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Mathematics LibreTexts

8.1: The Arctangent Function

  • Page ID
    22685
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    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 \(\arctan (-1)=-\frac{\pi}{4}\).

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