Welcome to inverse circular functions part 2

Having explored the definitions, graphs, and identities of arcsin, arccos, and arctan in Part 1, we now turn to their calculus. This stage derives the derivatives of each inverse circular function and uses them to evaluate a family of important integrals.

The chain rule and implicit differentiation are your main tools here—use what you know!

Dr Brian Brooks
Mathematics InSight

differentials of inverse circular functions

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

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

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

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

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

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

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

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

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 \(u = \arcsin x\) 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 \(x=\sin u\). Try it this way.

Use the subsitution \(x=\sin u\) 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 \(u=\arcsin x\) so that \(\frac{\mathrm{d}x}{\mathrm{d}u}>0\)

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

Use the subsitution \(x=\cos u\) 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 \(u=\arccos x\) so that \(\frac{\mathrm{d}x}{\mathrm{d}u}<0\)

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

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 \(\frac{\pi}{2}\). That is, the "c" is different in each case.

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

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

Use the substitution \(x = \tan u\) 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 \(u=\arctan x\), there is no difficulty with the signs as there was before.

Congratulations on completing this stage!

You've now mastered inverse circular functions! You understand how to define arcsin, arccos, and arctan properly, know their domains and ranges, and can work with them both algebraically and graphically. You've explored their derivatives and integrals, and seen how they relate to each other through beautiful identities.

Keep practising with inverse functions—they're essential tools for solving equations and understanding circular functions at a deeper level!

Dr Brian Brooks
Mathematics InSight

Welcome to inverse circular functions part 2

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Congratulations on completing this stage!