\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We must have \begin {align*} \DiffFrac {y}{x} & = \DiffOp {x} \cdot \frac {\arcsin x}{\sqrt {1 - x^2}} \\ & = \frac {1}{1 - x^2} \cdot \left (\frac {1}{\sqrt {1 - x^2}} \cdot \sqrt {1 - x^2} - \arcsin x \cdot (-2x) \cdot \left (\frac {1}{2}\right ) \cdot \frac {1}{\sqrt {1 - x^2}}\right ) \\ & = \frac {1}{1 - x^2} \cdot \left (1 + x \cdot \frac {\arcsin x}{\sqrt {1 - x^2}}\right ) \\ & = \frac {1}{1 - x^2} \cdot \left (1 + xy\right ), \end {align*}
which gives \[ (1 - x^2) \DiffFrac {y}{x} - xy - 1 = (1 + xy) - xy - 1 = 0 \] as desired.
Differentiating both sides of this equation w.r.t. \(x\) gives \[ \NdiffFrac {2}{y}{x} \cdot (1 - x^2) - 2x \cdot \DiffFrac {y}{x} - y - x \DiffFrac {y}{x} = 0, \] which combined gives \[ (1 - x^2) \cdot \NdiffFrac {2}{y}{x} - 3x \cdot \DiffFrac {y}{x} - y = 0. \]
If we extend the definition of the differentiation operator to \[ \NdiffFrac {0}{y}{x} = y, \] then this precisely proves the desired statement for the case \(n = 0\) since \(2n + 3 = 3\) and \((n + 1)^2 = 1\), and we will prove the desired statement for all non-negative integer \(n\). The base case is shown as above.
Now, assume the given holds for some \(n = k\) where \(k\) is a non-negative integer, i.e. \[ (1 - x^2) \cdot \NdiffFrac {k + 2}{y}{x} - (2k + 3) x \cdot \NdiffFrac {k + 1}{y}{x} - (k + 1)^2 \cdot \NdiffFrac {k}{y}{x} = 0, \] we aim to show that the same holds for \(n = k + 1\).
Differentiating both sides with respect to \(x\) gives \[ (-2x) \cdot \NdiffFrac {k + 2}{y}{x} + (1 - x^2) \cdot \NdiffFrac {k + 3}{y}{x} - (2k + 3) \cdot \NdiffFrac {k + 1}{y}{x} - (2k + 3)x \cdot \NdiffFrac {k + 2}{y}{x} - (k + 1)^2 \cdot \NdiffFrac {k + 1}{y}{x} = 0, \] which then simplifies to \[ (1 - x^2) \cdot \NdiffFrac {k + 3}{y}{x} - (2k + 5)x \cdot \NdiffFrac {k + 2}{y}{x} - (k^2 + 4k + 4) \cdot \NdiffFrac {k + 1}{y}{x} = 0. \]
But notice that \(n + 2 = (k + 1) + 2 = k + 3\), \(n + 1 = (k + 1) + 1 = k + 2\), \((n + 1)^2 = (k + 2)^2 = k^2 + 4k + 4\), \(2n + 3 = 2(k + 1) + 3 = 2k + 5\), so this is exactly the statement when \(n = k + 1\), which finishes our inductive step.
Hence, by the Principle of Mathematical Induction, we can conclude that the original statement holds for any non-negative integer \(n\), and hence for any positive integer \(n\).
We have that \[ \LEvalAt {y}{x = 0} = \frac {\arcsin 0}{\sqrt {1 - 0^2}} = \frac {0}{1} = 0, \] and evaluating the equation on the first derivative at \(x = 0\) gives \[ \LEvalAt {\DiffFrac {y}{x}}{x = 0} = 1. \]
Evaluating the proven equation at \(x = 0\) gives \[ \LEvalAt {\NdiffFrac {n + 2}{y}{x}}{x = 0} = (n + 1)^2 \LEvalAt {\NdiffFrac {n}{y}{x}}{x = 0}. \]
Using this, we can conclude that \[ \LEvalAt {\NdiffFrac {2r}{y}{x}}{x = 0} = 0 \] for all \(r \geq 0\) where \(r\) is an integer, since it is \(0\) when \(n = 0\), and that \[ \LEvalAt {\NdiffFrac {2r + 1}{y}{x}}{x = 0} = ((2r)!!)^2 = 2^{2r} \cdot (r!)^2 \] for all \(r \geq 0\) where \(r\) is an integer, by mathematical induction.
Hence, the MacLaurin Series for \(\frac {\arcsin x}{\sqrt {1 - x^2}}\), must be \begin {align*} \frac {\arcsin x}{\sqrt {1 - x^2}} & = \sum _{k = 0}^{\infty } \frac {\LEvalAt {\NdiffFrac {k}{y}{x}}{x = 0}}{k!} \cdot x^k \\ & = \sum _{r = 0}^{\infty } \frac {\LEvalAt {\NdiffFrac {2r}{y}{x}}{x = 0}}{(2r)!} \cdot x^{2r} + \sum _{r = 0}^{\infty } \frac {\LEvalAt {\NdiffFrac {2r + 1}{y}{x}}{x = 0}}{(2r + 1)!} \cdot x^{2r + 1} \\ & = 0 + \sum _{r = 0}^{\infty } \frac {2^{2r} \cdot (r!)^2}{(2r + 1)!} \cdot x^{2r + 1} \\ & = \sum _{r = 0}^{\infty } \frac {2^{2r} \cdot (r!)^2}{(2r + 1)!} \cdot x^{2r + 1}. \end {align*}
This means the general term for even powers of \(x\) is zero, and the general term for odd powers of \(x\) is \[ \frac {2^{2r} \cdot (r!)^2}{(2r + 1)!} \cdot x^{2r + 1} \] where \(r\) is any non-negative integer.
The infinite sum can be expressed as \[ \sum _{r = 0}^{\infty } \frac {(r!)^2}{(2r + 1)!} = 2 \cdot \sum _{r = 0}^{\infty } \frac {2^{2r} \cdot (r!)^2}{(2r + 1)!} \cdot \left (\frac {1}{2}\right )^{2r+1}, \] which is precisely double the value of \[ \SqEvalAt {\frac {\arcsin x}{\sqrt {1 - x^2}}}{x = \frac {1}{2}} = \frac {\arcsin \frac {1}{2}}{\sqrt {1 - \left (\frac {1}{2}\right )^2}} = \frac {\pi / 6}{\sqrt {3} / 2} = \frac {\pi }{3 \sqrt {3}}, \]
Hence, the sum evaluates to \(\frac {2\pi }{3 \sqrt {3}}\).