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% Document Info
\author{smithandrewa}
\pdfinfo{
  /Title (calc-ii.pdf)
  /Creator (Cheatography)
  /Author (smithandrewa)
  /Subject (Calc II Cheat Sheet)
}

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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Calc II Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{smithandrewa} via \textcolor{DarkBackground}{\uline{cheatography.com/71791/cs/18986/}}}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}smithandrewa \\
  \uline{cheatography.com/smithandrewa} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 27th February, 2019.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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% www.emerson.emory.edu/services/latex/latex_169.html
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{x{4.16 cm} x{3.84 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Trig Integrals}}  \tn
% Row 0
\SetRowColor{LightBackground}
∫sinx dx & = -cosx dx + C \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
∫cosx dx & = sinx dx + C \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
∫sec\textasciicircum{}2\textasciicircum{}x dx & = tanx dx + C \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
∫tanx dx & = ln|secx| + C \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
∫secx tanx dx & = secx + C \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
∫csc\textasciicircum{}2\textasciicircum{}x dx & = -cotx + C \tn 
% Row Count 6 (+ 1)
% Row 6
\SetRowColor{LightBackground}
∫cscx cotx dx & = -cscx + C \tn 
% Row Count 7 (+ 1)
% Row 7
\SetRowColor{white}
∫cotx dx & = ln|sinx| + C \tn 
% Row Count 8 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{4.32 cm} x{3.68 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Trig Identities}}  \tn
% Row 0
\SetRowColor{LightBackground}
∫(1/x\textasciicircum{}2\textasciicircum{} + a\textasciicircum{}2\textasciicircum{}) dx & = 1/a tan\textasciicircum{}-1\textasciicircum{}(x/a) + C \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
∫(1/Sqrt(a\textasciicircum{}2\textasciicircum{} - x\textasciicircum{}2\textasciicircum{}) dx & = sin\textasciicircum{}-1\textasciicircum{}(x/a) + C \tn 
% Row Count 4 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{(While a \textgreater{} 0)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Area Between Curves}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Area = ∫{[}Height{]} Width \newline A = ∫(f(x) - g(x){]} dx \newline  \newline 1. Graph Equasions \newline 2. Label \newline 3. Determine how to slice \newline 4. Set up dA \newline 5. dA = height*dx \newline 6. Get range a \& b from intersections \newline 7. Plug in and find area} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{There is never (-) area.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Volume by Disk}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{dV = A(x) dx \newline  \newline V = ∫A(x)dx \newline  \newline Volume = ∫Radius\textasciicircum{}2\textasciicircum{} * Thickness \newline  \newline V = ∫(pi(r)\textasciicircum{}2\textasciicircum{}) dx} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Volume by Washer}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{dV = A(x) dx \newline  \newline V = ∫A(x) dx \newline  \newline Volume = ∫{[}(pi r out\textasciicircum{}2\textasciicircum{})-(pi r in\textasciicircum{}2\textasciicircum{}){]} dx} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Slice Perpendicular to Axis of Rotation}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Volume by Shell}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{dVolume = Circumference * dArea \newline dV = (2 pi r) dArea \newline  \newline V = ∫(2 pi r)(Area)dx \newline  \newline 1. Write: dV = 2 pi r dA \newline 2. Find dA(height dx) \newline 3. Find Radius(x or y) \newline 4. Plug in  \newline 5. Take integral} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Slice Parallel to Axis of Rotation}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Average Value of a Function}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Average Value = 1/b-a * ∫f(x) dx \newline  \newline Symmetry: \newline If f(x) is EVEN, then ∫f(x)dx from -a to a = 2∫f(x) from 0 to a \newline  \newline If f(x) is ODD, then  ∫f(x)dx from -a to a = 0} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{3.44 cm} x{4.56 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Important Integrals}}  \tn
% Row 0
\SetRowColor{LightBackground}
∫c f(x) dx & = c ∫f(x) dx \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
∫{[}f(x) + g(x){]} dx & = ∫f(x) dx + ∫g(x) dx \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
∫ 1/x dx & = ln|x| + C \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
∫e\textasciicircum{}x\textasciicircum{} dx & = e\textasciicircum{}x\textasciicircum{} + C \tn 
% Row Count 5 (+ 1)
% Row 4
\SetRowColor{LightBackground}
∫b\textasciicircum{}x\textasciicircum{} dx & = (b\textasciicircum{}x\textasciicircum{} / lnb) + C \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{2.28 cm} x{3.42 cm} x{1.9 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Methods of Integration}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Method}} & {\bf{When to Use}} & {\bf{Example}} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\seqsplit{U-Substitution} & When a Polynomial is \{\{nl\}\}raised to a power \textgreater{} 1 & ∫(3x + 5)\textasciicircum{}5\textasciicircum{} \tn 
% Row Count 5 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Integration by Parts & When U-Sub will not work & ∫xe\textasciicircum{}x\textasciicircum{} \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
\seqsplit{Trigonometric} Integration & Only Trig raised to powers & ∫sin\textasciicircum{}6\textasciicircum{}x cos\textasciicircum{}3\textasciicircum{}xdx \tn 
% Row Count 10 (+ 3)
% Row 4
\SetRowColor{LightBackground}
\seqsplit{Trigonometric} \seqsplit{Substitution} & 3/2 powers or Sqrt(a\textasciicircum{}2\textasciicircum{}-x\textasciicircum{}2\textasciicircum{}) etc. & dx/(x\textasciicircum{}2\textasciicircum{}Sqrt(25-x\textasciicircum{}2\textasciicircum{})) \tn 
% Row Count 13 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{p{0.8 cm} p{0.8 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{U-Substitution}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{} \tn 
% Row Count 0 (+ 0)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Integration by Parts}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Logarithmic \newline Inverse trig \newline Algebraic \newline Trigonometric \newline Exponential \newline  \newline ∫u dv = u v - ∫v du \newline  \newline 1. Write u v - ∫v du \newline 2. Use LIATE to find u; the other term becomes dV \newline 3. Setup u=      dV=       du=      V= \newline 4. Solve} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Cyclical Functions will need to be split and substituted.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Trigonometric Integration}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{{\emph{Identities}} \newline sin\textasciicircum{}2\textasciicircum{}t+cos\textasciicircum{}2\textasciicircum{}t = 1 \newline sin\textasciicircum{}2\textasciicircum{}t = 1/2 {[}1-cos(2t){]} \newline cos\textasciicircum{}2\textasciicircum{}t = 1/2 {[}1+cos(2t){]} \newline  \newline Can use with U-Substitution} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Don't change all of the trig to the same form.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Trigonometric Integration}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{{\emph{Identities}} \newline sin\textasciicircum{}2\textasciicircum{}t+cos\textasciicircum{}2\textasciicircum{}t = 1 \newline sin\textasciicircum{}2\textasciicircum{}t = 1/2 {[}1-cos(2t){]} \newline cos\textasciicircum{}2\textasciicircum{}t = 1/2 {[}1+cos(2t){]} \newline  \newline Can use with U-Substitution} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Don't change all of the trig to the same form.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Trigonometric Substitution}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{{\emph{Pythag. Identities}} \newline sin\textasciicircum{}2\textasciicircum{} + cos\textasciicircum{}2\textasciicircum{} = 1 \newline 1 + tan\textasciicircum{}2\textasciicircum{} = sec\textasciicircum{}2\textasciicircum{} \newline 1 + cot\textasciicircum{}2\textasciicircum{} = csc\textasciicircum{}2\textasciicircum{} \newline  \newline 1. Identify a and u \newline 2. Sub in the trig \newline 3. Manipulate to simplify \newline 4. Get rid of trig with a triangle} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}