Thread: Vol of revolution: Integral of tan(x)

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


shows that ∫tan(x) = ln(sec(x))+c




shows the way I did it.

& my Stroud engineering maths says ∫tan(x) = sec²x
WolframAlpha gets the same answer as me, too.

So am I right?

Your answer is correct.

Originally Posted by froodles01

Having looked at several other places
... shows that ∫tan(x) = ln(sec(x))+c

... shows the way I did it.

Indeed, ln(sec(x)) = -ln(cos(x)).

Originally Posted by froodles01

& my Stroud engineering maths says ∫tan(x) = sec²x

This does not seem to be correct.

Stroud is wrong. (The derivative of tanx is sec^2x )

your title says "volume of revolution" ... how did the definite integral of tanx fit into the problem?

Title is wrong. I'm doing too many things at once & although I tried to edit this to make title correct, it still says the wrong thing.
As for vol of revolution I have been able to do this.
∏ ∫ (f(x))^2 dx

but the volume formula says ∏ ∫ (f(x))^2 dx.giving ∏ ∫ (tan(x))^2 dx.does that equate to ∏ ∫ (sin(x))^2/ (cos(x))^2 dx and integrate from there????

tan^2(x)=sec^2(x)-1 So integral of tan^2(x)=tanx-x