Let $f:[0,\infty)\to \Bbb R$ be defined by $f(x)=\int_{0}^x \sin^2(t^2)\mathrm dt$. Show that $f$ is uniformly continuous Let $f:[0,\infty)\to \Bbb R$ be defined by
$$f(x)=\int_{0}^x \sin^2(t^2)\mathrm dt$$
Then Show that the function is uniformly continuous on $[0,1)$ and $(0,\infty)$
Attempt: Differentiating with respect to $x$ we get,
$$f'(x)=\sin^2(x^2)$$ then $$|f'(x)|=|\sin^2(x^2)|\le1,\forall x\in(0,\infty)$$
this means that derivative of $f$ is bounded, hence f is uniformly continuous on given intervals. 
am I right? Also provide other methods to solve this. Thank you.
 A: Your solution is very good. By proving that the derivative is bounded and next using Lebesgue's mean value theorem, you can prove that $f$ is Lipschitz continuous (of Lipschitz constant $1$), hence in particular uniformly continuous not just on intervals, but even on the whole $\Bbb R$.
An alternative could be the following: assume $x \le y$; then $|f(x) - f(y)| = | \int \limits _x ^y \sin ^2 (t^2) \Bbb d x | \le \int \limits _x ^y |\sin ^2 (t^2)| \Bbb d x \le \int \limits _x ^y 1 \ \Bbb d x = y - x = |x - y|$. The advantage of this proof is that it uses integration without even touching differentiability or mean values theorems (and, in general, integrability is considered a more convenient setting than derivability, because it is weaker).
A: You're attempt is correct. In general if you're derivative is a priori bounded $\|f'\|_\infty\leq M$, then the function is uniformly continuous by the mean value theorem:
$$f(x)-f(y)=f'(\xi)(x-y)\Longrightarrow|f(x)-f(y)|=|f'(\xi)||x-y|\leq M|x-y|.$$
A: Let $\epsilon >0$. 
By MVT we have $\vert f(x)-f(y)\vert =\vert f'(c)(x-y)\vert \leq \vert x-y\vert ;\ y<c<x$.
Now set $\delta =\epsilon $ and the result follows. 
