# College Math Teaching

## May 20, 2016

### Student integral tricks…

Ok, classes ended last week and my brain is way out of math shape. Right now I am contemplating how to show that the complements of this object

and of the complement of the object depicted in figure 3, are NOT homeomorphic.

I can do this in this very specific case; I am interested in seeing what happens if the “tangle pattern” is changed. Are the complements of these two related objects *always* topologically different? I am reasonably sure yes, but my brain is rebelling at doing the hard work to nail it down.

Anyhow, finals are graded and I am usually treated to one unusual student trick. Here is one for the semester:

$\int x^2 \sqrt{x+1} dx =$

Now I was hoping that they would say $u = x +1 \rightarrow u-1 = x \rightarrow x^2 = u^2-2u+1$ at which case the integral is translated to: $\int u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} du$ which is easy to do.

Now those wanting to do it a more difficult (but still sort of standard) way could do two repetitions of integration by parts with the first set up being $x^2 = u, \sqrt{x+1}dx =dv \rightarrow du = 2xdx, v = \frac{2}{3} (x+1)^{\frac{3}{2}}$ and that works just fine.

But I did see this: $x =tan^2(u), dx = 2tan(u)sec^2(u)du, x+1 = tan^2(x)+1 = sec^2(u)$ (ok, there are some domain issues here but never mind that) and we end up with the transformed integral: $2\int tan^5(u)sec^3(u) du$ which can be transformed to $2\int (sec^6(u) - 2 sec^4(u) + sec^2(u)) tan(u)sec(u) du$ by elementary trig identities.

And yes, that leads to an answer of $\frac{2}{7}sec^7(u) +\frac{4}{5}sec^5(u) + \frac{2}{3}sec^3(u) + C$ which, upon using the triangle

Gives you an answer that is exactly in the same form as the desired “rationalization substitution” answer. Yeah, I gave full credit despite the “domain issues” (in the original integral, it is possible for $x \in (-1,0]$ ).

What can I say?