day One Fall 2012 « blueollie — December 7, 2012 @ 3:28 am

  2. […] 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 […]

    Pingback by Domains and Anti Derivatives (Indefinite Integration) « College Math Teaching — December 13, 2012 @ 2:50 am

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