TITLE: Show that $\int_{0}^{\pi/6} {\cos (x^2)}\mathrm{d}x\ge\frac12$.
QUESTION [7 upvotes]: 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.

REPLY [19 votes]: For $0 < x \le \frac \pi 6 < 1$ we have $x^2 < x$ and therefore
$$
 \int_{0}^{\pi/6} \cos (x^2) \, dx >  \int_{0}^{\pi/6} \cos (x) \ dx
 = \sin( \frac \pi 6) = \frac 12 
$$

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

REPLY [5 votes]: Since $\cos(x^2)\geq \cos((\pi/6)^2)$ for $0\leq x \leq \frac\pi6$, we get
$$
\int_0^{\pi/6}\cos(x^2)dx \geq \int_0^{\pi/6}\cos\left(\frac{\pi^2}{36}\right)dx\approx 0.504
$$