I gave my Calc II class the following problem on their first assignment of the quarter:

$int_{-frac{pi}{2}}^{frac{pi}{2}} (x^5+1)cos x : dx$,

which evaluates to 2 when one takes into account that the odd part ($x^5cos x $) evaluates to zero on a symmetric interval — something we covered a few lessons in.

I naively thought that no one would bother integrating by parts since we had not yet covered that technique in class (and it’s tedious to do even if one knows how), but of course several students did. More interestingly, I had a handful of students submit something that looks like

begin{align}

int_{-frac{pi}{2}}^{frac{pi}{2}} (x^5+1)cos x : dx &= int_{-frac{pi}{2}}^{frac{pi}{2}} frac{x^{10}-1}{x^5-1} cos x : dx \&= int_{-frac{pi}{2}}^{1} frac{x^{10}-1}{x^5-1} cos x : dx + int_{1}^{frac{pi}{2}} frac{x^{10}-1}{x^5-1} cos x : dx \

&= (102 sin(1) -60pi -frac{pi^5}{32} + frac{5 pi^3}{2} + 65 cos(1) + 1) + \ &~~~~~~(-102 sin(1) +60pi +frac{pi^5}{32} – frac{5 pi^3}{2} – 65 cos(1) + 1) \

&= 2

end{align}

This did not earn them much credit since there is a lot of work missing between steps 2 and 3, and to be honest I suspect this solution may have been spit out by some kind of integration software (though every one I have checked just does integration by parts). Ordinarily I would write this off as a student doing something weird, but the fact that several submitted the same method (possibly from the same source) has piqued my curiosity — is there an actual useful technique being applied here? Rewriting the expression in this way and then splitting the integral at the (artificially created) singularity does not seem particularly helpful to me.