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$$I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}x + \sin^{2n-3}x + ... + \sin x dx, n\in \mathbb{N}$$

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:

$J_{n} = \frac{2^{2n+1}n!(n+1)!}{(2n+2)!}$

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    \begin{eqnarray*} J_n= \frac{2^{2n} (n!)^2}{(2n+1)!} \end{eqnarray*} – 2017-02-25
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    I think that in this case for the product of the odd numbers, you use n+1 since if n=1, then the number of odd numbers is 2 – 2017-02-25

3 Answers 3

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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

$$\begin{align}\int_0^{\pi/2}\sum_{k=0}^n (\sin x)^{2k+1}\mathrm dx&=\frac{2 \sqrt{\pi } n \Gamma (n+2)+3 \sqrt{\pi } \Gamma (n+2)-2 \Gamma \left(\frac{1}{2} (2 n+5)\right)}{2 \Gamma \left(\frac{1}{2} (2 n+5)\right)}\\&=\frac{\sqrt\pi(n+1)!}{\Gamma(n+3/2)}-1\end{align}$$

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    Your formula for $I_{2k+1}$ seems to be incorrect, unless I'm making a mistake. – 2017-02-25
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    @mrnovice I think my formulas are correct. I checked it with mathematica from both approaches: the integral and the summation, and both coincide. – 2017-02-26
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    By your own formula $I_{3} = 4^{1}*1!*1^{-2} = 4$ But $ I_{3} = \frac{2}{3}$ – 2017-02-26
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    @mrnovice the elements $a^\underline k$ are falling factorials, not powers, then $I_3|_A=4/6=2/3$, because $1^\underline{-2}=\frac12\cdot\frac13=\frac16$ – 2017-02-26
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    Oh ok, there's some more factorial notation I have now learned lol, sorry for the confusion. – 2017-02-26
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    @mrnovice it was my fault, sorry... its too late now and I forget that not everyone know this notation. This notation was introduced many years ago by Donald Knuth and it is common nowadays. – 2017-02-26
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\pi/2}\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.

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    $\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 \\ $ Could you explain how you arrived at this? – 2017-02-26
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    @mrnovice It's equivalent to $t = \cos\left(x\right)$. Note that $\sin^{2}\left(x\right) = 1 - \cos^{2}\left(x\right) = 1 - t^{2}$ and $\sin^{2n}\left(x\right) = \left[1 - \cos^{2}\left(x\right)\right]^{n} = \left(1 - t^{2}\right)^{n}$. Moreover, $\mathrm{d}t = -\sin\left(x\right)\,\mathrm{d}x$. – 2017-02-27
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    Ah okay thanks, also what is $(-\frac{1}{2})!$? – 2017-02-27
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    @mrnovice In general, the factorial is defined as $z! = \Gamma\left(z +1\right)$. So, $\left(-\,{1 \over 2}\right)! = \Gamma\left(1 \over 2\right) = \,\sqrt{\,\pi\,}\,$. – 2017-03-02
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    Ah... I didnt noticed that $\sqrt\pi=\Gamma(1/2)$. Sorry but it is not correct to write $(-1/2)!$ because factorial is not defined for negative numbers, but the binomial is correct because the generalized binomial is defined from falling factorials, what are well defined for any $x\in\Bbb C$. – 2017-03-07
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    @Masacroso $z!$ is defined $\forall z \in \mathbf{C}\setminus\left\{-1,-2,-3,\ldots\right\}$. I already talk about it in a previous comment. [See this link](http://mathworld.wolfram.com/Factorial.html). Thanks for your remark. – 2017-03-08
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    @FelixMarin this is not the general consensus about the definition of the factorial. Just [one example](https://en.wikipedia.org/wiki/Factorial#Extension_of_factorial_to_non-integer_values_of_argument). The reason is that exists infinite possible extensions of the factorial function to other values. – 2017-03-08
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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*}

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    I don't think that formula is correct – 2017-02-25
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    Take n = 1, then $\int_{0}^{\frac{\pi}{2}}sin^{3}xdx = \frac{2}{3}$. But according to your formula it would be much much smaller than that. – 2017-02-25
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    Oh ... is it the double factorial notation ? $(2n)!! =(2n)(2n-2) \cdots 4 \times 2$ ... I do not mean $((2n)!)!$ – 2017-02-25
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    Oh I did not know that, then it probably is correct – 2017-02-25
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    phew ... I easily doubt myself ... I do think it is correct. ... goto Wolfram alpha & check ? integral_0^{pi/2} (sin(t))^11 dt – 2017-02-25