Recent content by mrworf

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...