Uniform continuity of $f(x)=\int_0^x \sin^2(t^2)\textrm{d}t.$

Question

Let $f: [0,\infty ) \to \Bbb{R}$ be defined by $$f(x)=\int_0^x \sin^2(t^2)\textrm{d}t.$$ Then how to show that $f$ is uniformly continuous on both $[0,1)$ and $(0,\infty)$. Is it uniformly continuous on $[0,\infty)$?

Answer 1

$$\left|\int_0^x \sin^2(t^2)dt-\int_0^y \sin^2(t^2)dt\right|=\left|\int_y^x \sin^2(t^2)dt\right|\le \left(\sup_{t \in (y,x)}\sin^2(t^2)\right) |x-y| $$