Integral :Complete the square and give a substitution (not necessarily trigonometric)

**highc1157** · September 30th 2010, 05:37 PM

I have tried multiple substitutions and cannot get this homework problem right xD

$\int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx$

everything in the denominator is under the square root

**skeeter** · September 30th 2010, 06:11 PM

Originally Posted by highc1157

I have tried multiple substitutions and cannot get this homework problem right xD

$\int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx$

everything in the denominator is under the square root

$\displaystyle \int \frac{x-1}{\sqrt{1 -(x^2-2x+1)}} \, dx$

$\displaystyle \int \frac{x-1}{\sqrt{1 -(x-1)^2}} \, dx$

can you finish?

**sa-ri-ga-ma** · September 30th 2010, 06:15 PM

After completing the square in the squareroot sing, the problem becomes

$\int\frac{1-x}{\sqrt{1 - (x-1)^2}}dx$

Substitue (x-1)^2 = t and proceed.

**highc1157** · September 30th 2010, 06:41 PM

I did that substitution and got (x-2)(x-1), but not same as answer in the book.

I screwed up somewhere

**highc1157** · September 30th 2010, 06:49 PM

hmmm the book answer is : $1-(1-x)^2$

I tried it again, and got : $\sqrt{1-(1-x)^2$

I dont know where i messed up, simplified out everything and this is the integral that I take
$1/2\int{(1-U)^(-(1/2))} du$ cant fix the exponent, supposed to be (1-U)^(-1/2)

**skeeter** · September 30th 2010, 06:59 PM

$\displaystyle \int \frac{x-1}{\sqrt{1 -(x-1)^2}} \, dx$

let $t = (x-1)^2$

$dt = 2(x-1) \, dx$

$\displaystyle \frac{1}{2} \int \frac{2(x-1)}{\sqrt{1 - (x-1)^2}} \, dx$

$\displaystyle \frac{1}{2} \int \frac{dt}{\sqrt{1 - t}}$

$\displaystyle -\int \frac{-1}{2\sqrt{1 - t}} \, dt$

$\displaystyle -\sqrt{1-t} + C$

$\displaystyle -\sqrt{1-(x-1)^2} + C$

$\displaystyle -\sqrt{2x - x^2} + C$

**highc1157** · September 30th 2010, 07:01 PM

that is not the answer in the book

, i think i posted the book answer above

**skeeter** · September 30th 2010, 07:06 PM

Originally Posted by highc1157

that is not the answer in the book

, i think i posted the book answer above

the "book" answer is incorrect ... they left off the radical and the constant of integration.

take the derivative of my solution ... what do you get?

**highc1157** · September 30th 2010, 07:10 PM

Hey Skeeter,

Thanks for you're time and help you put into this , very much appreciated

But as for your answer and my answer they are slightly different, but this isn't right, what I got? $\sqrt{1-(1-x)^2$

oh wow you are right, The derivative of my solution does not equal the original equation !! grrrr lol

**skeeter** · September 30th 2010, 07:17 PM

Originally Posted by highc1157

Hey Skeeter,

Thanks for you're time and help you put into this , very much appreciated

But as for your answer and my answer they are slightly different, but this isn't right, what I got? $\sqrt{1-(1-x)^2$

oh wow you are right, The derivative of my solution does not equal the original equation !! grrrr lol

your solution is correct ... note that $\sqrt{1-(x-1)^2} = \sqrt{1 - (1-x)^2}$

Thread: Integral :Complete the square and give a substitution (not necessarily trigonometric)

LinkBack

Thread Tools

Display

Integral :Complete the square and give a substitution (not necessarily trigonometric)

Similar Math Help Forum Discussions

Integral ( complete the square )

Integral - trigonometric substitution?

Trigonometric Substitution Integral

Transforming Integral for Trigonometric Substitution

Trigonometric Substitution Integral

Search Tags