" int_(1)^(3^(1/3)) z^2 dz cos(3pi/9) = ln e^(1/3)

(Note: Readers outside the US, this works better if you allow yourself the agony of pronouncing the letter “z” as “zee”.)

I will post this answer one week from today"

OK, the week has passed; here's the answer:

The integral of z squared dz

from 1 to the cube root of 3

times the cosine

of 3 pi over 9

equals log of the cube root of e

How did you do in solving it?

mike