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If $n$ is a positive integer, show that

$$ \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} \sin(t^2) dt = \frac{(-1)^n}{c}$$

for some $c \in [\sqrt{n\pi}, \sqrt{(n+1)\pi}]$

I have an idea that i can use Mean value Theorem, but I am not able to proceed.

Please give me hints, and to why the steps come to mind with intuition?

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Make the two substitutions $u=t^2$ and $y =u -n\pi$ to find

$$I =\int_{n\pi}^{(n+1)\pi} \frac{\sin(u)}{2\sqrt{u}}du = \int_{0}^{\pi} \frac{\sin(y)(-1)^n}{2\sqrt{y+n\pi}}dy$$

where we have used $\sin(y+n\pi) = \sin(y)(-1)^n$. Now, for $y\in[0,\pi]$ we have that

$$\frac{1}{\sqrt{(n+1)\pi}}\leq \frac{1}{\sqrt{n\pi +y}} \leq \frac{1}{\sqrt{n\pi}}$$

and also $\sin(y) \geq 0$ so

$$\frac{2}{\sqrt{(n+1)\pi}} = \int_{0}^{\pi} \frac{\sin(y)}{\sqrt{(n+1)\pi}}dy \leq \frac{2I}{(-1)^n} \leq \int_{0}^{\pi} \frac{\sin(y)}{\sqrt{n\pi}}dy = \frac{2}{\sqrt{n\pi}}$$

and the claim follows.

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