202211040914 Solution to 2009-CE-AMATH-II-1

Find

(a) $\displaystyle{\int (4x+1)^2\,\mathrm{d}x}$ ,
(b) $\displaystyle{\int \sin 3\theta\cos\theta\,\mathrm{d}\theta}$ .

Roughwork.

(a)

$\begin{aligned} & \quad\, \int (4x+1)^2\,\mathrm{d}x \\ \dots\enspace & \textrm{let }u=4x+1\textrm{ s.t. }\mathrm{d}x=\frac{\mathrm{d}u}{4}\enspace\dots \\ & = \int u^2\,\bigg(\frac{\mathrm{d}u}{4}\bigg) \\ & = \frac{u^3}{12}+C \textrm{ for some constant}\\ & = \frac{(4x+1)^3}{12} + C \\ \end{aligned}$

(b)

$\begin{aligned} & \quad\, \int \sin 3\theta\cos\theta\,\mathrm{d}\theta \\ & = \int \bigg(\frac{e^{3\mathrm{i}\theta}-e^{-3\mathrm{i}\theta}}{2\mathrm{i}}\bigg)\bigg(\frac{e^{\mathrm{i}\theta}+e^{-\mathrm{i}\theta}}{2}\bigg)\,\mathrm{d}\theta \\ & = \int \frac{e^{4\mathrm{i}\theta}+e^{2\mathrm{i}\theta}-e^{-2\mathrm{i}\theta}-e^{-4\mathrm{i}\theta}}{4\mathrm{i}} \,\mathrm{d}\theta \\ & = \frac{1}{2}\int \bigg( \frac{e^{4\mathrm{i}\theta}-e^{-4\mathrm{i}\theta}}{2\mathrm{i}} + \frac{e^{2\mathrm{i}\theta}-e^{-2\mathrm{i}\theta}}{2\mathrm{i}} \bigg) \,\mathrm{d}\theta \\ & = \frac{1}{2}\int (\sin 4\theta +\sin 2\theta )\,\mathrm{d}\theta \\ & = \frac{1}{8}\int \sin 4\theta\,\mathrm{d}(4\theta ) + \frac{1}{4}\int\sin 2\theta\,\mathrm{d}(2\theta ) \\ & = -\frac{1}{8}\cos 4\theta - \frac{1}{4}\cos 2\theta +C\textrm{ for some constant} \\ \end{aligned}$

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