Integrating Tangent by parts; 0 = -1

[NoOne0507]

Homework Statement

The question is what has gone wrong in this proof, it is worth noting this a definite integral between pi/6 and pi/4:

∫ tan(x) dx = ∫ sin(x)/cos(x) dx
Let u = 1/cos(x) and dv = sin(x) dx
So du= sec(x)tan(x) and v = -cos(x)

When we substitute back in we get:
∫ tan(x) dx = -1 + ∫ tan(x) dx subtract the integral:
0 = -1

Homework Equations

∫u dv = uv - ∫v du

The Attempt at a Solution

First off, I know how to integrate tangent, with just straight u substitution. The problem is why 0 = -1 shows up in this proof; I can't find the error.

The two things I can think of is the discontinuity of tangent on pi/2, 3pi/2, etc, but those aren't included in the bounds of integration. Even so, the indefinite integral doesn't work either.

The other is the constant of integration. An indefinite integral is related to its definite integral by some constant, C. But it seems like since ∫tan(x) dx shows up on both sides the same arbitrary constant should cancel itself out.

 

[micromass]

Homework Statement

The question is what has gone wrong in this proof, it is worth noting this a definite integral between pi/6 and pi/4:

∫ tan(x) dx = ∫ sin(x)/cos(x) dx
Let u = 1/cos(x) and dv = sin(x) dx
So du= sec(x)tan(x) and v = -cos(x)

When we substitute back in we get:
∫ tan(x) dx = -1 + ∫ tan(x) dx subtract the integral:
0 = -1

Homework Equations

∫u dv = uv - ∫v du

The Attempt at a Solution

First off, I know how to integrate tangent, with just straight u substitution. The problem is why 0 = -1 shows up in this proof; I can't find the error.

The two things I can think of is the discontinuity of tangent on pi/2, 3pi/2, etc, but those aren't included in the bounds of integration. Even so, the indefinite integral doesn't work either.

The other is the constant of integration. An indefinite integral is related to its definite integral by some constant, C. But it seems like since ∫tan(x) dx shows up on both sides the same arbitrary constant should cancel itself out.

It's the constant of integration that is the culprit. It is true that \int\tan(x)dx appears on both sides, but those two integrals do not necessarily have the same constant of integration!

 

[dynamicsolo]

The question is what has gone wrong in this proof, it is worth noting this a definite integral between pi/6 and pi/4:

∫ tan(x) dx = ∫ sin(x)/cos(x) dx
Let u = 1/cos(x) and dv = sin(x) dx
So du= sec(x)tan(x) dx and v = -cos(x)

When we substitute back in we get:
∫ tan(x) dx = -1 + ∫ tan(x) dx subtract the integral:

This is the deceptive step: the indefinite integral has an arbitrary constant, which can "absorb" the "-1". This is equivalent to saying that since the general antiderivative function is unaffected by a "vertical shift", you can easily "shift away" the numerical term.* You have conjured a finite number out of the infinite summation, but it can be made to disappear again. (In the definite integration, the numerical constant term will simply cancel out.)

*which sounds like a slightly "shifty" argument...In my own experience when I was learning integration-by-parts, obtaining an "equation" like this told me that this was a wrong technique of integration for the purpose...