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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    Got another fun one for you fellow math nerds out there.... much more straightforward than the last and this time no silly errors :D

  2. Astrophysics
    • one year ago
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    Try \[x = \tan \theta \]

  3. anonymous
    • one year ago
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    That 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

  4. xapproachesinfinity
    • one year ago
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    IBP works fine i got \(\pi\frac{\ln 2}{2}-\frac{\pi^2}{32}\)

  5. xapproachesinfinity
    • one year ago
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    unless i made ARITHMETIC mistake should be good value

  6. xapproachesinfinity
    • one year ago
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    if we did x=tanu we would have to integrate \[\int \ln (\tan u+1)du\]

  7. xapproachesinfinity
    • one year ago
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    lost complex to me to do that integral

  8. xapproachesinfinity
    • one year ago
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    looks*

  9. anonymous
    • one year ago
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    Close its \[\pi \frac{ln(2)}{8}\]

  10. xapproachesinfinity
    • one year ago
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    then i must have made arithmetic mistake somewhere

  11. anonymous
    • one year ago
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    Ok 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

  12. anonymous
    • one year ago
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    Im interested in your IBP solution though could you post it plz I didnt treat it like that myself

  13. xapproachesinfinity
    • one year ago
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    oh right wolfram gives that solution too

  14. xapproachesinfinity
    • one year ago
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    let me see my math

  15. anonymous
    • one year ago
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    Dinner 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

  16. xapproachesinfinity
    • one year ago
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    oh i see where i made a mistake

  17. xapproachesinfinity
    • one year ago
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    oh interesting it crossed my mind to proceed that way but felt unsafe

  18. anonymous
    • one year ago
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    Then, 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: \[ \int_0^{\frac{\pi}{4}} (\ln( \sqrt{2}( \cos \theta \cos \frac{\pi}{4}- \sin \theta \sin \frac{\pi}{4}))- \ln( \cos \theta)) d \theta \\ = \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos \theta -\frac{\pi}{4})- \ln( \cos \theta)) d \theta \]

  19. anonymous
    • one year ago
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    darn that should be: \[ \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos( \theta -\frac{\pi}{4}))- \ln( \cos \theta)) d \theta \]

  20. xapproachesinfinity
    • one year ago
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    oh IBP won't work actually had made a severe mistake

  21. anonymous
    • one year ago
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    Ok 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: \[ \int_0^{\frac{\pi}{4}} \frac{1}{2} \ln(2) d \theta=\frac{\pi \ln2}{8} \]

  22. xapproachesinfinity
    • one year ago
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    oh very nice

  23. xapproachesinfinity
    • one year ago
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    neat work!

  24. anonymous
    • one year ago
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    You 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

  25. anonymous
    • one year ago
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    Thank 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

  26. anonymous
    • one year ago
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    Anyways your IBP must not have been that off... you were pretty close except for that pi^2 correction term

  27. anonymous
    • one year ago
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    What was the flaw

  28. xapproachesinfinity
    • one year ago
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    i 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

  29. anonymous
    • one year ago
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    ahhh 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

  30. anonymous
    • one year ago
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    So 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

  31. xapproachesinfinity
    • one year ago
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    ah yes! some good tricks!

  32. anonymous
    • one year ago
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    Anyways 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

  33. anonymous
    • one year ago
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    @Astrophysics solution is posted if you wanted to have a look but I am closing the question now

  34. xapproachesinfinity
    • one year ago
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    yea see ya man! that was some a good problem i should have given it more thought hhhh good night

  35. anonymous
    • one year ago
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    You 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

  36. Astrophysics
    • one year ago
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    Hey thanks for the tag! Nice question for sure!

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