Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $

Question

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.

$$ u = x ^ n $$ $$ v = \arcsin x $$ $$ \mathrm d v = \frac { \mathrm d x } { \sqrt { 1 - x ^ 2 } } $$ $$ I _ n = \int u \ \mathrm d v = u v - \int v \ \mathrm d u = x ^ n \arcsin x - ( n - 1 ) \int x ^ { n - 1 } \arcsin x \ \mathrm d x $$

Answer 1

$$I_n=\int\frac{x^n}{\sqrt{1-x^2}}dx$$

Let $x=\sin y\implies dx=\cos y \,dy$

$$I_n=\int\sin^ny \,dy$$

which has a well known reduction formula

Answer 2

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

$$I_n = -\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n I_{n-2}$$