Is there a general formula for the integral $I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}x + \sin^{2n-3}x + ... + \sin x dx, n\in \mathbb{N}$. $$I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}x + \sin^{2n-3}x + ... + \sin x dx, n\in \mathbb{N}$$
I'm using this integral to form part of answer to someone's question, but I'm struggling to find a formula for this integral (if one exists).
Edit:
I believe the formula for $J_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n+1}xdx$ is:
$J_{n} = \frac{2^{2n+1}n!(n+1)!}{(2n+2)!}$
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{0}^{\pi/2}\sum_{k = 1}^{n}\sin^{2k - 1}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\sin\pars{x}\,{\sin^{2n}\pars{x} - 1 \over \sin^{2}\pars{x} - 1}
\,\dd x\,\,\, \stackrel{\cos\pars{x}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{1}{1 - \pars{1 - x^{2}}^{n} \over x^{2}}\,\dd x
\\[5mm] = &\
-1 + \int_{0}^{1}{1 \over x}\bracks{-n\pars{1 - x^{2}}^{n - 1}}\pars{-2x}\,\dd x =
-1 + 2n\int_{0}^{1}\pars{1 - x^{2}}^{n - 1}\,\dd x
\\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, &\
-1 + n\int_{0}^{1}x^{-1/2}\,\pars{1 - x}^{n - 1}\,\dd x =
-1 + n\,{\Gamma\pars{1/2}\Gamma\pars{n} \over \Gamma\pars{1/2 + n}} =
-1 + {\pars{-1/2}!\,n! \over \pars{n - 1/2}!}
\\[5mm] = &\
{1 \over \ds{{n - 1/2 \choose n}}} - 1=
\bbx{\ds{{2^{2n} \over \ds{{2n \choose n}}} - 1}}
\end{align}

A proof of the last identity can be seen in one of my previous answers.

A: The terms of your integral can be evaluated one by one
\begin{eqnarray*}
\int_{0}^{\pi/2} \sin^{2n+1}x dx =\frac{(2n)!!}{(2n+1)!!}
\end{eqnarray*}
So your integral can be written as a sum
\begin{eqnarray*}
I_n = \sum_{i=0}^{n-1}\frac{(2i)!!}{(2i+1)!!}
\end{eqnarray*}
A: HINT: by integration by parts we have that
$$I_n:=\int(\sin x)^n\mathrm dx=-\cos x(\sin x)^{n-1}+(n-1)\int (\cos x)^2(\sin x)^{n-2}=\\=-\cos x(\sin x)^{n-1}+(n-1)\int (1-\sin x)^2(\sin x)^{n-2}=\\=-\cos x(\sin x)^{n-1}+(n-1)(I_{n-2}-I_n)$$
hence
$$I_n=\frac1n((n-1)I_{n-2}-\cos x(\sin x)^{n-1})\tag{1}$$
where clearly $I_1=-\cos x$ and $I_0=x$. Evaluating in $A:=[0,\pi/2]$ we have that $I_1|_A=1$, then because (1) we have that $I_3|_A=2/3$, $I_5|_A=\frac23\cdot\frac45$, and in general $$I_{2k+1}|_A=\frac{(2k)!!}{(2k+1)!!}=2^k\frac{k!}{(2k+1)!!}=\frac{2^kk!(2k)!!}{(2k+1)!}=\\=\frac{4^k(k!)^2}{(2k+1)!}=\frac{4^kk!}{(2k+1)^\underline{k+1}}=4^kk!k^\underline{-k-1}$$
Hence
$$\int_0^{\pi/2}\sum_{k=0}^n (\sin x)^{2k+1}\mathrm dx=\sum_{k=0}^{n}I_{2k+1}\big|_A$$
what dont seems to have a closed form of elementary functions. From Wolfram Mathemathica I get a "closed" solution in terms of the gamma function
$$\begin{align}\int_0^{\pi/2}\sum_{k=0}^n (\sin x)^{2k+1}\mathrm dx&=\frac{2 \sqrt{\pi } n \Gamma (n+2)+3 \sqrt{\pi } \Gamma
   (n+2)-2 \Gamma \left(\frac{1}{2} (2 n+5)\right)}{2 \Gamma
   \left(\frac{1}{2} (2 n+5)\right)}\\&=\frac{\sqrt\pi(n+1)!}{\Gamma(n+3/2)}-1\end{align}$$
