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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.

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  • $\begingroup$ $\mathrm{arccot}$ 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 yesterday
  • $\begingroup$ @peek-a-boo I am sorry for the confusing writeup. I meant the sine function. $\endgroup$ Commented yesterday
  • $\begingroup$ @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 yesterday
  • $\begingroup$ $\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$ 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 yesterday
  • $\begingroup$ 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 yesterday

2 Answers 2

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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. :>

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    $\begingroup$ 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 yesterday
  • $\begingroup$ $\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$ 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 yesterday
  • $\begingroup$ @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 yesterday
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    $\begingroup$ @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 23 hours ago
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Your first method is not correct because the sine term is not entirely an odd function, in fact, for any $\,x\in\Bbb R\,,\,$

$\sin(\cot^{-1}(-x))=\sin(\pi-\cot^{-1}x)=\sin(\cot^{-1}x)$

and it means that actually it is entirely an even function.

For the same reason, your second method gives you an incorrect result, nevertheless it is true that, for any $\,x\in\Bbb R\,,$

$\cot^{-1}{x}+\tan^{-1}x=\pi/2\qquad\qquad$ and
$\cos(\tan^{-1}x) = \cos(\pi/2 - \cot^{-1}{x}) = \sin(\cot^{-1}{x})\;.\quad\;\color{blue}{(*)}$

Finally, your third method is correct, indeed, by using the equality $\,(*)\,,\,$ it follows that

$\begin{align}\displaystyle I&=\int_{-1}^{1}\frac{\sin(\cot^{-1}x)+\cos(\tan^{-1}x)}{x^2+1}\,\mathrm{d}x=\\[3pt]&=2\int_{-1}^{1}\frac{\cos(\tan^{-1}x)}{x^2+1}\,\mathrm{d}x\underset{\overbrace{\;u=\tan^{-1}\!x\;}}{=}\\[3pt]&=2\int_{-\pi/4}^{\pi/4}\cos u\,\mathrm{d}u=2\bigg[\sin u\bigg]_{-\pi/4}^{\pi/4}\!\!\!\!\!=2\sqrt2\;.\end{align}$

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