User:Raul654/proof
|
|
Proof that 0 = -1
I stumbled across this in my high school Calculus BC class. Here's a challenge for my fellow math geeks - what is wrong here?
Given:
<math> \int_ {} u\, dv = u * v - \int_ {} v\, du<math>
Let <math>\mathbf{dv} = \mathbf{sin(x) * dx}<math>
- Therefore, <math>\mathbf{v} = - \mathbf{ cos(x)}<math>
Let <math>\mathbf{u} = \mathbf{sec(x)}<math>
- Therefore, <math>\mathbf{du} = \mathbf{sec(x) * tan (x) * dx}<math>
<math> \int_ {} tan (x)\, dx = \int_ {} tan (x)\, dx <math>
<math> \int_ {} tan (x)\, dx = \int_ {} sec (x) * sin(x)\, dx <math>
- Substitute from above.
<math> \int_ {} tan (x)\, dx = sec (x) * (- cos (x)) - \int_ {} - cos (x) * sec(x) * tan(x) \, dx <math>
- Sec(x) * cos(x) = 1
<math> \int_ {} tan (x)\, dx = - 1 - \int_ {} - tan(x) \, dx <math>
- Group the minus signs.
<math> \int_ {} tan (x)\, dx = - 1 + \int_ {} tan(x) \, dx <math>
- (Subtract the integral from each side)
<math>\mathbf{0} = \mathbf{-1}<math>
- Voilą - a logical inconsistency.