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Welcome to Inverse Circular Functions: Part 3

This stage covers the derivatives of arcsin, arccos, and arctan, and how to use these results to evaluate integrals involving inverse circular functions. It concludes with a deeper look at differentiating arcsin directly from the definition of the derivative.

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Pay close attention to the substitution technique — it unlocks both derivatives and integrals!

Dr Brian Brooks
Mathematics InSight

differentials of inverse circular functions

If \(\displaystyle{y=\arcsin x}\), what is \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}}\) in terms of \(\displaystyle{y}\) ?

If \(\displaystyle{y=\arcsin x}\), what is \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}}\) in terms of \(\displaystyle{y}\) ?

\[x=\sin y\Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}=\cos y\]

If \(\displaystyle{y=\arcsin x}\), what is \(\displaystyle{\cos y}\)?

If \(\displaystyle{y=\arcsin x}\), what is \(\displaystyle{\cos y}\)?

\[\begin{aligned} &\sin^2y+\cos^2y=1\\[6pt] \Rightarrow &\cos^2y=1-x^2\\[12pt] &-\frac{\pi}{2}\leq \sin y\leq \frac{\pi}{2}\\[6pt] \Rightarrow &\cos y\geq 0\\[12pt] \Rightarrow &\cos y=\sqrt{1-x^2} \end{aligned}\]

What are \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}\text{ and }\frac{\mathrm{d}y}{\mathrm{d}x}}\) in terms of \(\displaystyle{x}\) ?

What are \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}\text{ and }\frac{\mathrm{d}y}{\mathrm{d}x}}\) in terms of \(\displaystyle{x}\) ?

\[\begin{aligned} &\frac{\mathrm{d}x}{\mathrm{d}y}=\cos y=\sqrt{1-x^2}\\[6pt] \Rightarrow &\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}}\\ \end{aligned}\]
Graph showing segment choices

Notice that the gradient of the pink segments on the graph are always positive, and our formula for \(\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}\) is also always positive.

How would \(\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}\) change if we had chosen one of the green segments as our graph of \(\arcsin x\)? Look back at the previous page for a clue.

unit circle for differentiation
\[\begin{aligned} \text{As }\delta\theta\to 0,\; \alpha\to\theta\\[6pt] \text{and }\delta x\to\sin\theta\times\delta\theta,\; \delta y\to\cos\theta\times\delta\theta\\[6pt] \Rightarrow \frac{\delta \theta}{\delta y}\to \frac{1}{\cos\theta}\\[6pt] =\frac{1}{\sqrt{1-y^2}}\text{ and }\theta=\arcsin y\\[6pt] \Rightarrow \frac{\mathrm{d}}{\mathrm{d}y}\arcsin y=\frac{1}{\sqrt{1-y^2}} \end{aligned}\]

Follow the same method to find \[\frac{\mathrm{d}}{\mathrm{d}y}\arccos x\]

Follow the same method to find \[\frac{\mathrm{d}}{\mathrm{d}y}\arccos x\]

\[\begin{aligned} y=\arccos x\Rightarrow &x=\cos y\\[6pt] \Rightarrow &\frac{\mathrm{d}x}{\mathrm{d}y}=-\sin y\\[12pt] &\sin^2y+\cos^2y=1\\[6pt] \Rightarrow &\sin^2y=1-x^2\\[12pt] &0\leq \cos y\leq \pi\\[6pt] \Rightarrow &\sin y\geq 0\\[12pt] \Rightarrow &\sin y=\sqrt{1-x^2}\\[6pt] \Rightarrow &\frac{\mathrm{d}x}{\mathrm{d}y}=-\sqrt{1-x^2}\\[6pt] \Rightarrow &\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-x^2}}\\ \end{aligned}\]
Graph showing different segments

Notice that the gradient of the graph \(y=\arccos x\) is always negative, as is our formula for \(\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}\).

Follow the same method to find \[\frac{\mathrm{d}}{\mathrm{d}y}\arctan x\]

Follow the same method to find \[\frac{\mathrm{d}}{\mathrm{d}y}\arctan x\]

\[\begin{aligned} y=\arctan x\Rightarrow &x=\tan y\\[6pt] \Rightarrow &\frac{\mathrm{d}x}{\mathrm{d}y}=\sec^2 y=1+\tan^2y=1+x^2\\[12pt] \Rightarrow &\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1+x^2}\\ \end{aligned}\]

integrals using inverse circular functions

Use the substitution \(\displaystyle{u = \arcsin x}\) for the integral \[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x\]

Use the substitution \(\displaystyle{u = \arcsin x}\) for the integral \[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x\]

\[\begin{aligned} u = \arcsin x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} & = \frac{1}{\sqrt{1 - x^{2}}} \end{aligned}\]

\[\begin{aligned} \int\frac{1}{\sqrt{1 - x^{2}}}\;\mathrm{d}x & = \int\frac{1}{\sqrt{1 - x^{2}}}\,\frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u \\ & = \int\frac{1}{\sqrt{1 - x^{2}}}\sqrt{1 - x^{2}}\;\mathrm{d}u \\ & = \int \mathrm{d}u \\ & = u + c \\ & = \arcsin x + c \end{aligned}\]

In practice, you might write the subsitution as \(\displaystyle{x=\sin u}\). Try it this way.

Use the subsitution \(\displaystyle{x=\sin u}\) for the integral \[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x\]

\[\begin{aligned} x = \sin u \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} & = \cos u \\ & = \pm \sqrt{1 - \sin^{2}u} \\ & = \pm \sqrt{1 - x^{2}} \end{aligned}\]

At this point, it is really important to remember that the actual substitution is \(\displaystyle{u=\arcsin x}\) so that \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}u}>0}\)

\[\begin{aligned} \int\frac{1}{\sqrt{1 - x^{2}}}\;\mathrm{d}x & = \int\frac{1}{\sqrt{1 - x^{2}}}\;\frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u \\ & = \int\frac{1}{\sqrt{1 - x^{2}}}\sqrt{1 - x^{2}}\;\mathrm{d}u \\ & = \int \mathrm{d}u \\ & = u + c \\ & = \arcsin x + c \end{aligned}\]

Use the subsitution \(\displaystyle{x=\cos u}\) for the integral \[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x\]

\[\begin{aligned} x = \cos u \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} & = -\sin u \\ & = \pm \sqrt{1 - \cos^{2}u} \\ & = \pm \sqrt{1 - x^{2}} \end{aligned}\]

At this point, it is really important to remember that the actual substitution is \(\displaystyle{u=\arccos x}\) so that \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}u}<0}\)

\[\begin{aligned} \int\frac{1}{\sqrt{1 - x^{2}}}\;\mathrm{d}x & = \int\frac{1}{\sqrt{1 - x^{2}}}\;\frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u \\ & = -\int\frac{1}{\sqrt{1 - x^{2}}}\sqrt{1 - x^{2}}\;\mathrm{d}u \\ & = -\int \mathrm{d}u \\ & = -u + c \\ & = -\arccos x + c \end{aligned}\]

Explain why it is possible that both

\[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x=\arcsin x+c\]

and

\[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x=-\arccos x+c\]

Explain why it is possible that both

\[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x=\arcsin x+c\]

and

\[\int \frac{1}{\sqrt{1-x^2}}\;\mathrm{d}x=-\arccos x+c\]

Remember that

\[\arcsin x+\arccos x=\frac{\pi}{2}\]

so the two integrals differ by \(\displaystyle{\frac{\pi}{2}}\). That is, the "c" is different in each case.

Use the substitution \(\displaystyle{x = \tan u}\) for the integral:

\[\begin{aligned} \int\frac{1}{1 + x^{2}}\;\mathrm{d}x \end{aligned}\]

Use the substitution \(\displaystyle{x = \tan u}\) for the integral:

\[\begin{aligned} \int\frac{1}{1 + x^{2}}\;\mathrm{d}x \end{aligned}\]

\[\begin{aligned} x = \tan u \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} & = \sec^{2}u \\ & = 1 + \tan^{2}u \\ & = 1 + x^{2} \end{aligned}\]

\[\begin{aligned} \int\frac{1}{1 + x^{2}}\;\mathrm{d}x & = \int\frac{1}{1 + x^{2}}\;\frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u \\ & = \int\frac{1}{1 + x^{2}}\left( 1 + x^{2} \right)\;\mathrm{d}u \\ & = \int \mathrm{d}u \\ & = u + c \\ & = \arctan x + c \end{aligned}\]

Here, although the substitution is actually \(\displaystyle{u=\arctan x}\), there is no difficulty with the signs as there was before.

The area under \(\displaystyle{y = \sqrt{1-x^2}}\) (the upper unit semicircle) from \(\displaystyle{x=0}\) to \(\displaystyle{x=a}\) (where \(\displaystyle{0 < a < 1}\)) can be split into a circular sector and a right-angled triangle.

Let \(\displaystyle{\theta = \arcsin a}\). Show geometrically that

\[\int_0^a\!\sqrt{1-x^2}\,\mathrm{d}x = \frac{1}{2}\arcsin a + \frac{1}{2}a\sqrt{1-a^2}\]

and hence write down \(\displaystyle{\displaystyle\int\sqrt{1-x^2}\,\mathrm{d}x}\).

Find \(\displaystyle{\displaystyle\int\sqrt{1-x^2}\,\mathrm{d}x}\) geometrically, then verify using the substitution \(\displaystyle{x = \sin u}\).

Geometric argument:

The region is composed of a circular sector (radius 1, angle \(\displaystyle{\theta = \arcsin a}\)) with area \(\displaystyle{\tfrac{1}{2}\theta}\), and a right-angled triangle (base \(\displaystyle{a}\), height \(\displaystyle{\sqrt{1-a^2}}\)) with area \(\displaystyle{\tfrac{1}{2}a\sqrt{1-a^2}}\).

\[\int_0^a\!\sqrt{1-x^2}\,\mathrm{d}x = \frac{\arcsin a}{2} + \frac{a\sqrt{1-a^2}}{2}\] \[\boxed{\int\sqrt{1-x^2}\,\mathrm{d}x = \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C}\]

Verification by substitution \(\displaystyle{x = \sin u}\):

\[\begin{aligned} &\mathrm{d}x = \cos u\,\mathrm{d}u, \quad \sqrt{1-x^2} = \cos u\\[6pt] &\int\sqrt{1-x^2}\,\mathrm{d}x = \int\cos^2 u\,\mathrm{d}u = \int\frac{1+\cos 2u}{2}\,\mathrm{d}u\\[6pt] &= \frac{u}{2} + \frac{\sin 2u}{4} + C = \frac{\arcsin x}{2} + \frac{x\sqrt{1-x^2}}{2} + C \;\checkmark \end{aligned}\]

Well done on completing Part 3!

You have derived the derivatives of arcsin, arccos, and arctan, used them to evaluate integrals, and differentiated arcsin directly from the definition of the derivative.

Continue to Part 4 for integrals of inverse circular functions.

Dr Brian Brooks
Mathematics InSight

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