Let $f(x): (0, +\infty)\rightarrow \mathbb{R}$ be a function defined as: $0, 0 \leq x < \frac{\pi}{2}$ and $\frac{{cos}^2x}{x},x\geq \frac{\pi}{2}$. Let $F(x): (0, +\infty)\rightarrow \mathbb{R}$ be a function defined as $F(x)=\int_{0}^{x} f(t)dt$. Show that $F(x)$ is uniformly continuous on $(0, +\infty)$.
To be honest, I have absolutely no idea how to even begin this problem. f is continuous on $(0, +\infty)$ so the function F is well defined. The most obvious way to do it for me was to show F(x) converges when $x\to\infty$($\frac{1}{t}\to0, {cos}^2t$ is bounded) and then evaluate the improper integral and show its derivative is bounded, but I can't really solve the integral $\int \frac{{cos}^2x}{x}dx$ so I'm not really sure what direction I'm supposed to take.
The improper integral is what's bothering me the most. I've also thought of using the fact $\frac{{cos}^2x}{x}$ is bounded (f too), so F' is bounded on $[0, a]$ where $a\in\mathbb{R}$. But I'm not sure if this is correct or not.
I would appreciate any hints!
Hint: If a function $F$ is differentiable in some interval $I$ and its derivative is bounded then $F$ is uniformly continuous in $I$.
Can you show that for your function $F$?