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shubhamsrg Group Title

integral [ log( sqrt(1+x) + sqrt(1-x) ) dx] from x=0 to 1

  • one year ago
  • one year ago

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  1. lgbasallote Group Title
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    \[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})\] ???

    • one year ago
  2. vikrantg4 Group Title
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    yep i think so

    • one year ago
  3. shubhamsrg Group Title
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    yep.. and a dx there,,ofcorse..

    • one year ago
  4. vikrantg4 Group Title
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    ah.. leave the typos.. concentrate on question. lol

    • one year ago
  5. hartnn Group Title
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    simple substitution of x=cos 2t

    • one year ago
  6. shubhamsrg Group Title
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    then our dx = -2sin 2t dt.. dont you think that will create a problem ? altough the sq. roots will be eliminated..

    • one year ago
  7. shubhamsrg Group Title
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    how'd you get that ? what i got was inside the integral (something) + log( cos t + sin t ) sin 2t dt are you missing the log there buddy ?

    • one year ago
  8. mukushla Group Title
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    is that \(\ln\)

    • one year ago
  9. shubhamsrg Group Title
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    yep..ln..

    • one year ago
  10. hartnn Group Title
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    i need to reconsider

    • one year ago
  11. mukushla Group Title
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    man this is interesting

    • one year ago
  12. mukushla Group Title
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    let\[t=\ln(\sqrt{1+x}+\sqrt{1-x})\]

    • one year ago
  13. hartnn Group Title
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    is there a standard formula for: \[\int\limits_{}^{}\ln(1+\sqrt{1-x^2})\] if yes,then this is so simple!

    • one year ago
  14. mukushla Group Title
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    i think there is

    • one year ago
  15. hartnn Group Title
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    \[=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1-x^2})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1-x^2})^2 dx\]

    • one year ago
  16. hartnn Group Title
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    \[=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1-x^2}) dx\]

    • one year ago
  17. mukushla Group Title
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    \[\ln(1+\sqrt{1-x^2})=\text{sech}^{-1}x+\ln x\]this is hard to get o.O

    • one year ago
  18. mukushla Group Title
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    http://www.wolframalpha.com/input/?i=%5Cint+%5Cln+%281%2B%5Csqrt%7B1-x%5E2%7D%29+dx

    • one year ago
  19. hartnn Group Title
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    http://www.wolframalpha.com/input/?i=integral+of+ln%281%2Bsqrt%7B1-x%5E2%7D%29&dataset=

    • one year ago
  20. hartnn Group Title
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    lol,i did same exact thing :P

    • one year ago
  21. hartnn Group Title
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    i guess,thats integration by parts........???

    • one year ago
  22. hartnn Group Title
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    taking u=ln(....),v=1

    • one year ago
  23. shubhamsrg Group Title
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    bingo,,got it by parts! @hartnn thanks a lot!

    • one year ago
  24. hartnn Group Title
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    belated welcome :)

    • one year ago
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