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aayushi.somani

  • 4 months ago

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

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  1. myininaya
    • 4 months ago
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    Have you tried completing the square inside the square root first?

  2. myininaya
    • 4 months ago
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    This might end up being a trig sub.

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

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

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

  6. aayushi.somani
    • 4 months ago
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    @ikram002p please help !

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

  8. aayushi.somani
    • 4 months ago
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    yeah exactly!

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

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

  11. myininaya
    • 4 months ago
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    to what integral have you transformed it?

  12. aayushi.somani
    • 4 months ago
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    i could not figure out anything

  13. myininaya
    • 4 months ago
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    can you tell me what substitutions you performed ?

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

  15. myininaya
    • 4 months ago
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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?

  16. myininaya
    • 4 months ago
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    I'm confused

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

  18. myininaya
    • 4 months ago
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    did you complete the square inside the square root part?

  19. aayushi.somani
    • 4 months ago
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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..

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

  21. myininaya
    • 4 months ago
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    So by nothing means absolutely nothing I get it

  22. aayushi.somani
    • 4 months ago
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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..

  23. myininaya
    • 4 months ago
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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

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

  25. myininaya
    • 4 months ago
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    oh mine what is all of that?

  26. aayushi.somani
    • 4 months ago
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    u mean i gotto Rationalise? @ikram002p

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

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

  29. aayushi.somani
    • 4 months ago
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    got it lol sorry :P

  30. aayushi.somani
    • 4 months ago
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    i got the answer.. anyways thanks (:

  31. ikram002p
    • 4 months ago
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    :o how did u get it ?

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

  33. ikram002p
    • 4 months ago
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    :* np girl

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

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

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