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Prove that $\displaystyle\int_{0}^{\frac\pi 6} {\cos ({x^2)}\mathrm{d}x\ge\dfrac12}$.

I know this is a Fresnel integral but without going into advanced calculus is there a way to show that this is true? using calculus 1 knowledge, I tried Riemann's sum to prove this and got stuck. Thanks for any help.

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For $0 < x \le \frac \pi 6 < 1$ we have $x^2 < x$ and therefore $$ \int_{0}^{\pi/6} \cos (x^2) \, dx > \int_{0}^{\pi/6} \cos (x) \ dx = \sin( \frac \pi 6) = \frac 12 $$

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Since $\cos(x^2)\geq \cos((\pi/6)^2)$ for $0\leq x \leq \frac\pi6$, we get $$ \int_0^{\pi/6}\cos(x^2)dx \geq \int_0^{\pi/6}\cos\left(\frac{\pi^2}{36}\right)dx\approx 0.504 $$

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Alternatively, there is a non-calculus proof that $\cos(x)\geq 1-\dfrac{x^2}{2}$ for all $x\in \mathbb{R}$. Use this to show that $$\int_0^{\frac{\pi}{6}}\,\cos(x^2)\,\text{d}x\geq \int_0^{\frac{\pi}{6}}\,\left(1-\frac{x^4}{2}\right)\,\text{d}x=\frac{\pi}{6}-\frac{\pi^5}{77760}\gtrsim 0.519663>\frac12\,.$$ This approximation looks to be very good: $$\int_0^{\frac{\pi}{6}}\,\cos(x^2)\,\text{d}x\approx0.519677\,.$$

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