tag:blogger.com,1999:blog-7136655045781905113.post7939187453780727787..comments2024-03-25T19:01:19.600+03:00Comments on Διασκεδαστικά Μαθηματικά : Putnam Training Problem [3]Σωκράτης Δ. Ρωμανίδηςhttp://www.blogger.com/profile/05364191669604847034noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-7136655045781905113.post-10408669489320191902022-05-02T18:27:55.038+03:002022-05-02T18:27:55.038+03:00χ=tanu , dχ=1/(cosu)^2 du . Άρα: [ π/4
...χ=tanu , dχ=1/(cosu)^2 du . Άρα: [ π/4<br /> Ι= / ln(1+ tan(u)) du<br /> ]0<br />φ=π/4 - u: [π/4<br /> I= / ln(2/(1+tan(φ))) dφ, <br /> ]0<br />αλλά ln(2/(1+tan(φ)))=ln2 - ln(1+tan(φ)). Επομένως <br /> ]π/4<br />2Ι=/ ln2 . Άρα Ι= ln(2)*π/8 <br /> ] 0 <br /> vimarkoulis@gmail.comhttps://www.blogger.com/profile/04321859530122488997noreply@blogger.com