Second degree function under root - integral

1. Feb 10, 2013

Chromosom

1. The problem statement, all variables and given/known data

Integral:

$$\int\sqrt{1+\frac{1}{a^2+x^2}}\,\text dx$$

2. Relevant equations

3. The attempt at a solution

I don't know any method at the moment. Maybe Euler substitution? But this integral is $$\int\sqrt{\frac{1+a^2+x^2}{a^2+x^2}}$$ after making some calculation, but it is still not similar to any other I have seen in the past.

2. Feb 10, 2013

Dick

That doesn't look like an elementary integral to me. More like an elliptic integral. I don't think the usual techniques will get you very far.

3. Feb 10, 2013

LCKurtz

Indeed. Maple gives a rather complicated expression in terms of ellipitic functions.

4. Feb 10, 2013

Chromosom

The original problem was: calculate y coordinate of mass center of line: $$y=\text{arsinh}\,\frac xa$$

I need to compute:

$$y_a=\frac{\int_Ly\,\text dl}{\int_L\text dl}$$

and I got this integral from first post. Any other method for mass center?

Thanks for answers by the way :)

5. Feb 10, 2013

SammyS

Staff Emeritus
What is the wording of the original problem?

6. Feb 10, 2013

Chromosom

It's other language... in translation, I need to compute y coordinate of mass center.

7. Feb 10, 2013

SammyS

Staff Emeritus
So, are you to compute the center of mass of a uniform wire lying along the curve, $\displaystyle \ \ y=\text{arcsinh}\,\frac xa\ ?$

What are the endpoints?

8. Feb 10, 2013

Chromosom

Yeah. $$x\in[0,b]$$

Thanks for help :)

9. Feb 10, 2013

SammyS

Staff Emeritus
If $\displaystyle \ \ y=\text{arcsinh}\,\frac xa\,,\$ then $\displaystyle \ \ x=a\, \sinh(y)\ .$

You can compute the line integral in terms of either x or y -- or some other parameter for that matter.

$\displaystyle d\ell^2=dx^2+dy^2\ .$

$\displaystyle d\ell=\sqrt{1+\left(dy/dx\right)^2\ }\ dx=\sqrt{1+\left(dx/dy\right)^2\ }\ dy$

10. Feb 11, 2013

Chromosom

Thanks for help:) but then I have $$\sqrt{1+\cosh^2x}$$ and nothing to do with it...

11. Feb 11, 2013

LCKurtz

As has been pointed out before, the original integral gives a formula in terms of an elliptic function. A simple change of variable isn't going to change the nature of the problem.