# 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

Staff Emeritus
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.