Recursive integration

Question

The integral I have is $$I_{n} = \int^{\pi/2}_{0} \cos^{2n+1}y \ \mathrm{d}y$$

And I have found $I_{n} = \frac{2n}{1+2n}I_{n-1}$ but I want to express $I_{n}$ in a form without $I_{n-1}$ how do I do this?

I know $$I_{0} = \int^{\pi/2}_{0} \cos{y} \ \mathrm{d}y = \sin{y}\left.\right|^{\pi/2}_{0}=1$$

Then $I_{1} = 2/3I_{0} = 2/3$,

$I_{2} = 4/5I_{1} = 4/5(2/3) = 8/15$

$I_{3} = 6/7I_{2} = 6/7(8/15) = 48/105$

and so on but how would I spot a pattern in this sequence, what's the best way to find one?

Answer 1

It seems that $$I_n=\prod_{k=1}^n \frac{2k}{2k+1}$$

For a proof, use induction.

Answer 2

$$\frac{2n}{2n+1}I_{n-1}=\frac{2n}{2n+1}\frac{2n-2}{2n-1}I_{n-2}=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}I_{n-3}...$$