We give an example of obtaining the Taylor series expansion of an elementary function by integrating the Taylor series of its derivative.

For $-1 < x < 1$ we have the derivative of the principal branch of the arcus sine function: $$\frac{d\,\arcsin{x}}{dx} = \frac{1}{\sqrt{1\!-\!x^2}} = (1\!-\!x^2)^{-\frac{1}{2}}.$$

Using the generalized binomial coefficients ${-\frac{1}{2} \choose r}$ we thus can form the Taylor series for it as Newton's binomial series: $$(1\!-\!x^2)^{-\frac{1}{2}} = \sum_{r = 0}^\infty{-\frac{1}{2} \choose r}(-x^2)^r = 1\!+\!{-\frac{1}{2}\choose 1}(-x^2)\!+\!{-\frac{1}{2}\choose 2}(-x^2)^2\!+\! {-\frac{1}{2}\choose 3}(-x^2)^3\!+\!\cdots =$$ $$ =1\!-\!\frac{-\frac{1}{2}}{1!}x^2\!+ \!\frac{-\frac{1}{2}(-\frac{1}{2}\!-\!1)}{2!}x^4\! -\!\frac{-\frac{1}{2}(-\frac{1}{2}\!-\!1)(-\frac{1}{2}\!-\!2)}{3!}x^6\!+-\cdots =$$ $$ = 1\!+\!\frac{1}{2}x^2\!+\!\frac{1\cdot 3}{2\cdot 4}x^4\!+\! \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6\!+\!\cdots \quad\quad\quad\quad \mathrm{for}\,\, -1 < x < 1$$

Because $\arcsin{0} = 0$ for the principal branch of the function, we get, by integrating the series termwise, the expansion $$\arcsin{x} = \int_0^x\frac{dx}{\sqrt{1\!-\!x^2}} = x\!+\!\frac{1}{2}\!\cdot\!\frac{x^3}{3}\!+ \!\frac{1\!\cdot\!3}{2\!\cdot\!4}\!\cdot\!\frac{x^5}{5}\!+\! \frac{1\!\cdot\!3\cdot\!5}{2\!\cdot\!4\cdot\!6}\!\cdot\!\frac{x^7}{7}\!+\!\cdots,$$ the validity of which is true for $|x| < 1$ . It can be proved, in addition, that it is true also when $x = \pm 1$ .