Show that $\int_{0}^{\pi/6} {\cos (x^2)}\mathrm{d}x\ge\frac12$.

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.

3 Answers

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

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

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: $$\int_0^{\frac{\pi}{6}}\,\cos(x^2)\,\text{d}x\approx0.519677\,.$$