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Viktor [21]
3 years ago
10

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">
Mathematics
1 answer:
Zina [86]3 years ago
5 0

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.)

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