1. ## Tricky Integral

I'm trying to evaluate the integral:

The integral seems to be challenging (computer algebra systems will have a long time dealing this integral). But, the book where I got this integral from came with a hint:

What is the value of: in the context of the problem?

I'm stuck. How should I continue? Thanks for any advice!

2. ## Re: Tricky Integral

Originally Posted by zachd77
I'm trying to evaluate the integral:

The integral seems to be challenging (computer algebra systems will have a long time dealing this integral). But, the book where I got this integral from came with a hint:

What is the value of: in the context of the problem?

I'm stuck. How should I continue? Thanks for any advice!
Have you multiplied out the hint? I think you'll recognize it when you do. What does that do with the big square root?

-Dan

3. ## Re: Tricky Integral

The hint gives you the expression under the square root. But how do I go from there?

4. ## Re: Tricky Integral

Well if it gives you the same thing as what's under the square root, surely you can substitute it in place of the stuff under the square root. Does that make things easier?

5. ## Re: Tricky Integral

Yes and no. So if you substitute that expression under the square root, won't you have the same thing? Or would the value be 1? I know I'm just missing that one thing and then I'll understand the substitution.

6. ## Re: Tricky Integral

Originally Posted by zachd77
I'm trying to evaluate the integral:

The integral seems to be challenging (computer algebra systems will have a long time dealing this integral). But, the book where I got this integral from came with a hint:

What is the value of: in the context of the problem?

I'm stuck. How should I continue? Thanks for any advice!
Take it step by step:
$\displaystyle \int \sqrt{x - \sqrt{x^2 - 4}}~dx = \int \sqrt{\frac{1}{2} \left ( \sqrt{x + 2} - \sqrt{x - 2} \right )^2 }~dx$

$\displaystyle = \frac{1}{\sqrt{2}} \int \left ( \sqrt{x + 2} - \sqrt{x - 2} \right )~dx$

I'm sure you can do the rest.

-Dan

7. ## Re: Tricky Integral

Apparanently the hint that came with the integral is somehow equal to 1. How did that come about?

8. ## Re: Tricky Integral

Why do you think it is equal to 1?

9. ## Re: Tricky Integral

Here's my argument. So if we call the expression under the square root I, and we get I again when we simplify the hint, I can set sqrt(I) = I and solve. I end up with I to be 0 or 1. I'm not sure this logic makes sense, but it's an effort. Does this seem legit?

10. ## Re: Tricky Integral

You can't set the sqrt(I)=I
that's like saying sqrt4=4
Topsquark has done most of the work for you.
You just have to make sense of it.
He has left you at a spot where you can integrate directly. just change the square roots to powers of a half.