Showing that $\int_0^{ \frac \pi 2} \cos^{2n+1}(x) dx \ = \frac{4^n(n!)²}{(2n+1)!}$?

Question

I don't succed to show that using induction proof however the first condition was satisfied $\int_0^{ \frac \pi 2} \cos^{2n+1}(x) dx\ = \frac{4^n(n!)²}{(2n+1)!}$ , probably it is well known reccurence formula which i missed to know it , Then is it possible to show that using induction proof or any simple way ?

Answer 1

We have\begin{align}\int_0^{\frac\pi2}\cos^{2n+1}(x)\,\mathrm dx&=\int_0^{\frac\pi2}\cos(x)\cos^{2n}(x)\,\mathrm dx\\&=\left[\sin(x)\cos^{2n}(x)\right]_0^{\frac\pi2}+\int_0^{\frac\pi2}2n\sin^2(x)\cos^{2n-1}(x)\,\mathrm dx\\&=2n\int_0^{\frac\pi2}(1-\cos^2 x)\cos^{2n-1}(x)\,\mathrm dx\\&=2n\left(\int_0^{\frac\pi2}\cos^{2n-1}(x)\,\mathrm dx-\int_0^{\frac\pi2}\cos^{2n+1}(x)\,\mathrm dx\right)\end{align}and therefore$$\int_0^{\frac\pi2}\cos^{2n+1}(x)\,\mathrm dx=\frac{2n}{2n+1}\int_0^{\frac\pi2}\cos^{2n-1}(x)\,\mathrm dx.$$Besides, $\int_0^{\frac\pi2}\cos(x)\,\mathrm dx=1$. So$$\int_0^{\frac\pi2}\cos^{2n+1}(x)\,\mathrm dx=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\cdots\times1.$$Can you take it from here?

Answer 2

Generalization:

$$\dfrac{d(\sin^mx\cos^nx)}{dx}=-n\sin^{m+1}\cos^{n-1}x+m\sin^{m-1}x\cos^{n+1} x$$

$$=\sin^{m-1}x[m\cos^{n+1}x-n(1-\cos^2x)\cos^{n-1}x]$$

$$\implies K+\sin^mx\cos^nx=(m+n)I_{m-1,n+1}-n I_{m-1,n-1}$$ where $\displaystyle I_{m-1,n+1}=\int\sin^{m-1}x\cos^{n+1}x\ dx$

$$\implies0=(m+n)J_{m-1,n+1}-nJ_{m-1,n-1}$$ $\displaystyle I_{m-1,n+1}=\int_0^{\pi/2}\sin^{m-1}x\cos^{n+1}x\ dx$

Answer 3

You can get a recurrence relation: $$\int_{0}^{\pi/2} (1-\sin^2 x)^n d\sin x.$$ $$I_n=\int_0^1 (1-t^2)^n dt=\underbrace{(1-t^2)^nt|_0^1}_{=0}-\int_0^1 tn(1-t^2)^{n-1} (-2t)dt=$$ $$-2n\int_0^1 (1-t^2)^ndt+2n\int_0^1 (1-t^2)^{n-1}dt=-2nI_n+2nI_{n-1} \Rightarrow I_n=\frac{2n}{2n+1}I_{n-1}, \ \ I_0=1.$$ $$I_n=\frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdots \frac67 \cdot \frac45 \cdot \frac23=\frac{(2^n\cdot n!)^2}{(2n+1)!}.$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\pi/2}\cos^{2n + 1}\pars{x}\,\dd x & = \int_{0}^{\pi/2}\bracks{1 - \sin^{2}\pars{x}}^{n}\cos\pars{x}\,\dd x \,\,\,\stackrel{t\ =\ \sin\pars{x}}{=}\,\,\, \int_{0}^{1}\pars{1 - t^{2}}^{n}\,\dd t \\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, {1 \over 2}\int_{0}^{1}t^{-1/2}\pars{1 - t}^{n}\,\dd t = {1 \over 2}\,{\Gamma\pars{1/2}\Gamma\pars{n + 1} \over \Gamma\pars{n + 3/2}} \\[5mm] & = {\root{\pi} \over 2}\,{n! \over \Gamma\pars{2n + 2}/ \bracks{\pars{2\pi}^{-1/2}\, 2^{2n + 3/2}\,\Gamma\pars{n + 1}}} \label{1}\tag{1} \\[5mm] & = {\root{\pi} \over 2}\,{n! \over \pars{2n + 1}!/ \bracks{\pi^{-1/2}\, 2^{2n + 1}\,n!}} = \bbx{{4^{n}\pars{n!}^{2} \over \pars{2n + 1}!}} \end{align}

In expression \eqref{1}, I used the Gamma Duplication Formula.