What would be the best/quickest way to integrate

squareroot(1-x^2) ?

Trigonometric or Hyperbolic substitution works well...

Let \(\displaystyle x = \sin{\theta}\) so that \(\displaystyle dx = \cos{\theta}\,d\theta\).

The integral becomes

\(\displaystyle \int{\sqrt{1 - x^2}\,dx} = \int{\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}\)

\(\displaystyle = \int{\cos{\theta}\cos{\theta}\,d\theta}\)

\(\displaystyle = \int{\cos^2{\theta}\,d\theta}\)

\(\displaystyle = \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}\)

\(\displaystyle = \frac{1}{4}\sin{2\theta} + \frac{1}{2}\theta + C\)

\(\displaystyle = \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C\)

\(\displaystyle = \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C\).

Remembering that \(\displaystyle x = \sin{\theta}\), that means \(\displaystyle \cos{\theta} = \sqrt{1 - x^2}\) and \(\displaystyle x = \arcsin{\theta}\) and substituting, we find

\(\displaystyle \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C\).