% (find-angg "LATEX/2015-2-C2-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2015-2-C2-P1.tex" :end))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2015-2-C2-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2015-2-C2-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2015-2-C2-P1"))
% (find-xpdfpage "~/LATEX/2015-2-C2-P1.pdf")
% (find-sh0 "cp -v  ~/LATEX/2015-2-C2-P1.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2015-2-C2-P1.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2015-2-C2-P1.pdf
%               file:///tmp/2015-2-C2-P1.pdf
%           file:///tmp/pen/2015-2-C2-P1.pdf
% http://angg.twu.net/LATEX/2015-2-C2-P1.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
%\usepackage{tikz}
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.tex")
\input istanbuldefs               % (find-LATEX "istanbuldefs.tex")
%
\begin{document}

\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
%\directlua{output(preamble1)}
\def\expr#1{\directlua{output(tostring(#1))}}
\def\eval#1{\directlua{#1}}
\def\pu{\directlua{pu()}}

%\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
%%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end

\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}}


% 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}}





%   ____      _                    _ _           
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___  
% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ 
%                                                

{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2015.2
\par P1 - 14/mar/2016 - Eduardo Ochs
% \par Versão: 14/mar/2016
\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}
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\bsk
\bsk

% (c2q 7 "integral de Riemann")
% (c2q 9 "integral de Riemann sem partição especificada")
% (c2q 11 "TFC")
% (c2q 13 "TFC 2")
% (c2q 15 "Substituição")
% (c2q 16 "Diferenciais")
% (c2q 22 "G(x,y) = x^2 + y^2")
% (c2q 24 "Derivada da função inversa")
% (c2q 27 "Integrando funções racionais")
% (c2q 30 "Método de Heaviside")
% (c2q 32 "Integrando funções racionais impróprias")
% (c2q 34 "Integração por partes")
% (c2q 35 "Truque do `onde'")
% (c2q 36 "Tabelas de integrais")
% (c2q 37 "Substituição trigonométrica")
% (c2q 43 "Série de Taylor")
% (c2q 45 "Plano complexo")
% (c2q 48 "Grande truque: E")
% (c2q 50 "Substituição trigonométrica")
% (c2q 51 "EDOs")
% (c2q 53 "EDOs: D")
% (c2q 55 "EDOs: sen e cos vezes exp")


% (find-es "ipython" "2015.2-C2-P1")

1) Seja $f_a(x) = x^2\cos ax$. Calcule:

$$a) \;\; (\text{2.0 pts}) \quad \intx {f_a(x)}$$

$$b) \;\; (\text{1.0 pts}) \quad \Intx {0} {π} {f_2(x)}$$


2) Calcule:

$$(\text{3.0 pts}) \quad \intx {\frac {x^3} {x^2 + x - 20}}$$

3) Calcule:

$$a) \;\; (\text{2.0 pts}) \quad \intx {\frac{\sqrt{x^2-1}}{x^3}}$$

$$b) \;\; (\text{2.0 pts}) \quad \intx {\frac{\sqrt{4x^2-9}}{x^3}}$$

% (find-xpdfpage "~/2015.2-C2/3c3-TrigonometSubstitu_Stu.pdf" 5 "ex.13")
% (find-xpdfpage "~/2015.2-C2/3c3-TrigonometSubstitu_Stu.pdf" 7 "answer")



\newpage



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

\bsk

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}
$



\end{document}


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