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shubhamsrg

  • 3 years ago

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

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  1. lgbasallote
    • 3 years ago
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    \[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})\] ???

  2. vikrantg4
    • 3 years ago
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    yep i think so

  3. shubhamsrg
    • 3 years ago
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    yep.. and a dx there,,ofcorse..

  4. vikrantg4
    • 3 years ago
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    ah.. leave the typos.. concentrate on question. lol

  5. hartnn
    • 3 years ago
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    simple substitution of x=cos 2t

  6. shubhamsrg
    • 3 years ago
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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..

  7. shubhamsrg
    • 3 years ago
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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 ?

  8. mukushla
    • 3 years ago
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    is that \(\ln\)

  9. shubhamsrg
    • 3 years ago
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    yep..ln..

  10. hartnn
    • 3 years ago
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    i need to reconsider

  11. mukushla
    • 3 years ago
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    man this is interesting

  12. mukushla
    • 3 years ago
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    let\[t=\ln(\sqrt{1+x}+\sqrt{1-x})\]

  13. hartnn
    • 3 years ago
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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!

  14. mukushla
    • 3 years ago
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    i think there is

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

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

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

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

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

  20. hartnn
    • 3 years ago
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    lol,i did same exact thing :P

  21. hartnn
    • 3 years ago
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    i guess,thats integration by parts........???

  22. hartnn
    • 3 years ago
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    taking u=ln(....),v=1

  23. shubhamsrg
    • 3 years ago
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    bingo,,got it by parts! @hartnn thanks a lot!

  24. hartnn
    • 3 years ago
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    belated welcome :)

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