For the integral
[tex]I=\displaystyle\int_1^3\frac{\mathrm dx}{x\sqrt{1-(\log x)^2}}[/tex]
(assuming [tex]\log x[/tex] is the natural logarithm with base [tex]e[/tex]) substitute [tex]u=\log x[/tex] and [tex]\mathrm du=\frac{\mathrm dx}x[/tex]. Then the integral is equivalent to
[tex]I=\displaystyle\int_{\log1}^{\log e}\frac{\mathrm du}{\sqrt{1-u^2}}=\int_0^1\frac{\mathrm du}{\sqrt{1-u^2}}[/tex]
Next, substitute [tex]u=\sin t[/tex] and [tex]\mathrm du=\cos t\,\mathrm dt[/tex]:
[tex]I=\displaystyle\int_{\sin^{-1}0}^{\sin^{-1}1}\frac{\cos t}{\sqrt{1-\sin^2t}}\,\mathrm dt=\int_0^{\frac\pi2}\frac{\cos t}{\sqrt{\cos^2t}}\,\mathrm dt[/tex]
We have [tex]\sqrt{x^2}=|x|[/tex] for all [tex]x[/tex], but in the given integration interval, [tex]\cos t\ge0[/tex], so
[tex]I=\displaystyle\int_0^{\frac\pi2}\frac{\cos t}{\cos t}\,\mathrm dt=\int_0^{\frac\pi2}\mathrm dt=\boxed{\dfrac\pi2}[/tex]
(Of course, with a little foresight, you could have immediately combined the two substitutions and started off with letting [tex]\sin u=\log x[/tex].)
