# Thread: integrating an expression

1. ## integrating an expression

hello again,

I am trying to integrate this expression and I can't seem to be able to do so. I would greatly appreciate it if you can help me. The expression is: 1/(1 + sqr(2x - x^2)).

I tried multiplying the denominator with 1- sqr(2x - x^2) but to no avail. I then tried to rewrite the numerator as 1 - x + x, still stuck. I also tried to rewrite the denominator as (u - 2)/2, but I failed to find a solution. anyone any hint, please?

2. ## Re: integrating an expression

Why do you believe you should be able to do it?

Have you noticed the symmetry about x = 1? And the limited Domain, 0 <= x <= 2?

3. ## Re: integrating an expression

Originally Posted by bibiki
hello again,

I am trying to integrate this expression and I can't seem to be able to do so. I would greatly appreciate it if you can help me. The expression is: 1/(1 + sqr(2x - x^2)).

I tried multiplying the denominator with 1- sqr(2x - x^2) but to no avail. I then tried to rewrite the numerator as 1 - x + x, still stuck. I also tried to rewrite the denominator as (u - 2)/2, but I failed to find a solution. anyone any hint, please?
looks like a messy one ...

$\displaystyle \frac{1}{1+\sqrt{2x-x^2}} =$

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

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

$\displaystyle \frac{1 - \sqrt{1-(x-1)^2}}{(x-1)^2} =$

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

I believe the last term can be integrated using a trig substitution ...

4. ## Re: integrating an expression

(Sqrt(-(x-2) * x) - (x-1) * sin^(-1) * (1 - x) - 1) / (x-1), where sin^(-1) is an inverse sin function.

5. ## Re: integrating an expression

Originally Posted by Will57
(Sqrt(-(x-2) * x) - (x-1) * sin^(-1) * (1 - x) - 1) / (x-1), where sin^(-1) is an inverse sin function.
Wolfram Alpha's solution ...

integral of 1&#47;&#40;1&#43;sqrt&#40;2x-x&#94;2&#41;&#41; dx - Wolfram|Alpha

6. ## Re: integrating an expression

You WILL need to worry about Domain issues. x = 1 seems to be a problem unless you can prove it isn't.

7. ## Re: integrating an expression

TKHunny,
I believe I should be able to evaluate the integral because it was on a test in the past at the school I go.

skeeter,
that is the expression I arrived at, but did not have the idea of reforming it into a trig expression. I guess I will write x - 1 = sin (or perhaps cos), and see what I get.

Thank you all for your reply.

8. ## Re: integrating an expression

Originally Posted by bibiki
TKHunny,
I believe I should be able to evaluate the integral because it was on a test in the past at the school I go.
No good. It is or it isn't. As soon as the denominator was "rationalized", the Domain issue was introduced unless it can be show that there is no Domain issue. You might impress your teacher if you take a look at it.

You must answer this question: Are the first and last expressions equivalent? $\displaystyle \frac{a}{b}=\frac{a}{b}\cdot\frac{0}{0}=\frac{a \cdot 0}{b \cdot 0}=\frac{0}{0}$. If they are, then you're done. If they are not, what are you going to do about the transformation?

9. ## Re: integrating an expression

TKHunny, impress my teacher? I'd rather not before I get into wining and complaining about all the wrong at school, I'll just ask you or skeeter whether I integrated correctly 'the second' expression that skeeter derived. otherwise, I understand you perfectly well, TKHunny. You do have a point.

$\displaystyle \int{\frac{{\sqrt{1 - (x - 1)^2}}}{(x - 1)^2}} == {\frac{1}{1 - x}} - \frac{(x - 1)}{\cos({\arcsin(x - 1)})} - \arcsin(x - 1)$

The substitutions I did are as follows:
$\displaystyle (x - 1) = \sin(t), \mathrm{d}x = \cos(t)\mathrm{d}t$