Upper bound for a Fresnel Integral

Question

I was wondering if there is an upper bound strictly less than 1 to the function $x \mapsto \int_{0}^{1} \sin (x^2 + t^2) dt$ for $x \in [0,1]$. And if it is how to derive it.

Thank you for your answer,

Answer 1

Use a trig identity to write

$$ \sin(x^2+t^2)=\sin(x^2)\cos(t^2)+\sin(t^2) \cos(x^2) $$

Now you have two Fresnel integrals; S and C, leading to

$$ I(x)=\cos(x^2)\operatorname{S}(1) +\sin(x^2)\operatorname{C}(1) $$

Over $0 \leq x \leq 1$, this is maximum at $x=1$, so

$$ I_{\text{max}}=I(1)=\cos(1)\operatorname{S}(1) +\sin(1)\operatorname{C}(1) \approx 0.93 $$