There is a little mistake in your typing, $b_{n}=\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}xdx$ but not $\int_{0}^{1}\cos^{2n+1}xdx$.
There are two methods in calculation:
The first method:
$\begin{align*} b_{n}&=\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}xdx\\ &=\int_{0}^{\frac{\pi}{2}}\cos^{2n}xd\sin x\\ &=2n\int_{0}^{\frac{\pi}{2}}\sin^{2}x\cos^{2n-1}xdx\\ &=2n\int_{0}^{\frac{\pi}{2}}\cos^{2n-1}x-\cos^{2n+1}xdx\\ &=2nb_{n-1}-2nb_{n} \end{align*}$
so $b_{n}=\frac{2n}{2n+1}b_{n-1}$, or $\frac{b_{n}}{b_{n-1}}=\frac{2n}{2n+1}$, as $b_{0}=1$, we can obtain that:
$b_{n}=b_{0}\prod_{k=0}^{n-1}\frac{b_{k+1}}{b_{k}}=\prod_{k=0}^{n-1}\frac{2k+2}{2k+3}$.
With the knowledge of $Gamma$ and $Beta$ functions, we can get the representation as Wolfram Alpha gave.