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Recently I stumbled upon someone who said he wanted to understand why

$\arctan x = \int\dfrac{dx}{1+x^2}$

At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it "intuitively". He wanted to see why it was in terms of arclength and addition and subtraction.

My question is: Is there an "intuitive" way to explain this identity?

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    I believe you mean $1+x^2$ in the denominator.2010-08-20
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    Just a thought: it's probably easier to find an intuitive explanation of the equivalent d/dx arctan x = 1/(1+x^2).2010-08-21
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    @Akjil Thank you. I knew I would make a typo somewhere on my first post.2010-08-22
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    As the waitress said, "plus a constant!". http://preposterousuniverse.blogspot.com/2004/07/is-that-riemann-or-lebesgue-integral.html2011-01-05

7 Answers 7

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The geometric picture is as follows. Let $O = (0, 0), A = (1, 0), X = (1, x)$. Then $\arctan x = \angle AOX$. We want to understand why $\angle AOY \approx \arctan x + \frac{h}{1 + x^2}$ where $Y = (1, x + h)$ and $h$ is small; equivalently, we want to understand why $\angle XOY \approx \frac{h}{1 + x^2}$. Since this angle is small, we equivalently want to understand why $\sin \angle XOY \approx \frac{h}{1 + x^2}$.

Now $\triangle XOY$ evidently has area $\frac{h}{2}$. On the other hand, it has area $\frac{1}{2} |OX| |OY| \sin \angle XOY$ where $|OX| = \sqrt{1 + x^2}$ and $|OY| \approx |OX|$. The result follows.

(The derivative follows, anyway. The integral follows by dividing up $AX$ into little pieces and drawing a bunch of lines to $O$, then summing up all of the contributions.)

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    +1 for the geometric explanation. Would you mind inserting the geometric figure?2010-08-25
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Stereographic projection offers a geometric interpretation of this formula. In fact the factor $(1+x^2)^{-1}$ is the metric of the circle in the stereographic projection coordinate. The construction is as follows:

Take a circle of radius 1/2, call the angle of (the radius vector of) a general point on the circle with respect to the radius to the south pole $2\theta$. Connect the point with the north pole and call the distance of the point of intersection of the extension of this chord with the tangent at the south pole $x$. One can parameterize the points on the circle by the coordinate $x$ which is called the stereographic projection coordinate. It is easy to see that $ x = tan(\theta)$. The length of an arc of the circle between $\theta_1$ and$\theta_2$ (having the stereographic projection coordinates $x_1$ and $x_2$) is given by:

$s = 2 \int_{\theta_1}^{\theta_2} d\theta = 2\int_{x_1}^{x_2} \frac{dx}{1+x^2}$

This construction has further nice properties

  1. The group of plane rotations $SO(2)$ acts on $x$ by a Möbius transformation.

  2. This construction generalizes to spheres in higher dimensions.

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Put $\arctan x=:\alpha$, let $n$ be large, choose points $t_k:=\tan(k\alpha/n)$ $\ (0\leq k\leq n)$ and put $\tau_k:=\sqrt{t_k t_{k-1}}$. Then $$\sum_{k=1}^n {t_k - t_{k-1} \over 1 + \tau_k^2}=\sum_{k=1}^n {t_k - t_{k-1} \over 1 + t_k t_{k-1}}= n \tan(\alpha/n).$$ Here the left side is a Riemann sum for the integral $$\int_0^{\tan\alpha}{dt \over 1+t^2}=\int_0^x{dt \over 1+t^2},$$ and the right side has limit $\alpha=\arctan x$ when $n\to\infty$.

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    how is the first equation found?2010-08-25
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    First equality because $\tau_k^2=\tau_k \tau_{k+1}$. Second by the angle addition formula for tangent http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Tangents_of_sums_of_finitely_many_terms2010-08-25
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    PS Nice answer!2010-08-25
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Since $D \tan(x) = 1 + \tan(x)^2$ and $\tan(\arctan(x)) = x$ we see that:

$$\begin{align} \tan(\arctan(x)) &= x \\ (1 + \tan(\arctan(x))^2 \cdot D \arctan(x) &= 1 \\ D \arctan(x) &= \frac{1}{(1 + x)^2} \\ \end{align}$$

So by the fundamental theorem of calculus we recover the integral.

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    Again, this is a proof, but I do not see how it directly provides any intuition.2010-08-21
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The real answer I think involves complex numbers. Factor $1+x^2 = (i-x)(-i-x)$, decompose in partial fractions, get an expression involving logarithms that ends up being $\arctan x$ as described in http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms.

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    This is a proof (one of my favorites), but I do not immediately see how it relates to the geometric definition of arctan, which I assume is what the OP is asking about.2010-08-21
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    "Intuitively" does not mean "geometrically", does it? For me, explaining integrals in terms in logarithms is intuitive, but I'm not a first-year calculus student. :-)2010-08-21
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    What I mean is, this is a general-purpose algorithm for obtaining the answer. But why would one expect the answer in this case to be arctan and not arctanh or some other expression involving logarithms?2010-08-21
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This was actually my question originally. To put the demand for intuitiveness in starker terms: what does the arc of a circle have to do with the area of a square and the operation of reciprocation?

If someone can show me a geometric figure with area $1 \over 1 + x^2$ that might be a good start.

Or, they can explain how this sequence

  1. Times a number by itself;
  2. Add one to it;
  3. Flip that result

leads to "a little bit of" circular arc.

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    @Lao Tzu: have you read my answer? The 1/(1+x^2) term basically comes from the Pythagorean theorem, and it doesn't measure an area but an angle.2010-08-25
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    I've read it, and it's a huge improvement but not all the way to where I want to go. Thanks for going into the geometric detail.2010-08-26
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    One problem for me is that $\sin x \approx x$ is still too Taylor-y for me to explain to a fifth-grader.2010-08-26
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    Also, why is it evident that $\Delta XOY$ has area $h \over 2$? Let $x=.0001$ for example.2010-08-26
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    @Lao Tzu: sin x being approximately x is just the statement that a circle is approximately linear. More precisely, compare the length of a very small arc from (cos x, sin x) to (1, 0) where x is small with the length of the perpendicular from (cos x, sin x) to (cos x, 0). As for the triangle, take XY as the base of the triangle. (PS: in the future to ensure that I am notified of these comments you should begin them with @Qiaochu Yuan.)2010-08-29
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    @Qiaochu Yuan This is all very good but I'm still not convicted. I think the fault lies with me for not describing well enough the specific intuitional apparatus I'm trying to link up with this result. I'm very picky in this respect so don't feel it's your fault I'm not satisfied. I couldn't even explain why $x^2 \approx 2x$ for small $x$ to a kid, so what I'm asking is unfair.2010-09-03
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    @Lao Tzu: then maybe you should ask about that apparatus as a separate question. (And you should revise the statement of that approximation...)2010-09-03
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    @Qiaochu Yuan Sorry, but the reason I didn't post my question here is that this community isn't the right place. Too many Real Mathematicians here and not people to whom formulae and symbols are incomprehensible.2010-12-11