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TuringTest

  • 3 years ago

Fun integral that seems to have gotten deleted, so I'm posting it again.\[\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\]

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  1. TuringTest
    • 3 years ago
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    @No-data this is from the cinvestav exam, and it was one I could remember how to do. @mukushla perhaps you wanna have a try?

  2. jk_16
    • 3 years ago
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    your killing me hahah

  3. mandonut
    • 3 years ago
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    i think i got it

  4. mandonut
    • 3 years ago
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    u integrate xsinx and sinx/cos^2x separetely

  5. TuringTest
    • 3 years ago
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    not how I did it, but there more than one way to skin a cat

  6. mandonut
    • 3 years ago
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    lol nvm

  7. mandonut
    • 3 years ago
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    i read it upside down

  8. TuringTest
    • 3 years ago
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    I'm gonna leave this up for quite a while before giving any hints :)

  9. TuringTest
    • 3 years ago
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    wow I totally didn't mean to close this question I'll repost it :S

  10. eigenschmeigen
    • 3 years ago
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    @TuringTest i might have it... although i might well have made a mistake. i have it coming out as a periodic integral after a substitution

  11. eigenschmeigen
    • 3 years ago
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    ill post what i have

  12. eigenschmeigen
    • 3 years ago
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    \[u = cosx \]\[\frac{du}{dx} = -sinx\]\[dx = \frac{du}{-sinx}\] let our integral = I \[I = - \int\limits^{-1}_{1} \frac{\cos^{-1}u}{1+u^2} du\]\[I = \int\limits^{1}_{-1} \frac{\cos^{-1}u}{1+u^2} du\]\[f'(u) = \frac{1}{1+u^2} , g(u) = \cos^{-1}(u)\]\[f(u) = \tan^{-1}u \text{ , }g'(u) = \frac{1}{\sqrt{1-u^2}}\] \[\int\limits{f' \times g } \text{ } du = f \times g - \int\limits{f \times g'} \text{ } du\] \[I = [\tan^{-1}(u) \times \cos^{-1}(u)]^{1}_{-1} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\]\[I = \frac{\pi^2}{4} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\] \[ \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du = [\tan^{-1}u \times \cos^{-1}]^{1}_{-1} - \int\limits{\frac{\cos^{-1}u}{1+u^2}du}\] therefore \[I = \frac{\pi^2}{4} - I\]\[I= \frac{\pi^2}{8}\] if this is completely wrong my excuse is tiredness!

  13. TuringTest
    • 3 years ago
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    oh you are sooooooo close! let me see if I can find your apparently minor mistake....

  14. eigenschmeigen
    • 3 years ago
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    ah pellet i think i see a mistake already

  15. eigenschmeigen
    • 3 years ago
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    last couple of lines i think i made a sign error?

  16. eigenschmeigen
    • 3 years ago
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    aaah crap i think i made a sign error, and without it i have I = I

  17. TuringTest
    • 3 years ago
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    you have it right up until \[\int{\tan^{-1}u\over\sqrt{1-u^2}}du\]though you did it very differently than me there is a trick to this integral, would you like a hint?

  18. TuringTest
    • 3 years ago
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    I personally think you over-complicated things by leaving them in terms of u by the way, so it took me a while to decipher your work, but it is correct :)

  19. eigenschmeigen
    • 3 years ago
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    so up until i tried to integrate by parts?

  20. TuringTest
    • 3 years ago
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    the integration by parts was correct this setup for the integral is correct, though it looks much differently than mine as you left it in terms of u and I kept mine in x

  21. TuringTest
    • 3 years ago
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    you have done everything right, but you need to integrate this last guy....

  22. eigenschmeigen
    • 3 years ago
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    ah ok, so i just evaluated that last guy wrong

  23. TuringTest
    • 3 years ago
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    yes

  24. eigenschmeigen
    • 3 years ago
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    ugh im tired, ill have one last crack at it :D

  25. eigenschmeigen
    • 3 years ago
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    also, here's one in return: \[\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx\]

  26. TuringTest
    • 3 years ago
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    thanks I'll write it down, I'm leaving soon

  27. TuringTest
    • 3 years ago
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    let me know if you want a hint on the other one...

  28. eigenschmeigen
    • 3 years ago
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    no hints, but i'll have to continue it tommorow :)

  29. TuringTest
    • 3 years ago
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    good spirit :) see ya!

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