right on, with sammyS' suggestion, my equation becomes:
2 p \int_{\tan^{-1}{\frac{p}{\alpha}}}^{\tan^{-1}{\frac{-p}{\alpha}}} \bigg[ \frac{\sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} \bigg] d\theta
Homework Statement
\int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{p^2 + v^2 +1 }} dv
Homework Equations
1 + \tan{\theta}^2 = \sec{\theta}^2
The Attempt at a Solution
I thought the best way to go about this was to rename some constants.
Let \alpha^2 = 1 + p^2 so that we are left with...