Definite Integration with limits

Question

Let

$$I_n := \int\limits_0^1 x^n\sqrt{1-x^2} \,\mathrm d x$$

What is the following limit equal to?

$$\lim_{n\to\infty} \dfrac{I_n}{I_{n-2}}$$

I tried using the integration by parts method, but couldn't come to any final answer.

Answer 1

Hint:

Set $x=\sin t,dx=\cos t\ dt$

$$I_n=\int_0^{\pi/2}\sin^nt\cos^2t\ dt=\int_0^{\pi/2}\sin^nt(1-\sin^2t)\ dt=J_n-J_{n+2}$$

where $\displaystyle J_m=\int_0^{\pi/2}\sin^mt\ dt$

Use this to find $$J_m=\dfrac{m-1}mJ_{m-2}$$

$I_n=J_n-J_{n+2}=J_n-\dfrac{m-1}{m+2}J_n=\dfrac{3J_n}{m+2}$

$I_{n-2}=?$

Answer 2

Integrate by parts $u=x^{n-1}, dv = x\sqrt{1-x^2}\;dx$. I chose this because I can easily integrate $dv$. Then we get $$ \int_{0}^{1}\!{x}^{n}\sqrt {1-x^2}\,{\rm d}x=-\frac{n-1}{3} \int_{0}^{1}\, { {{x}^{n-2} (x^2-1) \sqrt {1-x^2}}}\,{\rm d}x $$ Tne right-hand-side is a linear combination of $I_n$ and $I_{n-2}$.

Answer 3

Alternatively: $$\begin{align}I_n &= \int\limits_0^1 x^n\sqrt{1-x^2} \,\mathrm d x=\\ &= \int\limits_0^1 (1-(x^2-1))x^{n-2}\sqrt{1-x^2} \,\mathrm d x=\\ &=I_{n-2}-\int\limits_0^1 (1-x^2)^{3/2}dx=\\ &=I_{n-2}-\frac{3\pi}{16}.\end{align}$$ Hence: $$\lim_\limits{n\to \infty} \frac{I_n}{I_{n-2}}=1.$$

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Note: $$I_0=\frac{\pi}{4},I_1=\frac13.$$ The solution is: $$I_n = \frac16 ((-1)^{n+1} + 1) + \frac1{64} π (-6 n + 5 (-1)^n + 3 (-1)^{2n} + 8)$$ WA answer: 1, 2, 3, 4.