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

- aayushi.somani

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

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- schrodinger

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- myininaya

Have you tried completing the square inside the square root first?

- myininaya

This might end up being a trig sub.

- myininaya

i will be back if you have questions

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## More answers

- aayushi.somani

yeah i tried it.. but that didnt worked !

- anonymous

u can take (x-1) as u and the rest term as v and integrate by parts

- aayushi.somani

@ikram002p please help !

- ikram002p

addy :o
is it like this ?
\(\int (x-1) \sqrt (1-x-x^2)\)

- aayushi.somani

yeah exactly!

- myininaya

The trig sub seems to work
Not saying there aren't other ways though

- myininaya

@aayushi.somani may I ask how far you got with that method?

- myininaya

to what integral have you transformed it?

- aayushi.somani

i could not figure out anything

- myininaya

can you tell me what substitutions you performed ?

- myininaya

some people like to do two here
i just like one

- 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

I'm confused

- aayushi.somani

i tried getting a trig sub but unfortunatly i couldnt.. thats what it is..
@ikram002p could u figure out?

- myininaya

did you complete the square inside the square root part?

- 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

Oh I'm sorry I didn't know you couldn't do the algebra either

- myininaya

So by nothing means absolutely nothing
I get it

- 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

\[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{a})^2)+c-a(\frac{b}{a})^2 \\ =a(x+\frac{b}{a})^2+c-a \frac{b^2}{a^2}=a(x+\frac{b}{a})^2+c-\frac{b^2}{a}\]
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

|dw:1408601698600:dw|
i dont know if this work next to use tring mmm sin maybe ?

- myininaya

oh mine what is all of that?

- aayushi.somani

u mean i gotto Rationalise? @ikram002p

- myininaya

You don't have to rationalize...And tht wouldn't be the way to rationalize the top anyhow

- ikram002p

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

- aayushi.somani

got it lol sorry :P

- aayushi.somani

i got the answer.. anyways thanks (:

- ikram002p

:o
how did u get it ?

- aayushi.somani

I just made a small mistake.. I rechecked it.. and i could figure out the mistake! thankyou so much.. @ikram002p

- ikram002p

:*
np girl

- myininaya

I thought you didn't do any of it...

- 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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