How to show $\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\pi/2$ How to show that 
$$\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi}{2}$$
The problem is that I don't know what is 
$$\lim\limits_{x\to\infty}{\mathrm{arcsec}\ x}$$

The $\displaystyle \lim_{x \to \infty} \sec^{-1} x$ $\mathbf{is}$ $\frac{\pi}{2}$.
$$\sec^{-1} x = \cos^{-1}\frac{1}{x}$$
And $\cos^{-1} 0 = \frac{\pi}{2}$
