-1 = 0 ?? | Shuzak.com

-1 = 0 ?? - 22 May, 2008

			Seth says
			$<br /> int tan{x}dx = int sec{x}sin{x}dx<br />$ $<br /> u = sec{x} <br />$ $<br /> dv = sin{x}dx<br />$ $<br /> du = sec{x}tan{x}dx<br />$ $<br /> v = -cos{x}<br />$ $<br /> int udv = uv - int vdu<br />$ $<br /> int tan{x}dx = int sec{x}sin{x}dx = -sec{x}cos{x} - int -cos{x}sec{x}tan{x}dx<br />$ $<br /> int tan{x}dx = -1 + int tan{x}dx<br />$ $<br /> 0 = -1<br />$ I cant figure this out i dint even devide by zero [/math]
			Total Topic Karma: 5	- More by this Author

Ati says

+0 Karma

Sorry, not my area of expertise. Try God?

- Author's History - 22 May, 2008

Walter says

+3 Karma

Until the second to last line the process was correct. But an integral is an antiderative plus a constant, so you can't assume both
$\int tan{x}dx$
are equal. In fact that equation can be expressed as
$ln {|sec{x}|} + C_1 = -1 + (ln {|sec{x}|} + C_2)$
We can see -1 is a constant, and then the equation is merely expressed as
$ln {|sec{x}|} + C_1 = ln {|sec{x}|} + C_2$
From this, it is expected that
$C_1 = C_2$
given that those constants can assume any real value.

- Author's History - 25 May, 2008

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