$W_n=\int_0^{\pi/2}\sin^n(x)\,dx$ Find a relation between $W_{n+2}$ and $W_n$

Question

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?

Answer 1

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 $u=\cos x$.

Answer 2

Just integrate by parts, to get $$W_n=\int_0^{\pi/2} \sin^n x = \left. -\sin^{n-1}x\cos x \right|_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2} x (1- \sin^2 x) d x = (n-1) W_{n-2} - (n-1) W_n.$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}