# Finding $\int\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x$.

In Arnold's Mathematical Methods of Classical Mechanics he uses,

$$\int_0^1\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x = \left\{\frac{\pi}{2-b} : 0\leqslant b < 2\right\}$$ but he doesn't explain how to get it.

Via Mathematica (wolfram alpha works also), $$\int\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x = -\frac{2x^{b/2}\sqrt{1-x^{2-b}}\arcsin{\big(x^{(2-b)/2} }\big)}{(2-b)\sqrt{x^b-x^2}}.$$

(Besides using a CAS) I am not sure how to solve the general case or the specific improper integral.

Write it in the following way: $$\int_0^1\frac{dx}{\sqrt{x^b-x^2}}=\int_0^1 dx\,\frac{x^{-b/2}}{\sqrt{1-x^{2-b}}}.$$ Now make a trigonometric substitution $x^{1-\tfrac{b}{2}}=\sin\theta$ so that $\left(1-\frac{b}{2}\right)x^{-b/2}dx=\cos\theta\,d\theta$ and the integral becomes (ignore the bounds for a moment) $$=\int d\theta\,\frac{\cos\theta}{\left(1-\frac{b}{2}\right) \cos \theta}=\left.\frac{1}{1-\frac{b}{2}}\sin^{-1}\left(x^{1-b/2}\right)\right|_0^1=\frac{\pi}{2-b}.$$