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Question Number 141805 by mohammad17 last updated on 23/May/21

$$\int \frac{dx}{1-tanx}$$

Answered by MJS_new last updated on 23/May/21

$$\int \frac{dx}{1-\tan x}=$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=\int \frac{dt}{(1-t)(t^2+1)}=$$$$=\frac{1}{4}\int \frac{2t}{t^2+1}dt+\frac{1}{2}\int \frac{dt}{t^2+1}-\frac{1}{2}\int \frac{dt}{t-1}=$$$$=\frac{1}{4}\ln (t^2+1)+\frac{1}{2}\arctan t-\frac{1}{2}\ln (t-1)=$$$$=\frac{1}{4}\ln \frac{t^2+1}{{(t-1)}^2}+\frac{1}{2}\arctan t=$$$$=\frac{x}{2}-\frac{1}{4}\ln (1-\sin 2x)+C$$

Answered by mathmax by abdo last updated on 24/May/21

$$\mathrm{\Phi }=\int \frac{\mathrm{dx}}{1-\mathrm{tanx}}\Rightarrow \mathrm{\Phi }{=}_{\mathrm{tanx}=\mathrm{t}}\int \frac{\mathrm{dt}}{(1+{\mathrm{t}}^2)(1-\mathrm{t})}=-\int \frac{\mathrm{dt}}{(\mathrm{t}-1)({\mathrm{t}}^2+1)}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{1}{(\mathrm{t}-1)({\mathrm{t}}^2+1)}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{a}}{\mathrm{t}-1}+\frac{\mathrm{bt}+\mathrm{c}}{{\mathrm{t}}^2+1},\mathrm{a}=\frac{1}{2},{\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tF}\left(\mathrm{t}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{1}{2}$$$$\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=\frac{1}{2(\mathrm{t}-1)}+\frac{-\frac{1}{2}\mathrm{t}+\mathrm{c}}{{\mathrm{t}}^2+1}$$$$\mathrm{F}\left(\mathrm{o}\right)=-1=-\frac{1}{2}+\mathrm{c}\Rightarrow \mathrm{c}=-\frac{1}{2}\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=\frac{1}{2(\mathrm{t}-1)}-\frac{1}{2}\times \frac{\mathrm{t}+1}{{\mathrm{t}}^2+1}$$$$\Rightarrow \mathrm{\Phi }=-\int \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=-\frac{1}{2}\int \frac{\mathrm{dt}}{\mathrm{t}-1}+\frac{1}{2}\int \frac{\mathrm{t}+1}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=-\frac{1}{2}\log \mid \mathrm{t}-1\mid +\frac{1}{4}\log ({\mathrm{t}}^2+1)+\frac{1}{2}\mathrm{arctant}+\mathrm{C}$$$$=-\frac{1}{2}\log \mid \mathrm{tanx}-1\mid +\frac{1}{4}\log (1+{\tan }^2\mathrm{x})+\frac{\mathrm{x}}{2}+\mathrm{C}$$

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