Estimates of $f(x) = \int_x^{x+1}\sin(t^2)dt$

Question

$$f(x)=\int_x^{x+1}\sin(t^2)dt$$ I have to prove that $|f(x)| < \frac{1}{x}$ for $x>0$. then I have to show that $2xf(x) = \cos(x^2) - \cos(x+1)^2 + r(x)$ where $|r(x)| < \frac{c}{x}$ for some constant $c$. I used the fundamental theorem of calculus and calculated its derivative, but I don't know how to continue. Any hint?

Answer 1

In fact, by IBP, one has $$f(x)=\int_x^{x+1}\sin(t^2)dt=-\int_x^{x+1}\frac{1}{2t}d\cos(t^2)=-\frac{1}{2t}\cos(t^2)\bigg|_x^{x+1}+\frac12\int_x^{x+1}\frac{\cos(t^2)}{t^2}dt$$ which implies $$ |f(x)|\le \bigg|\frac{1}{2t}\cos(t^2)\bigg|_x^{x+1}\bigg|+\frac12\int_x^{x+1}\frac{1}{t^2}dt\le\frac12(\frac1x+\frac1{x+1})+\frac12(\frac{1}{x}-\frac1{x+1})=\frac1x.$$