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anonymous
 one year ago
Putnam Calculus Problem: \[\int\limits_0^1 \frac{\ln(x+1)dx}{x^2+1}= \ ?\]
anonymous
 one year ago
Putnam Calculus Problem: \[\int\limits_0^1 \frac{\ln(x+1)dx}{x^2+1}= \ ?\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Got another fun one for you fellow math nerds out there.... much more straightforward than the last and this time no silly errors :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Try \[x = \tan \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That works @Astrophysics but it takes more than that... I just put this up for a fun challenge to whoever wants to try... if you have a solution by all means post it please :D

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1IBP works fine i got \(\pi\frac{\ln 2}{2}\frac{\pi^2}{32}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1unless i made ARITHMETIC mistake should be good value

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1if we did x=tanu we would have to integrate \[\int \ln (\tan u+1)du\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1lost complex to me to do that integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Close its \[\pi \frac{ln(2)}{8}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1then i must have made arithmetic mistake somewhere

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok give me a sec Ill post my solution but I just put a microwave dinner in and the plastic has melted so now my apartment reeks of plastic hold on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im interested in your IBP solution though could you post it plz I didnt treat it like that myself

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh right wolfram gives that solution too

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1let me see my math

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dinner ruined :/ but at least I found another math's enthusiast :D... Ok so I started out with the same substitution yieldling: \[\int_0^{\frac{\pi}{4}} \ln( \tan \theta +1) d \theta= \int_0^{\frac{\pi}{4}} (\ln( \sin \theta + \cos \theta) \ln( \cos \theta)) d \theta\] via factoring out the cos and reduing the log

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh i see where i made a mistake

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh interesting it crossed my mind to proceed that way but felt unsafe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then, the thing that took me forever to see was this: \[ \cos \theta + \sin \theta = \sqrt{2}( \cos \theta \cos \frac{\pi}{4}+ \sin \theta \sin \frac{\pi}{4})\] Which yields: \[ \int_0^{\frac{\pi}{4}} (\ln( \sqrt{2}( \cos \theta \cos \frac{\pi}{4} \sin \theta \sin \frac{\pi}{4})) \ln( \cos \theta)) d \theta \\ = \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos \theta \frac{\pi}{4}) \ln( \cos \theta)) d \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0darn that should be: \[ \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos( \theta \frac{\pi}{4})) \ln( \cos \theta)) d \theta \]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh IBP won't work actually had made a severe mistake

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so since cosine is an even function and the integration region is [0,pi/4] for the last two terms... those integrals are equal and opposite so they cancel. Thus: \[ \int_0^{\frac{\pi}{4}} \frac{1}{2} \ln(2) d \theta=\frac{\pi \ln2}{8} \]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh very nice

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1neat work!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can show the cosine cancellation explicitly by simply breaking up the integral and just using a simple u sub to remove the  pi/4 from the argument of cosine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you :D.... I had to sleep on it to get it to work for the life of me I was stuck playing around with various trig identities for a while

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyways your IBP must not have been that off... you were pretty close except for that pi^2 correction term

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1i had int ( arctan/x+1)dx by blind error i made it (arctanx/x^2+1) but again arctanx/x+1 is not an easy one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahhh yes I went down that path also... I was getting really annoyed because these are supposed to be solvable without reference to integral tables and arctan(x)/(x+1) is not fun to work with an honestly I had to look up the integral of arctan its been that long

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that's when I just went to bed and I tried again tonight and sure enough product to sum trick to coax out that symmetry

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1ah yes! some good tricks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyways I got some more hw to do as is... so Im signing off for tonight.. I will probably find and post another as soon as I finish my tests this week cya later :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics solution is posted if you wanted to have a look but I am closing the question now

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1yea see ya man! that was some a good problem i should have given it more thought hhhh good night

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You too... dont sweat it the first one I posted I blew really badly... I solved it in about 5 mins and didnt look back and notice I failed to properly distribute a minus sign (facepalm) no wonder it was so easy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Hey thanks for the tag! Nice question for sure!
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