What is the indefinite integral $\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$? What is
$$\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$$
I have tried substituting everything and it doesn't seem to be working. Substituting trigonometry doesn't seem to work either. 

The primitive is quite hard to write down, but we can notice, by symmetry, that:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx = 2\int_{0}^{1/2}\sqrt{1-2\sqrt{x-x^2}}\,dx = 2\int_{0}^{1/2}\sqrt{1-2\sqrt{\frac{1}{4}-x^2}}\,dx$$
hence:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx=\int_{0}^{1}\sqrt{1-\sqrt{1-x^2}}\,dx=\int_{0}^{\pi/2}\cos\theta\sqrt{1-\cos\theta}\,d\theta$$
so:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx = \sqrt{2}\int_{0}^{\pi/2}\cos\theta \sin\frac{\theta}{2}\,d\theta = \color{red}{\frac{2}{3}(2-\sqrt{2})}.$$
By following the same steps for the indefinite integral, we have:

$$\int_{0}^{x} \sqrt{1-2\sqrt{z-z^2}}\,dz = \frac{2}{3}+\frac{2(2x-1)\left(1-\sqrt{x-x^2}\right)}{3\sqrt{1-2\sqrt{x-x^2}}}$$

for any $x\in\left[0,\frac{1}{2}\right]$. In order to compute the integral for $x\in\left[\frac{1}{2},1\right]$ it is sufficient to notice that the integrand function is symmetric around $z=\frac{1}{2}$.
