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Question Number 42407 by Tawa1 last updated on 25/Aug/18

$$\int \frac{1}{1+\mathrm{tanx}}\mathrm{dx}$$

Commented by maxmathsup by imad last updated on 25/Aug/18

$$letI=\int \frac{dx}{1+tanx}changementtanx=tgive$$$$I=\int \frac{1}{(1+t)(t^2+1)}dtletdecomposeF\left(t\right)=\frac{1}{(t+1)(t^2+1)}$$$$F\left(t\right)=\frac{a}{t+1}+\frac{bt+c}{t^2+1}wehavea={lim}_{t\to -1}(t+1)F\left(t\right)=\frac{1}{2}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+b\Rightarrow b=-\frac{1}{2}$$$$F\left(0\right)=1=a+c\Rightarrow c=1-a=\frac{1}{2}\Rightarrow F\left(t\right)=\frac{1}{2(t+1)}-\frac{1}{2}\frac{t-1}{t^2+1}\Rightarrow$$$$I=\int F\left(t\right)dt=\frac{1}{2}ln\mid t+1\mid -\frac{1}{4}\int \frac{2t-2}{t^2+1}dt$$$$=\frac{1}{2}ln\mid t+1\mid -\frac{1}{4}ln(t^2+1)+\frac{1}{2}arctan\left(t\right)+c$$$$=\frac{x}{2}+\frac{1}{2}ln\mid 1+tanx\mid -\frac{1}{4}ln(1+{tan}^{2}x)+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

$$\frac{1}{2}\int \frac{2cosx}{sinx+cosx}dx$$$$\frac{1}{2}\int \frac{sinx+cosx+cosx-sinx}{sinx+cosx}dx$$$$\frac{1}{2}\int dx+\frac{1}{2}\int \frac{d(sinx+cosx)}{sinx+cosx}$$$$=\frac{1}{2}\{x+ln(sinx+cosx)\}+c$$

Commented by Tawa1 last updated on 25/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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