Calculate $\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x$ 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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\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
${\bf\mbox{6.1.18}}$.
A: Substitute $x=\sin{t}$. Then $t=\arcsin{x}$, $dt=\frac{dx}{\sqrt{1-x^2}}$, and:
$$I(n)=\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x = \int_{0}^{\pi/2}\sin^{2n}{t}\,\mathrm{d}t.$$
