A trigonometric integral inequality $$\displaystyle\frac{4\sin 1}{\pi }<\int_{0}^{1}{\frac{\cos x}{\sqrt{1-{{x}^{2}}}}}\text{d}x\le \frac{\pi }{2}\ln \left( \sec 1+\tan 1 \right)$$
I've got no ideas for this one.

Numerically, the integral evaluates to $\approx 1.201969715$. Your RHS is $1.926096588$ which seems like a fairly bad estimate. Here's one way to do better:
Partial integration gives
\begin{align}
\int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\
&\le \frac{\pi}{2} \cos 1 + \sin 1 \int_0^1 \arcsin x\,dx \\
&= \frac{\pi}2 \cos 1 + \sin 1\left(\frac{\pi}2 -1\right) \approx 1.329013425
\end{align}
(we use that $\sin x$ is increasing on $[0,1]$).
For the other direction, your LHS is $\approx 1.071394134$. The same partial integration trick and the inequality $\arcsin x \ge x$ gives
\begin{align}
\int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\
&\ge \frac{\pi}{2} \cos 1 + \int_0^1 x \sin x\,dx \\
&= \frac{\pi}2 \cos 1 + \sin 1 - \cos 1 \approx 1.149873556
\end{align}
