I was playing with integrals, and came up with
$$L=\lim_{n\to\infty}\int_0^1 \frac{1}{(1+|\sin (nx)|)^2}dx.$$
Conjecture: $L=\dfrac{4}{3\pi}$
Is my conjecture true?
Remarks on numerical investigation:
Desmos and Wolfram don't do a good job with numerical investigation of this limit, but we can consider the series $f(n)=\dfrac{1}{n}\sum\limits_{k=1}^n \dfrac{1}{(1+|\sin{k}|)^2}$.
$f(10^3)\approx0.999568\left(\frac{4}{3\pi}\right)$
$f(10^6)\approx0.999999635\left(\frac{4}{3\pi}\right)$
$f(10^9)\approx0.999999999807\left(\frac{4}{3\pi}\right)$
This suggests that $\lim\limits_{n\to\infty}f(n)=\frac{4}{3\pi}$.
Using Riemann sums, we have $\lim\limits_{n\to\infty}f(n)=L$.
My attempt:
I tried to use $\,\sin nx = \dfrac{1}{2i}(e^{nxi}-e^{-nxi})\,$ in $\;\displaystyle\int_0^1 \dfrac{1}{(1+|\sin (nx)|)^2}\,\mathrm dx\;,\;\;$ to no avail.
I also tried to use complex numbers in the series $f(n)$, as in answers to a question about $\sum_{n=1}^{\infty} \frac{\cos (n)}{n}$, to no avail.
Best Answer
Use the substitution $y = nx$:
$$I = \lim_{n\to\infty}\frac{1}{n}\int_0^n\frac{dy}{(1+|\sin y|)^2} = \lim_{n\to\infty}\frac{1}{n}\int_0^{\pi\lfloor\frac{n}{\pi}\rfloor}\frac{dy}{(1+|\sin y|)^2}+\frac{1}{n}\int_{\pi\lfloor\frac{n}{\pi}\rfloor}^n\frac{dy}{(1+|\sin y|)^2}$$
Since the integrand is $\pi$-periodic, the second piece is bounded by $\frac{\pi}{n}$ and goes to zero by squeeze theorem. The first piece can be further divided into
$$I = \lim_{n\to\infty}\frac{1}{\pi}\int_0^{\pi}\frac{dy}{(1+|\sin y|)^2} - \frac{\left\{\frac{n}{\pi}\right\}}{n}\int_0^{\pi}\frac{dy}{(1+|\sin y|)^2}$$
where again the second piece is bounded by $\frac{\pi}{n}$ and goes to zero by squeeze theorem. To evaluate the integral, from symmetry considerations we have the equivalent
$$I = \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\frac{dy}{(1+\cos y)^2} = \frac{1}{\pi}\int_0^{\frac{\pi}{2}}\frac{1}{2}\sec^2\left(\frac{y}{2}\right)\Bigr(1+\tan^2\left(\frac{y}{2}\right)\Bigr)\:dy$$
$$= \frac{1}{\pi}\Biggr[\tan\left(\frac{y}{2}\right)+\frac{1}{3}\tan^3\left(\frac{y}{2}\right)\Biggr]_0^{\frac{\pi}{2}} = \boxed{\frac{4}{3\pi}}$$