The value of ${\lim }_{x\to 0}\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}$ is:

The form is $0/0$. Use L'Hopital's Rule and Leibniz Rule for differentiating integrals.Numerator derivative: $\frac{d}{dx}{\int }_0^{x^2}{\sec }^{2}tdt={\sec }^2(x^2)\cdot 2x$.Denominator derivative: $\frac{d}{dx}(x\sin x)=\sin x+x\cos x$.Limit becomes:
$$\underset{x\to 0}{\lim }\frac{2x{\sec }^2(x^2)}{\sin x+x\cos x}$$
Divide numerator and denominator by $x$:
$$\underset{x\to 0}{\lim }\frac{2{\sec }^2(x^2)}{\frac{\sin x}{x}+\cos x}$$
Substitute $x=0$:
$$\frac{2{\sec }^2(0)}{1+\cos (0)}=\frac{2(1)}{1+1}=1$$