Finding length!

Adding $x^2+y^2$ to both sides of the given equation $x^4+y^4+2x^2y^2+2xy^2+2x^3+x^2=x^2+y^2.$ By factoring the right side we get $(x^2+y^2+x)^2=x^2+y^2$ . Now we can convert this equation to polar coordinates, so we get $(r^2+r \cos \theta)^2= r^2$ . Then taking square root of both sides we get $r^2+r \cos \theta= \pm r$ . Dividing both sides of these equations by $r$ we obtain two equations yielding the same polar curve; indeed, the equations are $r+ \cos \theta= \pm 1$ . We can use one of this two equations to find the length. Let us pick $r+\cos \theta =1$ or equivalently, $r =1-\cos \theta$ . Then the arc length will be $L=\int_{0}^{2\pi} \sqrt{r^2+(r')^2}d\theta=\int_{0}^{2\pi} \sqrt{(1-\cos \theta)^2+(\sin \theta)^2}d \theta=\int_{0}^{2\pi} \sqrt{2- 2\cos \theta}d\theta=$ $=2\int_{0}^{2\pi} |\sin \frac{\theta}{2}| d\theta=8.$