# Show that if $c_n = \int_{-1}^1 (1-t^2)^\frac{n-1}{2}dt$ then $c_n = \frac{n-1}{n}c_{n-2}$

Question: Show that if $c_n = \int_{-1}^1 (1-t^2)^\frac{n-1}{2}dt$ then $c_n = \frac{n-1}{n}c_{n-2}$. I have tried two things both of which I didn't get super far with. First, I tried dividing n into two cases odd and even and applying the binomial theorem and second I tried integration by parts with $u = (1-t^2)^\frac{n-1}{2}$ and $dv = dt$. Maybe one of these approaches are right and I just screwed up somewhere but some help would be appreciated.

• $1= t^2+(1-t^2)$ and integration by parts. – Jack D'Aurizio Mar 1 '17 at 1:26
• You can rearrange this to show that $c_n +\frac{1}{n} c_{n-2} = 1$ is equivalent to your identity. It could potentially be easier to work with this. – Mark Mar 1 '17 at 1:29
• This can be inspiring: math.stackexchange.com/questions/2162309/… – Olivier Oloa Mar 1 '17 at 1:31

Observe that $$c_n = \int_{-1}^{0}(1-t^2)^{\frac{n-1}{2}}dt + \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = \\ =2\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt,$$

where you make the substitution $t\mapsto -t$ in the first integral (or just observe that the function being integrated is even). Then, we do partial integration:

$$\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = t(1-t^2)^{\frac{n-1}{2}}\big|_{0}^{1} + (n-1)\int_{0}^{1}t^2(1-t^2)^{\frac{n-3}{2}}dt = \\ =(n-1) \int_{0}^{1}t^2(1-t^2)^{\frac{n-3}{2}}dt.$$

Then, we conclude that $$(n-1)\int_{0}^{1}(1-t^2)^{\frac{n-3}{2}}dt - \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = (n-1)\int_{0}^{1}(1-t^{2})^{\frac{n-1}{2}}dt,$$

which means $$\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = \frac{n-1}{n}\int_{0}^{1}(1-t^2)^{\frac{n-3}{2}}dt.$$

Then, it follows that $$\frac{c_n}{2}=\frac{n-1}{n}\frac{c_{n-2}}{2} \implies c_n = \frac{n-1}{n}c_{n-2}.$$

Let $t=\cos x$ with $x\in [0,\pi].$

For $n\geq 1$ we have $$c_n=\int_{\pi}^0 (\sin^2 x)^{(n-1)/2}\;d\cos x=\int_{\pi}^0(\sin x)^{n-1}(-\sin x)\;dx=$$ $$=\int_0^{\pi}\sin^n x\;dx.$$ Now for $n\geq 2$ we have, integrating by parts, $$c_n=\int_0^{\pi}\sin^{n-1}x\;d (-\cos x)=$$ $$=(\sin^{n-1}x)(-\cos x)|_{x=0}^{x=\pi}\;-\int_0^{\pi}(-\cos x)(n-1)(\sin^{n-2}x) (\cos x)\;dx.$$ And since $0=(\sin^{n-1}0)(\cos 0)=(\sin^{n-1}\pi)(\cos \pi)$ when $n\geq 2$, we have $$c_n= \int_0^{\pi}(n-1)(\cos^2 x)(\sin^{n-2}x)\; dx=\int_0^{\pi}(n-1)(1-\sin^2x)(\sin^{n-2}x)\;dx=$$ $$=\int_0^{\pi} (n-1)(\sin^{n-2}x-\sin^nx)\;dx=$$ $$=(n-1)c_{n-2}-(n-1)c_n.$$