I have worked out ∫ tan(x) dx -∏/6<x<∏/3
to 0.5493 to 4d.p.
But I'm checking things through & would like some clarification, please.
Using trig definitions I let tan(x) = sin(x)/cos(x)
then I used substitution
let u = cos(x) so du/dx = -sin(x)
du = -sin(x) dx
& - du/sin(x) = dx
-∫ sin(x)/u * du/sin(x) = du/u as sin(x) cancels.
[-ln(u)] -∏/6<x<∏/3
=0.5493 to 4d.p.
Having looked at several other places
MathDoctorBob
shows that ∫tan(x) = ln(sec(x))+c
clindelof
shows the way I did it.
& my Stroud engineering maths says ∫tan(x) = sec^{2}x
WolframAlpha gets the same answer as me, too.
So am I right?