Represent the function $f(x)=\sin ^{-1}x$ by a Maclaurin series.

Solution Recall that $$\sin ^{-1}x=\int_{0}^{x}\frac{dt}{\sqrt{1-t^{2}}}=\int_{0}^{x}(1-t^{2})^{-1/2}dt$$

We write the integrand as a binomial series, with $x=-t^{2}$ and $m=-\frac{1}{2}$. $$\begin{eqnarray*} (1-t^{2})^{-1/2} &=&1+\left( -\frac{1}{2}\right) (-t^{2}) + \frac{\left( -\frac{1}{2}\right) \left( -\frac{3}{2}\right) }{2!}(-t^{2}) ^{2}\\[8pt] && +\frac{\left( -\frac{1}{2}\right) \left( -\frac{3}{2} \right) \left( -\frac{5}{2}\right) }{3!}(-t^{2}) ^{3}+\cdots \\[8pt] &=&1+\frac{1}{2}t^{2}+\frac{3}{8}\,t^{4}+\frac{5}{16}t^{6}+\cdots \end{eqnarray*}$$

Now we use the integration property of a power series to obtain $$\begin{eqnarray*} \int_{0}^{x}\frac{dt}{\sqrt{1-t^{2}}} &=&\int_{0}^{x}\left( 1+\frac{1}{2} t^{2}+\frac{3}{8}\,t^{4}+\frac{5}{16}t^{6}+\cdots \right)\,dt \\ \sin ^{-1}x &=& x+\left( \frac{1}{2}\right) \left( \frac{x^{3}}{3}\right) +\left( \frac{3}{8}\right) \left( \frac{x^{5}}{5}\right) +\left( \frac{5}{16} \right) \left( \frac{x^{7}}{7}\right) +\cdots \\ &=& x+\frac{x^{3}}{6}+\frac{3}{40}x^{5}+\frac{5}{112}x^{7}+\cdots \end{eqnarray*}$$

In Problem 57, you are asked to show that the interval of convergence is $[-1,1].$