Question

what is intigration of tanx

Answer 1

integration  sinx/cosx
let cosx=z   so   dz/dx=-sinx    orr   dx=-dz/sinx
  putting value of dx   we get
integration  (sinx/cosx)-dz/sinx= -dz/z
hence integration of 1/z= log z
so integration tanx= log|cosx|+c

Answer 2

we know that
tan x=(sinx/cosx)
∫ tan(x) dx = ∫ sin(x)/cos(x) dx


Let cos(x) = u
-sin(x) dx = du
sin(x) dx = -du


The integral becomes

- ∫ du/u = -ln|u| = - ln |cos(x)| + C

tanx=-ln cos(x)