
# 8.1: The Arctangent Function


## Definition

For any $$x \in \mathbb{R},$$ we call

$\arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t$

the arctangent of $$x .$$

## Proposition $$\PageIndex{1}$$

The arctangent function is differentiable at every $$x \in \mathbb{R} .$$

Moreover, if $$f(x)=\arctan (x),$$ then

$f^{\prime}(x)=\frac{1}{1+x^{2}}.$

Proof

The result follows immediately from Theorem $$7.5 .4 .$$ $$\quad$$ Q.E.D.

## Proposition $$\PageIndex{2}$$

The arctangent is increasing on $$\mathbb{R}$$.

Proof

The result follows immediately from the previous proposition and the fact that

$\frac{1}{1+x^{2}}>0$

for every $$x \in \mathbb{R}$$. $$\quad$$ Q.E.D.

## Definition

$$\pi=2 \lim _{x \rightarrow+\infty} \arctan (x)=2 \int_{0}^{+\infty} \frac{1}{1+t^{2}} d t$$

Note that $$0<\pi<4$$ by Example $$7.7 .1 .$$

The following proposition says that the arctangent function is an odd function.

## Theorem $$\PageIndex{3}$$

For any $$x \in \mathbb{R},$$ arctan $$(x)=-\arctan (-x)$$.

Proof

Using the substitution $$t=-u,$$ we have

\begin{aligned} \arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t=-\int_{0}^{-x} \frac{1}{1+u^{2}} d u=-\arctan (-x). & \end{aligned}

Q.E.D.

It now follows that

$\lim _{x \rightarrow-\infty} \arctan (x)=-\lim _{x \rightarrow-\infty} \arctan (-x)=-\frac{\pi}{2}.$

Hence the range of the arctangent function is $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

## Proposition $$\PageIndex{4}$$

If $$x>0,$$ then

$\arctan (x)+\arctan \left(\frac{1}{x}\right)=\frac{\pi}{2}.$

Proof

Using the substitution $$t=\frac{1}{u},$$ we have

\begin{aligned} \arctan \left(\frac{1}{x}\right) &=\int_{0}^{\frac{1}{x}} \frac{1}{1+t^{2}} d t \\ &=\int_{+\infty}^{x} \frac{1}{1+\frac{1}{u^{2}}}\left(-\frac{1}{u^{2}}\right) d u \\ &=-\int_{+\infty}^{x} \frac{1}{1+u^{2}} d u \\ &=\int_{x}^{+\infty} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\int_{0}^{x} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\arctan (x). \end{aligned}

Q.E.D.

## Proposition $$\PageIndex{5}$$

If $$x<0$$, then

$\arctan (x)+\arctan \left(\frac{1}{x}\right)=-\frac{\pi}{2}.$

Proof

The result follows immediately from the preceding proposition and the fact that arctangent is an odd function.

## Exercise $$\PageIndex{1}$$

Show that arctan $$(1)=\frac{\pi}{4}$$ and $$\arctan (-1)=-\frac{\pi}{4}$$.