# $\lim\limits_{n\rightarrow +\infty}\int_0^{2k\pi}\sqrt{\sin^{2n}(x)+\cos^{2n}(x)}dx$

I think the limit value of the function $$\sin^{2n}(x) + \cos^{2n}(x)$$ when $$n$$ tends to infinity is $$0$$ for all real values of $$x$$ except when $$x$$ is an integral multiple of $$\pi$$, where it comes out to be one.

Is my thinking correct?

(Keep in mind that I am a high school student, so please bear with me if I've used incorrect mathematical terminologies anywhere, I'm apologizing in advance... Also, kindly try to explain as lucidly as possible.) If this question is way above my level, then let me know that too in the comments.

If my approach is correct, then that would mean the function is discrete (I'm not sure if its the right term to use here), so how would we find the integral of it? If not, please guide me to the right approach to solve this question.

• Is the $k$ in the upper limit of integration supposed to be $n$? Nov 18 '20 at 2:53
• Right, as $n$ increases, the integrand goes to zero except at $x$ a multiple of $\pi/2,$ where the integrand is $1.$
– mjw
Nov 18 '20 at 3:36
• @peterag Your font isnt legible in my device... Nov 18 '20 at 3:48
• Without measure theory, use the fact that the integrand is $\pi/2$-periodic and is $\leqslant 1$ if $n>0$. So you just have to show the claim for $\int_0^{\pi/2}$. Pick any "small" $\epsilon>0$, write $\int_0^{\pi/2}=\int_0^{\epsilon/2}+\int_{\epsilon/2}^{(\pi-\epsilon)/2}+\int_{(\pi-\epsilon)/2}^{\pi/2}$, see that the middle integral tends to $0$ as $n\to\infty$, and the sum of the other two is $\leqslant\epsilon$. As $\epsilon$ is arbitrary, you get 0 as the answer. Nov 18 '20 at 16:00
• (BTW, the integrand is $1$ when $x$ is an integral multiple of $\color{red}{\pi/2}$, not just $\pi$.) Nov 18 '20 at 16:06