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….)
Ok, there are two standard methods. The first (and easiest) is to do the change of variable which transforms this to which is very easy to do. The second method: parts, let 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: which leads to 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. . of course, I thought that I was testing “partial fractions” which leads to an answer of . Fair enough. But what did one of my students do? Well, this looked like trig substitution to him so: so this was transformed to which transforms back to 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 but I wasn’t being that picky. Besides, this trig substitution is really setting and we are really just choosing a convenient “branch” (meaning: viewing the domain “mod (-1,1)”) of the function.
3. . Easy, you say? Why not let etc. Yes, most did it that way. But then we had a couple do the following: which lead to which transforms to 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. 🙂