how can we prove that $\displaystyle \int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}$

I think the proof starts by letting $\displaystyle x=cos(u)$ but I didn't reach anything

Thanks

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- Jun 2nd 2010, 09:53 AMmohammadfawazProving Inequality
how can we prove that $\displaystyle \int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}$

I think the proof starts by letting $\displaystyle x=cos(u)$ but I didn't reach anything

Thanks - Jun 3rd 2010, 04:03 AMOpalg
I don't know where the $\displaystyle \pi/2$ comes from, but there is a fairly simple way to prove a stronger inequality.

Let $\displaystyle I_n = \int_0^1(1-x^2)^n\,dx$. Integrate by parts to get $\displaystyle I_n = \Bigl[x(1-x^2)^n\Bigr]_0^1 + 2n\int_0^1x^2(1-x^2)^{n-1}dx = 2n(I_{n-1}-I_n)$. Therefore $\displaystyle I_n = \frac{2n}{2n+1}I_{n-1}$.

Using that recurrence relation, you can show by induction that $\displaystyle I_n\leqslant \frac1{\sqrt{n+1}}$. In fact, the base case n=0 is obvious. The inductive step amounts to the inequality $\displaystyle \frac{2n+2}{2n+3}\,\frac1{\sqrt{n+1}}\leqslant \frac1{\sqrt{n+2}}$, which you can check by squaring both sides.

Since $\displaystyle \frac1{\sqrt{n+1}}$ is clearly less than $\displaystyle \frac\pi{2\sqrt n}$ , that is a better inequality than what was asked for.