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Question Number 145064 by akolade last updated on 02/Jul/21

$$\int \cos 2\mathrm{xln}(1+\tan \mathrm{x})\mathrm{dx}$$$$$$

Answered by mathmax by abdo last updated on 02/Jul/21

$$\mathrm{\Psi }=\int \cos \left(2\mathrm{x}\right)\log (1+\mathrm{tanx})\mathrm{dx}\mathrm{by}\mathrm{parts}$$$$\mathrm{\Psi }=\frac{\sin \left(2\mathrm{x}\right)}{2}\log (1+\mathrm{tanx})-\int \frac{\sin \left(2\mathrm{x}\right)}{2}\times \frac{(1+{\tan }^2\mathrm{x})}{1+\mathrm{tanx}}\mathrm{dx}$$$$\int \sin \left(2\mathrm{x}\right).\frac{1+{\tan }^2\mathrm{x}}{1+\mathrm{tanx}}\mathrm{dx}{=}_{\mathrm{tanx}=\mathrm{t}}\int \frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\times \frac{1+{\mathrm{t}}^2}{1+\mathrm{t}}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$=2\int \frac{\mathrm{t}}{(\mathrm{t}+1)({\mathrm{t}}^2+1)}\mathrm{dt}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{(\mathrm{t}+1)({\mathrm{t}}^2+1)}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{a}}{\mathrm{t}+1}+\frac{\mathrm{bt}+\mathrm{c}}{{\mathrm{t}}^2+1}\Rightarrow \mathrm{a}=-\frac{1}{2}$$$${\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tF}\left(\mathrm{t}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=\frac{1}{2}$$$$\mathrm{F}\left(0\right)=0=\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=\frac{1}{2}\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=\frac{-1}{2(\mathrm{t}+1)}+\frac{1}{2}\times \frac{\mathrm{t}+1}{{\mathrm{t}}^2+1}\Rightarrow$$$$\int \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=-\frac{1}{2}\log \mid \mathrm{t}+1\mid +\frac{1}{4}\log ({\mathrm{t}}^2+1)+\frac{1}{2}\mathrm{arctant}+\mathrm{K}\Rightarrow$$$$\mathrm{\Psi }=\frac{1}{2}\sin \left(2\mathrm{x}\right)\log (1+\mathrm{tanx})+\frac{1}{2}\log \mid \mathrm{tanx}+1\mid -\frac{1}{4}\log (1+{\tan }^2\mathrm{x})+\frac{\mathrm{x}}{2}+\mathrm{K}$$

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