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.
 A: $$I_n-I_{n+2}=\int_0^{1/2}\frac{x^n(1-x^2)}{\sqrt{1-x^2}}\,dx=\int_0^{1/2}x^n\sqrt{1-x^2}\,dx\\$$Now do integration by parts on this, by integrating $x^n$ and differentiating the other term.
This gives:
$$I_n-I_{n+2}=\left[\frac{1}{n+1} x^{n+1}\sqrt{1-x^2}\right]_0^{1/2}+\frac{1}{n+1}\int_0^{1/2}\frac{x^{n+2}}{\sqrt{1-x^2}}\,dx\\I_n=\frac{\sqrt3}{2^{n+2}(n+1)}+\frac{n+2}{n+1}I_{n+2}\\I_{n+2}=\frac{n+1}{n+2}I_n-\frac{\sqrt 3}{2^{n+2}(n+2)}$$
A: $$I_n=\int_0^{1/2}\frac{x^n}{\sqrt{1-x^2}}dx\\
=-x^{n-1}\sqrt{1-x^2}\,\Big\vert_0^{1/2}+(n-1)\int_0^{1/2}x^{n-2}\sqrt{1-x^2}dx\\
-(1/2)^{n-1}(3/4)^{1/2}+(n-1)\int_0^{1/2}\frac{x^{n-2}-x^n}{\sqrt{1-x^2}}dx\\
=-(1/2)^{n-1}(3/4)^{1/2}+(n-1)(I_{n-2}-I_n).
$$
A: Let's replace the variable $x=\sin t$  and integration by part becomes more pleasant 
$$I_n=\int\limits_0^{\frac\pi6}\sin^nt\,dt=-\cos t\sin^{n-1}t\bigg|_0^{\frac\pi6}+
(n-1)\int\limits_0^{\frac\pi6}\cos^2t\sin^{n-2}t\,dt=$$
$$-\frac{\sqrt3}{2^n}+(n-1)(I_{n-2}-I_n)$$
So $$\boxed{I_n=-\frac{\sqrt3}{2^nn}+\frac{n-1}nI_{n-2}}$$
