Question Number 201348 by sonukgindia last updated on 05/Dec/23
Answered by aleks041103 last updated on 05/Dec/23
$${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right)\right){dx} \\ $$$${u}=\pi/\mathrm{2}−{x}\Rightarrow{du}=−{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({u}\right)\right){du} \\ $$$$\Rightarrow\mathrm{2}{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({ln}\left({sin}\left({x}\right)\right)+{ln}\left({cos}\left({x}\right)\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\mathrm{2}{x}\right)\right){d}\left(\mathrm{2}{x}\right)−{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$${x}=\frac{\pi}{\mathrm{2}}+{u}\Rightarrow{dx}={du} \\ $$$$\Rightarrow\mathrm{2}{I}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{2}}+{u}\right)\right){du}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({u}\right)\right){du}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+{I}=\mathrm{2}{I} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$