# $\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2)\; dt = \frac{(-1)^n}{c}, \text{ where } \sqrt{n\pi} \leq c \leq \sqrt{(n+1)\pi}.$

The following is a problem from Apostol Vol 1 Calculus from the section: Continuity. Since Differentiation hasn't been introduced yet, the objective is to solve it without direct reference to differentiation.

My approach is to show $\displaystyle\int_{a}^{b} f(x^2)\; dx = \int_{a^2}^{b^2} \frac{f(x)}{2x}\; dx$ using intuitive step functions transformation principles. I am struggling towards the end of this approach. Please let me know if there is an intuitive solution without direct reference to differentiation.
Many thanks!
If $n$ is a positive integer, use "weighted mean value theorem for integrals" to show : $$\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2)\; dt = \frac{(-1)^n}{c}, \text{ where } \sqrt{n\pi} \leq c \leq \sqrt{(n+1)\pi}.$$

After the substitution $x=t^2$ ($t=\sqrt x$), we obtain $$I_n:=\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2)\; dt =\int_{n\pi}^{(n+1)\pi}\frac{\sin(x)}{2\sqrt x}dx.$$ Then use the transformation $x=s+n\pi$: this gives $$I_n=(-1)^n\int_0^{\pi}\frac{\sin(s)}{2\sqrt{s+n\pi}}\mathrm ds.$$ Notice that $\sin s$ is non-negative; this implies that
$$\frac 1{2\sqrt{(n+1)\pi}}\int_0^{\pi}\sin s\mathrm{d}s\leqslant \int_0^{\pi}\frac{\sin(s)}{2\sqrt{s+n\pi}}\mathrm ds\leqslant \frac 1{2\sqrt{n\pi}}\int_0^{\pi}\sin s\mathrm{d}s.$$