ITT we confuse the mathsfags [math] -20 = -20 [/math] [math]

>>7789768
>±

[math] \int tan(x)dx = \int tan(x)dx [/math]
[math] \int tan(x)dx = \int sin(x)sec(x)dx [/math]
[math] \int tan(x)dx = -sec(x)cos(x) + \int cos(x)tan(x)sec(x)dx [/math]
[math] \int tan(x)dx = -1 + \int tan(x) dx [/math]
[math] 0 = -1 [/math]

Line 5 does not imply line 6

>>7789777
>comparing indefinite integrals

[math] x = 0 [/math]
[math] 1 = 0 [/math]

[math] \frac{1}{3} = 0.\overline{3} [/math]
[math] \frac{2}{3} = 0.\overline{6} [/math]
[math] \frac{3}{3} = 0.\overline{9} [/math]

[math] \frac{3}{3} = 1 = 0.\overline{9} [/math]

eat shit infintesimalfags

>>7789768

In the transition from line 5 to line 6 of the op, we are supposed to be rused into letting root extraction take place without comment, or consideration as to signing; note that the bases are variously +-0.5.

Since it may happen that (-1)^2 = 1^2, etc, it is not valid to assume that x^2 = y^2 -> x = y. As in, say, the derivation of the quadratic formula, there is a possibility of +- signing, which due to the above requires a bit more investigation.

By quick re-arrangement, one legitimate re-phrase which causes the two to be equal to each other is when the RHS is 9/2 - 4. this forces (correctly), say,

5 = 9 - 4 , or, 5 - 9 = -4.