Question Number 26357 by abdo imad last updated on 24/Dec/17
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\theta\right){d}\theta\:\:{and}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:\:. \\ $$
Commented by prakash jain last updated on 24/Dec/17
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\right){dx} \\ $$$${u}=\frac{\pi}{\mathrm{2}}−{x} \\ $$$${du}=−{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{sin}\:{x}\right){dx}=−\int_{\pi/\mathrm{2}} ^{\mathrm{0}} \mathrm{ln}\:\left(\mathrm{cos}\:{u}\right){du} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\mathrm{sin}\:{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\mathrm{sin}\:\mathrm{2}{xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\mathrm{2}{dx} \\ $$$$={I}−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${I}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$
Answered by prakash jain last updated on 24/Dec/17
$$−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$