Question Number 152161 by peter frank last updated on 26/Aug/21
$$\int\mathrm{cos}\:\left(\mathrm{log}\:\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 26/Aug/21
$$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{cos}\left(\mathrm{log}{x}\right)\:{dx} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\int\mathrm{cos}{u}\:{e}^{{u}} {du} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\mathrm{Re}\int{e}^{{iu}} \:{e}^{{u}} {du} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\mathrm{Re}\left(\frac{{e}^{{iu}} {e}^{{u}} }{\mathrm{1}+{i}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\left(\mathrm{1}−{i}\right){e}^{{iu}} {e}^{{u}} \right) \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{cos}{u}+\mathrm{sin}{u}\right){e}^{{u}} \right) \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{{x}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{log}{x}\right)+\mathrm{sin}\left(\mathrm{log}{x}\right)\right)\:\:\left(+\mathrm{C}\right) \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Paradoxical last updated on 26/Aug/21
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by puissant last updated on 26/Aug/21
$${K}=\int{cos}\left({lnx}\right){dx} \\ $$$$ \\ $$$${u}={cos}\left({lnx}\right)\:;\:{v}'=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{K}=\:{xcos}\left({lnx}\right)+\int{sin}\left({lnx}\right){dx} \\ $$$$ \\ $$$${u}={sin}\left({lnx}\right)\:;\:{v}'=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{K}={xcos}\left({lnx}\right)+{xsin}\left({lnx}\right)−\int{cos}\left({lnx}\right){dx} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{2}{K}={xcos}\left({lnx}\right)+{xsin}\left({lnx}\right) \\ $$$$ \\ $$$$\therefore\because\:\:{Q}=\frac{{x}}{\mathrm{2}}{cos}\left({lnx}\right)+\frac{{x}}{\mathrm{2}}{sin}\left({lnx}\right)+{C} \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by qaz last updated on 26/Aug/21
$$\int\mathrm{x}^{\mathrm{i}} \mathrm{dx}=\int\mathrm{e}^{\mathrm{ilnx}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{i}+\mathrm{1}} }{\mathrm{i}+\mathrm{1}}+\mathrm{C}=\frac{\mathrm{1}−\mathrm{i}}{\mathrm{2}}\mathrm{x}\centerdot\mathrm{e}^{\mathrm{ilnx}} +\mathrm{C} \\ $$$$\Rightarrow\int\mathrm{cos}\:\left(\mathrm{lnx}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xcos}\:\left(\mathrm{lnx}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xsin}\:\left(\mathrm{lnx}\right)+\mathrm{C} \\ $$