Integration of complicated equation I was looking at a Wolfram Mathworld article about bean curves: https://mathworld.wolfram.com/BeanCurve.html
And it states that the area enclosed by the curve of
$$x^4+x^2y^2+y^4=ax(x^2+y^2)$$ is
$$A=\sqrt{2}a^2\int_0^1\sqrt{x\left(1-x+\sqrt{1+(2-3x)x}\right)}\text{d}x=\frac{7\pi a^2}{12\sqrt{3}}$$
Although a brief calculation shows that the area enclosed by the curve is $$A=\sqrt{2}a^2\int_0^1\sqrt{x\left(1-x+\sqrt{1+(2-3x)x}\right)}\text{d}x$$ I have no clue how the integration, in which $$A=\frac{7\pi a^2}{12\sqrt{3}}$$
, works.
I would like to know how!
 A: use polar coordinates to evaluate the area.
$$r^4(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))=ar^3\cos(\theta)$$
$$r=\frac{a\cos(\theta)}{\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta)}$$
$$\frac{A}{a^2}=\frac{1}{2a^2}\int_{-\pi/2}^{\pi/2}r^2 d\theta$$
$$=\frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\cos^2(\theta)}{(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))^2} d\theta$$
$$=\int_{0}^{\pi/2} \frac{\cos^2(\theta)}{(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))^2} d\theta$$
$$=\int_{0}^{\pi/2} \frac{\cos^2(\theta)}{(1-\frac{1}{4}\sin^2(2\theta))^2} d\theta$$
$$=8\int_{0}^{\pi/2} \frac{2\cos^2(\theta)}{(4-sin^2(2\theta))^2} d\theta$$
$$=8\int_{0}^{\pi/2} \frac{1+\cos(2\theta)}{(4-\sin^2(2\theta))^2} d\theta$$
$$=4\int_{0}^{\pi} \frac{1+\cos(\phi)}{(4-\sin^2(\phi))^2} d\phi$$
$$=4\int_{0}^{\pi} \frac{1}{(4-\sin^2(\phi))^2} d\phi$$
$$=16\int_{0}^{\pi} \frac{1}{(8-2\sin^2(\phi))^2} d\phi$$
$$=16\int_{0}^{\pi} \frac{1}{(7+\cos(2\phi))^2} d\phi$$
$$=8\int_{0}^{2\pi} \frac{1}{(7+\cos(\alpha))^2} d\alpha$$
$$=16\int_{0}^{\pi} \frac{1}{(7+\cos(\alpha))^2} d\alpha$$
$$=16*\frac{7\sqrt3}{288}*\tan^{-1}(\frac{\sqrt3}{2}\tan(\frac{\alpha}{2}))|_0^{\pi}$$
$$=\frac{7\sqrt3}{18}*\frac{\pi}{2}$$
$$A=\frac{7\pi a^2}{12\sqrt3}$$
change of variables: $\phi=2\theta$ and $\alpha=2\phi$
