To find area enclosed within curve $x^4+y^4=x^2+y^2$ Find area enclosed by the curve 
$x^4+y^4=x^2+y^2$
My attempt:
How to describe $y$ as function of $x$.
Only after that we can integrate.
Or can this curve be described parametrically.

$$y^4-y^2+x^4-x^2=0$$
$$r^2(\cos^4(\theta)+\sin^4(\theta))=1$$
$$r^2=\frac 1 {\cos^4(\theta)+\sin^4(\theta)}$$
$$A/4=\frac 12\int_0^{\pi/2}\frac 1 {\cos^4(\theta)+\sin^4(\theta)}d\theta$$
$$A/4=\frac 12\int_0^{\pi/2} \frac 1{\cos^4 \theta}\frac{1}{1+\tan^4\theta}d\theta $$
Let $s=\tan(\theta)$ then $$A/4=\frac 12\int_0^{\infty}\frac{s^2+1}{s^4+1}ds$$
Decomposing the rational function:
$$A/4=\frac 14\int_0^{\infty}\frac1{s^2+\sqrt 2 s +1}-\frac1{-s^2+\sqrt 2 s -1}ds$$
$$A/4=\frac 14\int_0^{\infty}\frac1{(s+1/\sqrt2)^2+1/2}-\frac1{-(s+1/\sqrt2)^2-1/2}ds$$
And keeping in mind that $$\int \frac{1}{a^2+(x+b)^2}dx=\frac 1a\tan^{-1}\left(\frac{x+b}a\right)$$
You find that $$\frac A 4=\frac {\pi}{2\sqrt 2}$$
EDIT: Another, more interesting way of solving the integral: $$\frac{1}{\cos^4(\theta)+\sin^4(\theta)}=\frac{1}{1-2\sin^2(\theta)+2\sin^4(\theta)}=\frac{1}{1-\sin^2(2\theta)/2}=\frac{1}{1-\dfrac{1-\cos^2(2\theta)}{2}}=\frac{1}{\dfrac{2-1+\cos^2(2\theta)}{2}}=\frac{2}{1+\cos^2(\theta)}=\frac{1+\tan^2(2\theta)}{1+\dfrac{\tan^2(2\theta)}2}$$
If $u=\dfrac{\tan(2\theta)}{\sqrt2}$ then $du=\sqrt2(1+\tan^2(2\theta)) d\theta$, so that the integral becomes:$$\frac 1{2\sqrt2}\int \frac{du}{1+u^2}=\frac 1{2\sqrt2}\tan^{-1}(u)=\frac 1{2\sqrt2}\tan^{-1}\left(\dfrac{\tan(2\theta)}{\sqrt2}\right)$$
