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 
$$\lim\limits_{x\to\infty}{\mathrm{arcsec}\ x}$$
 A: The $\displaystyle \lim_{x \to \infty} \sec^{-1} x$ $\mathbf{is}$ $\frac{\pi}{2}$.
$$\sec^{-1} x = \cos^{-1}\frac{1}{x}$$
And $\cos^{-1} 0 = \frac{\pi}{2}$
A: $$\int \frac{dx}{x\sqrt{x^2-1} }= -\tan^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)+C$$
A: Begin by writing
\begin{eqnarray} 
y=\sqrt{x^2-1}\\ y^2=x^2-1, \ x=\sqrt{1+y^2}\\ ydy=xdx\\ y(1)=0,\ y(\infty)=\infty. 
\end{eqnarray}
Then re-write as
\begin{eqnarray}
I &=& \int_{0}^{\infty}\frac{dy}{1+y^{2}} \\
  &=& \tan^{-1}y|_{0}^{\infty} \\
  &=& \frac{\pi}{2}
\end{eqnarray}
A: Assuming that you intended to write 
$$ 
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi}{2}
$$
First note that
$$ 
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx= \int_1^\infty \frac{x}{x^2\sqrt{x^2-1}}dx
$$
Let $u=\sqrt{x^2-1}$, then $du=\frac{x}{\sqrt{x^2-1}}dx$. So now
$$ 
\int_0^\infty \frac{1}{u^2+1}du
$$
$$=\lim\limits_{\beta\to \infty} \arctan \beta- \lim\limits_{\alpha\to 0^+} \arctan \alpha
$$
$$=\frac{\pi}{2}- 0= \frac{\pi}{2}
$$
As already mentioned, your original integral does not converge.
A: When evaluating integrals containing $\sqrt{x^2\pm a^2}$, one of the basic substitutions is $x=a\sinh t$ $($for the case with $+)$ or $x=a\cosh t$ $($for the case with $-)$. Then use $\cosh^2t-\sinh^2t=1$, together with $\sinh't=\cosh t$ and $\cosh't=\sinh t$. Also, a second substitution, $u=e^t$ might sometimes also be in order, to reduce the integrand to a rational function.
