Without Integration III

$\begin{aligned} C & = \int_0^\frac \pi 2 \cos (\cos x) \ dx & \small \color{#3D99F6} \text{By reflections }\int_a^b f(x)\ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 2 \cos (\sin x) \ dx \\ & = \int_0^\frac \pi 2 \sin \left(\frac \pi 2 - \sin x \right) \ dx \end{aligned}$

For $x \in \left[0, \frac \pi 2\right]$ , the integrand of $S$ , $\sin (\sin x) \in [0, \sin 1]$ and that of $C$ , $\sin \left(\frac \pi 2 - \sin x\right) \in \left[\sin \left(\frac \pi 2- 1 \right), 1\right]$ , this implies that $\boxed {S< C}$ .

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Reference: Reflections

Since $\displaystyle x > \sin x, \, \, \sin x > \sin(\sin x)$ , so $\displaystyle S < \int_0^{\frac{\pi}{2}} \sin x \, dx = 1$ .

And since $\displaystyle 1-\frac{x^2}{2} < \cos x, \, \, 1-\frac{\cos^2x}{2} = \frac{3}{4}-\frac{\cos(2x)}{4} < \cos(\cos x)$ , so $C > \displaystyle \int_0^{\frac{\pi}{2}} \frac{3}{4}-\frac{\cos(2x)}{4} dx = \frac{3}{4} \cdot \frac{\pi}{2} \approx 1.17$

We have $S < 1 < 1.17 < C$ , so $\boxed{S < C}$

On the interval $[0,\pi/2] sin(sin(x))<cos(cos(x))$ and furthermore that these two functions are always positive and monotonically increasing (this could be shown by differentiating the two function. Because of this, one could see that the area under sin(sin(x)) must be less than cos(cos(x))

$J_0(1)>\pmb{H}_0(1)$