Calculate the limit $\lim_{n \rightarrow \infty} \int_0^{2022\pi} \sin^n(x) dx$. What did I do wrong? I had to calculate the limit
$$\lim_{n \rightarrow \infty} \int_0^{2022\pi} \sin^n(x) dx$$
I massively overcomplicated things and landed on $2022\pi / \sqrt{e}$, while my friend simply got $0$. Now that I looked at the graph and played around with high values of $n$, $0$ does make a lot more sense than whatever I got. However, I can't seem to find what is wrong with my approach, other than it being unnecessarily complicated. I'll try to keep it short, but essentially what I did is the following:
I applied the reduction formula
$$\int \sin^n(x) dx = - \frac{\sin^{n-1}(x) \cos (x)}{n} + \frac{n-1}{n}\int \sin^{n-2}(x) dx$$
But
$$\left[- \frac{\sin^{n-1}(x) \cos (x)}{n}\right]_0^{2022\pi} = 0$$
Therefore
$$\int_0^{2022\pi} \sin^n(x) dx = \frac{n-1}{n}\int_0^{2022\pi} \sin^{n-2}(x) dx$$
If we now apply this same reduction formula over and over again, $n/2$ times, we will end up with:
\begin{align}
\int_0^{2022\pi} \sin^n(x) dx & = \left(\frac{n-1}{n}\right)^{\frac{n}{2}}\int_0^{2022\pi} \sin^{n-n}(x) dx \\
 & = \left(\frac{n-1}{n}\right)^{\frac{n}{2}}\int_0^{2022\pi} 1 dx \\
 & = \left(\frac{n-1}{n}\right)^{\frac{n}{2}}2022\pi
\end{align}
Then I took the limit, and if my calc skills didn't fail me, I got:
$$\lim_{n\rightarrow \infty}\left(\frac{n-1}{n}\right)^{\frac{n}{2}}2022\pi = \frac{1}{\sqrt{e}}2022\pi$$
Where is the mistake? (I can already feel it, it's gonna be something incredibly simple that I did wrong)
I'm not looking for the correct solution, the one my friend came up with makes sense to me. I am simply asking why I got a different result.
 A: The reduction step is correct, but the repeated application is wrong. Also, you assumed $n$ was even. That might seem trivial, but if you apply the reduction on odd $n$ you finish with: $$\int_0^{2022\pi}\sin=0$$So what went wrong for even $n$? Well, the first coefficient is $(n-1)/n$. The next one happens with $n\mapsto n-2$, so the multiplier is $(n-2-1)/(n-2)=(n-3)/(n-2)$, and the recurrence looks like: $$\int\sin^n=\frac{n-1}{n}\frac{n-3}{n-2}\int\sin^{n-4}$$And so on. You do not have $((n-1)/n)^{n/2}$.
A: It is possible to determine the limit without any explicit calculations. We will consider
$$I_n={1\over 4044}\int\limits_0^{2022\pi}|\sin x|^n\,dx$$ The sequence $I_n$ is positive and nonincreasing, therefore convergent. As  $\sin(\pi -x)=\sin x$ and $|\sin (x+\pi)|=|\sin x|$ we get
$$I_n= \int\limits_0^{\pi/2}\sin^nx\,dx$$
Fix $0<a<\pi/2.$ Then
$$I_n= \int\limits_0^{a}\sin^nx\,dx+\int\limits_{a}^{\pi/2}\sin^nx\,dx$$
Thus
$$I_n\le a\sin^na+\left ({\pi\over 2}-a\right )$$
As $0<\sin a<1$ we obtain
$$\lim_nI_n\le {\pi\over 2}-a,\qquad 0<a<{\pi\over 2}$$ Since $a$ is arbitrary we get
$$\lim_nI_n=0$$
