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AbdulRehman

  • 3 years ago

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

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  1. Atooty-S
    • 3 years ago
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    \[\int\limits 1/(1+x ^{2})^{1/2}\] it will be \[\ln (1 + x ^{2})^{1/2}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  17. AbdulRehman
    • 3 years ago
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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)

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

  19. AbdulRehman
    • 3 years ago
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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

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

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

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

  23. uzma
    • 3 years ago
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    yes :)

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

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

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

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

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