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Welcome to Inverse Circular Functions: Part 5

This extension stage covers the graphs, domains, ranges, and derivatives of arcsec, arccosec, and arccot, together with compositions and identities involving these functions.

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These extension functions build directly on arcsin, arccos, and arctan — and the same implicit differentiation technique from Part 3 applies to their derivatives!

Dr Brian Brooks
Mathematics InSight

We've worked through the basic circular functions: sin, cos, and tan. We've worked through the reciprocal circular functions: cosec, sec, and cot. We've worked on the inverses of the basic functions: arcsin, arccos, and arctan. Surely that's plenty? Yes it is, especially when we've worked through them all in such depth. But if you are still just a little bit interested to see more, we could work on the inverses of the reciprocal functions: arccosec, arcsec, arccot. Strictly for interest only, though!
Here is the graph \(y=\sec x\). Draw this graph, and, on the same set of axes, draw the graph \(x=\sec y\).
Graph of sec x
Graphs of sec x and sec inverse x
Why is the graph \(x=\sec y\) not the same as the graph \(y=\mathrm{arcsec} \,x\)?
Graphs of sec x and sec inverse x
Graphs of sec x and sec inverse x
For the graph to represent a function, the vertical line must only ever intersect the graph at a single point.

How can we choose part of the graph \(x=\sec y\) to create a sensible graph of the function

\(y=\operatorname{arcsec}x\)?

Graphs of sec x and sec inverse x
If this is the graph \(y=\mathrm{arcsec}\,x\), what are the domain and the range of the function arcsec?
\[\begin{aligned} &\text{Domain: } (-\infty,-1]\cup[1,\infty)\\[6pt] &\text{Range: } \left[0,\, \pi\right]\setminus\left\{\frac{\pi}{2}\right\} \end{aligned}\]
Graphs of sec x and sec inverse x

We know that \(\sec x=\displaystyle\frac{1}{\cos x}\).

Now we'll look for a similar relationship between \(\operatorname{arccos}x\) and \(\operatorname{arcsec}x\), the pink and blue graphs in the diagram.

Putting \(x=\operatorname{arcsec}A\), what are \(\sec x\) and \(\cos x\) in terms of \(A\)?

\[\sec x=A\qquad \cos x=\frac{1}{\sec x}=\frac{1}{A} \]

Now make \(x\) the subject of \(\cos x=\displaystyle\frac{1}{A}\).

\(x=\operatorname{arccos}\displaystyle\frac{1}{A}\).

What do we now know about \(\operatorname{arcsec}A\) and \(\operatorname{arccos}\displaystyle\frac{1}{A}\)?

\[\operatorname{arcsec}A=\operatorname{arccos}\displaystyle\frac{1}{A}\]
Draw the graphs \(y=\mathrm{cosec}\,x\) and \(x=\mathrm{cosec}\,y\) on the same set of axes.
Graphs of cosec x and cosec inverse x
Graphs of cosec x and arccosec x
How can we choose part of the graph \(x=\mathrm{cosec}\, y\) to create a sensible graph of the function \(y=\mathrm{arccosec}\,x\)?
What are the domain and range of the function arccosec?
Graphs of sec x and sec inverse x
\[\begin{aligned} &\text{Domain: } (-\infty,-1]\cup[1,\infty)\\[6pt] &\text{Range: } \left[-\tfrac{\pi}{2},\, \tfrac{\pi}{2}\right]\setminus\{0\} \end{aligned}\]

Draw the graphs \(y=\cot x\) and \(x=\cot y\).

Graph of cot x and cot inverse x

Use this to draw the graph \(y=\mathrm{arccot}\,x\)

Graphs of cot x and cot inverse x
\[\begin{aligned} &\text{Domain: } \mathbb{R}\\[6pt] &\text{Range: } (0,\,\pi) \end{aligned}\]

The graphs of \(y = \operatorname{arccot} x\) and \(y = \arctan x\) are shown together below.

What symmetry do you observe? Use it to state the value of \(\operatorname{arccot} x + \arctan x\).

Graphs of cot inverse x and tan inverse x

The average of \(\operatorname{arccot} x\) and \(\arctan x\) is\(\displaystyle{\frac{\pi}{4}}\). So

\[\operatorname{arccot} x + \arctan x=\frac{\pi}{2} \]

The graphs of \(y = \operatorname{arcsec} x\) and \(y = \operatorname{arccosec} x\) are shown together below.

What symmetry do you observe? Use it to state the value of \(\operatorname{arcsec} x + \operatorname{arccosec} x\).

Graphs of sec inverse x and cosec inverse x

The average of \(\operatorname{arcsec} x\) and \(\operatorname{arccosec} x\) is \(\displaystyle{\frac{\pi}{4}}\) So

\[\operatorname{arcsec} x + \operatorname{arccosec} x= \frac{\pi}{2} \]
Find the differential of \(\operatorname{arcsec} x\).
\(y = \operatorname{arcsec}x \Rightarrow x\) \(=\) \(\sec y\)
\(\Rightarrow\dfrac{\mathrm{d}x}{\mathrm{d}y}\) \(=\) \(\sec y\tan y\)
\(\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(=\) \(\pm\dfrac{1}{x\sqrt{1-x^2}}\)
but gradient is always positive
\(\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(=\) \(\dfrac{1}{|x|\sqrt{1-x^2}}=\dfrac{1}{\sqrt{x^2(1-x^2)}}\)
Find the differential of \(\operatorname{arccsc} x\).
\(y = \operatorname{arccosec}x \Rightarrow x\) \(=\) \(\operatorname{cosec} y\)
\(\Rightarrow\dfrac{\mathrm{d}x}{\mathrm{d}y}\) \(=\) \(-\operatorname{cosec} y\cot y\)
\(\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(=\) \(\pm\dfrac{1}{x\sqrt{1-x^2}}\)
but gradient is always negative
\(\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(=\) \(-\dfrac{1}{|x|\sqrt{1-x^2}}=-\dfrac{1}{\sqrt{x^2(1-x^2)}}\)
Find the differential of \(\operatorname{arccot} x\).
\(y = \operatorname{arccot}x \Rightarrow x\) \(=\) \(\cot y\)
\(\Rightarrow\dfrac{\mathrm{d}x}{\mathrm{d}y}\) \(=\) \(-\operatorname{cosec}^2 y\)
\(\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(=\) \(-\dfrac{1}{\operatorname{cosec}^2 y}=-\dfrac{1}{1+x^2}\)

Find (using integration by parts) \[\int_1^2 \operatorname{arcsec}x\,\mathrm{d}x\]

\(u=\operatorname{arcsec}x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{x\sqrt{x^2-1}}\qquad v=x\)
\(\displaystyle\int_1^2 \operatorname{arcsec}x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v \,\mathrm{d}u\)
\(=\) \(\Big[x\operatorname{arcsec}x\Big]_1^2-\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)

Use the substitution \(x=\sec u\) to find \[\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\]

\(\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(x=\sec u\qquad \dfrac{\mathrm{d}x}{\mathrm{d}u}=\sec u\tan u\)
\(\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\) \(=\) \(\displaystyle\int_0^\frac{\pi}{3} \frac{1}{\sqrt{x^2-1}}\frac{\mathrm{d}x}{\mathrm{d}u}\,\mathrm{d}u\)
\(=\) \(\displaystyle\int_0^\frac{\pi}{3} \frac{1}{\sqrt{x^2-1}}\sec u\tan u\,\mathrm{d}u\)
\(=\) \(\displaystyle\int_0^\frac{\pi}{3} \frac{1}{\tan u}\sec u\tan u\,\mathrm{d}u\qquad\) as \(\tan u \geqslant 0\) in this range
\(=\) \(\displaystyle\int_0^\frac{\pi}{3} \sec u\,\mathrm{d}u\)
\(=\) \(\Big[\ln|\sec u+\tan u|\Big]_0^\frac{\pi}{3}=\ln\left(2+\sqrt{3}\right)\)
\(=\) \(\Bigg[\ln\left|x+\sqrt{x^2-1}\right|\Bigg]_1^2=\ln\left(2+\sqrt{3}\right)\)

Use this result to find\[\int_1^2 \operatorname{arcsec}x\,\mathrm{d}x\]

\(\displaystyle\int_1^2 \operatorname{arcsec}x\,\mathrm{d}x\) \(=\) \(\Big[x\operatorname{arcsec}x\Big]_1^2-\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(=\) \(\displaystyle\frac{2\pi}{3}-\ln\left(2+\sqrt 3\right)\)

Find (using integration by parts) \[\int_{-2}^{-1} \operatorname{arcsec}x\,\mathrm{d}x\]

\(u=\operatorname{arcsec}x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{x\sqrt{x^2-1}}\qquad v=x\)
\(\displaystyle\int_{-2}^{-1} \operatorname{arcsec}x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v \,\mathrm{d}u\)
\(=\) \(\Big[x\operatorname{arcsec}x\Big]_{-2}^{-1}-\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)

Use the substitution \(x=\sec u\) to find \[\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\]

\(\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(x=\sec u\qquad \dfrac{\mathrm{d}x}{\mathrm{d}u}=\sec u\tan u\)
\(\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\) \(=\) \(\displaystyle\int_\frac{2\pi}{3}^\pi \frac{1}{\sqrt{x^2-1}}\frac{\mathrm{d}x}{\mathrm{d}u}\,\mathrm{d}u\)
\(=\) \(\displaystyle\int_\frac{2\pi}{3}^\pi \frac{1}{\sqrt{x^2-1}}\sec u\tan u\,\mathrm{d}u\)
\(=\) \(-\displaystyle\int_\frac{2\pi}{3}^\pi \frac{1}{\tan u}\sec u\tan u\,\mathrm{d}u\qquad\) as \(\tan u \leqslant 0\) in this range
\(=\) \(-\displaystyle\int_\frac{2\pi}{3}^\pi \sec u\,\mathrm{d}u\)
\(=\) \(-\Big[\ln|\sec u+\tan u|\Big]_\frac{2\pi}{3}^\pi=\ln\left(2+\sqrt{3}\right)\)
\(=\) \(-\Bigg[\ln\left|x-\sqrt{x^2-1}\right|\Bigg]_{-2}^{-1}=\ln\left(2+\sqrt{3}\right)\)

Use this result to find\[\int_{-2}^{-1} \operatorname{arcsec}x\,\mathrm{d}x\]

\(\displaystyle\int_{-2}^{-1} \operatorname{arcsec}x\,\mathrm{d}x\) \(=\) \(\Big[x\operatorname{arcsec}x\Big]_{-2}^{-1}-\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(=\) \(\displaystyle\frac{4\pi}{3}-\ln\left(2+\sqrt 3\right)\)

Find (using integration by parts) \[\int_1^2 \operatorname{arccosec}x\,\mathrm{d}x\]

\(u=\operatorname{arccosec}x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=-\dfrac{1}{x\sqrt{x^2-1}}\qquad v=x\)
\(\displaystyle\int_1^2 \operatorname{arccosec}x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v \,\mathrm{d}u\)
\(=\) \(\Big[x\operatorname{arccosec}x\Big]_1^2+\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)

Using the result \[\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x=\ln\left(2+\sqrt{3}\right)\] (found earlier by the substitution \(x=\sec u\)), find \[\int_1^2 \operatorname{arccosec}x\,\mathrm{d}x\]

\(\displaystyle\int_1^2 \operatorname{arccosec}x\,\mathrm{d}x\) \(=\) \(\Big[x\operatorname{arccosec}x\Big]_1^2+\displaystyle\int_1^2 \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(=\) \(\ln\left(2+\sqrt 3\right)-\displaystyle\frac{\pi}{6}\)

Find (using integration by parts) \[\int_{-2}^{-1} \operatorname{arccosec}x\,\mathrm{d}x\]

\(u=\operatorname{arccosec}x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=-\dfrac{1}{x\sqrt{x^2-1}}\qquad v=x\)
\(\displaystyle\int_{-2}^{-1} \operatorname{arccosec}x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v \,\mathrm{d}u\)
\(=\) \(\Big[x\operatorname{arccosec}x\Big]_{-2}^{-1}+\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)

Using the result \[\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x=\ln\left(2+\sqrt{3}\right)\] (found earlier by the substitution \(x=\sec u\)), find \[\int_{-2}^{-1} \operatorname{arccosec}x\,\mathrm{d}x\]

\(\displaystyle\int_{-2}^{-1} \operatorname{arccosec}x\,\mathrm{d}x\) \(=\) \(\Big[x\operatorname{arccosec}x\Big]_{-2}^{-1}+\displaystyle\int_{-2}^{-1} \frac{1}{\sqrt{x^2-1}}\,\mathrm{d}x\)
\(=\) \(\ln\left(2+\sqrt 3\right)+\displaystyle\frac{\pi}{6}\)

Find (using integration by parts) \[\int \operatorname{arccot}x\,\mathrm{d}x\]

\(u=\operatorname{arccot}x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=-\dfrac{1}{1+x^2}\qquad v=x\)
\(\displaystyle\int \operatorname{arccot}x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v \,\mathrm{d}u\)
\(=\) \(x\operatorname{arccot}x+\displaystyle\int x\frac{1}{1+x^2}\,\mathrm{d}x\)
\(=\) \(x\operatorname{arccot}x+\frac{1}{2}\ln\left(1+x^2\right)+c\)

Well done on completing Part 5!

You have explored the graphs, domains, ranges, and compositions of arcsec, arccosec, and arccot, and derived their derivatives using implicit differentiation.

Dr Brian Brooks
Mathematics InSight

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