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Question Number 60240 by ANTARES VY last updated on 19/May/19

∫_0 ^2 ((ln(x))/(√(4−x^2 )))dx

$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}} \\ $$

Commented by maxmathsup by imad last updated on 19/May/19

let I =∫_0 ^2  ((ln(x))/(√(4−x^2 ))) dx  chang. x=2sint give  I=∫_0 ^(π/2)  ((ln(2)+ln(sint))/(2cost)) 2costdt  =(π/2)ln(2) +∫_0 ^(π/2) ln(sint)dt  but ∫_0 ^(π/2)   ln(sint)dt =−(π/2)ln(2)(result proved) ⇒  I =0 .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left({x}\right)}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:{dx}\:\:{chang}.\:{x}=\mathrm{2}{sint}\:{give}\:\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{2}\right)+{ln}\left({sint}\right)}{\mathrm{2}{cost}}\:\mathrm{2}{costdt} \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sint}\right){dt}\:\:{but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{ln}\left({sint}\right){dt}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\left({result}\:{proved}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{0}\:. \\ $$

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