I'm also learning line integration, but I think I got this one right.
You want to find
$\sqrt {{{\left( {\frac{{dx}}{{d\phi }}} \right)}^2} + {{\left( {\frac{{dy}}{{d\phi }}} \right)}^2}} $
We have that
$\eqalign{ & \frac{{dx}}{{d\phi }} = \frac{{dr}}{{d\phi }}\cos \phi - r\sin \phi \cr & \frac{{dy}}{{d\phi }} = \frac{{dr}}{{d\phi }}\sin \phi + r\cos \phi \cr} $
So we get
$\eqalign{ & {\left( {\frac{{dx}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2}{\cos ^2}\phi - 2\frac{{dr}}{{d\phi }}r\cos \phi \sin \phi + {r^2}{\sin ^2}\phi \cr & {\left( {\frac{{dy}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2}{\sin ^2}\phi + 2\frac{{dr}}{{d\phi }}r\cos \phi \sin \phi + {r^2}{\sin ^2}\phi \cr} $
This means that
${\left( {\frac{{dx}}{{d\phi }}} \right)^2} + {\left( {\frac{{dy}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2} + {r^2}$
(Actually, this is a known result, namely that in polar coordinates one has ds = \sqrt{r^2+r'^2}d\phi
Moving on, we have
$\eqalign{ & {r^2} = {a^2}\cos 2\phi \cr & {\left( {\frac{{dr}}{{d\phi }}} \right)^2} = {a^2}\frac{{\sin 2\phi }}{{\cos 2\phi }}\sin 2\phi \cr} $
So
$\sqrt {{{\left( {\frac{{dr}}{{d\phi }}} \right)}^2} + {r^2}} = \sqrt {\frac{{{a^2}\left( {{{\sin }^2}2\phi + {{\cos }^2}2\phi } \right)}}{{\cos 2\phi }}} = \frac{a}{{\sqrt {\cos 2\phi } }}$
The integral ends up being rather "user friendly"
$\eqalign{ & {a^2}\int\limits_{ - \pi /4}^{\pi /4} {\frac{{\sqrt {\cos 2\phi } \cos \phi - \sqrt {\cos 2\phi } \sin \phi }}{{\sqrt {\cos 2\phi } }}} d\phi \cr & {a^2}\int\limits_{ - \pi /4}^{\pi /4} {\left( {\cos \phi - \sin \phi } \right)} d\phi = {a^2}\sqrt 2 \cr} $