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Question Number 201509 by Calculusboy last updated on 07/Dec/23

Answered by witcher3 last updated on 09/Dec/23

$$\mathrm{tack}\mathrm{principal}\mathrm{definition}\mathrm{of}\mathrm{Log}\left(\mathrm{z}\right)=\ln \mid \mathrm{z}\mid +\mathrm{iarg}\left(\mathrm{z}\right)$$$$\mathrm{z}\in ]-\frac{\pi }{2},\frac{3\pi }{2}[$$$$\ln (\mathrm{ix}+\ln \left(\cos \left(\mathrm{x}\right)\right)=\ln \mid \sqrt{{\mathrm{x}}^2+{\ln }^2(\cos \left(\mathrm{x}\right)}\mid +\mathrm{i}\arg (\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix})$$$$\arg (\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix})\in [\frac{\pi }{2},\pi [$$$$\arg (\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix})={\tan }^{-1}\left(\frac{\mathrm{x}}{\ln (\cos \left(\mathrm{x}\right)}\right)+\pi$$$$\cos (\ln \left(\cos \left(\mathrm{x}\right)\right)+\mathrm{ix})=\frac{1}{2}\ln ({\mathrm{x}}^2+{\ln }^2\left(\cos \left(\mathrm{x}\right)\right)+\mathrm{i}({\tan }^{-1}\left(\frac{\mathrm{x}}{\ln (\cos \left(\mathrm{x}\right)}\right)+\pi )$$$$\Rightarrow 2\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix})=\ln ({\mathrm{x}}^2+{\ln }^2\left(\cos \left(\mathrm{x}\right)\right)+{\mathrm{itan}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{lncos}\left(\mathrm{x}\right)}\right)+2\pi$$$$\mathrm{principal}\mathrm{definition}$$$$\Rightarrow 2\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix})=\ln ({\mathrm{x}}^2+{\ln }^2\left(\cos \left(\mathrm{x}\right)\right)+{2\mathrm{itan}}^{-1}\left(\frac{\mathrm{x}}{\ln (\cos \left(\mathrm{x}\right)}\right)$$$$\mid 2\ln (\cos \left(\mathrm{x}\right)+\mathrm{ix}){\mid }^2={\ln }^2({\mathrm{x}}^2+{\ln }^2\left(\cos \left(\mathrm{x}\right)\right))+{4\arctan }^2\left(\frac{\mathrm{x}}{\ln (\cos \left(\mathrm{x}\right)}\right)$$$$\mathrm{easy}\mathrm{from}\mathrm{here}$$$$$$$$$$

Commented by Calculusboy last updated on 10/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by witcher3 last updated on 10/Dec/23

$$\mathrm{withe}\mathrm{Pleasur}$$$$$$

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