I want to prove equation (1). Note that the equation (2) can be proven by the reduction formula for $\displaystyle \int sin^n(x)dx$, but I'll just write the formula down since I know it's true.

I want to show that:

----------------------------------------------------

$\displaystyle \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\frac{(2)(4)(6 )...(2n)}{(3)(5)(7)...(2n+1)}$

----------------------------------------------------

This seems rather difficult, and I'm not entirely sure where to begin. I'm an amateur, so I don't have much experience with proofs like this. I'll write the equation like this:

(1) $\displaystyle \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\prod_{i=1}^n\ frac{2i}{2i+1}$

Using the following reduction formula:

(2) $\displaystyle \int_0^{\frac{\pi}{2}}sin^n(x)dx=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}sin^{n-2}xdx$

I can write the hypothesis for $\displaystyle n=k$

$\displaystyle \int_0^{\frac{\pi}{2}}sin^{2k+1}(x)dx=\frac{(2k+1)-1}{2k+1}\int_0^{\frac{\pi}{2}}sin^{(2k+1)-2}(x)dx$

So now I have:

$\displaystyle \frac{2k}{2k+1}\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=1}^k\frac{2i}{2i+1}$

Factoring the left side:

$\displaystyle \prod_{i=1}^k\frac{2i}{2i+1}=\frac{2k}{2k+1}\prod_ {i=2}^k\frac{2(i-1)}{2i-1}$

(3) $\displaystyle \int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=2}^k\frac{2(i-1)}{2i-1}$

All I've managed to do is write the inductive hypothesis. It's really frustrating that I can't proceed. I can't figure out how to connect one side to the other. Wring the next term for n=k+1 doesn't seem to get me anywhere. Can anyone offer some insight?