Stuck on Double Integral $\int_{0}^{\sqrt{2}/2} \int_{0}^{u} \frac{1}{1- \frac{u^2 - v^2}{2}} dv du $

Question

I was stuck on this double integral below and wondered if anyone could help me out on it.

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

I know it needs to be in arctan integration form; however, I am stuck on how to proceed.

Answer 1

Proceed as follows

\begin{align} & \int_{0}^{\sqrt{2}/2} \int_{0}^{u} \frac{1}{1- \frac{u^2 - v^2}{2}} dv \>du \\ =& \int_{0}^{\sqrt{2}/2} \int_{0}^{u} \frac{2}{2- u^2 + v^2} dv \>du =\int_{0}^{\sqrt{2}/2} \frac{2}{\sqrt{2- u^2}} \arctan\frac u{\sqrt{2-u^2} }\>du \\ =& \int_{0}^{\sqrt{2}/2} \frac{2}{\sqrt{2- u^2}} \arcsin\frac u{\sqrt2 }\>du =\int_{0}^{\sqrt{2}/2} d\left(\arcsin^2\frac u{\sqrt2}\right) =\frac{\pi^2}{36} \end{align}