2
$\begingroup$

The integral is:

$$I = \int_{-1}^{1} \frac{\sin(\cot^{-1}x) + \cos(\tan^{-1}x)}{x^2 + 1}\,\mathrm{d}x$$

First method: We notice that the sine term is entirely an odd function so its contribution to the integral is $0$. Working with a triangle and after a few steps, we get $I = \sqrt2$. This result seems consistent with WolframAlpha.

Second method: We use the fact that $\cot^{-1}{x} + \tan^{-1}x = \pi/2$. Then, $\cos(\tan^{-1}x) = \cos(\pi/2 - \cot^{-1}{x}) = \sin(\cot^{-1}{x})$. And the whole integral becomes $0$ after noticing that the function becomes odd.

Third method: We use the same fact but we make $\sin(\cot^{-1}{x}) = \cos(\tan^{-1}x)$ and we get $I = 2\sqrt2$.

What am I missing? I suspect that the identity might not hold true in this case but I fail to see why.

$\endgroup$
7
  • $\begingroup$ $\mathrm{arccot}$ is not an odd function; its value at $x=0$ is $\pi/2$, but for odd functions it should be $0$; so assuming I’m not making silly mistakes, it is $\mathrm{arccot}(x)-\frac{\pi}{2}$ which is an odd function of $x$. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ @peek-a-boo I am sorry for the confusing writeup. I meant the sine function. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ @peek-a-boo: $\operatorname{arccot}-\frac{\pi}2$ isn't odd either. One way to fix that issue is to just ignore it, since changing the value of a function at a point does not affect the integral. So, for the purpose of integration, that function can be treated as an odd function. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ $\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$ is standard where I live. Of course then you don't have $\operatorname{arccot}(-x)=-\operatorname{arccot}(x)$ or $\operatorname{arccot}(x)=\arctan\frac1x$, which would be stupid anyways. You gain, of course, in the fact that then $\operatorname{arccot}$ is differentiable. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ FWIW, being consistent with WolframAlpha isn't always the best litmus test. Their definition of ArcCot deviates from the common convention of being continuous. $\endgroup$ Commented 13 hours ago

1 Answer 1

Reset to default
3
$\begingroup$

Note. The original answer used the convention of WolframAlpha which is not the usual way one defines $\operatorname{arccot}$. See the edit below.

The identity is not correct. What does hold is: $$\arctan(x)+\operatorname{arccot}(x)=\begin{cases}\pi/2,&\mathrm{if}\ x>0\\-\pi/2,&\mathrm{if}\ x<0\end{cases}$$ Working with this identity (after breaking the integral at $0$) will lead you to the correct answer.

With the usual definition, the identity is indeed correct. What is not correct in that case, is that $\operatorname{arccot}$ is not an odd function anymore. Working with that definition, indeed $$\sin(\operatorname{arccot}(x))=\cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}$$ The integral then becomes $$\int_{-1}^1\frac{2}{(1+x^2)^{3/2}}\,\mathrm{d}x$$ which can be computed to be $2\sqrt{2}$.

Hope this helps. :>

$\endgroup$
4
  • $\begingroup$ This definitely resolved my confusion. Thank you. Of course, as Sassatelli Giulio said above, only the first part is the widely taught standard form. Is there a reason for that? $\endgroup$ Commented 13 hours ago
  • $\begingroup$ $\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$ is standard where I live. Of course then you don't have $\operatorname{arccot}(-x)=-\operatorname{arccot}(x)$ or $\operatorname{arccot}(x)=\arctan\frac1x$, which were stupid anyways. You gain in the fact that then $\operatorname{arccot}(x)$ is differentiable. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ @SassatelliGiulio: Now that you say it, yeah you are right. I went and checked and realised that the discrepancy comes from the fact that WolframAlpha does not follow that (the usual) convention. $\endgroup$ Commented 13 hours ago
  • 1
    $\begingroup$ @ultralegend5385, $$\arctan(x)+\operatorname{arccot}(x)=\begin{cases}\pi/2,&\mathrm{if}\ x>0\\-\pi/2,&\mathrm{if}\ x<0\end{cases}$$ is wrong, in fact it results that $$\arctan(x)+\operatorname{arccot}(x)=\pi/2$$ for any $x\in\Bbb R$. $\endgroup$ Commented 11 hours ago

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.