aayushi.somani
  • aayushi.somani
Integrate : (x-1) root(1-x-x^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
Have you tried completing the square inside the square root first?
myininaya
  • myininaya
This might end up being a trig sub.
myininaya
  • myininaya
i will be back if you have questions

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More answers

aayushi.somani
  • aayushi.somani
yeah i tried it.. but that didnt worked !
anonymous
  • anonymous
u can take (x-1) as u and the rest term as v and integrate by parts
aayushi.somani
  • aayushi.somani
@ikram002p please help !
ikram002p
  • ikram002p
addy :o is it like this ? \(\int (x-1) \sqrt (1-x-x^2)\)
aayushi.somani
  • aayushi.somani
yeah exactly!
myininaya
  • myininaya
The trig sub seems to work Not saying there aren't other ways though
myininaya
  • myininaya
@aayushi.somani may I ask how far you got with that method?
myininaya
  • myininaya
to what integral have you transformed it?
aayushi.somani
  • aayushi.somani
i could not figure out anything
myininaya
  • myininaya
can you tell me what substitutions you performed ?
myininaya
  • myininaya
some people like to do two here i just like one
myininaya
  • myininaya
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?
myininaya
  • myininaya
I'm confused
aayushi.somani
  • aayushi.somani
i tried getting a trig sub but unfortunatly i couldnt.. thats what it is.. @ikram002p could u figure out?
myininaya
  • myininaya
did you complete the square inside the square root part?
aayushi.somani
  • aayushi.somani
if i could have done it then y would i ask this ques..??? @myininaya i asked so that anyone could help me..
myininaya
  • myininaya
Oh I'm sorry I didn't know you couldn't do the algebra either
myininaya
  • myininaya
So by nothing means absolutely nothing I get it
aayushi.somani
  • aayushi.somani
from what i did m getting a wrong answer so consider i did nothing and now please help me from the very begnning..
myininaya
  • myininaya
\[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{a})^2)+c-a(\frac{b}{a})^2 \\ =a(x+\frac{b}{a})^2+c-a \frac{b^2}{a^2}=a(x+\frac{b}{a})^2+c-\frac{b^2}{a}\] 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
ikram002p
  • ikram002p
|dw:1408601698600:dw| i dont know if this work next to use tring mmm sin maybe ?
myininaya
  • myininaya
oh mine what is all of that?
aayushi.somani
  • aayushi.somani
u mean i gotto Rationalise? @ikram002p
myininaya
  • myininaya
You don't have to rationalize...And tht wouldn't be the way to rationalize the top anyhow
ikram002p
  • ikram002p
:P ok we dont need conjucate xD just hmm 1-x-^2=(5/2-(x+1/2)^2 >.<
aayushi.somani
  • aayushi.somani
got it lol sorry :P
aayushi.somani
  • aayushi.somani
i got the answer.. anyways thanks (:
ikram002p
  • ikram002p
:o how did u get it ?
aayushi.somani
  • aayushi.somani
I just made a small mistake.. I rechecked it.. and i could figure out the mistake! thankyou so much.. @ikram002p
ikram002p
  • ikram002p
:* np girl
myininaya
  • myininaya
I thought you didn't do any of it...
myininaya
  • myininaya
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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