# Parametric integration problem

If $$I_n = \int_0^1{x^n\sqrt{1-x^2}}\, \mathrm{d}x,$$ then find $$\lim_{n\to\infty}\frac{I_n}{I_{n-2}}.$$

• Are you familiar with the Beta Function? A substitution $x = \sin u$ puts this into a classic form, at which point we find that $I_n = \frac{\sqrt{\pi} \,\Gamma(\frac{n+1}{2})}{4 \Gamma(\frac{n}{2}+2)}$. Clearly the constants are irrelevant here, and the Gamma Functions can be simplified using well known properties to evaluate the limit Apr 28, 2017 at 4:55
• Using a trigonometric substitution, the expression reduces to a Wallis integral. Sep 1, 2017 at 14:15

Putting $x=\sin t$ we get $$I_{n} =\int_{0}^{\pi/2}\sin^{n}t\cos^{2}t\,dt$$ and using integration by parts we can show that $$(n+2)I_{n}=(n-1)I_{n-2}$$ and the desired limit is $1$.
More generally we can show that if $$J_{m, n} = \int_{0}^{\pi/2}\cos^{m}x\sin^{n}x\,dx$$ then $$(m + n)J_{m, n} = (m - 1)J_{m - 2, n} = (n - 1)J_{m, n - 2}$$ We have \begin{align} (m + 1)J_{m, n} &=-\int_{0}^{\pi/2}\sin^{n - 1}\frac{d}{dx}(\cos^{m + 1}x)\,dx\notag\\ &= -[\sin^{n - 1}x\cos^{m + 1}x]_{0}^{\pi/2} + \int_{0}^{\pi/2}(n - 1)\sin^{n - 2}x\cos^{m + 2}x\,dx\notag\\ &= (n - 1)\int_{0}^{\pi/2}\sin^{n - 2}x\cos^{m}x(1 - \sin^{2}x)\,dx\notag\\ &= (n - 1)J_{m, n - 2} - (n - 1)J_{m, n}\notag \end{align} And thus we get $$(m + n)J_{m, n} = (n - 1)J_{m, n - 2}$$ The other reduction formula is available by interchanging the roles of $m, n$ and noting that this does not affect $J_{m,n}$. For the current question we can see that $I_{n} = J_{2, n}$.
• @AjaySubramanian: I hope you know how I arrived at the reduction formula for $I_{n}$ using integration by parts. This is pretty standard and available in any introductory calculus text. Apr 28, 2017 at 5:18
• @Stephan: Note that $\sqrt{1-\sin^{2}t}=\cos t$ and the other $\cos t$ comes from $dx=\cos t\, dt$. Remember the rule of substitution $$\int f(x) \, dx=\int f(g(t)) g'(t) \, dt$$ where $x=g(t)$. Apr 28, 2017 at 10:19