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/
 A: Use the formula:
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
So we have:
$$B\left(\frac{11}2,\frac{1}2\right)=\frac{\Gamma\left(\frac{11}2\right)\Gamma\left(\frac{1}2\right)}{\Gamma\left(\frac{11}2+\frac{1}2\right)}=\frac{\Gamma\left(\frac{11}2\right)\Gamma\left(\frac{1}2\right)}{\Gamma(6)}$$
Use the formula:
$$\Gamma(z)=(z-1)\Gamma(z-1)$$
So we get:
$$\Gamma\left(\frac{11}2\right)=\frac{9}{2}\cdot \frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot\Gamma\left(\frac{1}2\right),~~~~\Gamma(6)=5!$$
Finally,
$$B\left(\frac{11}2,\frac{1}2\right)=\frac{\frac{9}{2}\cdot \frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot\Gamma\left(\frac{1}2\right)\cdot\Gamma\left(\frac{1}2\right)}{5!}=\frac{63\pi}{256}$$
A: The Beta function is defined as;
$$B(x,y)={\Gamma(x)\Gamma(y)\over \Gamma(x+y)}$$
So, for your specific example we have;
$$B(11/2, 1/2)= {\Gamma(11/2)\sqrt{\pi}\over 5!}$$
We can then use the recursive property of the gamma function simplify our answer;
$$\Gamma(n)=(n-1)\cdot \Gamma(n-1) \implies \Gamma\left({11\over2}\right)=\left({9\over2}\right)\left({7\over2}\right)\left({5\over2}\right)\left({3\over2}\right)\left({1\over2}\right)\sqrt{\pi}$$
So, our final answer becomes;
$$B(11/2, 1/2)= {\Gamma(11/2)\sqrt{\pi}\over 5!}={1\cdot 3 \cdot 5 \cdot 7 \cdot 9 \over2^5 \cdot5!}\pi={63\pi\over256}$$
A: 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
$$\int_0^{\pi/2}\sin(x)^{10}\mathrm dx=I_5 \\ =\frac{9}{8}I_4 \\ =\frac{9\cdot 7}{8\cdot 6}I_3 \\ =\frac{9\cdot 7\cdot 5}{8\cdot 6\cdot 4}I_2 \\ =\frac{9\cdot 7\cdot 5\cdot 3}{10\cdot 8\cdot 6\cdot 4}I_1 \\ =\frac{9\cdot 7\cdot 5\cdot 3\cdot 1}{10\cdot 8\cdot 6\cdot 4\cdot 2}I_0 \\ =\frac{9\cdot 7\cdot 5\cdot 3\cdot 1}{10\cdot 8\cdot 6\cdot 4\cdot 2}\frac{\pi}{2}  \\ =\frac{63}{512}\pi$$
A: Using the power reduction formula for sine (7), we have
$$\int_0^{\frac\pi2} \sin^{10}(x) \, dx = \frac1{2^{10}}\binom{10}5 \int_0^{\frac\pi2}\,dx - \frac1{2^9} \sum_{k=0}^4 \binom{10}k \int_0^{\frac\pi2} \cos((10-2k)x) \, dx$$
Observe that for integer $k$,
$$\int_0^{\frac\pi2} \cos((10-2k)x) \, dx = \frac1{10-2k} \sin\left(\frac{(10-2k)\pi}2\right) = \frac1{10-2k} \sin((5-2k)\pi) = 0$$
So we have
$$\int_0^{\frac\pi2} \sin^{10}(x) \, dx = \frac\pi{2^{11}} \binom{10}5 = \frac{63\pi}{512}$$
Note that
$$\begin{align*}
\frac\pi{2^{11}} \binom{10}5 &= \frac{\pi \cdot 10!}{2^{11} \cdot 5! \cdot 5!} \\[1ex]
&= \Gamma\left(\frac12\right)^2 \cdot \frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2^{11}\cdot5\cdot4\cdot3\cdot2\cdot1\cdot5\cdot4\cdot3\cdot2\cdot1} \\[1ex]
&= \Gamma\left(\frac12\right) \cdot \Gamma\left(\frac{11}2\right) \cdot \frac{10\cdot8\cdot6\cdot4\cdot2}{2^6\cdot5\cdot4\cdot3\cdot2\cdot5\cdot4\cdot3\cdot2} \\[1ex]
&= \Gamma\left(\frac12\right) \cdot \Gamma\left(\frac{11}2\right) \cdot \frac{1}{2\cdot5!} \\[1ex]
&= \frac{\Gamma\left(\frac12\right) \Gamma\left(\frac{11}2\right)}{2\Gamma(6)} \\[1ex]
&= \frac12 \operatorname{B}\left(\frac12,\frac{11}2\right)
\end{align*}$$
