<s>[INST] evaluate beta function I know:
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1}{2}\cdot \mathrm B\left(11/2, 1/2\right)$$
and
$$\Gamma(1/2)=\sqrt{\pi}$$
But what is the following calculation based on?
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1\cdot3\cdot5\cdot7\cdot9\cdot\pi}{2\cdot4\cdot6\cdot8\cdot10\cdot2}=\frac{63\pi}{512}$$
https://www.goseeko.com/blog/what-are-beta-and-gamma-functions/
 [/INST] If you let
$$I_n=\int_{0}^{\pi/2}\sin(x)^{2n}\mathrm dx$$
And integrate by parts the formula for $I_{n+1}$, you will find (exercise!) that
$$I_{n+1}=\frac{2n+1}{2n+2}I_n$$
Using $I_0=\pi/2$, you get
$$\int_0^{\pi/2}\sin(x)^{10}\mathrm dx=I_5 \\ =\frac{9}{8}I_4 \\ =\frac{9\cdot 7}{8\cdot 6}I_3 \\ =\frac{9\cdot 7\cdot 5}{8\cdot 6\cdot 4}I_2 \\ =\frac{9\cdot 7\cdot 5\cdot 3}{10\cdot 8\cdot 6\cdot 4}I_1 \\ =\frac{9\cdot 7\cdot 5\cdot 3\cdot 1}{10\cdot 8\cdot 6\cdot 4\cdot 2}I_0 \\ =\frac{9\cdot 7\cdot 5\cdot 3\cdot 1}{10\cdot 8\cdot 6\cdot 4\cdot 2}\frac{\pi}{2}  \\ =\frac{63}{512}\pi$$
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