- #1

- 16

- 0

## 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.