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Welcome to Inverse Circular Functions: Part 4

This stage explores integrals of the inverse circular functions — \(\displaystyle{\int \arcsin x\,\mathrm{d}x}\), \(\displaystyle{\int \arccos x\,\mathrm{d}x}\), and \(\displaystyle{\int \arctan x\,\mathrm{d}x}\) — using geometric area arguments and integration by parts.

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The area argument gives geometric insight; IBP then confirms the result algebraically!

Dr Brian Brooks
Mathematics InSight

We can find \(\displaystyle{\int \arcsin x\,\mathrm{d}x}\) using an area argument. The region under \(\displaystyle{y = \arcsin x}\) from \(\displaystyle{x=0}\) to \(\displaystyle{x=a}\) is the area of a rectangle minus a complementary curved region.
Graph showing different segments

What is the area of the shaded rectangle (pink and green together)?

What is the pink shaded area?

What is the green shaded area?

\(\text{rectangle area}\) \(=\) \(a\arcsin a\)
\(\text{pink area}\) \(=\) \(\displaystyle\int_0^{\arcsin a}\!\sin y\,\mathrm{d}y = \Big[-\cos y\Big]_0^{\arcsin a}\)
\(=\) \(-\cos(\arcsin a) + 1 = 1 - \sqrt{1-a^2}\)
\(\implies\text{green area}\) \(=\) \(a\arcsin a - \left(1 - \sqrt{1-a^2}\right)\)

Use this result to find

\[\int_0^a \arcsin x\,\mathrm{d}x\]

Since the integral is the green area,

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

Find \(\displaystyle{\displaystyle\int \arccos x\,\mathrm{d}x}\) using an area argument.

Graph showing different segments
\(\text{pink rectangle area}\) \(=\) \(x\arccos x\)
\(\text{green area}\) \(=\) \(\displaystyle\int_{\arccos x}^{\pi/2}\cos y\,\mathrm{d}y = \Big[\sin y\Big]_{\arccos x}^{\pi/2} = 1 - \sin(\arccos x) = 1 - \sqrt{1-x^2}\)
\(\implies\displaystyle\int_0^x \arccos x\,\mathrm{d}x\) \(=\) \(x \arccos x+ \left(1 - \sqrt{1-x^2}\right)\)
\(\implies\displaystyle\int \arccos x\,\mathrm{d}x\) \(=\) \(x\arccos x - \sqrt{1-x^2} + c\)

Use the area argument to find \(\displaystyle{\displaystyle\int \arctan x\,\mathrm{d}x}\).

arctan integral
\(\text{rectangle area}\) \(=\) \(x\arctan x\)
\(\text{pink area}\) \(=\) \(\displaystyle\int_0^{\arctan x}\!\tan y\,\mathrm{d}y = \Big[\ln\sec y\Big]_0^{\arctan x} = \ln\sec(\arctan x)\)
\(\sec(\arctan x) = \sqrt{1+x^2}, \;\text{so } \ln\sec(\arctan x) = \tfrac{1}{2}\ln(1+x^2)\)
\(\text{green area}\) \(=\) \(\displaystyle\int_0^x \arctan x\,\mathrm{d}x = x\arctan x - \tfrac{1}{2}\ln(1+x^2)\)

Use integration by parts to find \[\displaystyle{\int\arcsin x\,\mathrm{d}x}\]

\(u = \arcsin x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{\sqrt{1-x^2}}\qquad v=x\)
\(\displaystyle\int\arcsin x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v\frac{\mathrm{d}u}{\mathrm{d}x}\;\mathrm{d}x\)
\(=\) \(x\arcsin x-\displaystyle\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d} x\)
\(=\) \(x\arcsin x + \sqrt{1-x^2} + c\)

Use integration by parts to find \[\displaystyle{\int\arccos x\,\mathrm{d}x}\].

\(u = \arccos x\qquad \dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x}=-\dfrac{1}{\sqrt{1-x^2}}\qquad v=x\)
\(\displaystyle\int\arccos x\,\mathrm{d}x\) \(=\) \(uv-\displaystyle\int v\frac{\mathrm{d}u}{\mathrm{d}x}\;\mathrm{d}x\)
\(=\) \(x\arccos x+\displaystyle\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d} x\)
\(=\) \(x\arccos x -\sqrt{1-x^2} + c\)

Use integration by parts to find \[\displaystyle{\int\arctan x\,\mathrm{d}x}\].

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

Well done on completing Part 4!

You have evaluated \(\displaystyle{\int \arcsin x\,\mathrm{d}x}\), \(\displaystyle{\int \arccos x\,\mathrm{d}x}\), and \(\displaystyle{\int \arctan x\,\mathrm{d}x}\) using geometric arguments and integration by parts.

Continue to the extension material in Part 5 (graphs of arcsec, arccosec, arccot).

Dr Brian Brooks
Mathematics InSight

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