Proving that $\int_0^{\pi/2} \sin^{2n + 1}(x) dx = \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot ... \cdot\frac{2n}{2n + 1}$ I'm having some trouble with the following exercice:

Prove that, for all $n \in \Bbb N$: $$\int_0^{\pi/2} \sin^{2n + 1}(x) dx = \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot ... \cdot\frac{2n}{2n + 1}$$

So, the first thing I did was write the rhs using the product notation no make things a little easier:
$$\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot ... \cdot\frac{2n}{2n + 1} = \prod_{k=1}^n \frac{2k}{2k + 1}$$
And then I tried to use induction to prove this and I easily proved that if $n = 1$ this is true, But I'm having some trouble with the induction step of the proof. How can I prove this?
Note: In the last exercice the I proved that:
$$\int \sin^n(x) dx = - \frac{1}{n}(\sin^{n - 1}(x))\cos(x) + \frac{n - 1}{n} \int \sin^{n - 2}(x) dx$$
I don't know if this helps or if it's related to this exercise.

You need only prove$$\begin{align}0&=\int_0^{\pi/2}\sin^{2n-1}x(2n-1-2n\sin^2x)dx\\&=\int_0^{\pi/2}\sin^{2n-1}x(2n\cos^2x-1)dx\\&=[\sin^{2n}x\cos x]_0^{\pi/2}.\end{align}$$
