The question is as follows:

Given that $\displaystyle I_{n} = \int_0^\frac{\pi}{2} \sin^n (x) dx$ where $\displaystyle n \epsilon N$ (used capital N to show the sign for natural number or normal number, not too sure what it is)

a) show that $\displaystyle I_{2n} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2}$

b) show further that $\displaystyle I_{2n-1} I_{2n} = \frac{\pi}{4n}$

really need help with this one, its bugging me so much that i cant get it