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I was surprised to discover that

$$\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}\arccos(x)=\frac{-1}{\sqrt{1-x^2}}$$

This would seem to imply $\arcsin=-\arccos$ up to a constant, but then I discovered the identity

$$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$

and then it all made sense.

As a consequenc, if I am integrating $\frac{\pm 1}{\sqrt{1-x^2}}$ I can arbitrariy pick either $\arcsin$ or $\arccos$ as an antiderivative. So far so good.

I then discovered that, similarly,

$$\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}=-\frac{d}{dx}\mathrm{arccot}(x)$$

and assumed the explanation would be the same. But I sanity-checked it, and it turns out

$$\arctan(x)+\mathrm{arccot}(x)=\pm\frac{\pi}{2}$$

is not a constant.

So, if I am integrating $\frac{1}{1+x^2}$, how do I pick an antiderivative?

spraff
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  • https://www.quora.com/What-is-the-fault-in-this-proof-Arccot-x-+Arctan-x-Pi-2 – lab bhattacharjee Oct 01 '18 at 14:53
  • What is $$\int \frac {1}{x^2} dx ?$$ Don't tell me it is $$\frac {-1}{x}+\color{red}{0}$$. Or is it $$\frac {-1}{x}+\color{red}{1}?$$ Or could it be $$\frac {-1}{x}+\color{red}{5}?$$ But I think it can be $$\frac {-1}{x}+\color{red}{\zeta (4)}$$. I guess you should have got your answer. Cheers :-)))))) – Rohan Shinde Oct 01 '18 at 14:55
  • Pick any of the antiderivative of the two and just add a constant $\color{red}{C}$ at the end – Rohan Shinde Oct 01 '18 at 14:58
  • I'm asking how to handle the discontinuity. It seems like the choice of constant won't be valid because it isn't constant. – spraff Oct 01 '18 at 15:10
  • "and it turns out that $\arctan(x)+\operatorname{arccot}(x)=\pm\frac{\pi}{2}$"

    Desmos begs to differ: it says that on $\mathbb R,;,\arctan(x)+\operatorname{arccot}(x)=\frac{\pi}{2}$ (entirely positive). My explanation is here.

     – ryang Feb 21 '22 at 12:00

3 Answers3

1

Since the arccot function (at least in some conventions, like in Mathematica) is only defined for $x \neq 0$, the function $f(x)=\arctan x + \operatorname{arccot} x$ (defined for $x \neq 0$) has the derivative $f'(x)=0$ (defined for $x \neq 0$). A function which has derivative zero is locally constant, so $f(x)=C_1$ for $x<0$ and $f(x)=C_2$ for $x>0$, but $C_1$ doesn't have to be equal to $C_2$.

However, if you use the convention from Wikipedia, then $f(x)$ (defined on the whole real line) is indeed constant on $\mathbf{R}$.

Hans Lundmark
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0

The derivatives are $$\DeclareMathOperator{\arccot}{arccot} \arctan'x=\frac{1}{1+x^2} \qquad \arccot'x=-\frac{1}{1+x^2} $$ This of course implies that $$ f(x)=\arctan x+\arccot x $$ is constant. Since $f(0)=\arctan0+\arccot0=\pi/2$, we can conclude that $$ \arctan x+\arccot x=\frac{\pi}{2} $$

egreg
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$$\frac d{dx}\arctan(x)+\frac d{dx}\mathrm{arccot}(x)=\frac d{dx}\left(\pm\frac{\pi}{2}\right).$$

Now up to you to see when there is a switch between $+$ and $-$.