How do I evaluate this definite integral in Calc 1?

[opticaltempest]

I am trying to evaluate this integral,

$$\[<br /> \int_{ - r}^r {\left( {s\sqrt {1 + \frac{{x^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx\]$$

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,

$$\[<br /> \int_{ - r}^r {\left( {s\sqrt {1 + \frac{{r^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx\]$$

Thanks

 

[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.

 

[HallsofIvy]

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

 

[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

$$\[<br /> - \sqrt r \cdot \sqrt { - r} \cdot s\left( {\ln ( - r) - \ln (r)} \right)<br /> \]$$

Which simplifies to,

$$\[<br /> - r \cdot s \cdot i(\pi \cdot i)<br /> \]$$

Which simplified to,

$$\[<br /> r \cdot s \cdot \pi <br /> \]$$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.

 

[GCT]

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