202211041147 Solution to 2003-CE-AMATH-II-1

Find $\displaystyle{\int \cos^2\theta\,\mathrm{d}\theta}$ .

Roughwork.

We need the following

Lemma. (Double-angle formulae)

$\begin{aligned} \cos 2\theta & = \cos^2\theta -\sin^2\theta \\ & = \cos^2\theta - (1-\cos^2\theta )\\ & = 2\cos^2\theta -1 \\ \cos^2\theta & = \frac{1+\cos 2\theta}{2} \\ \end{aligned}$

Hence,

$\begin{aligned} &\quad\, \int\cos^2\theta\,\mathrm{d}\theta \\ & = \int \bigg( \frac{1+\cos 2\theta}{2}\bigg) \,\mathrm{d}\theta \\ & = \frac{1}{2}\int \mathrm{d}\theta + \frac{1}{2}\int \cos 2\theta\,\mathrm{d}\theta \\ & = \frac{\theta}{2} + \frac{1}{4}\sin 2\theta + C\textrm{ for some constant} \\ \end{aligned}$

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