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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
0 0 0
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)

1 5 1
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