- #1
jketts
- 4
- 0
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:
[itex]\int_{0}^{1-v} du \int_{0}^{\frac{1}{2}} dv \frac{1}{1+u^2-v^2} [/itex]
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 [itex]\int dx \frac{1}{c^2+x^2} = \frac{1}{c}tan^ {-1}\frac{x}{2}[/itex]
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?
[itex]\int_{0}^{1-v} du \int_{0}^{\frac{1}{2}} dv \frac{1}{1+u^2-v^2} [/itex]
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 [itex]\int dx \frac{1}{c^2+x^2} = \frac{1}{c}tan^ {-1}\frac{x}{2}[/itex]
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?