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. 🙂

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. 🙂