The task is to find $\int_L {x\sqrt{x^2-y^2}}$ds, where $L: (x^2+y^2)^2 = a^2 (x^2 - y^2),\ x\ge0$.
The curve is a right loop of a lemniscate. By tranistioning to polar coordinates I found:
$$r^2 = a^2 (cos^2(\theta)-sin^2(\theta)),\ r'=\frac{-2cos(\theta)sin(\theta)}{\sqrt{cos^2(\theta)-sin^2(\theta)}}$$
Then the integral becomes $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^2 cos(\theta) \sqrt{cos^2(\theta)-sin^2(\theta)}\sqrt{r^2+(r')^2} d\theta$.
The problem is that after plugging in the value of $r$, I get something increadibly messy and most likely not integrable in elementary functions. And the task is supposed to have an elementary simple answer and actually is aimed at people familiar only with some basics of integration, by far not assuming the knowledge of non-elementary functions and curves like lemniscate.
Maybe I'm missing something obvious? Or it is really that horrible...
Edit: yeah, now I see that my calculations aren't correct
Letting $x=r\cos\theta, y=r\sin\theta$ we have that the relation between $r$ and $\theta$ is given by
$$ r^2 = a^2\cos(2\theta) $$ or $r=a\sqrt{\cos(2\theta)}$, for $\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Given $$ x=a\cos\theta\sqrt{\cos(2\theta)},\quad y=a\sin\theta\sqrt{\cos(2\theta)} $$ we have $$ ds = \sqrt{\dot{x}^2+\dot{y}^2}\,d\theta =\frac{a}{\sqrt{\cos(2\theta)}}\,d\theta$$ hence the wanted integral equals $$ \int_{-\pi/4}^{\pi/4}a\cos\theta\sqrt{\cos(2\theta)}\cdot a\cos(2\theta)\cdot \frac{a}{\sqrt{\cos(2\theta)}}\,d\theta $$ and I guess you can see the nice simplification here. The outcome is $$ a^3 \int_{-\pi/4}^{\pi/4}\cos(\theta)\cos(2\theta)\,d\theta = \frac{2\sqrt{2}}{3}a^3.$$