Writing $$\int\limits_0^n\sin^2(t)dt=\int\limits_0^{\pi E(\frac{n}{\pi})}\sin^2(t)dt + \int\limits_{\pi E(\frac{n}{\pi})}^n\sin^2(t)dt$$
Where E(x) designates the floor function of x
Use the squeeze theorem to find $\lim\limits_{n\to+\infty}U_n$
I tried to evaluate the Integral but it's specifically asked to use $\pi E(\frac{n}{\pi})$
Best Answer
As we have:
$$\sin^2(t)=\frac12(1-\cos(2t))$$
It is clear that the function being integrated has a periodicity of $\pi$. Hence every integral through a whole $\pi$ period will have the same value. So, we have that:
$$\int\limits_0^\pi\sin^2(t)dt=\int\limits_0^\pi\frac12(1-\cos(2t))dt=\frac\pi2$$
Then, if we break down the integration as the function is suggesting:
$$\int\limits_0^n\sin^2(t)dt=\int\limits_0^{\pi \lfloor\frac{n}{\pi}\rfloor}\sin^2(t)dt + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt=\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt$$
Notice that the integrand is always positive, so we can easily find an lower and upper bound by excluding the left term or letting it complete another cycle, which means that:
$$\frac\pi2 \lfloor\frac{n}\pi\rfloor<\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt<\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2$$
$$\frac1n \frac\pi2 \lfloor\frac{n}\pi\rfloor<U_n<\frac1n\left(\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2 \right)$$
So, if we let $n \to \infty$, as both sides of the inequality tend to $1/2$, we have that
$$\lim_{n\to\infty}U_n=\frac12$$