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Integrate Sin Squared - Y12-C6-Q5b.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Year 12 Chapter 6B Q5 b)} \begin{align*} &\text{The following illustration contains many useful techniques and things to watch out. Read all the notes carefully.}\\ \\ &\because\quad\cos2x=1-2\sin^2x\\ &\therefore\quad\sin^2 x=\frac{1-\cos2x}{2}\\ &\text{Likewise,}\quad\sin^2 2x=\frac{1-\cos4x}{2}\qquad\text{(by replacing $x$ with $2x$)}\\ \\ I &=\int\sin^2~2x~dx\\ &=\int\frac{1-\cos4x}{2}~dx\\ &=\int\frac{1}{2}-\frac{\cos4x}{2}~dx\\ &=\int\frac{1}{2}~dx-\frac{1}{2}\int\cos4x~dx\\ &=\frac{1}{2}~x-\frac{1}{2}\left(\frac{1}{4}\sin4x\right)+C\qquad\text{(Must remove all $\int$ signs and add $C$ in the same step.)}\\ &=\frac{1}{2}~x-\frac{1}{8}\sin4x+C.\\ \\ \\ &\text{Notes:}\\ &\text{In order to use the formula}\quad\int\cos x~dx=\sin x+C,\quad\text{the variable in $\cos$ and the one in $d$}\\ &\text{must be the same, so we have to make up the factor $4$.}\\ \\ &\text{You can image $dx$ being replaced by $\frac{d(4x)}{4}$, so}\\ &\int\cos4x~dx=\int\cos4x~\frac{d(4x)}{4}=\frac{1}{4}\int\cos 4x~d(4x)=\frac{1}{4}\sin4x+C\quad\text{(now both $\cos$ and $d$ carry $4x$).}\\ \\ &\text{However, you are not allowed to write $d(4x)$, even it is correct in concept.}\\ &\text{Therefore, in your work you need to skip the steps in between and write:}\\ &\int\cos4x~dx=\frac{1}{4}\sin4x+C.\\ \\ &\text{Alternatively, you can introduce a new variable $u=4x$, so $du=4dx$ and $dx=\frac{du}{4}$.}\\ &\int\cos4x~dx=\int\cos u~\frac{du}{4}=\frac{1}{4}\int\cos u~du\qquad\text{(now the variable in $\cos$ and the one in $d$ are the same)}\\ &=\frac{1}{4}\sin u+C=\frac{1}{4}\sin 4x+C.\\ \\ &\text{IMPORTANT:}\\ &\text{The answer can only contain $x$, not $u$. So you need to change $u$ back to the equivalent expression of $x$.}\\ \end{align*} \end{document}