# Don't the derivatives of $\arctan x$ and $\operatorname{arccot} x$ imply they're negatives of each other?

I think I'm misunderstanding something, but I'm not sure exactly what.

So we know: $$\int - \frac{1}{x^2+1}dx=\operatorname{arccot}(x)+C$$

If we then bring the negative sign out of the integrand, we get:

$$-\int \frac{1}{x^2+1}dx=-\arctan(x)+C$$

Since the integrals are the same, doesn't that imply that $$\operatorname{arccot}(x) = -\arctan(x)$$?

I've repeated this process using $$\sin(x)$$ and $$\cos(x)$$, but they check out as the negatives cancel to produce the same results.

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• Did you try plugging in some numbers for $x$ ? A word to the wise: computation is an excellent cure for headache. Jun 25, 2019 at 4:28

If two functions $$f$$ and $$g$$ has the same derivative, $$f'=g'$$, then we can say $$f=g+c$$ for some constant $$c$$. In this case $$-\arctan(x)=\operatorname{arccot}(x)+c$$ and, since $$\arctan(1)=\operatorname{arccot}(1)=\frac{\pi}{4}$$ this constant $$c$$ must be equal to $$-\pi/2$$. Then we have a nice identity: $$\arctan(x)+\operatorname{arccot}(x)=\frac{\pi}{2}$$
The confusion seems to stem from the fact that the same symbol is used for the constant of integration in both integrals. If we instead write the general antiderivatives using the (a priori distinct) arbitrary constants $$C, D$$, our equality is then $$\operatorname{arccot} x + C = -\arctan x + D ,$$ and rearranging gives that $$\arctan x + \operatorname{arccot} x = E,$$ where $$E := D - C$$ is some constant. Evaluating at, e.g., $$x = 0$$, gives $$E = \frac{\pi}{2},$$ so $$\arctan x$$ and $$\operatorname{arccot} x$$ are not negatives of one another, and instead they together satisfy the identity $$\color{#df0000}{\boxed{\arctan x + \operatorname{arccot} x = \frac{\pi}{2}}} .$$