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I encountered the following integral when dealing with Legendre polynomials, trying to derive their orthonormality relation starting from the Rodriguez Formula. $$\int_0^{\pi/2}\sin(\theta)^{2l+1}d\theta$$ According to my book, the result is: $$\int_0^{\pi/2}\sin(\theta)^{2l+1}d\theta=\frac{(2^ll!)^2}{(2l+1)!}$$

Yet this result seems completely unintuitive for me. Where does this come from?

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    $\begingroup$ For $n \geq 1$, you may think along the following lines:$$\int_0^{\frac{\pi}{2}}\sin^{2n}x \sin x \, dx=\int_0^{\frac{\pi}{2}}(1-\cos^2x)^{n} \sin x \, dx$$ Now use the substitution $t=\cos x$ and binomial theorem. $\endgroup$
    – Anurag A
    Dec 28, 2022 at 19:00

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I will give a method of solving this integral that is not in the linked answer. It's just iterated by-parts.

Let $\displaystyle I_{n}=\int_0^{\pi/2}\sin^{2n+1}(\theta)\; d\theta$ and $\displaystyle u=\sin^{2n}\theta, v'=\sin\theta$ which gives $u'=2n\cos\theta\sin^{2n-1}\theta, v=-\cos\theta$.

$\displaystyle\therefore I_{n}=\left[-\cos\theta\sin^{2n}\theta\right]^{\pi/2}_{0}+\int^{\pi/2}_0 2n\cos^2\theta\sin^{2n-1}\theta \;dx=2n\int^{\pi/2}_0 \left(1-\sin^2\theta\right)\sin^{2n-1}\theta \;dx$

$\displaystyle =2n\left(\int^{\pi/2}_0 \theta\sin^{2n-1}\theta \;dx\int^{\pi/2}_0 \theta\sin^{2n+1}\theta \;dx\right)$

$\displaystyle \therefore I_{n}=2n\left(I_{n-1}-I_{n}\right)\Longrightarrow I_{n}=\frac{2n}{2n+1}I_{n-1}$.

This formula allows us to retrieve $I_{n}$ rather quickly since we just apply it $n$ times like so:

$\displaystyle I_{n}=\frac{2n}{2n+1}I_{n-1}=\frac{2n}{2n+1}\frac{2n-2}{2n-1}I_{n-2}=...=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}...\frac{2}{3}I_0$.

We know $\displaystyle I_0=\int_0^{\pi/2}\sin^{1}(\theta) \ d\theta=1$ and so what we do to get the factorials is this:

$$\displaystyle \frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}...\frac{2}{3}=\frac{2n}{2n+1}\left(\frac{2n}{2n}\right)\frac{2n-2}{2n-1}\left(\frac{2n-2}{2n-2}\right)\frac{2n-4}{2n-3}...\frac{2}{3}\left(\frac{2}{2}\right)$$

Factorise a $2$ from each term in the top (noting that there at $2n$ terms) to get $(2n)^2(2n-2)^2...2^2=2^{2n}(n!)^2$.

$$\displaystyle \therefore I_n=\frac{2^{2n}(n!)^2}{(2n+1)(2n)(2n-1)...2}=\frac{(2^nn!)^2}{(2n+1)!}$$

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