$W_n=\int_0^{\pi/2}\sin^n(x)\,dx$ Find a relation between $W_{n+2}$ and $W_n$ 
Set $$W_n=\int_0^{\pi/2}\sin^n(x)\,dx.$$ Compute $W_0$ and $W_1$. Find
  a relation between $W_n$ and $W_{n+2}$ and deduce a formula for $W_n$.

What I have so far is: $$W_{2k}=\frac{1}{2^k}\left( \sum_{\substack{0\leq j \leq k \\ j \text{ even }}} \binom{k}{j} W_j \right)\\[30pt]W_{2k+1}= \sum_{j=1}^k \binom{k}{j}\frac{(-1)^j}{2j+1}.$$
How I got this is:
$$\int_0^{\pi/2}\sin^{2k}x\,dx=\int_0^{\pi/2}\frac{1}{2^k}\left(1-\cos2x\right)^k = \frac{1}{2^{k+1}}\int_0^{\pi}(1-\cos x)^k$$ the odd powers evaluate to zero, so
$$=\frac{1}{2^{k+1}}\int_0^{\pi}\sum_{\substack{0\leq j \leq k \\ j \text{ even }}}\binom{k}{j}\cos^jx=\frac{1}{2^k}\int_0^{\pi/2}\sum_{\substack{0\leq j \leq k \\ j \text{ even }}}\binom{k}{j}\cos^j x = \frac{1}{2^k}\left( \sum_{\substack{0\leq j \leq k \\ j \text{ even }}} \binom{k}{j} W_j \right).\\[40pt]\int_0^{\pi/2}\sin^{2k+1} x\,dx=\int_0^1(1-u^2)^kdu = \sum_{j=1}^k \binom{k}{j}\frac{(-1)^j}{2j+1}.$$
But I haven't found a relation between $W_n$ and $W_{n+2}$ as such. Any ideas?
 A: Just integrate by parts, to get $$W_n=\int_0^{\pi/2} \sin^n x = \left. -\sin^{n-1}x\cos x \right|_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2} x (1- \sin^2 x) d x = (n-1) W_{n-2} - (n-1) W_n.$$
A: Note $W_{n+2}=\int_0^{\pi/2}\sin^n x\sin^2 x\,dx$. Use $\sin^2x=1-\cos^2x$ and integrate by parts appropriately on the integral that has a $\cos^2x$ term, namely, let $dv=\sin^n x\cos x$ and $u=\cos x$.
A: $\newcommand{\+}{^{\dagger}}%
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With $\ds{t = \sin\pars{x}}$:
\begin{align}
W_{n} &= \int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x
= \int_{0}^{1}t^{n}\,{\dd t \over \root{1 - t^{2}}}
= \int_{0}^{1}t^{n/2}\pars{1 - t}^{-1/2}\,\half\,t^{-1/2}\dd t
\\[3mm]&=\half\int_{0}^{1}t^{\pars{n - 1}/2}\pars{1 - t}^{-1/2}\,\dd t
=\half\,{\rm B}\pars{n + \half,\half}\,,\qquad \Re n > -\,\half
\end{align}
where ${\rm B}\pars{x,y}$ is the Beta function which satisfies
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$.
$\Gamma\pars{z}$ is the Gamma function.

\begin{align}
W_{n} &= \int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x
=\half\,{\Gamma\pars{n + 1/2}\,\Gamma\pars{1/2} \over \Gamma\pars{n + 1}}
=\half\,\root{\pi}\,{\Gamma\pars{n + 1/2} \over \Gamma\pars{n + 1}}
\end{align}
where we used $\ds{\Gamma\pars{\half} = \root{\pi}}$.

Also $\pars{~\mbox{with the property}\ \Gamma\pars{z + 1} = z\,\Gamma\pars{z}~}$:
\begin{align}
W_{n + 1} &= \half\,\root{\pi}\,{\Gamma\pars{n + 3/2} \over \Gamma\pars{n + 2}}
=\half\,\root{\pi}\,{\pars{n + 1/2}\Gamma\pars{n + 1/2} \over \pars{n + 1}\Gamma\pars{n + 1}}
\\[3mm]&\imp\quad 
\color{#00f}{\large W_{n + 1} = \half\,{2n + 1 \over n + 1}\,W_{n}\,,\qquad
W_{0} = {\pi \over 2}}
\end{align}
