Question

D%20%5C%2C%20dx" id="TexFormula1" title="\int\limits^e_1 {\frac{1}{x\sqrt{1-(logx)^{2} } } } \, dx" alt="\int\limits^e_1 {\frac{1}{x\sqrt{1-(logx)^{2} } } } \, dx" align="absmiddle" class="latex-formula">

Answer 1

For the integral

$I=\displaystyle\int_1^3\frac{\mathrm dx}{x\sqrt{1-(\log x)^2}}$

(assuming $\log x$ is the natural logarithm with base $e$) substitute $u=\log x$ and $\mathrm du=\frac{\mathrm dx}x$. Then the integral is equivalent to

$I=\displaystyle\int_{\log1}^{\log e}\frac{\mathrm du}{\sqrt{1-u^2}}=\int_0^1\frac{\mathrm du}{\sqrt{1-u^2}}$

Next, substitute $u=\sin t$ and $\mathrm du=\cos t\,\mathrm dt$:

$I=\displaystyle\int_{\sin^{-1}0}^{\sin^{-1}1}\frac{\cos t}{\sqrt{1-\sin^2t}}\,\mathrm dt=\int_0^{\frac\pi2}\frac{\cos t}{\sqrt{\cos^2t}}\,\mathrm dt$

We have $\sqrt{x^2}=|x|$ for all $x$, but in the given integration interval, $\cos t\ge0$, so

$I=\displaystyle\int_0^{\frac\pi2}\frac{\cos t}{\cos t}\,\mathrm dt=\int_0^{\frac\pi2}\mathrm dt=\boxed{\dfrac\pi2}$

(Of course, with a little foresight, you could have immediately combined the two substitutions and started off with letting $\sin u=\log x$.)