Without Integration II

$\begin{aligned} C & = \int_0^1 \frac {\cos x}{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{Since }\cos \theta = \sin \left(\frac \pi 2 - \theta\right) \\ & = \int_0^1 \frac {\sin \left(\frac \pi 2-x\right)}{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{and }\sin (\pi - \theta) = \sin \theta \\ & = \int_0^1 \frac {\sin \left(\frac \pi 2+x\right)}{\sqrt{1-x^2}} dx \\ & = \int_\frac \pi 2^{\frac \pi 2 + 1} \frac {\sin x}{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{Since }\sin x \text{ is symmetrical about }\frac \pi 2 \\ & = \int^\frac \pi 2_{\frac \pi 2 - 1} \frac {\sin x}{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{and }\sin \left(\frac \pi 2 -1 + x\right) > \sin x \ \forall x \in [0, 1] \\ & > \int_0^1 \frac {\sin x}{\sqrt{1-x^2}} dx \end{aligned}$

Therefore, $\boxed{C > S}$

Sine starts from zero and grows in that range. Cosine starts from 1 and shrinks in that range and stays larger than sine over most of that range.

$\int_0^1 \frac{\sin (x)}{\sqrt{1-x^2}} \, dx= \frac{\pi \bm{H}_0(1)}{2}\approx 0.893243740975026$

$\int_0^1 \frac{\cos (x)}{\sqrt{1-x^2}} \, dx=\frac{\pi J_0(1)}{2} \approx 1.20196971531721$