202212211715 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\int\frac{\mathrm{d}x}{\sqrt{9-4x^2}}}$

i. by using the substitution $\displaystyle{x=\frac{3}{2}\cos\theta}$ ;
ii. by using the substitution $\displaystyle{x=\frac{3}{2}\sin\theta}$ .

Explain why you appear to get two different answers.

Roughwork.

i.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\cos\theta \textrm{ s.t. }\mathrm{d}x=-\frac{3}{2}\sin\theta\,\mathrm{d}\theta \,\dots \\ & = -\int \frac{3\sin\theta}{2\sqrt{9-4(\frac{3}{2}\cos\theta )^2}}\,\mathrm{d}\theta \\ & = -\int \frac{3\sin\theta}{6\sin\theta}\,\mathrm{d}\theta \\ & = -\frac{\theta}{2} +C_1\textrm{ for some constant}\\ & = -\frac{1}{2}\arccos \bigg(\frac{2x}{3}\bigg) +C_1 \\ \end{aligned}$

ii.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\sin\theta \textrm{ s.t. }\mathrm{d}x=\frac{3}{2}\cos\theta\,\mathrm{d}\theta \,\dots \\ & = \int \frac{3\cos\theta}{2\sqrt{9-4(\frac{3}{2}\sin\theta )^2}}\,\mathrm{d}\theta \\ & = \int \frac{3\cos\theta}{6\cos\theta}\,\mathrm{d}\theta \\ & = \frac{\theta}{2} +C_2\textrm{ for some constant} \\ & = \frac{1}{2}\arcsin \bigg(\frac{2x}{3}\bigg) +C_2 \\ \end{aligned}$

Let’s see if

$\displaystyle{-\frac{1}{2}\arccos\bigg(\frac{2x}{3}\bigg) +C_1 \stackrel{\textrm{?}}{\equiv}\frac{1}{2}\arcsin\bigg(\frac{2x}{3}\bigg)} + C_2$

or, to be rephrased, whether

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\arcsin \bigg(\frac{2x}{3}\bigg) + \arccos\bigg(\frac{2x}{3}\bigg) \bigg]\stackrel{\textrm{?}}{\equiv} 0}$

This is left an exercise for the reader.

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