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Question Number 41675 by Raj Singh last updated on 11/Aug/18

Commented by math khazana by abdo last updated on 12/Aug/18

$$I=\int \frac{dx}{1+\frac{cosx}{sinx}}=\int \frac{sinx}{sinx+cosx}dx$$$${=}_{tan\left(\frac{x}{2}\right)=t}\int \frac{\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{2t}{2t+1-t^2}\frac{2dt}{1+t^2}$$$$=\int \frac{4t}{(1+t^2)(-t^2+2t+1)}dt=-\int \frac{4tdt}{(t^2+1)(t^2-2t-1)}$$$$letdecomposeF\left(t\right)=\frac{4t}{(t^2+1)(t^2-2t-1)}$$$$rootsof{t}^2-2t-1$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-(-1)=2\Rightarrow t_1=1+\sqrt{2}and{t}_2=1-\sqrt{2}$$$$F\left(t\right)=\frac{4t}{(t^2+1)(t-t_1)(t-t_2)}$$$$=\frac{a}{t-t_1}+\frac{b}{t-t_2}+\frac{ct+d}{t^2+1}$$$$a=\frac{4t_1}{(t_1^2+1)\left(2\sqrt{2}\right)}=\frac{2t_1}{\sqrt{2}(t_1^2+1)}=\frac{2+2\sqrt{2}}{\sqrt{2}(4+2\sqrt{2})}$$$$=\frac{1+\sqrt{2}}{\sqrt{2}(2+\sqrt{2})}$$$$b=\frac{4t_2}{(t_2^2+1)(-2\sqrt{2})}=\frac{4(1-\sqrt{2})}{(4-2\sqrt{2})(-2\sqrt{2})}$$$$=\frac{1-\sqrt{2}}{(2-\sqrt{2})(-\sqrt{2})}=\frac{\sqrt{2}-1}{\sqrt{2}(2-\sqrt{2})}...becontinued...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

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

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