To evaluate this integral, it is best to remember that `tanx=sinx/cosx` , which suggests the substitution `u=cosx` . This means that `du=-sinxdx` , so the integral becomes:

`int tanxdx`

`=int sinx/cosx dx`

`=-int 1/udu` now integrate using logarithm rule

`=-ln|u|+C` where C is a constant of integration

`=-ln|cosx|+C` now switch negative 1 to exponent

`=ln|cosx|^{-1}+C` use trig definition of `secx` .

`=ln|secx|+C`

**The integral evaluates to `ln|secx|+C` .**

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