# 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:

${\displaystyle \int _{}u\,dv=u*v-\int _{}v\,du}$

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

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

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

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

${\displaystyle \int _{}\tan(x)\,dx=\int _{}\tan(x)\,dx}$

${\displaystyle \int _{}\tan(x)\,dx=\int _{}\sec(x)*\sin(x)\,dx}$

Substitute from above.

${\displaystyle \int _{}\tan(x)\,dx=\sec(x)*(-\cos(x))-\int _{}-\cos(x)*\sec(x)*\tan(x)\,dx}$

Sec(x) * cos(x) = 1

${\displaystyle \int _{}\tan(x)\,dx=-1-\int _{}-\tan(x)\,dx}$

Group the minus signs.

${\displaystyle \int _{}\tan(x)\,dx=-1+\int _{}\tan(x)\,dx}$

(Subtract the integral from each side)

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

Voilà - a logical inconsistency.