The answer to the second semi-mathematical puzzle - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: The answer to the second semi-mathematical puzzle (/thread-239067.html) The answer to the second semi-mathematical puzzle - Mike Reed - 02-14-2013 " 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 Re: The answer to the second semi-mathematical puzzle - Les Koller - 02-16-2013 Not sure how to read it in Lymerick meter...is this close? The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three Re: The answer to the second semi-mathematical puzzle - Marcus von Cube, Germany - 02-17-2013 ```The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three``` The formum software has wrapped the text. What you see here is enclosed in [pre]...[/pre] tags. As an alternative use [nl] at each line end. Re: The answer to the second semi-mathematical puzzle - Derek Walker (UK) - 02-18-2013 Shouldn't 'cosine' and 'nine' be at the ends of lines, to rhyme: The integral of z squared dzfrom 1 to the cube root of 3 times the cosine of 3 pi over 9 equals log of the cube root of e