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

$$Bonjourbesoin{d}^{\prime }aide$$$$Calculer\int ln\left(cosx\right)dx$$

Answered by Olaf last updated on 14/Sep/20

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

Answered by MJS_new last updated on 18/Sep/20

$$\int \ln \cos xdx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=x\ln \cos x+\int x\tan xdx$$$$$$$$\int x\tan xdx=-\mathrm{i}\int x\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{{\mathrm{e}}^{\mathrm{i}x}+{\mathrm{e}}^{\mathrm{i}x}}dx=$$$$=-\mathrm{i}\int x\frac{{\mathrm{e}}^{2\mathrm{i}x}-1}{{\mathrm{e}}^{2\mathrm{i}x}+1}dx=$$$$[t={\mathrm{e}}^{2\mathrm{i}x}\iff x=-\frac{\mathrm{i}}{2}\ln t\to dx=-\frac{\mathrm{i}}{2t}dt]$$$$=\frac{\mathrm{i}}{4}\int \ln t\frac{t-1}{t(t+1)}dt=$$$$\frac{\mathrm{i}}{2}\int \frac{\ln t}{t+1}dt-\frac{\mathrm{i}}{4}\int \frac{\ln t}{t}dt$$$$$$$$\frac{\mathrm{i}}{2}\int \frac{\ln t}{t+1}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{\mathrm{i}}{2}\ln t\ln (t+1)-\frac{\mathrm{i}}{2}\int \frac{\ln (t+1)}{t}dt=$$$$=\frac{\mathrm{i}}{2}\ln t\ln (t+1)+\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(-t)$$$$$$$$-\frac{\mathrm{i}}{4}\int \frac{\ln t}{t}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{\mathrm{i}}{8}{\left(\ln t\right)}^2$$$$$$$$\Rightarrow$$$$\int x\tan xdx=\frac{\mathrm{i}}{2}\ln t\ln (t+1)+\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(-t)+\frac{\mathrm{i}}{8}{\left(\ln t\right)}^2=$$$$=\frac{\mathrm{i}}{2}\ln {\mathrm{e}}^{2\mathrm{i}x}\ln ({\mathrm{e}}^{2\mathrm{i}x}+1)+\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(-{\mathrm{e}}^{2\mathrm{i}x})+\frac{\mathrm{i}}{8}{\left(\ln {\mathrm{e}}^{2\mathrm{i}x}\right)}^2=$$$$=-\frac{\mathrm{i}}{2}{x}^2-x\ln ({\mathrm{e}}^{2\mathrm{i}x}+1)+\frac{\mathrm{i}}{2}\mathrm{L}{i}_2(-{\mathrm{e}}^{2\mathrm{i}x})$$$$\Rightarrow$$$$\int \ln \cos xdx=$$$$=x\ln \cos x-\frac{\mathrm{i}}{2}{x}^2-x\ln ({\mathrm{e}}^{2\mathrm{i}x}+1)+\frac{\mathrm{i}}{2}\mathrm{L}{i}_2(-{\mathrm{e}}^{2\mathrm{i}x})+C$$$$$$$$\mathrm{hopefully}\mathrm{I}\mathrm{made}\mathrm{no}\mathrm{errors},\mathrm{please}\mathrm{check}!$$

Commented by Tawa11 last updated on 06/Sep/21

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

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