Question Number 41514 by maxmathsup by imad last updated on 08/Aug/18
$${find}\:\:\:\int\:\:{cos}\left({lnx}\right){dx}\: \\ $$
Answered by alex041103 last updated on 09/Aug/18
$${We}\:{know}\:{that}\:{cos}\left(\theta\right)={Re}\left({e}^{{i}\theta} \right) \\ $$$${We}\:{will}\:{assume}\:{x}\epsilon\mathbb{R}. \\ $$$$\Rightarrow\int{cos}\left({ln}\left({x}\right)\right){dx}={Re}\left(\int{e}^{{iln}\left({x}\right)} {dx}\right)= \\ $$$$={Re}\left(\int{x}^{{i}} {dx}\right)={Re}\left(\frac{{x}^{\mathrm{1}+{i}} }{\mathrm{1}+{i}}\right)=\frac{{x}}{\mathrm{2}}{Re}\left({x}^{{i}} \left(\mathrm{1}−{i}\right)\right)= \\ $$$$=\frac{{x}}{\mathrm{2}}{Re}\left(\left({cos}\left({ln}\left({x}\right)\right)+{isin}\left({ln}\left({x}\right)\right)\right)\left(\mathrm{1}−{i}\right)\right)= \\ $$$$=\frac{{x}}{\mathrm{2}}{Re}\left({cos}\left({ln}\:{x}\right)+{sin}\left({ln}\:{x}\right)+{i}\left(...\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}\left({sin}\left({ln}\:{x}\right)+{cos}\left({ln}\:{x}\right)\right) \\ $$$${We}\:{can}\:{simplify}\:{further}\:{by}\:{using} \\ $$$${sin}\:\alpha\:+\:{cos}\:\alpha\:=\:\sqrt{\mathrm{2}}{sin}\left(\alpha+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\int{cos}\left({ln}\:{x}\right){dx}=\frac{{xsin}\left(\frac{\pi}{\mathrm{4}}+{ln}\left({x}\right)\right)}{\sqrt{\mathrm{2}}}\:+\:{C} \\ $$