202211041133 Solution to 2004-CE-AMATH-II-1

Find

(a) $\displaystyle{\int\cos (3x+1)\,\mathrm{d}x}$ ,
(b) $\displaystyle{\int (2-x)^{2004}\,\mathrm{d}x}$ .

Roughwork.

(a)

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

(b)

$\begin{aligned} &\quad\, \int (2-x)^{2004}\,\mathrm{d}x \\ \dots\enspace &\textrm{let }u=2-x\textrm{ s.t. }\mathrm{d}x=-\mathrm{d}u \\ & = -\int u^{2004}\,\mathrm{d}u \\ & = -\frac{u^{2005}}{2005} + C\textrm{ for some constant} \\ & = -\frac{(2-x)^{2005}}{2005} + C \\ \end{aligned}$

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