TITLE: $\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{x^2+y^2}}\,dx\,dy=$?
QUESTION [5 upvotes]: I'm having difficulties thinking of a good variable change for 
$$\int_0^1 \int_0^1 \frac{1}{\sqrt{x^2+y^2}}\,dx\,dy=?$$
the most natural choice would be something like $x=r\cos\theta$ and $y=r\sin\theta$ since that would make it a very simple integral to calculate, but the bounds of $r$ and $\theta$ are all mixed up. This is a square. not a circle.

REPLY [8 votes]: Here is another way to perform the integral without using polar coordinates.
Notice the integrand and the domain to integrate are both symmetric with respect to $x \leftrightarrow y$.
This means we can split the integral into two equal pieces. One piece for $x \le y$ and another piece for $x \ge y$. On one of the piece, say $x \le y$, introduce variable $t$ such that $x = yt$, we have:
$$
\int_0^1\int_0^1 \frac{1}{\sqrt{x^2+y^2}} dx dy 
= 2 \int_0^1 \int_0^y \frac{1}{\sqrt{x^2+y^2}} dx dy
= 2 \int_0^1 \left(\int_0^1 \frac{1}{y\sqrt{1+t^2}} ydt\right) dy\\
= 2\int_0^1\frac{1}{\sqrt{1+t^2}}dt
= 2\sinh^{-1}(1)
= 2\log(\sqrt{1^2+1}+1)
= 2\log(\sqrt{2}+1)$$