The convergence speed of $ \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x $? I have already known how to prove
\begin{equation*}
        \lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \sqrt{\frac{\pi}{2n}}
    \end{equation*}
with Wallis's formula
\begin{equation*}
 \quad \frac{\pi}{2}=\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 8 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 9 \cdots} 
\end{equation*}
But the method I used was considered not to be universal.
How to prove that
\begin{equation*}
        \lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+\frac{\sqrt{2 \pi}}{64} \cdot \frac{1}{n^{\frac{5}{2}}}
    \end{equation*}
And is
\begin{equation*}
         \lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+ \dots +
         (-1)^{k}\cdot\frac{\sqrt{2 \pi}}{2^{\frac {k(k+1)}{2}}} \cdot \frac{1}{n^{\frac{2k+1}{2}}}
    \end{equation*}
true? Are there any more powerful tools, like numerical methods to calculate the integration?
 A: If you are familiar with the Gaussian hypergeometric function, if $0 \leq x \leq \frac \pi 2$
$$\int \sin^n(x)\,dx=-\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^2(x)\right)$$ and the definite integral just becomes
$$I_n=\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt{\pi }} 2\,\,\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n+2}{2}\right)}$$ Using Stirling approximation, then
$$\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n+2}{2}\right)}=\sqrt{\frac{2}{n}}\left(1-\frac{1}{4 n}+\frac{1}{32 n^2}+\frac{5}{128
   n^3}+O\left(\frac{1}{n^4}\right) \right)$$
$$I_n= \sqrt{\frac{\pi}{2n}}\left(1-\frac{1}{4 n}+\frac{1}{32 n^2}+\frac{5}{128
   n^3}+O\left(\frac{1}{n^4}\right) \right)$$
If you want an even more accurate formula, use more terms in the expansion and make it a $[n,n]$ Padé approximant to get
$$I_n \sim \sqrt{\frac{\pi}{2n}}\,\,\,\frac{64 n^2-8 n+11}{64 n^2+8 n+11}$$ whose error is $\frac{709}{16384 n^5}$.
Using it, the relative error is $0.0035$% for $n=4$
A: You need to have an equivalent of the factorial, or equivalently, of the gamma function. A well known is the Stirling's approximation but you can go further as $$ {\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\left(1+{\frac {1}{12n}}+{\frac {1}{288n^{2}}}-{\frac {139}{51840n^{3}}}-{\frac {571}{2488320n^{4}}}+\cdots \right).}$$ where the coefficients corresponds to the "Stirling series".
A: 
Are there any more powerful tools, like numerical methods to calculate the intergration?

Yes. There are.
Interval Integration
e.g.
We can take the sine to the n power of x intervals and then integrate those intervals.
This would probably be the easiest way to solve it:
$$
\begin{align*}
z &= \int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \mathrm{e}^{\ln(\sin(x)^{n})}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \mathrm{e}^{n \cdot \ln(\sin(x))}\\
\ln(\sin(x)) &= \begin{cases}
\ln(\sin(x)) = 0, \text{ if } x = \frac{\pi}{2} + 2 \cdot k \cdot \pi\\ \ln(\sin(x)) < 1, \text{ if } x \ne \frac{\pi}{2} + 2 \cdot k \cdot \pi
\end{cases}\\
\ln(\sin(x)) &= \begin{cases}
\ln(\sin(x)) \leq 0, \text{ if } x \in \mathbb{R}\\ 
\end{cases}\\
\ln(\sin(x)) &\leq \begin{cases}
0, \text{ if } x \in \mathbb{R}\\ 
\end{cases}\\
\lim_{{n} \to {\infty}} n \cdot \ln(\sin(x)) &= \begin{cases}
\lim_{{n} \to {\infty}} n \cdot 0, \text{ if } x \in \mathbb{R}\\ \end{cases}\\
\lim_{{n} \to {\infty}} n \cdot \ln(\sin(x)) &= \begin{cases}
0, \text{ if } x \in \mathbb{R}\\ 
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \begin{cases}
\lim_{{n} \to {\infty}} \mathrm{e}^{n \cdot \ln(\sin(x))}, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq \begin{cases}
\mathrm{e}^{0}, \text{ if } x \in \mathbb{R}\\
\end{cases}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq 
\mathrm{e}^{0}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &\leq 
1\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= 
0\\
F(x) = \int_{0}^{x} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x &= \int_{0}^{x} \lim_{{n} \to {\infty}} 0 ~\operatorname{d}x \\
F(x) = \int_{0}^{x} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x &= 0\\
\\
\int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} &= F(\frac{\pi}{2}) - F(0) = 0 - 0 = 0\\
z &= 0\\
\end{align*}
$$
Complex Numbers Way
$$\begin{align*}
z &= \int_{0}^{\frac{\pi}{2}} \lim_{{n} \to {\infty}} \sin(x)^{n} ~\operatorname{d}x\\
\sin(x) &= \frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} (\frac{\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}}}{2} \cdot -\mathrm{i})^{n}\\
\lim_{{n} \to {\infty}} \sin(x)^{n} &= \lim_{{n} \to {\infty}} \frac{(\mathrm{e}^{x \cdot \mathrm{i}} - \mathrm{e}^{-x \cdot \mathrm{i}})^{n}}{2^{n}} \cdot (-\mathrm{i})^{n}\\
&= \dots
\end{align*}
$$
This way is method is way harder...
It may be some nice taskt to train solving integrals.
