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aayushi.somani Group Title

Integrate : (x-1) root(1-x-x^2)

  • 3 months ago
  • 3 months ago

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  1. myininaya Group Title
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    Have you tried completing the square inside the square root first?

    • 3 months ago
  2. myininaya Group Title
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    This might end up being a trig sub.

    • 3 months ago
  3. myininaya Group Title
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    i will be back if you have questions

    • 3 months ago
  4. aayushi.somani Group Title
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    yeah i tried it.. but that didnt worked !

    • 3 months ago
  5. vaibhavthacker95 Group Title
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    u can take (x-1) as u and the rest term as v and integrate by parts

    • 3 months ago
  6. aayushi.somani Group Title
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    @ikram002p please help !

    • 3 months ago
  7. ikram002p Group Title
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    addy :o is it like this ? \(\int (x-1) \sqrt (1-x-x^2)\)

    • 3 months ago
  8. aayushi.somani Group Title
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    yeah exactly!

    • 3 months ago
  9. myininaya Group Title
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    The trig sub seems to work Not saying there aren't other ways though

    • 3 months ago
  10. myininaya Group Title
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    @aayushi.somani may I ask how far you got with that method?

    • 3 months ago
  11. myininaya Group Title
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    to what integral have you transformed it?

    • 3 months ago
  12. aayushi.somani Group Title
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    i could not figure out anything

    • 3 months ago
  13. myininaya Group Title
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    can you tell me what substitutions you performed ?

    • 3 months ago
  14. myininaya Group Title
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    some people like to do two here i just like one

    • 3 months ago
  15. myininaya Group Title
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    When you say you could not figure out anything does that mean you did not do anything with the trig sub even though you said you did?

    • 3 months ago
  16. myininaya Group Title
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    I'm confused

    • 3 months ago
  17. aayushi.somani Group Title
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    i tried getting a trig sub but unfortunatly i couldnt.. thats what it is.. @ikram002p could u figure out?

    • 3 months ago
  18. myininaya Group Title
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    did you complete the square inside the square root part?

    • 3 months ago
  19. aayushi.somani Group Title
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    if i could have done it then y would i ask this ques..??? @myininaya i asked so that anyone could help me..

    • 3 months ago
  20. myininaya Group Title
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    Oh I'm sorry I didn't know you couldn't do the algebra either

    • 3 months ago
  21. myininaya Group Title
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    So by nothing means absolutely nothing I get it

    • 3 months ago
  22. aayushi.somani Group Title
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    from what i did m getting a wrong answer so consider i did nothing and now please help me from the very begnning..

    • 3 months ago
  23. myininaya Group Title
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    \[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{a})^2)+c-a(\frac{b}{a})^2 \\ =a(x+\frac{b}{a})^2+c-a \frac{b^2}{a^2}=a(x+\frac{b}{a})^2+c-\frac{b^2}{a}\] I hope this helps you complete the square Please follow these steps I prefer you ask me questions on how to do this than just using the end result there

    • 3 months ago
  24. ikram002p Group Title
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    |dw:1408601698600:dw| i dont know if this work next to use tring mmm sin maybe ?

    • 3 months ago
  25. myininaya Group Title
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    oh mine what is all of that?

    • 3 months ago
  26. aayushi.somani Group Title
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    u mean i gotto Rationalise? @ikram002p

    • 3 months ago
  27. myininaya Group Title
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    You don't have to rationalize...And tht wouldn't be the way to rationalize the top anyhow

    • 3 months ago
  28. ikram002p Group Title
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    :P ok we dont need conjucate xD just hmm 1-x-^2=(5/2-(x+1/2)^2 >.<

    • 3 months ago
  29. aayushi.somani Group Title
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    got it lol sorry :P

    • 3 months ago
  30. aayushi.somani Group Title
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    i got the answer.. anyways thanks (:

    • 3 months ago
  31. ikram002p Group Title
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    :o how did u get it ?

    • 3 months ago
  32. aayushi.somani Group Title
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    I just made a small mistake.. I rechecked it.. and i could figure out the mistake! thankyou so much.. @ikram002p

    • 3 months ago
  33. ikram002p Group Title
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    :* np girl

    • 3 months ago
  34. myininaya Group Title
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    I thought you didn't do any of it...

    • 3 months ago
  35. myininaya Group Title
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    I want to go back to my completing square thingy... (i made some mistakes) \[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ =a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2}=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \]

    • 3 months ago
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