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Question Number 152161 by peter frank last updated on 26/Aug/21

∫cos (log x)dx

$$\int\mathrm{cos}\:\left(\mathrm{log}\:\mathrm{x}\right)\mathrm{dx} \\ $$

Answered by Olaf_Thorendsen last updated on 26/Aug/21

F(x) = ∫cos(logx) dx  F(e^u ) = ∫cosu e^u du  F(e^u ) = Re∫e^(iu)  e^u du  F(e^u ) = Re(((e^(iu) e^u )/(1+i))) = (1/2)Re((1−i)e^(iu) e^u )  F(e^u ) = (1/2)((cosu+sinu)e^u )  F(x) = (x/2)(cos(logx)+sin(logx))  (+C)

$$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{cos}\left(\mathrm{log}{x}\right)\:{dx} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\int\mathrm{cos}{u}\:{e}^{{u}} {du} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\mathrm{Re}\int{e}^{{iu}} \:{e}^{{u}} {du} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\mathrm{Re}\left(\frac{{e}^{{iu}} {e}^{{u}} }{\mathrm{1}+{i}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\left(\mathrm{1}−{i}\right){e}^{{iu}} {e}^{{u}} \right) \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{cos}{u}+\mathrm{sin}{u}\right){e}^{{u}} \right) \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{{x}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{log}{x}\right)+\mathrm{sin}\left(\mathrm{log}{x}\right)\right)\:\:\left(+\mathrm{C}\right) \\ $$

Commented by peter frank last updated on 26/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Paradoxical last updated on 26/Aug/21

Commented by peter frank last updated on 26/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by puissant last updated on 26/Aug/21

K=∫cos(lnx)dx    u=cos(lnx) ; v′=1    ⇒ K= xcos(lnx)+∫sin(lnx)dx    u=sin(lnx) ; v′=1    ⇒ K=xcos(lnx)+xsin(lnx)−∫cos(lnx)dx    ⇒ 2K=xcos(lnx)+xsin(lnx)    ∴∵  Q=(x/2)cos(lnx)+(x/2)sin(lnx)+C

$${K}=\int{cos}\left({lnx}\right){dx} \\ $$$$ \\ $$$${u}={cos}\left({lnx}\right)\:;\:{v}'=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{K}=\:{xcos}\left({lnx}\right)+\int{sin}\left({lnx}\right){dx} \\ $$$$ \\ $$$${u}={sin}\left({lnx}\right)\:;\:{v}'=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{K}={xcos}\left({lnx}\right)+{xsin}\left({lnx}\right)−\int{cos}\left({lnx}\right){dx} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{2}{K}={xcos}\left({lnx}\right)+{xsin}\left({lnx}\right) \\ $$$$ \\ $$$$\therefore\because\:\:{Q}=\frac{{x}}{\mathrm{2}}{cos}\left({lnx}\right)+\frac{{x}}{\mathrm{2}}{sin}\left({lnx}\right)+{C} \\ $$

Commented by peter frank last updated on 26/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by qaz last updated on 26/Aug/21

∫x^i dx=∫e^(ilnx) dx=(x^(i+1) /(i+1))+C=((1−i)/2)x∙e^(ilnx) +C  ⇒∫cos (lnx)dx=(1/2)xcos (lnx)+(1/2)xsin (lnx)+C

$$\int\mathrm{x}^{\mathrm{i}} \mathrm{dx}=\int\mathrm{e}^{\mathrm{ilnx}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{i}+\mathrm{1}} }{\mathrm{i}+\mathrm{1}}+\mathrm{C}=\frac{\mathrm{1}−\mathrm{i}}{\mathrm{2}}\mathrm{x}\centerdot\mathrm{e}^{\mathrm{ilnx}} +\mathrm{C} \\ $$$$\Rightarrow\int\mathrm{cos}\:\left(\mathrm{lnx}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xcos}\:\left(\mathrm{lnx}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xsin}\:\left(\mathrm{lnx}\right)+\mathrm{C} \\ $$

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