General case for sine integral: $ I = \displaystyle\int_{0}^{\infty}\frac{\sin^{2n+1}{x}}{x}dx $ where $n \in \Bbb N $ This integral was a exercise in a calculus book called "Advanced Calculus Explored" I have tried many different techniques and the closest one i got to an answer was using feynamn's technique . I have taken calc 1-3 and some other lower level math classes and a proof class but this integral has stumped me for months(working on it here and there). No elementary technique has got me close to  a solution.
$ I = \displaystyle\int_{0}^{\infty}\frac{\sin^{2n+1}{x}}{x}dx    $
where $n \in \ N $
The approach I found the most success with was
$I(a) = \displaystyle\int_{0}^{\infty}\frac{\sin^{2n+1}{x}}{x}e^{-ax}dx \Rightarrow I'(a) = -\displaystyle\int_{0}^{\infty}\sin^{2n+1}{x}e^{-ax}dx$
with initial condition $ \displaystyle\lim_{a \to \infty}I(a) = 0$.
After integration by parts I have got it down to
$I'(a) = -\frac{2n(2n+1)}{a^2}(I'(a) +\displaystyle\int_{0}^{\infty}\sin^{2n-1}{x}e^{-ax}dx)+ (2n+1)I'(a)dx$
$I'(a) = \frac{2n+1}{a^2+2n+1}\displaystyle\int_{0}^{\infty}\sin^{2n-1}{x}e^{-ax}dx$
I am unsure how to proceed from here to get  a closed form for $I$.
 A: Apply binomial expansion to express
$$\sin^{2n+1}x=\left(\frac{e^{i x}-e^{-i x}}{2i}\right)^{2n+1}
=\frac{1}{2^{2n}}
\sum_{k=0}^{n} \begin{pmatrix}2n+1\\k\end{pmatrix}(-1)^{n+k} \sin(2n+1-2k)x
$$
Then, note that $\int_{0}^{\infty}\frac{\sin(a{x})}{x}dx  =\frac\pi2  $ and integrate to obtain
$$I_n= \int_{0}^{\infty}\frac{\sin^{2n+1}{x}}{x}dx    
= \frac{\pi}{2^{2n+1}}
\sum_{k=0}^{n} \begin{pmatrix}2n+1\\k\end{pmatrix}(-1)^{n+k} $$
Listed below are a few sample results of $I_n$
$$I_1=\frac\pi4,\>\>\>
 I_2=\frac{3\pi}{16},\>\>\>
 I_3=\frac{5\pi}{32},\>\>\>
 I_4=\frac{35\pi}{256},\>\>\>
 I_5=\frac{63\pi}{512},\>\>\>\cdots
$$
A: Another short formula.
Starting from @Quanto'answer
$$I_n= \int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x}dx    
= \frac{\pi}{2^{2n+1}}\sum_{k=0}^{n} (-1)^{k+n} \binom{2 n+1}{k}$$
$$\sum_{k=0}^{n} (-1)^{k+n} \binom{2 n+1}{k}=\frac{n+1}{2n+1}\binom{2 n+1}{n+1}=\frac{2^{2 n}\, \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\,\, \Gamma (n+1)}$$
$$I_n=\frac{\sqrt \pi} 2\,\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ When $n$ starts to be large
$$I_n=\frac 12\sqrt{\frac{\pi}{n}}\Bigg[1-\frac{1}{8 n}+\frac{1}{128 n^2}+\frac{5}{1024 n^3}-\frac{21}{32768
   n^4}+O\left(\frac{1}{n^5}\right) \Bigg]$$
