F Ferocity Fanatic Jun 9, 2019 #1 Pa help dito mga sir, iproprove daw using rational functions of sinx and cosx, or weierstrass.
R Roby Robyn Eternal Poster Jun 9, 2019 #3 Rewrite mo to tapos check ko csc udu= csc u (csc u+cot u / csc u+cot u) du Let u=csc u + cot u So du= (-csc u cot u - csc^2 u)du = csc u + cot u csc udu= - du / u Integral nito -ln |u| + c Tapos balik mo lang -ln |csc u + cot u| + C
Rewrite mo to tapos check ko csc udu= csc u (csc u+cot u / csc u+cot u) du Let u=csc u + cot u So du= (-csc u cot u - csc^2 u)du = csc u + cot u csc udu= - du / u Integral nito -ln |u| + c Tapos balik mo lang -ln |csc u + cot u| + C
P PHC-BlackNight Eternal Poster Jun 9, 2019 #4 RobyRobyn00 said: Rewrite mo to tapos check ko csc udu= csc u (csc u+cot u / csc u+cot u) du Let u=csc u + cot u So du= (-csc u cot u - csc^2 u)du = csc u + cot u csc udu= - du / u Integral nito -ln |u| + c Tapos balik mo lang -ln |csc u + cot u| + C Click to expand... Talino ni paps
RobyRobyn00 said: Rewrite mo to tapos check ko csc udu= csc u (csc u+cot u / csc u+cot u) du Let u=csc u + cot u So du= (-csc u cot u - csc^2 u)du = csc u + cot u csc udu= - du / u Integral nito -ln |u| + c Tapos balik mo lang -ln |csc u + cot u| + C Click to expand... Talino ni paps
R Roby Robyn Eternal Poster Jun 9, 2019 #5 Eto rin cot udu= (cos u / sin u ) du Let u=sin u So du= cos u du Sub mo na lang yung du=cos u, tapos u =sin u (cos u / sin u) du= u / du Integral nito =ln |u| + C Balik mo lang cot udu= ln |sin u| + C
Eto rin cot udu= (cos u / sin u ) du Let u=sin u So du= cos u du Sub mo na lang yung du=cos u, tapos u =sin u (cos u / sin u) du= u / du Integral nito =ln |u| + C Balik mo lang cot udu= ln |sin u| + C
R Roby Robyn Eternal Poster Jun 9, 2019 #6 BlackNight3226 said: Talino ni paps Click to expand... Haha di naman napag aralan lang