Sigue tu idea: Sinx/(Sinx+Cosx)= 1/(1+COTX);
Supongamos: u=cotx, entonces x = arccotu, dx =-[ 1 /(1+u ^ 2)]du;
i=-∫du/[(1+u)(1+u^2)]=-(1/2)∫[1/ ( 1+u)+1/(1+u^2)]du=-(1/2)[ln|1+u|-arccotu]+c
= x/2-(1 / 2)ln | 1+cotx |+C .
Sigue tu idea: Sinx/(Sinx+Cosx)= 1/(1+COTX);
Supongamos: u=cotx, entonces x = arccotu, dx =-[ 1 /(1+u ^ 2)]du;
i=-∫du/[(1+u)(1+u^2)]=-(1/2)∫[1/ ( 1+u)+1/(1+u^2)]du=-(1/2)[ln|1+u|-arccotu]+c
= x/2-(1 / 2)ln | 1+cotx |+C .