$\int_0^1 x^n \sqrt{1-x^2}$ I am trying to solve the indeterminate integral first.By multiplying integrand with $\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}$ we get
$\int \frac{x^n-x^{n+2}}{\sqrt{1-x^2}}$
I then used ostrogradsky method thinking that I would get bunch of zeroes in the places of the coefficients but this isn't the case.
Is there some other way to approach this?
Any tips?
 A: Under $x^2\to x$ and by using Beta and Gamma functions, one has
\begin{eqnarray}
\int_0^1 x^n \sqrt{1-x^2}dx&=&\int_0^1 x^{\frac{n-1}{2}} \sqrt{1-x}dx\\
&=&\int_0^1 x^{\frac{n+1}{2}-1} (1-x)^{\frac32-1}dx\\
&=&B(\frac{n+1}{2},\frac32)\\
&=&\frac{\Gamma(\frac{n+1}{2})\Gamma(\frac32)}{\Gamma(\frac{n+4}{2})}\\
&=&\frac{\sqrt{\pi } \Gamma (\frac{n+1}{2})}{4 \Gamma (\frac{n}{2}+2)}
\end{eqnarray}
where
$$ \Gamma(\frac32)=\frac12\Gamma(\frac12)=\frac{\sqrt\pi}{2}.$$
A: You could try a recurrence method such as:
$$I_n=\int_0^1x^n\sqrt{1-x^2}dx$$
By parts, with $u=x^{n-1}$ and $v'=x((1-x^2)^{\frac12}$
$$\implies u'=(n-1)x^{n-2}$$ and $$v=-\frac13(1-x^2)^{\frac32}$$
$$\implies I_n=\left[-\frac13x^{n-1}(1-x^2)^{\frac32}\right]_0^1+\frac{n-1}{3}\int x^{n-2}(1-x^2)(1-x^2)^{\frac12}dx$$
$$\implies I_n=0+\frac{n-1}{3}\left[I_{n-2}-I_n\right]$$
$$\implies I_n=\frac{n-1}{n+2}I_{n-2}$$
...and take it from there, perhaps?
A: $\newcommand{\d}{\,\mathrm{d}}$Substitute $u=1-x^2$ to get $x=\sqrt{1-u},\d x=-\frac{1}{2}(1-u)^{-1/2}$ and note that $[0,1]\mapsto[1,0]$ under this transformation: $$\large{\begin{align}\int_0^1 x^n\sqrt{1-x^2}\d x&=\frac{1}{2}\int_0^1(1-u)^{\frac{1}{2}(n-1)}u^{1/2}\d u\\&=\frac{1}{2}\mathcal{B}\left(\frac{n+1}{2},\frac{3}{2}\right)\\&=\frac{1}{4}\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+2\right)}\quad\text{Valid for any real $n\gt0$}\\&=\frac{\sqrt{\pi}}{4}\cdot\begin{cases}\frac{(k-1/2)(k-3/2)\cdots(1/2)}{(k+1)!}\sqrt{\pi}&n=2k\in\Bbb N\\\frac{(k-1)!}{(k+1/2)(k-1/2)(k-3/2)\cdots(1/2)\sqrt{\pi}}&n+1=2k\in\Bbb N\end{cases}\\&=\begin{cases}\pi\cdot 2^{-\left(\frac{n}{2}+2\right)}\cdot\frac{(n-1)!!}{\left(\frac{n}{2}+1\right)!}&n\text{ is even}\\2^{\frac{1}{2}\left(n-1\right)}\cdot\frac{\left(\frac{1}{2}(n-1)\right)!}{(n+2)!!}&n\text{ is odd}\end{cases}\end{align}}$$
Where used was $x\Gamma(x)=\Gamma(x+1)$, $\Gamma(1/2)=\sqrt{\pi}$, and:
$$\int_0^1x^a(1-x)^b\d x=\mathcal{B}(a+1,b+1)=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$
And $\Gamma,\mathcal{B}$ denote the respective Gamma and Beta functions.
I appreciate this looks like a copy of xpaul's answer but I was typing this long beforehand :) I just got into a muddle and spent a while trying to convert the closed Gamma forms into (double) factorial form for odd/even $n$.
