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Question Number 144109 by henderson last updated on 21/Jun/21

$$\mathbf{hi},\mathbf{everybody}!$$$$1.\mathbf{calculate}:\mathbf{I}={\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}\mathit{l}\mathit{n}\left(\mathit{t}\mathit{a}\mathit{n}\mathit{x}\right)\mathit{d}\mathit{x}.$$$$2.\mathbf{calculate}:\underset{\mathit{x}\to \mathit{e}}{\mathit{l}\mathit{i}\mathit{m}}\frac{\mathit{x}\sqrt{1-\mathit{l}\mathit{n}\mathit{x}}}{\mathit{x}-\mathit{e}}.$$

Commented by bobhans last updated on 22/Jun/21

$$\left(2\right)\underset{x\to \mathrm{e}}{\lim }\frac{\mathrm{e}\sqrt{1-\ln \mathrm{x}}}{\mathrm{x}-\mathrm{e}}=$$$$\underset{x\to \mathrm{e}}{\lim }\frac{\mathrm{e}\left(\frac{-\frac{1}{\mathrm{x}}}{2\sqrt{1-\ln \mathrm{x}}}\right)}{1}=-\mathrm{\infty }$$$$$$

Answered by Olaf_Thorendsen last updated on 21/Jun/21

$$\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\ln \left(\tan x\right)dx$$$$\mathrm{Let}u=\frac{\pi }{2}-x$$$$\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\ln \left(\cot u\right)du=-{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\ln \left(\tan x\right)dx$$$$\mathrm{I}=-\mathrm{I}\Rightarrow \mathrm{I}=0$$

Answered by mathmax by abdo last updated on 21/Jun/21

$$1)\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\log \left(\mathrm{tanx}\right)\mathrm{dx}\mathrm{changement}\mathrm{tanx}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\mathrm{logt}}{1+{\mathrm{t}}^2}\mathrm{dt}{=}_{\mathrm{t}=\frac{1}{\mathrm{u}}}-{\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{-\mathrm{logu}}{1+\frac{1}{{\mathrm{u}}^2}}(-\frac{\mathrm{du}}{{\mathrm{u}}^2})$$$$=-{\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\mathrm{logu}}{1+{\mathrm{u}}^2}=-\mathrm{I}\Rightarrow 2\mathrm{I}=0\Rightarrow \mathrm{I}=0$$$$$$

Answered by mathmax by abdo last updated on 21/Jun/21

$$2)1-\mathrm{logx}>0\Rightarrow \mathrm{logx}<1\Rightarrow \mathrm{x}<\mathrm{e}$$$$\mathrm{f}\left(\mathrm{x}\right)=-\frac{\mathrm{x}\sqrt{1-\mathrm{logx}}}{\mathrm{e}-\mathrm{x}}\mathrm{changement}\mathrm{e}-\mathrm{x}=\mathrm{y}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=-\frac{(\mathrm{e}-\mathrm{y})\sqrt{1-\log (\mathrm{e}-\mathrm{y})}}{\mathrm{y}}=\frac{(\mathrm{y}-\mathrm{e})\sqrt{1-\log (\mathrm{e}-\mathrm{y})}}{\mathrm{y}}=\mathrm{g}\left(\mathrm{y}\right)(\mathrm{y}\to {\mathrm{o}}^{+})$$$$\mathrm{we}\mathrm{have}\log (\mathrm{e}-\mathrm{y})=1+\log (1-\frac{\mathrm{y}}{\mathrm{e}})\sim 1-\frac{\mathrm{y}}{\mathrm{e}}\Rightarrow 1-\log (\mathrm{e}-\mathrm{y})\sim \frac{\mathrm{y}}{\mathrm{e}}$$$$\Rightarrow \sqrt{1-\log (\mathrm{e}-\mathrm{y})}\sim \frac{\sqrt{\mathrm{y}}}{\sqrt{\mathrm{e}}}\Rightarrow$$$$\mathrm{g}\left(\mathrm{y}\right)\sim \frac{\sqrt{\mathrm{y}}}{\sqrt{\mathrm{e}}\mathrm{y}}(\mathrm{y}-\mathrm{e})\Rightarrow \mathrm{g}\left(\mathrm{y}\right)\sim \frac{\mathrm{y}-\mathrm{e}}{\sqrt{\mathrm{e}}\sqrt{\mathrm{y}}}\to -\mathrm{\infty }(\mathrm{y}\to 0^{+})\Rightarrow$$$${\lim }_{\mathrm{x}\to {\mathrm{e}}^{-}}\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{\infty }$$

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