Integrate with multiple variables in denominator?

1. Mar 20, 2013

jketts

Hey everyone, I need to do the following integral. I just need a little help getting this started, I'm not sure where I need to go. Here is the problem:

$\int_{0}^{1-v} du \int_{0}^{\frac{1}{2}} dv \frac{1}{1+u^2-v^2}$

I think I have the boundaries for the integral set up correctly, {0≤v≤1/2, 0≤u≤1-v}.
I know that I will have to use $\int dx \frac{1}{c^2+x^2} = \frac{1}{c}tan^ {-1}\frac{x}{2}$

I began by trying to substitute 1-v^2 as a variable, but then I had to try to integrate tan^-1 with a bunch of square roots in it and that got really bad looking pretty quickly. Thoughts?

2. Mar 20, 2013

haruspex

I would try a change of coordinates, like x = u+v, y = u-v. The region gets a little more complicated but it's not too bad. Don't forget the Jacobian.

3. Mar 20, 2013

jketts

The original problem wants to change it from x and y to terms of u and v. I.e., x=u-v, y=u+v.

4. Mar 20, 2013

Zondrina

Could you give us the original question with the original region?

5. Mar 20, 2013

jketts

Sure, $\int\int_{D}{} \frac{1}{1+xy} dxdy$ D={(x,y);0≤x≤1,0≤y≤1}, x=u-v, y=u+v. That's all I am given.

6. Mar 20, 2013

Zondrina

$dxdy = \frac{∂(x, y)}{∂(u, v)}dudv$

7. Mar 20, 2013

haruspex

Ok, then you need to look at what you can substitute for v after the first integration. You have $\frac1{\sqrt{1-v^2}} atan(\sqrt\frac{1-v}{{1+v}})$, right? Let $\theta = atan(\sqrt\frac{1-v}{{1+v}})$, so $tan^2(\theta) =\frac{1-v}{1+v}$. When you see tan-squared, what do you think of?

8. Mar 20, 2013

haruspex

That integration range doesn't match what you have in u, v coordinates in the OP.

9. Mar 20, 2013

jketts

$\int_{v}^{1-v}$

I think I had that part wrong.

10. Mar 21, 2013

haruspex

It gets a bit messy. Something like v from -1/2 to +1/2, u from |v| to 1-|v|.
Did you figure out what to do with the tan-squared?