\documentclass[a4paper]{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage{multicol}
\usepackage{multirow}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{harpoon}
\usepackage{tabularx}
\usepackage[dvipsnames, table]{xcolor}
\usepackage{blindtext}
\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usepackage{pgfplots}
\usetikzlibrary{calc}
\usetikzlibrary{angles}
\usetikzlibrary{datavisualization.formats.functions}
\usetikzlibrary{decorations.markings}
\usepgflibrary{arrows.meta}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhead[LO,LE]{Year 12 Specialist}
\fancyhead[CO,CE]{Andrew Lorimer}

\usepackage{mathtools}
\usepackage{xcolor} % used only to show the phantomed stuff
\renewcommand\hphantom[1]{{\color[gray]{.6}#1}} % comment out!
\setlength\fboxsep{0pt} \setlength\fboxrule{.2pt} % for the \fboxes
\newcommand*\leftlap[3][\,]{#1\hphantom{#2}\mathllap{#3}}
\newcommand*\rightlap[2]{\mathrlap{#2}\hphantom{#1}}
\newcolumntype{L}[1]{>{\hsize=#1\hsize\raggedright\arraybackslash}X}%
\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
\definecolor{cas}{HTML}{e6f0fe}
\linespread{1.5}
\newcommand{\midarrow}{\tikz \draw[-triangle 90] (0,0) -- +(.1,0);}

\begin{document}

\begin{multicols}{2}

  \section{Complex numbers}

  \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]

  \begin{align*}
    \text{Cartesian form: } & a+bi\\
    \text{Polar form: } & r\operatorname{cis}\theta
  \end{align*}

  \subsection*{Operations}

  \definecolor{shade1}{HTML}{ffffff}
  \definecolor{shade2}{HTML}{e6f2ff}
  \definecolor{shade3}{HTML}{cce2ff}
  \begin{tabularx}{\columnwidth}{r|X|X}
    & \textbf{Cartesian} & \textbf{Polar} \\
    \hline
    \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
    \hline
    \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
    \cline{1-1}\cline{3-3}
    \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
    \hline
    \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
    \hline
    \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
  \end{tabularx}

  \subsubsection*{Scalar multiplication in polar form}

  For \(k \in \mathbb{R}^+\):
  \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]

  \noindent For \(k \in \mathbb{R}^-\):
  \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]

    \subsection*{Conjugate}

    \begin{align*}
      \overline{z} &= a \mp bi\\
      &= r \operatorname{cis}(-\theta)
    \end{align*}

    \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}

    \subsubsection*{Properties}

    \begin{align*}
      \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
      \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
      \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
      z\overline{z} &= (a+bi)(a-bi)\\
      &= a^2 + b^2\\
      &= |z|^2
    \end{align*}

    \subsection*{Modulus}

    \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]

    \subsubsection*{Properties}

    \begin{align*}
      |z_1z_2|&=|z_1||z_2|\\
      \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
      |z_1+z_2|&\le|z_1|+|z_2|
    \end{align*}

    \subsection*{Multiplicative inverse}

    \begin{align*}
      z^{-1}&=\frac{a-bi}{a^2+b^2}\\
      &=\frac{\overline{z}}{|z|^2}a\\
      &=r \operatorname{cis}(-\theta)
    \end{align*}

    \subsection*{Dividing over \(\mathbb{C}\)}

    \begin{align*}
      \frac{z_1}{z_2}&=z_1z_2^{-1}\\
      &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
      &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
      & \qquad \text{(rationalise denominator)}
    \end{align*}

    \subsection*{Polar form}

    \begin{align*}
      z&=r\operatorname{cis}\theta\\
      &=r(\cos \theta + i \sin \theta)
    \end{align*}

    \begin{itemize}
      \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
      \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
      \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
      \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
      \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
      \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
    \end{itemize}

    \subsection*{de Moivres' theorem}

    \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]

    \subsection*{Complex polynomials}

    Include \(\pm\) for all solutions, incl. imaginary

    \begin{tabularx}{\columnwidth}{ R{0.55} X  }
      \hline
      Sum of squares & \(\begin{aligned} 
        z^2 + a^2 &= z^2-(ai)^2\\
      &= (z+ai)(z-ai) \end{aligned}\) \\
      \hline
      Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
      \hline
      Division & \(P(z)=D(z)Q(z)+R(z)\) \\
      \hline
      Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
      \hline
      Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
      \hline
      Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)\\
      \hline
    \end{tabularx}

    \subsection*{\(n\)th roots}

    \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:

    \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]

    \begin{itemize}

      \item{Same modulus for all solutions}
      \item{Arguments separated by \(\frac{2\pi}{n} \therefore\) there are \(n\) roots}
      \item{If one square root is \(a+bi\), the other is \(-a-bi\)}
      \item{Give one implicit \(n\)th root \(z_1\), function is \(z=z_1^n\)}
      \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
    \end{itemize}

    \noindent For \(0=az^2+bz+c\), use quadratic formula:

    \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    \subsection*{Fundamental theorem of algebra}

    A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):

    \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
    \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]

    \subsection*{Argand planes}

    \begin{center}\begin{tikzpicture}[scale=2]
      \draw [->] (-0.2,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
      \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
      \coordinate (P) at (1,1);
      \coordinate (a) at (1,0);
      \coordinate (b) at (0,1);
      \coordinate (O) at (0,0);
      \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
        \draw [gray, dashed] (1,1) -- (1,0) node[black, pos=1, below]{\(a\)};
        \draw [gray, dashed] (1,1) -- (0,1) node[black, pos=1, left]{\(b\)};
        \begin{scope}
          \path[clip] (O) -- (P) -- (a);
          \fill[red, opacity=0.5, draw=black] (O) circle (2mm);
          \node at ($(O)+(20:3mm)$) {$\theta$};
        \end{scope}
        \filldraw (P) circle (0.5pt);
    \end{tikzpicture}\end{center}

    \begin{itemize}
      \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
      \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
    \end{itemize}

    \subsection*{Sketching complex graphs}

    \subsubsection*{Linear}

    \begin{itemize}
      \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
      \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
      \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
    \end{itemize}

    \subsubsection*{Circles}

    \begin{itemize}
      \item \(|z-z_1|^2=c^2|z_2+2|^2\)
      \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
    \end{itemize}

    \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)

    \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
      \draw [->] (0,0) -- (1,0) node [right]  {$\operatorname{Re}(z)$};
      \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
      \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
      \draw [->, thick, blue] (0,0) -- (1,1);
      \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
      \begin{scope}
        \path[clip] (0,0) -- (1,1) -- (1,0);
        \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
        \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
      \end{scope}
      \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
      \node [blue, mydot] {};
    \end{tikzpicture}\end{center}

    \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)

    \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
      \draw [->] (-0.75,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
      \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
      \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
      \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
      \begin{scope}
        \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
        \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
      \end{scope}
      \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
      \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
      \node [brown, mydot] at (-0.25,0) {};
      \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
      \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
      \node [below, font=\footnotesize] at (1,0) {\(1\)};
    \end{tikzpicture}\end{center}

    \section{Vectors}
    \begin{center}\begin{tikzpicture}
      \draw [->] (-0.5,0) -- (3,0) node [right]  {\(x\)};
      \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
      \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
      \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
        \draw [gray, dashed, thick] (0.5,0.5) -- (2.5,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, below]{\(x\vec{i}\)};
        \draw [gray, dashed, thick] (2.5,0.5) -- (2.5,2.5) node [pos=0.5] {\midarrow};
      \end{scope}
      \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
    \end{tikzpicture}\end{center}
    \subsection*{Column notation}

    \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
      \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)

        \subsection*{Scalar multiplication}

        \[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]

        \noindent For \(k \in \mathbb{R}^-\), direction is reversed

        \subsection*{Vector addition}
        \begin{center}\begin{tikzpicture}[scale=1]
          \coordinate (A) at (0,0);
          \coordinate (B) at (2,2);
          \draw [->, thick, red] (0,0) -- (2,2) node [pos=0.5, below right] {\(\vec{u}=2\vec{i}+2\vec{j}\)};
          \draw [->, thick, blue] (2,2) -- (1,4) node [pos=0.5, above right] {\(\vec{v}=-\vec{i}+2\vec{j}\)};
          \draw [->, thick, orange] (0,0) -- (1,4) node [pos=0.5, left] {\(\vec{u}+\vec{v}=\vec{i}+4\vec{j}\)};
        \end{tikzpicture}\end{center}

        \[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]

        \begin{itemize}
          \item Draw each vector head to tail then join lines
          \item Addition is commutative (parallelogram)
          \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
        \end{itemize}

        \subsection*{Magnitude}

        \[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]

        \subsection*{Parallel vectors}

        \[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]

        For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
        \[\boldsymbol{a \cdot b}=\begin{cases}
          |\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
          -|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
        \end{cases}\]
        %\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
        %\includegraphics[width=1]{graphics/vector-subtraction.jpg}

        \subsection*{Perpendicular vectors}

        \[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]

        \subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
        \[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]

          \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}


          \begin{center}\begin{tikzpicture}[scale=2]
            \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
            \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
            \begin{scope}
              \path[clip] (1,0.5) -- (1,0) -- (0,0);
              \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
              \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
            \end{scope}
          \end{tikzpicture}\end{center}
          \begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\  &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
            \noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}

            \subsubsection*{Properties}

            \begin{enumerate}
              \item
                \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
              \item
                \(\boldsymbol{a \cdot 0}=0\)
              \item
                \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
              \item
                \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
              \item
                \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
              \item
                \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
            \end{enumerate}

            \subsection*{Angle between vectors}

            \[\cos \theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}\]

            \noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}

            (Action \(\rightarrow\) Vector \(\rightarrow\)Angle)

            \subsection*{Angle between vector and axis}

            \noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
            which makes angles \(\alpha, \beta, \gamma\) with positive side of
            \(x, y, z\) axes:
            \[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]

            \noindent \colorbox{cas}{On CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})}\\for angle
            between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
            \(x\)-axis

            \subsection*{Projections \& resolutes}

            \begin{tikzpicture}[scale=3]
              \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
              \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
              \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
              \begin{scope}
                \path[clip] (1,0.5) -- (1,0) -- (0,0);
                \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
                \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
              \end{scope}
              \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
                \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
              \end{scope}
              \draw (0,0) coordinate (O)
              (1,0) coordinate (A)
              (1,0.5) coordinate (B)
              pic [draw,red,angle radius=2mm] {right angle = O--A--B};
            \end{tikzpicture}

            \subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}

            \begin{align*}
              \boldsymbol{u} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
              & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
              & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
            \end{align*}

            \subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
            \[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]

            \subsubsection*{\(|\boldsymbol{u}|\) (scalar projection/resolute)}
            \begin{align*}
              s &= |\boldsymbol{u}|\\
              &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
              &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\\
              &= |\boldsymbol{a}| \cos \theta
            \end{align*}

            \subsubsection*{Rectangular (\(\parallel,\perp\)) components}

            \[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]


            \subsection*{Vector proofs}

            \textbf{Concurrent:} intersection of \(\ge\) 3 lines

            \begin{tikzpicture}
              \draw [blue] (0,0) -- (1,1);
              \draw [red] (1,0) -- (0,1);
              \draw [brown] (0.4,0) -- (0.6,1);
              \filldraw (0.5,0.5) circle (2pt);
            \end{tikzpicture}

            \subsubsection*{Collinear points}

            \(\ge\) 3 points lie on the same line

            \begin{tikzpicture}
              \draw [purple] (0,0) -- (4,1);
              \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
              \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
              \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
              \coordinate (O) at (2.8,-0.2);
              \node at (O) [below] {\(O\)}; 
              \begin{scope}[->, orange, thick] 
                \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
                \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
                \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
              \end{scope}
            \end{tikzpicture}

            \begin{align*}
              \text{e.g. Prove that}\\
              \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
              \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
              &= \overrightharp{OA} + m\overrightharp{AB}\\
              &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
              &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
              &=(1-m)\boldsymbol{a}+m{b}
            \end{align*}
            \begin{align*}
              \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
              \text{where } \lambda + \mu &= 1\\
              \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
            \end{align*}


            \subsubsection*{Parallelograms}

            \begin{center}\begin{tikzpicture}
              \coordinate (O) at (0,0) node [below left] {\(O\)};
              \coordinate (A) at (4,0);
              \coordinate (B) at (6,2);
              \coordinate (C) at (2,2);
              \coordinate (D) at (6,0);

              \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
              \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
              \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
              \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);

              \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
              \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
              \begin{scope}
                \path[clip] (C) -- (A) -- (O);
                \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
                \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
              \end{scope}
              \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
              \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
            \end{tikzpicture}\end{center}

            \begin{itemize}
              \item
                Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
              \item
                If diagonals are equal length, it is a rectangle
              \item
                \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
              \item
                Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
            \end{itemize}

            \subsubsection*{Useful vector properties}

            \begin{itemize}
              \item
                \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
                \(k \in \mathbb{R} \setminus \{0\}\)
              \item
                If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
                least one point in common, then they lie on the same straight line
              \item
                \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
              \item
                \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
            \end{itemize}

            \subsection*{Linear dependence}

            \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
            \begin{align*}
              0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
              \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
            \end{align*}

            \noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
            independent if no vector in the set is expressible as a linear
            combination of other vectors in set, or if they are parallel.

            \subsection*{Three-dimensional vectors}

            Right-hand rule for axes: \(z\) is up or out of page.

            \tdplotsetmaincoords{60}{120} 
            \begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick}, 
              vector/.style={-stealth,red,very thick}, 
              vector guide/.style={dashed,gray,thick}]

              %standard tikz coordinate definition using x, y, z coords
              \coordinate (O) at (0,0,0);

              %tikz-3dplot coordinate definition using x, y, z coords

              \pgfmathsetmacro{\ax}{1}
              \pgfmathsetmacro{\ay}{1}
              \pgfmathsetmacro{\az}{1}

              \coordinate (P) at (\ax,\ay,\az);

              %draw axes
              \draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
              \draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
              \draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};

              %draw a vector from O to P
              \draw[vector] (O) -- (P);

              %draw guide lines to components
              \draw[vector guide]         (O) -- (\ax,\ay,0);
              \draw[vector guide] (\ax,\ay,0) -- (P);
              \draw[vector guide]         (P) -- (0,0,\az);
              \draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
              \draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
              \draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
              \node[tdplot_main_coords,above right]
              at (\ax,\ay,\az){(\ax, \ay, \az)};
            \end{tikzpicture}\end{center}

            \subsection*{Parametric vectors}

            Parametric equation of line through point \((x_0, y_0, z_0)\) and
            parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:

            \[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]

              \section{Circular functions}

              \(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)

              \noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
              \indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)

              \subsection*{Reciprocal functions}

              \subsubsection*{Cosecant}

              \[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]

              \begin{itemize}
                \item
                  \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
                \item
                  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
                \item
                  \textbf{Turning points} at
                  \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
                \item
                  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
              \end{itemize}

              \subsubsection*{Secant}
           \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center}

                \[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]

                \begin{itemize}

                  \item
                    \textbf{Domain}
                    \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
                  \item
                    \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
                  \item
                    \textbf{Turning points} at
                    \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
                  \item
                    \textbf{Asymptotes} at
                    \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
                \end{itemize}

                \subsubsection*{Cotangent}

                \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center}

                  \[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]

                  \begin{itemize}

                    \item
                      \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
                    \item
                      \textbf{Range} \(= \mathbb{R}\)
                    \item
                      \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
                  \end{itemize}

                  \subsubsection*{Symmetry properties}

                  \[\begin{split}
                    \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
                    \operatorname{sec} (-x) & = \operatorname{sec} x \\
                    \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
                    \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
                    \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
                    \operatorname{cot} (-x) & = - \operatorname{cot} x
                  \end{split}\]

                  \subsubsection*{Complementary properties}

                  \[\begin{split}
                    \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
                    \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
                    \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
                    \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
                  \end{split}\]

                  \subsubsection*{Pythagorean identities}

                  \[\begin{split}
                    1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
                    1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
                  \end{split}\]

                  \subsection*{Compound angle formulas}

                  \[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
                  \[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
                  \[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]

                  \subsection*{Double angle formulas}

                  \[\begin{split}
                    \cos 2x &= \cos^2 x - \sin^2 x \\
                    & = 1 - 2\sin^2 x \\
                    & = 2 \cos^2 x -1
                  \end{split}\]

                  \[\sin 2x = 2 \sin x \cos x\]

                  \[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]

                  \subsection*{Inverse circular functions}

                  \pgfplotsset{every axis/.append style={
                    axis x line=middle,    % put the x axis in the middle
                    axis y line=middle,    % put the y axis in the middle
                    axis line style={<->}, % arrows on the axis
                    xlabel={$x$},          % default put x on x-axis
                    ylabel={$y$},          % default put y on y-axis
                    }}

% arrows as stealth fighters
\tikzset{>=stealth}

\begin{tikzpicture}
  \begin{axis}[domain = -1:1, samples = 500]
    \addplot[color = red]  {rad(asin(x))} node [pos=0.25, below right] {\(\sin^{-1}x\)};
    \addplot[color = blue] {rad(acos(x))} node [pos=0.25, below left] {\(\cos^{-1}x\)};
  \end{axis}
\end{tikzpicture}

                  Inverse functions: \(f(f^{-1}(x)) = x\) (restrict domain)

                  \[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
                  \hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)

                  \[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
                  \hfill where \(\cos y = x, \> y \in [0, \pi]\)

                  \[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
                  \hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)


                  \section{Differential calculus}

                  \subsection*{Limits}

                  \[\lim_{x \rightarrow a}f(x)\]
                  \(L^-,\quad L^+\) \qquad limit from below/above\\
                  \(\lim_{x \to a} f(x)\) \quad limit of a point\\

                  \noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
                  e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]

                  \subsubsection*{Limit theorems}

                  \begin{enumerate}
                    \item
                      For constant function \(f(x)=k\), \(\lim_{x \rightarrow a} f(x) = k\)
                    \item
                      \(\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G\)
                    \item
                      \(\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G\)
                    \item
                      \(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
                    \item
                      \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
                    \item
                      \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
                  \end{enumerate}

                  \subsection*{Gradients of secants and tangents}

                  \textbf{Secant (chord)} - line joining two points on curve\\
                  \textbf{Tangent} - line that intersects curve at one point

                  \subsection*{First principles derivative}

                  \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]

                  \subsubsection*{Logarithmic identities}

                  \(\log_b (xy)=\log_b x + \log_b y\)\\
                  \(\log_b x^n = n \log_b x\)\\
                  \(\log_b y^{x^n} = x^n \log_b y\)

                  \subsubsection*{Index identities}

                  \(b^{m+n}=b^m \cdot b^n\)\\
                  \((b^m)^n=b^{m \cdot n}\)\\
                  \((b \cdot c)^n = b^n \cdot c^n\)\\
                  \({a^m \div a^n} = {a^{m-n}}\)

                  \subsection*{Derivative rules}

                  \renewcommand{\arraystretch}{1.4}
                  \begin{tabularx}{\columnwidth}{rX}
                    \hline
                    \(f(x)\) & \(f^\prime(x)\)\\
                    \hline
                    \(\sin x\) & \(\cos x\)\\
                    \(\sin ax\) & \(a\cos ax\)\\
                    \(\cos x\) & \(-\sin x\)\\
                    \(\cos ax\) & \(-a \sin ax\)\\
                    \(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
                    \(e^x\) & \(e^x\)\\
                    \(e^{ax}\) & \(ae^{ax}\)\\
                    \(ax^{nx}\) & \(an \cdot e^{nx}\)\\
                    \(\log_e x\) & \(\dfrac{1}{x}\)\\
                    \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
                    \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
                    \(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
                    \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
                    \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
                    \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
                    \(\frac{d}{dy}f(y)\) & \(\dfrac{1}{\frac{dx}{dy}}\) (reciprocal)\\
                    \(uv\) & \(u \frac{dv}{dx}+v\frac{du}{dx} (product rule)\)\\
                    \(\dfrac{u}{v}\) & \(\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\) (quotient rule)\\
                    \(f(g(x))\) & \(f^\prime(g(x))\cdot g^\prime(x)\)\\
                    \hline
                  \end{tabularx}

                  \subsection*{Reciprocal derivatives}

                  \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]

                  \subsection*{Differentiating \(x=f(y)\)}
                  \begin{align*}
                    \text{Find }& \frac{dx}{dy}\\
                    \text{Then, }\frac{dx}{dy} &= \frac{1}{\frac{dy}{dx}} \\
                    \implies {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}\\
                    \therefore {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}
                  \end{align*}

                  \subsubsection*{Second derivative}
                  \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
                  \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}

                  \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken

                  \subsubsection*{Points of Inflection}

                  \emph{Stationary point} - i.e.
                  \(f^\prime(x)=0\)\\
                  \emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
                  \(f^{\prime\prime} = 0\))
                  %\begin{table*}[ht]
                  %\centering
                  %  \begin{tabularx}{\textwidth}{XXXX}
                  %\hline
                  %    \rowcolor{shade2}
                  %    & \(\dfrac{d^2 y}{dx^2} > 0\)  & \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
                  %\hline
                  %    \(\frac{dy}{dx}>0\) & \begin{tikzpicture} \draw[domain=1:2,smooth,variable=\x,blue] plot ({\x},{(1/10)*\x*\x*\x}) plot ({\x},{0.675*\x-0.677}); \end{tikzpicture} & cell 3\\
                  %cell 1 & cell 2 & cell 3\\
                  %\hline
                  %\end{tabularx}
                  %\end{table*}


\begin{itemize}
  \item
                      if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
                      \((a, f(a))\) is a local min (curve is concave up)
                    \item
                      if \(f^\prime (a) = 0\) and \(f^{\prime\prime} (a) < 0\), then point
                      \((a, f(a))\) is local max (curve is concave down)
                    \item
                      if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
                      inflection
                    \item
                      if also \(f^\prime(a)=0\), then it is a stationary point of inflection
                  \end{itemize}

                  \subsection*{Implicit Differentiation}

                  \noindent Used for differentiating circles etc.

                  If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
                  for all \(x\) and \(y\), then:

                  \[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]

                  \noindent \colorbox{cas}{\textbf{On CAS:}}\\
                  Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
                  Returns \(y^\prime= \dots\).

                  \subsection*{Integration}

                  \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]

                  \subsection*{Integral laws}

                  \renewcommand{\arraystretch}{1.4}
                  \begin{tabularx}{\columnwidth}{rX}
                    \hline
                    \(f(x)\) & \(\int f(x) \cdot dx\) \\
                    \hline
                    \(k\) (constant) & \(kx + c\)\\
                    \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
                    \(a x^{-n}\) &\(a \cdot \log_e |x| + c\)\\
                    \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
                    \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\>|\>n\ne 1\)\\
                    \((ax+b)^{-1}\) & \(\dfrac{1}{a}\log_e |ax+b|+c\)\\
                    \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
                    \(e^k\) & \(e^kx + c\)\\
                    \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
                    \(\cos kx\) & \(\dfrac{1}{k} \sin (kx) + c\)\\
                    \(\sec^2 kx\) & \(\dfrac{1}{k} \tan(kx) + c\)\\
                    \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
                    \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
                    \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
                    \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
                    \(\int f(u) \cdot \frac{du}{dx} \cdot dx\) & \(\int f(u) \cdot du\) (substitution)\\
                    \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
                    \hline
                  \end{tabularx}

                  Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)

                  \subsection*{Definite integrals}

                  \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]

                  \begin{itemize}

                    \item
                      Signed area enclosed by\\
                      \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
                    \item
                      \emph{Integrand} is \(f\).
                  \end{itemize}

                  \subsubsection*{Properties}

                  \[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]

                  \[\int^a_a f(x) \> dx = 0\]

                  \[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]

                  \[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]

                  \[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]

                  \subsection*{Integration by substitution}

                  \[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]

                  \noindent Note \(f(u)\) must be 1:1 \(\implies\) one \(x\) for each \(y\)
                  \begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
                    \text{let } u&=x+4\\
                    \implies& {\frac{du}{dx}} = 1\\
                    \implies& x = u - 4\\
                    \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
                    &\text{(solve as  normal integral)}
                  \end{align*}

                  \subsubsection*{Definite integrals by substitution}

                  For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
                  \(b\) for \(f(u) \cdot du\).

                  \subsubsection*{Trigonometric integration}

                  \[\sin^m x \cos^n x \cdot dx\]

                  \paragraph{\textbf{\(m\) is odd:}}
                  \(m=2k+1\) where \(k \in \mathbb{Z}\)\\
                  \(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
                  Substitute \(u=\cos x\)

                  \paragraph{\textbf{\(n\) is odd:}}
                  \(n=2k+1\) where \(k \in \mathbb{Z}\)\\
                  \(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
                  Substitute \(u=\sin x\)

                  \paragraph{\textbf{\(m\) and \(n\) are even:}}
                  use identities...

                  \begin{itemize}

                    \item
                      \(\sin^2x={1 \over 2}(1-\cos 2x)\)
                    \item
                      \(\cos^2x={1 \over 2}(1+\cos 2x)\)
                    \item
                      \(\sin 2x = 2 \sin x \cos x\)
                  \end{itemize}

                  \subsection*{Partial fractions}

                  \colorbox{cas}{On CAS:}\\
                  \indent Action \(\rightarrow\) Transformation \(\rightarrow\)
                  \texttt{expand/combine}\\
                  \indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
                  Expand \(\rightarrow\) Partial

                  \subsection*{Graphing integrals on CAS}

                  \colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
                  \(\int\) (\(\rightarrow\) Definite)\\
                  Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}

                  \subsection*{Applications of antidifferentiation}

                  \begin{itemize}

                    \item
                      \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
                      stationary points on \(y=F(x)\)
                    \item
                      nature of stationary points is determined by sign of \(y=f(x)\) on
                      either side of its \(x\)-intercepts
                    \item
                      if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
                      \(n+1\)
                  \end{itemize}

                  To find stationary points of a function, substitute \(x\) value of given
                  point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
                  original function.

                  \subsection*{Solids of revolution}

                  Approximate as sum of infinitesimally-thick cylinders

                  \subsubsection*{Rotation about \(x\)-axis}

                  \begin{align*}
                    V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
                    &= \pi \int^b_a (f(x))^2 \> dx
                  \end{align*}

                  \subsubsection*{Rotation about \(y\)-axis}

                  \begin{align*}
                    V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
                    &= \pi \int^b_a (f(y))^2 \> dy
                  \end{align*}

                  \subsubsection*{Regions not bound by \(y=0\)}

                  \[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]
                  \hfill where \(f(x) > g(x)\)

                  \subsection*{Length of a curve}

                  \[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]

                  \[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]

                  \noindent \colorbox{cas}{On CAS:}\\
                  \indent Evaluate formula,\\
                  \indent or Interactive \(\rightarrow\) Calculation
                  \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}

                  \subsection*{Rates}

                  \subsubsection*{Gradient at a point on parametric curve}

                  \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]

                  \[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]

                  \subsection*{Rational functions}

                  \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]

                  \subsubsection*{Addition of ordinates}

                  \begin{itemize}

                    \item
                      when two graphs have the same ordinate, \(y\)-coordinate is double the
                      ordinate
                    \item
                      when two graphs have opposite ordinates, \(y\)-coordinate is 0 i.e.
                      (\(x\)-intercept)
                    \item
                      when one of the ordinates is 0, the resulting ordinate is equal to the
                      other ordinate
                  \end{itemize}

                  \subsection*{Fundamental theorem of calculus}

                  If \(f\) is continuous on \([a, b]\), then

                  \[\int^b_a f(x) \> dx = F(b) - F(a)\]
                  \hfill where \(F = \int f \> dx\)

                  \subsection*{Differential equations}

                  \noindent\textbf{Order} - highest power inside derivative\\
                  \textbf{Degree} - highest power of highest derivative\\
                  e.g. \({\left(\dfrac{dy^2}{d^2} x\right)}^3\) \qquad order 2, degree 3

                  \subsubsection*{Verifying solutions}

                  Start with \(y=\dots\), and differentiate. Substitute into original
                  equation.

                  \subsubsection*{Function of the dependent
                  variable}

                  If \({\frac{dy}{dx}}=g(y)\), then
                  \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
                  \(e^c\) as \(A\).

                  \begin{table*}[ht]
                    \centering
                    \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png}
                  \end{table*}

                  \subsubsection*{Mixing problems}

                  \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]

                  \subsubsection*{Separation of variables}

                  If \({\frac{dy}{dx}}=f(x)g(y)\), then:

                  \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]

                  \subsubsection*{Euler's method for solving DEs}

                  \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]

                  \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]

                \end{multicols}
              \end{document}