<s>[INST] Definite Integral of $\int_{\frac{-1}{2}}^\frac{1}{2}\int_{\frac{-1}{2}}^\frac{1}{2} \sqrt{x^2+y^2} dxdy$ Here's my attempt at trying to evaluate the integral.
Let $x = y tan\theta$
$$\frac{dx}{d\theta} = \frac{y}{cos^2\theta}$$
$$dx = \frac{y}{cos^2\theta}d\theta$$
The new bounds of inner integral would be $\theta = tan^-(\frac{1}{2y}) $ and $\theta = tan^-(\frac{-1}{2y})$
$$\int_{\frac{-1}{2}}^\frac{1}{2}\int_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})} y^2 sec^3{\theta} dy$$
Evaluating the innermost integral
$$\int_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})} sec^3{\theta} dy =  {\huge|} \frac{sec\theta tan\theta + ln|sec\theta + tan\theta|}{2}{\huge|}_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})}$$
$$\begin{multline} = \left( \frac{sec(tan^-(\frac{1}{2y})) tan(tan^-(\frac{1}{2y})) + ln|sec(tan^-(\frac{1}{2y})) + tan(tan^-(\frac{1}{2y}))|}{2}\right)  
 - \\ \left( \frac{sec(tan^-(\frac{-1}{2y})) tan(tan^-(\frac{-1}{2y})) + ln|sec(tan^-(\frac{-1}{2y})) + tan(tan^-(\frac{-1}{2y}))|}{2}\right)  \end{multline}$$
$$\begin{multline} = \left( \frac{\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{2y}|}{2}\right)  
 - \left( \frac{-\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} - 1}{2y}|}{2}\right)  \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{2y}|\right)  
 - \left(-\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} - 1}{2y}|\right) \right] \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{2y^2} + ln|\sqrt{4y^2 + 1} + 1| 
 - ln|\sqrt{4y^2 + 1} - 1| \right) \right] \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{2y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{\sqrt{4y^2 + 1} - 1}| \right) \right] \end{multline}$$
Evaluating the outermost integral
$$\begin{multline} \int_{\frac{-1}{2}}^\frac{1}{2}\frac{\sqrt{4y^2 + 1}}{4} + y^2 \frac{ln{\large|}\frac{\sqrt{4y^2 + 1} + 1}{\sqrt{4y^2 + 1} - 1}{\large|}}{2} dy \end{multline}$$
 I am kinda stuck at this point, any help is greatly appreciated -:). I may have done something wrong in the steps above.
 [/INST] We can use symmetry to say that 
$$I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^2+y^2}\:dxdy = 8 \int_{0}^{\frac{1}{2}}\int_{0}^{x}\sqrt{x^2+y^2}\:dydx$$
From here, we can use polar coordinates:
$$I = 8 \int_{0}^{\frac{\pi}{4}} \int_0^{\frac{1}{2}\sec\theta}r^2\:drd\theta = \frac{1}{3}\int_{0}^{\frac{\pi}{4}}\sec^3\theta\:d\theta$$
Now, we could use integration by parts and trig identities to solve this integral, but instead let's use the substitution $\tan \theta = \sinh \tau$:
$$I = \frac{1}{3}\int_0^{\sinh^{-1}(1)}\cosh^2 \tau\:d\tau = \frac{1}{6}\int_0^{\sinh^{-1}(1)} 1 + \cosh 2\tau \:d\tau$$
$$= \frac{1}{6}+\frac{1}{6}\sinh\tau\cosh\tau \Biggr|_0^{\sinh^{-1}(1)} = \frac{\sinh^{-1}(1)+\sqrt{2}}{6}$$
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