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AbdulRehman Group Title

what will be the integration of 1/(1+x²)½ along x?

  • 3 years ago
  • 3 years ago

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  1. Atooty-S Group Title
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    \[\int\limits 1/(1+x ^{2})^{1/2}\] it will be \[\ln (1 + x ^{2})^{1/2}\]

    • 3 years ago
  2. Atooty-S Group Title
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    you know that when the numerator is the derivative of the dominator then the integration is ln "dominator" :)

    • 3 years ago
  3. uzma Group Title
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    but the derivative of denominator should be there in the numerator

    • 3 years ago
  4. uzma Group Title
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    \[x= \tan \theta\]

    • 3 years ago
  5. Atooty-S Group Title
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    ok , the derivative of (x^2 )^1/2 is 1/2 (2x) so 2 cancel the other 2 :)

    • 3 years ago
  6. uzma Group Title
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    and for natural log, exponent should be 1

    • 3 years ago
  7. uzma Group Title
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    \[\int\limits 1/x= \ln x\]

    • 3 years ago
  8. uzma Group Title
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    if the exponent is other than 1, we use power rule

    • 3 years ago
  9. AbdulRehman Group Title
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    yeah but we can't multiply it with 2x so ln(1+x²)½ is not the right answer

    • 3 years ago
  10. uzma Group Title
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    \[d x= \sec^2 \theta d \theta\]

    • 3 years ago
  11. Atooty-S Group Title
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    i think its correct .. anyways good luck ..

    • 3 years ago
  12. uzma Group Title
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    \[\int\limits \sec^2 \theta/(\sqrt{1+\tan^2 \theta} )\]

    • 3 years ago
  13. uzma Group Title
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    \[\int\limits \sec^2\theta/\sec \theta d \theta\]

    • 3 years ago
  14. uzma Group Title
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    \[\int\limits \sec \theta d \theta\]

    • 3 years ago
  15. uzma Group Title
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    \[\ln (\sec \theta + \tan \theta) +C\]

    • 3 years ago
  16. uzma Group Title
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    by backward substitution \[\tan \theta =x\]

    • 3 years ago
  17. AbdulRehman Group Title
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    yeah that will give us integration of secθ that is ln(secθ+tanθ) and by substituting and we will get probably get ln(sec(tan^-x)+x)

    • 3 years ago
  18. uzma Group Title
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    \[\sec \theta =\sqrt{1+x^2}\]

    • 3 years ago
  19. AbdulRehman Group Title
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    thanks for the help i have some questions on multiple integral usage in finding area and volume. Anyone please help me

    • 3 years ago
  20. uzma Group Title
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    post it as a new Q,some one would surly look at it :)

    • 3 years ago
  21. AbdulRehman Group Title
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    Uzma are you from Pakistan?

    • 3 years ago
  22. Atooty-S Group Title
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    ahaa ! ok waiting ur new Qs :)

    • 3 years ago
  23. uzma Group Title
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    yes :)

    • 3 years ago
  24. AbdulRehman Group Title
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    Can you please tell me your qualification and city

    • 3 years ago
  25. uzma Group Title
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    what would u do with that :)

    • 3 years ago
  26. AbdulRehman Group Title
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    nothing just to guess the level of your understanding mathematics

    • 3 years ago
  27. uzma Group Title
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    mathematics is as vast, that one can hardly grasp even one area

    • 3 years ago
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