# 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.

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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}
+1: I agree with your answer. Bad estimates are simple to give but hard to proof. E.g., show that $\int_0^1\frac{\cos x}{\sqrt{1-x^2}}< \zeta(2)$ – Fabian Jan 22 '13 at 14:04
You can push the RHS estimate to 1.241403959 by using $\int_0^1 \sin x \arcsin x\,dx \le \int_0^1 x\arcsin x\,dx$. – mrf Jan 22 '13 at 16:07