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:

$\int_ {} u\, dv = u * v - \int_ {} v\, du$

Let $\mathbf{dv} = \mathbf{sin(x) * dx}$

Therefore, $\mathbf{v} = - \mathbf{ cos(x)}$

Let $\mathbf{u} = \mathbf{sec(x)}$

Therefore, $\mathbf{du} = \mathbf{sec(x) * tan (x) * dx}$

$\int_ {} \tan (x)\, dx = \int_ {} \tan (x)\, dx$

$\int_ {} \tan (x)\, dx = \int_ {} \sec (x) * \sin(x)\, dx$

Substitute from above.

$\int_ {} \tan (x)\, dx = \sec (x) * (- \cos (x)) - \int_ {} - \cos (x) * \sec(x) * \tan(x) \, dx$

Sec(x) * cos(x) = 1

$\int_ {} \tan (x)\, dx = - 1 - \int_ {} - \tan(x) \, dx$

Group the minus signs.

$\int_ {} \tan (x)\, dx = - 1 + \int_ {} \tan(x) \, dx$

(Subtract the integral from each side)

$\mathbf{0} = \mathbf{-1}$

Voilà - a logical inconsistency.