Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $ I'm struggling to go any further. Anyone have any hints?
$$ I _ n = \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $$
I used integration by parts where I differentiated $ x ^ n $, but it resulted in an $ \arcsin x $ term which didn't get me anywhere.
$$ u = x ^ n $$
$$ v = \arcsin x $$
$$ \mathrm d v = \frac { \mathrm d x } { \sqrt { 1 - x ^ 2 } } $$
$$ I _ n = \int u \ \mathrm d v = u v - \int v \ \mathrm d u = x ^ n \arcsin x - ( n - 1 ) \int x ^ { n - 1 } \arcsin x \ \mathrm d x $$

Integrate by parts as follows
\begin{align}
 \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } d x 
&= -\int \frac { x ^{n-1} } { (1 - x ^ 2 )^{\frac{n-1}2}} d \left(\frac{(1-x^2)^{\frac n2}}n\right)\\
&=-\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n\int \frac { x ^ {n-2} } { \sqrt { 1 - x ^ 2 } } d x \\
\end{align}
hence the reduction formula
$$I_n = -\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n I_{n-2}$$
