How do I evaluate this definite integral in Calc 1?

1. Nov 25, 2005

opticaltempest

I am trying to evaluate this integral,

$$$\int_{ - r}^r {\left( {s\sqrt {1 + \frac{{x^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx$$$

Is it a valid integral?

If I evaluate it by plugging in r and -r, it becomes undefined.
How else can I evaluate this integral? I'm in calculus 1 and I am guessing
this may be a topic covered in later calculus classes?

EDIT: I had the integral wrong. It should be,

$$$\int_{ - r}^r {\left( {s\sqrt {1 + \frac{{r^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx$$$

Thanks

Last edited: Nov 25, 2005
2. Nov 25, 2005

Physics Monkey

The integral is well defined (the singularity you observed is called integrable because it doesn't bother the integral), and you can calculate it really easily. Hint: think about symmetry.

Edit: Haha. Ok Halls, fairly easily.

Last edited: Nov 25, 2005
3. Nov 25, 2005

HallsofIvy

Well, not all that "really easily". Opticaltempest, have you covered "trigonometric substitutions" yet?

4. Nov 25, 2005

opticaltempest

I got the original integral wrong. I updated it in an edit. Sorry

I am allowed to use technology right now to evaluate the above integral.

Using Maple 10 I find the above integral evaluates to

$$$- \sqrt r \cdot \sqrt { - r} \cdot s\left( {\ln ( - r) - \ln (r)} \right)$$$

Which simplifies to,

$$$- r \cdot s \cdot i(\pi \cdot i)$$$

Which simplified to,

$$$r \cdot s \cdot \pi$$$

This was what I was looking for :) I haven't covered using trigonometric substitution yet.
I was just curious how I would integrate that without using a CAS.

5. Nov 26, 2005

GCT

The form seems familiar, from what I remember it may have something to do with the arc length derivation.