<s>[INST] Find recurrence for $I_n$. Let $I_n=\int_{0}^{1/2} \frac {x^n}{\sqrt{1-x^2}}dx.$ I must find a recurrence for this so I just started using interation by parts:
Let $$f'(x)=x^n\to f(x)=\frac{x^{n+1}}{n+1}$$ and
$$g(x)=\frac 1{\sqrt{1-x^2}}\to g'(x)=\frac x{\sqrt{(1-x^2)^3}}$$
Therefore:
$$I_n=\frac {x^{n+1}}{(n+1)\sqrt{1-x^2}} - \int_{0}^{\frac 12}\frac{x^{n+2}}{(1-x^2)\sqrt{1-x^2}}dx$$
And I can't continue from here..
I tried rewriting $x^{n+2}=x^2x^n=(1-x^2+1)x^n$ but it's no good.
 [/INST] $$I_n=\int_0^{1/2}\frac{x^n}{\sqrt{1-x^2}}dx\\
=-x^{n-1}\sqrt{1-x^2}\,\Big\vert_0^{1/2}+(n-1)\int_0^{1/2}x^{n-2}\sqrt{1-x^2}dx\\
-(1/2)^{n-1}(3/4)^{1/2}+(n-1)\int_0^{1/2}\frac{x^{n-2}-x^n}{\sqrt{1-x^2}}dx\\
=-(1/2)^{n-1}(3/4)^{1/2}+(n-1)(I_{n-2}-I_n).
$$
</s>