Answer
We show that
${\sin ^{ - 1}}x = x + \mathop \sum \limits_{n = 1}^\infty \dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot4\cdot6\cdot\cdot\cdot\left( {2n} \right)}}\dfrac{{{x^{2n + 1}}}}{{2n + 1}}$,
for $\left| x \right| \lt 1$.
Work Step by Step
Write $f\left( x \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }} = {\left( {1 + \left( { - {x^2}} \right)} \right)^{ - 1/2}}$.
We have from Table 2:
${\left( {1 + x} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} = 1 + ax + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot$
converges for $\left| x \right| \lt 1$.
Let $a = - \dfrac{1}{2}$. Substituting $ - {x^2}$ for $x$ in the series above, we get
${\left( {1 + \left( { - {x^2}} \right)} \right)^{ - 1/2}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - {x^2}} \right)^n} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}{x^{2n}}$, ${\ \ \ }$ for $\left| { - {x^2}} \right| \lt 1$, or $\left| x \right| \lt 1$.
From Theorem 1 in Section 7.8, we know that
${\sin ^{ - 1}}x = \smallint \dfrac{1}{{\sqrt {1 - {x^2}} }}{\rm{d}}x$
Therefore,
${\sin ^{ - 1}}x = \smallint \dfrac{1}{{\sqrt {1 - {x^2}} }}{\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\smallint {x^{2n}}{\rm{d}}x$
${\sin ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{{{{\left( { - 1} \right)}^n}}}{{2n + 1}}{x^{2n + 1}} + C$, ${\ \ \ }$ where $C$ is a constant of integration
Setting $x=0$, we get ${\sin ^{ - 1}}0 = 0 = C$. Thus
${\sin ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{{{{\left( { - 1} \right)}^n}}}{{2n + 1}}{x^{2n + 1}}$
Since $\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}$, $\left( {\begin{array}{*{20}{c}}
a\\
0
\end{array}} \right) = 1$, so
${\sin ^{ - 1}}x = x + \mathop \sum \limits_{n = 1}^\infty \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot}}{{n!}}{\left( { - 1} \right)^n}\dfrac{{{x^{2n + 1}}}}{{2n + 1}}$
$ = x + \dfrac{1}{{2\cdot1!}}\dfrac{{{x^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot2\cdot2!}}\dfrac{{{x^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot2\cdot2\cdot3!}}\dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$
$ = x + \dfrac{1}{2}\dfrac{{{x^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot2\cdot\left( {1\cdot2} \right)}}\dfrac{{{x^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot2\cdot2\cdot\left( {1\cdot2\cdot3} \right)}}\dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$
$ = x + \dfrac{1}{2}\dfrac{{{x^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot4}}\dfrac{{{x^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}\dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$
Hence,
${\sin ^{ - 1}}x = x + \mathop \sum \limits_{n = 1}^\infty \dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot4\cdot6\cdot\cdot\cdot\left( {2n} \right)}}\dfrac{{{x^{2n + 1}}}}{{2n + 1}}$
