Proving $-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$

I have encountered a question:

Prove $$-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$$

There are plenty of solutions to $\int_0^{\infty} \sin(x^2) dx$ online, but there seems to be no solution to the boundary of $\int_a^b \sin(x^2) dx$.

Could anyone help me with this please? I tried to calculate the integral directly, but I cannot cancel out b and get a boundary only with $a$.

Applying inequality to the integrand $\sin(x^2)<x^2$ does not work either.

By change of variables $u=x^2$, we have $\displaystyle{\int_{a}^{b}\sin(x^2)\,{\rm d}x=\frac{1}{2}\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u}$.

Now, by integration by parts we have $$\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u=-\int_{a^2}^{b^2}\frac{1}{\sqrt{u}}{\rm d}(\cos u)=\frac{\cos(a^2)}{a}-\frac{\cos(b^2)}{b}-\frac{1}{2}\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u$$

Now, note that $$\left|\frac{\cos(a^2)}{a}\right|\leq \frac{1}{a},$$ $$\left|\frac{\cos(b^2)}{b}\right|\leq \frac{1}{b}$$ and $$\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u\right|< \frac{1}{2}\int_{a^2}^{b^2}\frac{1}{u^{3/2}}{\rm d}u=\frac{1}{a}-\frac{1}{b}.$$

Hence, $\displaystyle{\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a}+\frac{1}{b}+\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{2}{a}}$. We conclude $$\left|\int_{a}^{b}\sin(x^2)\,{\rm d}x\right|=\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a} $$