A reduction formula for $\int_0^1 x^n/\sqrt{9 - x^2}\,\mathrm dx$

Question

Let $$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx$$

Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2\sqrt2$$

I've found that $\displaystyle I_0 = \sin^{-1}\left(\frac{1}{3}\right)$, but that's it.

I'm struggling to go any further. Anyone have any hints?

Answer 1

$$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx= \int_0^1 x^{n-1}\cdot\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx$$ Using Integration By Parts $$\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx=\,\mathrm dv\iff-\sqrt{9-x^2}=v$$

$$x^{n-1}=u\iff (n-1)x^{n-2}\,\mathrm dx=\,\mathrm du$$

$$\begin{align} I_n &= -x^{n-1}\cdot\sqrt{9-x^2}\Bigg|_0^1+(n-1)\int_0^1x^{n-2}\sqrt{9-x^2}\,\mathrm dx\tag{1}\\ &= -2\sqrt2+(n-1)\int_0^1\frac{x^{n-2}(9-x^2)}{\sqrt{9-x^2}}\,\mathrm dx\tag{2}\\ &= -2\sqrt2+9(n-1)\int_0^1\frac{x^{n-2}}{\sqrt{9-x^2}}\,\mathrm dx-(n-1)\int_0^1\frac{x^{n}}{\sqrt{9-x^2}}\,\mathrm dx\tag{3}\\ &= -2\sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}\tag{4}\\ \end{align}$$

After rearrangement we get

$$(n)I_{n} =9(n-1)I_{n-2} -2\sqrt2$$

Answer 2

Hint:-

$x=3\sin \theta \implies dx=3\cos\theta\ d\theta$

$\therefore \displaystyle\int\dfrac{x^n}{\sqrt{9-x^2}}dx=\displaystyle\int\dfrac{3^n\sin^n\theta \cos \theta}{\cos \theta}d\theta=3^n\displaystyle\int{\sin^n\theta }\ d\theta$