Need help with integration by substitution for $\int_x^{x+1}\left(\sin\ t^2\right)dt$ I am doing exer 13 of baby Rudin Ch 6. I learned integration by substitution and by parts only recently, hence having trouble doing it.

Exercise: Define $$f(x)=\int_x^{x+1}\left(\sin\ t^2\right)dt\ .$$ Prove that  $|f(x)|<\frac{1}{x}$ if $x>0.$

Hint says to substitute $t^2=u$, which, I think means that substitute $t=\sqrt u$ or $t=-\sqrt u$. Which one should I substitute, and why?
Also, the solution I have says that $|f(x)|<\frac1x$ is obvious for $0<x\le 1$. I do not see how it is obvious, or why we have to consider this as separate case. 

Since $\vert\sin t^2\vert\le 1$ you always have that $\vert f(x)\vert\le\int_x^{x+1}1\,dt=1$. To get strict inequality, take any $t_0\in (x,x+1)$ such that $\vert\sin t^2_0\vert<1$. Then by continuity you can find that $\vert\sin t^2\vert<1-\varepsilon$ in a small interval $(t_0-\delta,t_0+\delta)$ contained in $(x,x+1)$. So now
\begin{align}\vert f(x)\vert&\le\int_x^{x+1}\vert \sin t^2\vert\,dt
=\int_{(t_0-\delta,t_0+\delta)}\vert \sin t^2\vert\,dt+\int_{(x,x+1)\setminus(t_0-\delta,t_0+\delta)}\vert \sin t^2\vert\,dt\\&\le (1-\varepsilon)\int_{(t_0-\delta,t_0+\delta)} 1\,dt+\int_{(x,x+1)\setminus(t_0-\delta,t_0+\delta)}1\,dt<1.
\end{align}
So if $0<x\le 1$ you have $1\le \frac1x$.
As for the change of variables, either one is admissible, but the one with the plus is simpler.
