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.

The proof you have done in your previous exercise, as you mentioned is exactly what you can use to get you answer for your question. A small step and you reach you answer. In you answer, if you take the integral on left hand side I{n} the integrals with exponential powers (n-1) and (n-2) get simplified into I{n-1} and I_{n-2} the you can form a series which will give you the value you require.
Forgive my typing as I am new to stackexchange.
Hope this  helps!
