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Question Number 87461 by hamdhan last updated on 04/Apr/20

$$\underset{e^{-1}}{\stackrel{\mathrm{e}}{\int }}\frac{\sqrt{1-{\left(\ln x\right)}^2}}{x}dx$$

Commented by ajfour last updated on 04/Apr/20

$$let\ln x=t\Rightarrow \left(dx\right)/x=dt$$$$I=2{\int }_0^1\sqrt{1-t^2}dt$$$$={[t\sqrt{1-t^2}+{\sin }^{-1}t]}_0^1$$$$=\pi /2.$$

Answered by Ar Brandon last updated on 04/Apr/20

$$\mathit{L}\mathit{e}\mathit{t}\mathit{x}={\mathit{e}}^{\mathit{y}}\Rightarrow \mathit{d}\mathit{x}={\mathit{e}}^{\mathit{y}}\mathit{d}\mathit{y}$$$$\mathit{I}={\int }_{-1}^1\frac{\sqrt{1-{\mathit{y}}^2}}{{\mathit{e}}^{\mathit{y}}}\cdot {\mathit{e}}^{\mathit{y}}\mathit{d}\mathit{y}={\int }_{-1}^1\sqrt{1-{\mathit{y}}^2}\mathit{d}\mathit{y}$$$$I=\sqrt{1-y^2}{\int }_{-1}^{1}dy-{\int }_{-1}^1[\frac{d}{dy}\left(\sqrt{1-y^2}\right)\int dy]dy$$$$={\left[y\sqrt{1-y^2}\right]}_{-1}^1+{\int }_{-1}^1\left[\frac{y^2}{\sqrt{1-y^2}}\right]dy$$$$lety=sin\theta \Rightarrow dy=cos\theta d\theta$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{sin}^2\theta d\theta =\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1-cos\left(2\theta \right))d\theta$$$$I=\frac{1}{2}{[\theta -\frac{1}{2}sin\left(2\theta \right)]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$$$$I=\frac{1}{2}[\frac{\pi }{2}+\frac{\pi }{2}]=\frac{\pi }{2}$$$$$$$$$$

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