Question Number 168429 by mathocean1 last updated on 10/Apr/22
$${calculate}: \\ $$$${P}=\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \frac{{cos}\left({lnx}\right)}{{x}}{dx} \\ $$
Answered by qaz last updated on 10/Apr/22
$$\int_{\mathrm{1}} ^{\mathrm{e}^{\pi/\mathrm{2}} } \frac{\mathrm{cos}\:\left(\mathrm{lnx}\right)}{\mathrm{x}}\mathrm{dx}=\Re\int_{\mathrm{1}} ^{\mathrm{e}^{\pi/\mathrm{2}} } \mathrm{x}^{\mathrm{i}−\mathrm{1}} \mathrm{dx}=\Re\frac{\mathrm{x}^{\mathrm{i}} }{\mathrm{i}}\mid_{\mathrm{1}} ^{\mathrm{e}^{\pi/\mathrm{2}} } =\Re\left(\frac{\mathrm{e}^{\frac{\pi}{\mathrm{2}}\mathrm{i}} }{\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{i}}\right)=\mathrm{1} \\ $$
Answered by FelipeLz last updated on 10/Apr/22
$$\mathrm{ln}\left({x}\right)\:=\:{u}\:\rightarrow\:\frac{\mathrm{1}}{{x}}{dx}\:=\:{du} \\ $$$${P}\:=\:\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \frac{\mathrm{cos}\left(\mathrm{ln}\:\left({x}\right)\right)}{{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left({u}\right){du}\:=\:\mathrm{sin}\left({u}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\:\mathrm{1} \\ $$