For integer $n\ge0$, Calculate: $$\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x.$$
I would like to get suggestions on how to calculate it? Should I expand $(1-x^2)^{-1/2}$ as a series?
Thanks.
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1trigonometric substitution should work just fine for you – mm-aops May 25 '14 at 23:11
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I am going in circle. I substitute $x=\sin\theta$ and now I get: $$\int_{0}^{\pi/2}\dfrac{\sin^{2n}\theta}{\cos\theta}\mathrm{d}\theta.$$ – Jika May 25 '14 at 23:19
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2$dx=\cos\theta d\theta$ – Math.StackExchange May 25 '14 at 23:19
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Thanks Aran. I get: $$\int_{0}^{\pi/2}\sin^{2n}\theta\mathrm{d}\theta.$$ – Jika May 25 '14 at 23:21
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \int_{0}^{1}{x^{2n} \over \root{1 - x^{2}}}\,\dd x& =\int_{0}^{1}{x^{n} \over \root{1 - x}}\,\half\,x^{-1/2}\,\dd x =\half\int_{0}^{1}x^{n - 1/2}\pars{1 - x}^{-1/2}\,\dd x \\[3mm]&=\half\,{\Gamma\pars{n + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{n + 1}} ={\root{\pi} \over 2\,n!}\,\Gamma\pars{n + \half}\tag{1} \end{align} $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$.
It's somehow related to Wallis Formula ${\bf\mbox{6.1.49}}$ since $$ \int_{0}^{1}{x^{2n} \over \root{1 - x^{2}}}\,\dd x =\int_{0}^{\pi/2}\sin^{2n}\pars{\theta}\,\dd\theta = {\pi \over 2^{2n + 1}} {2n \choose n}\tag{2} $$
$\pars{1}$ and $\pars{2}$ are related via Gamma Duplication Formula ${\bf\mbox{6.1.18}}$.
Felix Marin
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Substitute $x=\sin{t}$. Then $t=\arcsin{x}$, $dt=\frac{dx}{\sqrt{1-x^2}}$, and:
$$I(n)=\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x = \int_{0}^{\pi/2}\sin^{2n}{t}\,\mathrm{d}t.$$
David H
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