Showing that $\int_0^{ \frac \pi 2} \cos^{2n+1}(x) dx \ = \frac{4^n(n!)²}{(2n+1)!}$? 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 ?
 A: 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?
A: 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 $\displaystyle I_{m-1,n+1}=\int\sin^{m-1}x\cos^{n+1}x\ dx$
$$\implies0=(m+n)J_{m-1,n+1}-nJ_{m-1,n-1}$$ $\displaystyle I_{m-1,n+1}=\int_0^{\pi/2}\sin^{m-1}x\cos^{n+1}x\ dx$
A: You can get a recurrence relation:
$$\int_{0}^{\pi/2} (1-\sin^2 x)^n d\sin x.$$
$$I_n=\int_0^1 (1-t^2)^n dt=\underbrace{(1-t^2)^nt|_0^1}_{=0}-\int_0^1 tn(1-t^2)^{n-1} (-2t)dt=$$
$$-2n\int_0^1 (1-t^2)^ndt+2n\int_0^1 (1-t^2)^{n-1}dt=-2nI_n+2nI_{n-1} \Rightarrow I_n=\frac{2n}{2n+1}I_{n-1}, \ \ I_0=1.$$
$$I_n=\frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdots \frac67 \cdot \frac45 \cdot \frac23=\frac{(2^n\cdot n!)^2}{(2n+1)!}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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.

