Question-201348

Question Number 201348 by sonukgindia last updated on 05/Dec/23

Answered by aleks041103 last updated on 05/Dec/23

I=∫_0 ^( π/2) ln(sin(x))dx u=π/2−x⇒du=−dx ⇒I=∫_0 ^( π/2) ln(cos(u))du ⇒2I=∫_0 ^( π/2) (ln(sin(x))+ln(cos(x)))dx= =∫_0 ^( π/2) ln(((sin(2x))/2))dx= =(1/2)∫_0 ^( π/2) ln(sin(2x))d(2x)−ln(2)∫_0 ^( π/2) dx= =(1/2)∫_0 ^( π) ln(sin(x))dx−((πln(2))/2) x=(π/2)+u⇒dx=du ⇒2I=−((πln(2))/2)+(1/2)∫_(−π/2) ^( π/2) ln(sin((π/2)+u))du= =−((πln(2))/2)+(1/2)∫_(−π/2) ^( π/2) ln(cos(u))du= =−((πln(2))/2)+∫_0 ^( π/2) ln(cos(x))dx= =−((πln(2))/2)+I=2I ⇒I=∫_0 ^( π/2) ln(sin(x))dx=∫_0 ^( π/2) ln(cos(x))dx=−((πln(2))/2)

$${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right)\right){dx} \\ $$$${u}=\pi/\mathrm{2}−{x}\Rightarrow{du}=−{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({u}\right)\right){du} \\ $$$$\Rightarrow\mathrm{2}{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({ln}\left({sin}\left({x}\right)\right)+{ln}\left({cos}\left({x}\right)\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\mathrm{2}{x}\right)\right){d}\left(\mathrm{2}{x}\right)−{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$${x}=\frac{\pi}{\mathrm{2}}+{u}\Rightarrow{dx}={du} \\ $$$$\Rightarrow\mathrm{2}{I}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{2}}+{u}\right)\right){du}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({u}\right)\right){du}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+{I}=\mathrm{2}{I} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$

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