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quakeash

what is the integral of 2/ (2x-x^2)^1/2 my guesses is sin^-1 (whatever) +c maybe ?

  • one year ago
  • one year ago

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  1. experimentX
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    ??

    • one year ago
  2. experimentX
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    integral of ??

    • one year ago
  3. quakeash
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    the integral of x/ (2x-x^2)^1/2

    • one year ago
  4. experimentX
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    try completing squares .. 2x - x^2 = 1 - (1-x)^2

    • one year ago
  5. experimentX
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    and let 1 - x = sin u, x = 1 + sin u

    • one year ago
  6. experimentX
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    not sure though ... haven't checked!!

    • one year ago
  7. quakeash
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    the complete square u gave me is wrong its 1-(x-1)^2 i guess right ?

    • one year ago
  8. shivam_bhalla
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    @quakeash , would you mind writing your question on drawing board again? Click the draw button below and rewrite your question on that

    • one year ago
  9. shivam_bhalla
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    |dw:1337271739470:dw| Is this your question ??

    • one year ago
  10. quakeash
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    exactly

    • one year ago
  11. experimentX
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    LOL ... it wasn't trig substitution, http://www.wolframalpha.com/input/?i=integrate+x%2F%28sqrt%282x+-+x^2%29%29

    • one year ago
  12. shivam_bhalla
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    It is pretty simple. First follow what @experimentX said |dw:1337271981227:dw|

    • one year ago
  13. shivam_bhalla
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    |dw:1337272062537:dw|

    • one year ago
  14. shivam_bhalla
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    soory mistake in the lat step

    • one year ago
  15. shivam_bhalla
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    It should be |dw:1337272154314:dw|

    • one year ago
  16. quakeash
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    the solution . since the integral of dx/(a^2 - x^2) = sin^-1 (x/a) + c then my question after completing square would be .. 2/(1-(x-1)^2)^1/2 and its sin ^-1 (x-1/1) +c /////////////// case closed :D

    • one year ago
  17. shivam_bhalla
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    Now substitute back x -1 in place of t . You will get your answer

    • one year ago
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