- #1

- 181

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Consider the identity x

^{2}= - 2, where x = ± i√2.

Now rearrange this identity to x

^{2}+ 1 = -1

Invert the equation to get the reciprocals 1/(x

^{2}+ 1) = - 1

Multiply both sides by arctan(x)

arctan(x)(1/(x

^{2}+ 1)) = - arctan(x)

Let y = arctan(x)

Then we have the differential equation yy' = - y (Because 1/(x^2 + 1) is the derivative of arctan(x))

Divide both sides by y to get y' = - 1

Now integrate both sides ∫dy = - ∫dx

The result is y = - x

But since y = arctan(x), it follows that arctan(x) = - x

And since we know that x = ± i√2, it follows that arctan(i√2) = - i√2 and arctan(-i√2) = i√2

I know I must have done something wrong here. Could someone please point out the errors (which I'm sure are many ) in the above?