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Question Number 113634 by eric last updated on 14/Sep/20

Bonjour besoin d′aide  Calculer ∫ln(cosx)dx

$${Bonjour}\:{besoin}\:{d}'{aide} \\ $$$${Calculer}\:\int{ln}\left({cosx}\right){dx} \\ $$

Answered by Olaf last updated on 14/Sep/20

cosx = Σ_(k=0) ^∞ (((−1)^k x^(2k) )/((2k)!))...

$$\mathrm{cos}{x}\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!}... \\ $$$$ \\ $$

Answered by MJS_new last updated on 18/Sep/20

∫ln cos x dx=       [by parts]  =xln cos x +∫xtan x dx    ∫xtan x dx=−i∫x((e^(ix) −e^(−ix) )/(e^(ix) +e^(ix) ))dx=  =−i∫x((e^(2ix) −1)/(e^(2ix) +1))dx=       [t=e^(2ix)  ⇔ x=−(i/2)ln t → dx=−(i/(2t))dt]  =(i/4)∫ln t ((t−1)/(t(t+1)))dt=  (i/2)∫((ln t)/(t+1))dt−(i/4)∫((ln t)/t)dt    (i/2)∫((ln t)/(t+1))dt=       [by parts]  =(i/2)ln t ln (t+1) −(i/2)∫((ln (t+1))/t)dt=  =(i/2)ln t ln (t+1) +(i/2)Li_2  (−t)    −(i/4)∫((ln t)/t)dt=       [by parts]  =(i/8)(ln t)^2     ⇒  ∫xtan x dx=(i/2)ln t ln (t+1) +(i/2)Li_2  (−t) +(i/8)(ln t)^2 =  =(i/2)ln e^(2ix)  ln (e^(2ix) +1) +(i/2)Li_2  (−e^(2ix) ) +(i/8)(ln e^(2ix) )^2 =  =−(i/2)x^2 −xln (e^(2ix) +1) +(i/2)Li_2  (−e^(2ix) )  ⇒  ∫ln cos x dx=  =xln cos x −(i/2)x^2 −xln (e^(2ix) +1) +(i/2)Li_2  (−e^(2ix) ) +C    hopefully I made no errors, please check!

$$\int\mathrm{ln}\:\mathrm{cos}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$={x}\mathrm{ln}\:\mathrm{cos}\:{x}\:+\int{x}\mathrm{tan}\:{x}\:{dx} \\ $$$$ \\ $$$$\int{x}\mathrm{tan}\:{x}\:{dx}=−\mathrm{i}\int{x}\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{e}^{\mathrm{i}{x}} +\mathrm{e}^{\mathrm{i}{x}} }{dx}= \\ $$$$=−\mathrm{i}\int{x}\frac{\mathrm{e}^{\mathrm{2i}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{e}^{\mathrm{2i}{x}} \:\Leftrightarrow\:{x}=−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:{t}\:\rightarrow\:{dx}=−\frac{\mathrm{i}}{\mathrm{2}{t}}{dt}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{4}}\int\mathrm{ln}\:{t}\:\frac{{t}−\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:{t}}{{t}+\mathrm{1}}{dt}−\frac{\mathrm{i}}{\mathrm{4}}\int\frac{\mathrm{ln}\:{t}}{{t}}{dt} \\ $$$$ \\ $$$$\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:{t}}{{t}+\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:{t}\:\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}}{dt}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:{t}\:\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(−{t}\right) \\ $$$$ \\ $$$$−\frac{\mathrm{i}}{\mathrm{4}}\int\frac{\mathrm{ln}\:{t}}{{t}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{8}}\left(\mathrm{ln}\:{t}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\int{x}\mathrm{tan}\:{x}\:{dx}=\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:{t}\:\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(−{t}\right)\:+\frac{\mathrm{i}}{\mathrm{8}}\left(\mathrm{ln}\:{t}\right)^{\mathrm{2}} = \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:\mathrm{e}^{\mathrm{2i}{x}} \:\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}\right)\:+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(−\mathrm{e}^{\mathrm{2i}{x}} \right)\:+\frac{\mathrm{i}}{\mathrm{8}}\left(\mathrm{ln}\:\mathrm{e}^{\mathrm{2i}{x}} \right)^{\mathrm{2}} = \\ $$$$=−\frac{\mathrm{i}}{\mathrm{2}}{x}^{\mathrm{2}} −{x}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}\right)\:+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{L}{i}_{\mathrm{2}} \:\left(−\mathrm{e}^{\mathrm{2i}{x}} \right) \\ $$$$\Rightarrow \\ $$$$\int\mathrm{ln}\:\mathrm{cos}\:{x}\:{dx}= \\ $$$$={x}\mathrm{ln}\:\mathrm{cos}\:{x}\:−\frac{\mathrm{i}}{\mathrm{2}}{x}^{\mathrm{2}} −{x}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}\right)\:+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{L}{i}_{\mathrm{2}} \:\left(−\mathrm{e}^{\mathrm{2i}{x}} \right)\:+{C} \\ $$$$ \\ $$$$\mathrm{hopefully}\:\mathrm{I}\:\mathrm{made}\:\mathrm{no}\:\mathrm{errors},\:\mathrm{please}\:\mathrm{check}! \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

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