% (find-angg "LATEX/2015-2-C2-material.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2015-2-C2-material.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2015-2-C2-material.pdf"))
% (defun e () (interactive) (find-LATEX "2015-2-C2-material.tex"))
% (defun l () (interactive) (find-LATEX "2015-2-C2-material.lua"))
% (defun u () (interactive) (find-latex-upload-links "2015-2-C2-material"))
% (find-xpdfpage "~/LATEX/2015-2-C2-material.pdf")
% (find-lualatex-links   "2015-2-C2-material")
% (find-LATEX "falta-misandria-a5.tex")
% (find-LATEXfile "2014-1-GA-P2-gab.tex")
% (find-LATEXfile "2015-1-GA-P2-gabarito.tex" "\\catcode")
% (find-LATEXfile "2015-1-GA-P2-gabarito.tex" "dednat6dir =")
% (find-angg "LATEX/2015-1-C2-lista-edrx-1.tex")
% file:///home/edrx/LATEX/2015-2-C2-material.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
% \usepackage{luacode}
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.tex")
\input istanbuldefs.tex           % (find-istfile "defs.tex")
\def\Diag#1{\directlua{tf:processuntil()}\diag{#1}}
\def\Ded #1{\directlua{tf:processuntil()}\ded{#1}}
\def\Exec#1{\directlua{tf:processuntil() #1}}
\def\Expr#1{\directlua{tf:processuntil() output(#1)}}
\def\Expr#1{\directlua{tf:processuntil() output(tostring(#1))}}
%
\begin{document}

% dednat6
%
\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
% \directlua{output(preamble1)}
\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
\def\pu{\directlua{pu()}}

% tikz
% mygrid
% (find-angg ".emacs.papers" "tikz")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
\tikzset{grid/.style=gray!20,very thin}
\tikzset{anydot/.style={circle,inner sep=0pt,minimum size=1mm}}
\tikzset{opdot/.style={anydot, draw=black,fill=white}}
\tikzset{cldot/.style={anydot, draw=black,fill=black}}
%
\def\mygrid(#1,#2) (#3,#4){
  \clip              (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
  \draw[step=1,grid] (#1-0.2, #2-0.2) grid      (#3+0.2, #4+0.2);
  \draw[axis] (-10,0) -- (10,0);
  \draw[axis] (0,-10) -- (0,10);
  \foreach \x in {-10,...,10} \draw[tick] (\x,-0.2) -- (\x,0.2);
  \foreach \y in {-10,...,10} \draw[tick] (-0.2,\y) -- (0.2,\y);
}

\def\tikzp#1{\mat{\begin{tikzpicture}#1\end{tikzpicture}}}

% (find-LATEX "edrxtikz.lua" "drawdots0")
\def\drawdots#1{\directlua{output(drawdots0("#1"))}}









{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2015.2
\par Material para exercícios - Eduardo Ochs
\par Versão: 7/dez/2015
\par Links importantes:
\par \url{http://angg.twu.net/2015.2-C2.html} (página do curso)
\par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
\par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
     (lista, atualizada)
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\bsk
\bsk




% (find-LATEXgrep "grep --color -nH -e rigonom *.tex")

\def\ddx{\frac{d}{dx}}
\def\ddth{\frac{d}{d\theta}}
\def\arcsen{\operatorname{arcsen}}
\def\sen{\operatorname{sen}}
\def\sec{\operatorname{sec}}
\def\ln{\operatorname{ln}}

\def\subst#1{\left[\sm{#1}\right]}

\def\prims #1{∫#1\,ds}
\def\primth#1{∫#1\,dθ}

\def\ints #1#2#3{∫_{s=#1}^{s=#2}#3\,ds}
\def\intu #1#2#3{∫_{u=#1}^{u=#2}#3\,du}
\def\intx #1#2#3{∫_{x=#1}^{x=#2}#3\,dx}
\def\intth#1#2#3{∫_{θ=#1}^{θ=#2}#3\,dθ}

\def\difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}



Substituição:

$\begin{array}{l}
 \difx a b {g(h(x))} = \intx a b {g'(h(x))h'(x)} \\
 \phantom{mmm}|\,| \\
 \difu {h(a)} {h(b)} {g(u))} = \intu {h(a)} {h(b)} {g'(u)} \\
 \end{array}
$

\msk

$\intx a b {f(g(x))g'(x)} = \intu {g(a)} {g(b)} {f(u)}$

% Substituição inversa:
% 
% $
%  \begin{array}{rcl}
%  \intx {g¹(α)} {g¹(β)} {f(g(x))g'(x)} = \intu {α} {β} {f(u)} \\ \\
%  \int {f(g(x))g'(x)} \,dx = \int {f(u)} \,du \quad \subst{u = g(x)} \\ \\
%  \int {f(u)\frac{du}{dx}} \,dx = \int {f(u)} \,du \quad \subst{u = g(x)} \\
%  \end{array}
% $




\bsk

Um exemplo de substituição:

$\begin{array}[t]{l}
 \primth {s^3 c^3} \\
 = \primth {s^3 c^2 c} \\
 = \primth {s^3 (1-s^2) \frac{ds}{dθ}} \\
 = \primth {(s^3 - s^5) \frac{ds}{dθ}} \\
 = \prims  {s^3 - s^5} \\
 = \frac{s^4}{4} - \frac{s^6}{6} \\
 \end{array}
 %
 \qquad
 %
 \def\intthab{\intthαβ◻}
 \def\intsab {\ints{\senα}{\senβ}◻}
 \def\difthab{\difthαβ◻}
 \def\difsab {\difs{\senα}{\senβ}◻}
 %
 \def\intthab{(\intthαβ◻)}
 \def\intsab {(\ints{\senα}{\senβ}◻)}
 \def\difthab{(\difthαβ◻)}
 \def\difsab {(\difs{\senα}{\senβ}◻)}
 %
 \begin{array}[t]{lll}
 \primth {\sen^3θ \cos^3θ}                         & \intthab \\
 = \primth {\sen^3θ \cos^2θ \cosθ}                 & \intthab \\
 = \primth {\sen^3θ (1-\sen^2θ) \frac{d\senθ}{dθ}} & \intthab \\
 = \primth {(\sen^3θ - \sen^5θ) \frac{d\senθ}{dθ}}   & \intthab \\
 = \prims  {s^3 - s^5}                             & \intsab \\
 = \frac{s^4}{4} - \frac{s^6}{6}                   & \difsab \\
 = \frac{\sen^4θ}{4} - \frac{\sen^6θ}{6}           & \difthab \\
 \end{array}
$


\bsk
\bsk



\newpage


% Indefinite integrals
\def\ints #1{∫#1\,ds}
\def\intt #1{∫#1\,dt}
\def\intu #1{∫#1\,du}
\def\intx #1{∫#1\,dx}
\def\intz #1{∫#1\,dz}
\def\intth#1{∫#1\,dθ}

% Definite integrals
\def\Ints #1#2#3{∫_{s=#1}^{s=#2}#3\,ds}
\def\Intt #1#2#3{∫_{t=#1}^{t=#2}#3\,dt}
\def\Intu #1#2#3{∫_{u=#1}^{u=#2}#3\,du}
\def\Intx #1#2#3{∫_{x=#1}^{x=#2}#3\,dx}
\def\Intz #1#2#3{∫_{z=#1}^{z=#2}#3\,dz}
\def\Intth#1#2#3{∫_{θ=#1}^{θ=#2}#3\,dθ}

% Difference
\def\Difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\Difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\Difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\Difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}



Substituição:

\msk

$\begin{array}{l}
 \Difx a b {g(h(x))} = \Intx a b {g'(h(x))\frac{d\,h(x)}{dx}} \\
 \phantom{mmm}|\,| \\
 \Difu {h(a)} {h(b)} {g(u))} = \Intu {h(a)} {h(b)} {g'(u)} \\
 \end{array}
$

\msk

Fórmulas:

\msk

$\begin{array}{l}
 \Intx   a b           {f(g(x))\frac{d\,g(x)}{dx}} \\
 = \Intx a b           {f(u)\frac{du}{dx}} \\
 = \Intu {g(a)} {g(b)} {f(u)}
 \end{array}
 \qquad
 \begin{array}{ll}
 \intx   {f(g(x))\frac{d\,g(x)}{dx}} \\
 = \intx {f(u)\frac{du}{dx}}         & \subst{u=g(x)} \\
 = \intu {f(u)}                      & \subst{u=g(x)}
 \end{array}
$

\bsk

Substituição inversa:

$\def\a{{h¹(α)}}
 \def\b{{h¹(β)}}
 \begin{array}{l}
 \Difx \a \b {g(h(x))} = \Intx \a \b {g'(h(x))\frac{d\,h(x)}{dx}} \\
 \phantom{mmm}|\,| \\
 \Difu {h(\a)} {h(\b)} {g(u))} = \Intu {h(\a)} {h(\b)} {g'(u)} \\
 \phantom{mmm}|\,| \\
 \Difu α β {g(u))} = \Intu α β {g'(u)} \\
 \end{array}
$

\msk

Fórmulas:

\msk

$\def\a{{g¹(α)}}
 \def\b{{g¹(β)}}
 \begin{array}{l}
 \Intu   α β  {f(u)} \\
 = \Intx \a \b {f(u)\frac{du}{dx}} \\
 = \Intx \a \b {f(g(x))\frac{d\,g(x)}{dx}}
 \end{array}
 \qquad
 \begin{array}{ll}
 \intu   {f(u)} \\
 = \intx {f(u)\frac{du}{dx}}         & \subst{u=g(x)\\x=g¹(u)} \\
 = \intx {f(g(x))\frac{d\,g(x)}{dx}} & \subst{x=g¹(u)} \\
 \end{array}
$

\bsk

Substituição trigonométrica:

\msk

$
 \begin{array}{ll}
 \Ints a b {F(s, \sqrt{1-s^2})} \\
 = \Intth {\arcsen a} {\arcsen b} {F(\senθ, \sqrt{1-\sen^2θ}) \frac{d\senθ}{dθ}} \\
 = \Intth {\arcsen a} {\arcsen b} {F(\senθ, \cosθ) \cosθ}                        \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \ints    {F(s, \sqrt{1-s^2})} \\
 = \intth {F(s, \sqrt{1-s^2}) \frac{ds}{dθ}} & \subst{s=\senθ \\ θ=\arcsenθ} \\
 = \intth {F(s, c) c}                        & \subst{s=\senθ \\ c=\cosθ \\ θ=\arcsenθ} \\
 \end{array}
$

\msk

$
 \begin{array}{ll}
 \Intz a b {F(z, \sqrt{z^2-1})} \\
 = \Intth {\arcsec a} {\arcsec b} {F(\secθ, \sqrt{\sec^2θ-1}) \frac{d\secθ}{dθ}} \\
 = \Intth {\arcsec a} {\arcsec b} {F(\secθ, \tanθ) \secθ \tanθ}                   \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \intz    {F(z, \sqrt{z^2-1})} \\
 = \intth {F(z, \sqrt{z^2-1}) \frac{dz}{dθ}} & \subst{z=\secθ \\ θ=\arcsec z} \\
 = \intth {F(z, t) zt}                       & \subst{z=\secθ \\ θ=\arcsec z \\ t=\tanθ} \\
 \end{array}
$

\msk

$
 \begin{array}{ll}
 \Intt a b {F(t, \sqrt{1+t^2})} \\
 = \Intth {\arctan a} {\arctan b} {F(\tanθ, \sqrt{1+\tan^2θ}) \frac{d\tanθ}{dθ}} \\
 = \Intth {\arctan a} {\arctan b} {F(\tanθ, \secθ) \sec^2θ}                      \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \intt    {F(t, \sqrt{1+t^2})} \\
 = \intth {F(t, \sqrt{1+t^2}) \frac{dt}{dθ}} & \subst{t=\tanθ \\ θ=\arctan t} \\
 = \intth {F(t, z) z^2}                      & \subst{t=\tanθ \\ θ=\arctan t \\ z=\secθ} \\
 \end{array}
$




% & \subst{u = g(x)}



\newpage




% (find-LATEX "2010-1-C2-prova-1.tex")

Algumas fórmulas:

Integração por partes:

$\int_{x=a}^{x=b} f'(x)g(x)\,dx = f(x)g(x) \big|_{x=a}^{x=b} - \int_{x=a}^{x=b} f(x)g'(x)\,dx$

\msk

% Mudança de variável:
% 
% $\int_{x=a}^{y=b} \frac{dg}{du} \frac{du}{dx} \,dx = \int_{u=u(a)}^{u=u(b)} \frac{dg}{du} \, du$ 
% 
% $\int_{x=a}^{y=b} g'(u(x))u'(x)\,dx = \int_{u=u(a)}^{u=u(b)} g'(u) \, du$
% 
% $\int_{x=a}^{y=b} f(u(x))u'(x)\,dx = \int_{u=u(a)}^{u=u(b)} f(u) \, du$
% 
% \msk

Integrais de $(\sen θ)^m (\cos θ)^n$ com um expoente ímpar ($s = \sen θ$, $c= \cos θ$):

$\int s^n c^{2k+1} dθ = \int s^n c^{2k} · c \,dθ =
  \subst{\sen θ = s \\
         \cos^2 θ = 1 - s^2 \\
         \cos θ\,dθ = ds \\
         θ = \arcsen s}
  \int s^n (1-s^2)^k \, ds$

$\int c^n s^{2k+1} dθ = \int c^n s^{2k} · s \,dθ =
  \subst{\cos θ = c \\
         \sen^2 θ = 1 - c^2 \\
         - \sen θ\,dθ = dc \\
         θ = \arccos s}
  - \int c^n (1-c^2)^k \, dc$

\msk

Substituição trigonométrica:

$\int F(s, \sqrt{1 - s^2})\,ds =
 \subst{s = \sen θ \\
        \sqrt{1-s^2} = \cos θ \\
        ds = \cos θ \, dθ \\
        θ = \arcsen s}
 \int F(\sen θ, \cos θ) \cos θ \, dθ$

$\int F(t, \sqrt{1 + t^2})\,dt =
 \subst{t = \tan θ \\
        \sqrt{1+t^2} = \sec θ \\
        dt = \sec^2 θ \, dθ \\
        θ = \arctan t}
  \int F(\tan θ, \sec θ) \sec^2 θ \, dθ$

$\int F(z, \sqrt{z^2 - 1})\,dz =
 \subst{z = \sec θ \\
        \sqrt{z^2-1} = \tan θ \\
        dz = \tan θ \sec θ\, dθ \\
        θ = \arcsec z}
  \int F(\sec θ, \tan θ) \tan θ \sec θ \, dθ$

\ssk

$\int\sqrt{1-x^2}\,dx = \frac{\arcsen x}{2} + \frac{x\,\sqrt{1-x^2}}{2}$


\msk

Método de Heaviside:

Se $f(x) = \frac{\aa}{x-a} + \frac{\bb}{x-b} + \frac{\cc}{x-c} = \frac{p(x)}{(x-a)(x-b)(x-c)}$,

então $\lim_{x \to a} f(x)(x-a) = \aa = \frac{p(a)}{(a-b)(a-c)}$.



% \end{document}
\newpage



Funções usadas nas aulas de 30/nov e 2/dez/2015:

\msk

% (find-djvupage "~/2015.2-C2/2015.2-C2.djvu")
$f_1(x) = \tikzp{[scale=0.3,auto]
    \mygrid (-1,-2) (5,2);
    \drawdots{ (-2,1)--(2,1)c (2,-1)o--(3,-1)o (3,0)c--(6,0) };
  }
$

% (find-fline "~/2015.2-C2/20151130_C23.jpg")
$f_2(x) = \tikzp{[scale=0.3,auto]
    \mygrid (-1,-2) (4,3);
    \drawdots{ (-3,0)--(0,0)o (0,2)c--(1,0)c (1,-1)o--((5,-1) };
  }
$

% (find-fline "~/2015.2-C2/20151130_C23.jpg")
$f_3(x) = \tikzp{[scale=0.3,auto]
    \mygrid (-1,-2) (5,3);
    \drawdots{ (-2,0)--(1,0)c--(2,2)c--(6,2) };
  }
$

$f_4(x) = \tikzp{[scale=0.3,auto]
    \mygrid (-1,-1) (6,5);
    \drawdots{ (-2,0)--(0,0)c--(1,3)c--(2,4)c--(3,3)c--(4,0)c--(7,0) };
  }
$

$f_6(x) = \tikzp{[scale=0.3,auto]
    \mygrid (-1,-2) (8,3);
    \drawdots{ (-2,0)--(1,0)o (1,1)c--(2,1)c (2,2)o--(3,2)o (3,1)c--(4,1)c
               (4,0)o--(5,0)o (5,-1)c--(6,-1)c (6,0)o--(10,0) };
  }
$

\end{document}


% Local Variables:
% coding: utf-8-unix
% modes: (fundamental-mode emacs-lisp-mode lua-mode)
% End:
