Evaluate: `int{1+tanxtan(x+alpha)}dx`

Explanation

We have given

`int{1+tanxtan(x+alpha)}dx`

`I=int{1+tanxtan(x+alpha)}dx`

`I=int((1+tanxtan(x+alpha))/(tan(x+alpha)−tanx))(tan(x+alpha)−tanx)dx`

`I=int(tan(x+alpha)−tanx)/(tanalpha) dx`

`I=(1/(tan alpha))int(tan(x+alpha)−tanx)dx`

`I=(1/(tan alpha))(−logcos(x+alpha)1]+log∣cosx∣)+c`

`I=(1/(tan alpha))log|cosx/cos(x+alpha)|+c`

...