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?


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.
