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quakeash

  • 2 years ago

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

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  1. experimentX
    • 2 years ago
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    ??

  2. experimentX
    • 2 years ago
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    integral of ??

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

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

  5. experimentX
    • 2 years ago
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    and let 1 - x = sin u, x = 1 + sin u

  6. experimentX
    • 2 years ago
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    not sure though ... haven't checked!!

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

  8. shivam_bhalla
    • 2 years ago
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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

  9. shivam_bhalla
    • 2 years ago
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    |dw:1337271739470:dw| Is this your question ??

  10. quakeash
    • 2 years ago
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    exactly

  11. experimentX
    • 2 years ago
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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

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

  13. shivam_bhalla
    • 2 years ago
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    |dw:1337272062537:dw|

  14. shivam_bhalla
    • 2 years ago
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    soory mistake in the lat step

  15. shivam_bhalla
    • 2 years ago
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    It should be |dw:1337272154314:dw|

  16. quakeash
    • 2 years ago
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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

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

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