Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
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% (find-angg "LATEX/2016-1-C2-P1.lua")
% (find-es "ipython" "2016.1-C2-P1")
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% (find-sh0 "cp -v  ~/LATEX/2016-1-C2-P1.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2016-1-C2-P1.pdf
%               file:///tmp/2016-1-C2-P1.pdf
%           file:///tmp/pen/2016-1-C2-P1.pdf
% http://angg.twu.net/LATEX/2016-1-C2-P1.pdf
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\usepackage{amssymb}
%\usepackage{tikz}
%
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\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxchars.tex              % (find-LATEX "edrxchars.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.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()}}

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% %L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end


\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}



% (find-LATEX "2015-2-C2-P1.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{â«#1\,ds}
\def\primth#1{â«#1\,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{â«#1\,ds}
\def\intc #1{â«#1\,dc}
\def\intt #1{â«#1\,dt}
\def\intu #1{â«#1\,du}
\def\intx #1{â«#1\,dx}
\def\intz #1{â«#1\,dz}
\def\intth#1{â«#1\,dθ}

% Definite integrals
\def\Ints #1#2#3{â«_{s=#1}^{s=#2}#3\,ds}
\def\Intt #1#2#3{â«_{t=#1}^{t=#2}#3\,dt}
\def\Intu #1#2#3{â«_{u=#1}^{u=#2}#3\,du}
\def\Intx #1#2#3{â«_{x=#1}^{x=#2}#3\,dx}
\def\Intz #1#2#3{â«_{z=#1}^{z=#2}#3\,dz}
\def\Intth#1#2#3{â«_{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 CÃlculo 2
\par PURO-UFF - 2016.1
\par P1 - 4/jul/2016 - Eduardo Ochs
% \par VersÃo: 14/mar/2016
\par Links importantes:
\par \url{http://angg.twu.net/2016.1-C2.html} (pÃgina do curso)
\par \url{http://angg.twu.net/2016.1-C2/2016.1-C2.pdf} (quadros)
\par \url{http://angg.twu.net/LATEX/2016-1-C2-material.pdf}
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\bsk
\bsk

% (c2q 7 "integral de Riemann")
% (c2q 9 "integral de Riemann sem partiÃÃo especificada")
% (c2q 11 "TFC")
% (c2q 13 "TFC 2")
% (c2q 15 "SubstituiÃÃo")
% (c2q 16 "Diferenciais")
% (c2q 22 "G(x,y) = x^2 + y^2")
% (c2q 24 "Derivada da funÃÃo inversa")
% (c2q 27 "Integrando funÃÃes racionais")
% (c2q 30 "MÃtodo de Heaviside")
% (c2q 32 "Integrando funÃÃes racionais imprÃprias")
% (c2q 34 "IntegraÃÃo por partes")
% (c2q 35 "Truque do `onde'")
% (c2q 36 "Tabelas de integrais")
% (c2q 37 "SubstituiÃÃo trigonomÃtrica")
% (c2q 43 "SÃrie de Taylor")
% (c2q 45 "Plano complexo")
% (c2q 48 "Grande truque: E")
% (c2q 50 "SubstituiÃÃo trigonomÃtrica")
% (c2q 51 "EDOs")
% (c2q 53 "EDOs: D")
% (c2q 55 "EDOs: sen e cos vezes exp")


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


1) \T(Total: 1.0 pts) Calcule $\intx {\tan x}$.

\bsk

2) \T(Total: 1.0 pts) Calcule $\Intx {-1} {2} {x^{-4}}$.

\bsk

3) \T(Total: 1.5 pts) Calcule $\intx {\cos^4 x}$.

\bsk

4) \T(Total: 1.5 pts) Calcule $\intx {\frac {x^2} {x^2+x-2}}$.

\bsk

5) \T(Total: 1.5 pts) Calcule $\intx {x \, e^x \cos x}$.

\bsk

6) \T(Total: 2.0 pts)

6a) \B(1.5 pts) Calcule $\Intx 0 1 {\sqrt{4-x^2}}$.

6b) \B(0.1 pts) Represente $\Intx 0 1 {\sqrt{4-x^2}}$ graficamente.

6c) \B(0.4 pts) Mostre como calcular $\Intx 0 1 {\sqrt{4-x^2}}$ pelo grÃfico.

\bsk

7) \T(Total: 2.5 pts) Calcule $\Intx {-1} {2} {|e^x-1|}$.





\newpage



MÃtodo 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

SubstituiÃÃo:

\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

FÃrmulas:

\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

SubstituiÃÃo inversa:

$\def\a{{hÂ(α)}}
 \def\b{{hÂ(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

FÃrmulas:

\msk

$\def\a{{gÂ(α)}}
 \def\b{{gÂ(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=gÂ(u)} \\
 = \intx {f(g(x))\frac{d\,g(x)}{dx}} & \subst{x=gÂ(u)} \\
 \end{array}
$

\bsk

SubstituiÃÃo trigonomÃtrica:

\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}
$




\newpage

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

Gabarito (nÃo revisado, contÃm erros de vÃrios tipos!):

\bsk


%  _ 
% / |
% | |
% | |
% |_|
%    


1)
%
$\begin{array}[t]{rcl}
 \intth {\tanθ} &=& \intth {\frac{s}{c}} \\
                &=& \intth {\frac{1}{c} s} \\
                &=& - \intc {\frac{1}{c}} \\
                &=& - \ln |c| \\
                &=& - \ln |\cosθ| \\
 \end{array}
$

%  ____  
% |___ \ 
%   __) |
%  / __/ 
% |_____|
%        

{

\def\lima{\lim\limits_{aâ0^-}}
\def\limb{\lim\limits_{bâ0^+}}
\def\lima{\lim_{aâ0^-}}
\def\limb{\lim_{bâ0^+}}
\def\limap#1{\lim_{aâ0^-} \left(#1\right)}
\def\limbp#1{\lim_{bâ0^+} \left(#1\right)}
\def\p#1{\left(#1\right)}

\bsk

2)
%
$\begin{array}[t]{rcl}
 \intx {x^{-4}} &=& \frac{x^{-3}}{-3} \\
 \Intx{-1}{2} {x^{-4}} &=& \Intx{-1}{0} {x^{-4}} + \Intx{0}{2} {x^{-4}} \\
                       &=& \lima \Intx{-1}{a} {x^{-4}} +
                           \limb \Intx{b}{2}  {x^{-4}} \\
                       &=& \limap {\difx{-1}{a} {\frac{x^{-3}}{-3}}} +
                           \limbp {\difx{b}{2}  {\frac{x^{-3}}{-3}}} \\
                       &=& \limap {\frac{a^{-3}}{-3} - \frac{(-1)^{-3}}{-3}} +
                           \limbp {\frac{2^{-3}}{-3} - \frac{b^{-3}}{-3}} \\
                       &=&     \p {\frac{-â}{-3} - \frac{-1}{-3}} +
                               \p {\frac{1/8}{-3} - \frac{+â}{-3}} \\
 \end{array}
$

}

\bsk

%  _____ 
% |___ / 
%   |_ \ 
%  ___) |
% |____/ 
%        

3) Seja $E = e^{iθ}$. EntÃo

$\begin{array}[t]{rcl}
 \cosθ     &=& \frac {E + EÂ} {2}, \\
 (\cosθ)^2 &=& \frac {E^2 + 2 + E^{-2}} {4}, \\
 (\cosθ)^4 &=& \frac {E^4 + 4E^2 + 6 + 4E^{-2} + E^{-4}} {16} \\
           &=& \frac 1 8 \frac{E^4 + E^{-4}}{2} + \frac 1 2 \frac{E^2 + E^{-2}}{2} + \frac 3 8 \\
           &=& \frac 1 8 \cos 4θ + \frac 1 2 \cos 2θ + \frac 3 8 \\
 \intth {(\cosθ)^4} &=& \intth {\frac 1 8 \cos 4θ + \frac 1 2 \cos 2θ + \frac 3 8} \\
                    &=& \frac 1 8 \frac{\sen 4θ}{4} + \frac 1 2 \frac{\sen 2θ}{2} + \frac 3 8 θ \\
                    &=& \frac{\sen 4θ}{32} + \frac{\sen 2θ}{4} + \frac 3 8 θ \\
 \end{array}
$

\bsk

%  _  _   
% | || |  
% | || |_ 
% |__   _|
%    |_|  
%         

4) $\frac {x^2} {x^2 + x - 2} =
    \frac {(x^2 + x - 2) - x + 2} {x^2 + x - 2} =
    1 + \frac {- x + 2} {x^2 + x - 2} =
    1 + \frac {- x + 2} {(x+2)(x-1)}
   $

Se $\frac {- x + 2} {(x+2)(x-1)} = \frac A {x+2} + \frac B {x-1}$ entÃo

$\lim_{xâ-2} \frac {-x+2} {x-1} = A = \frac 4 {-3}$ e 

$\lim_{xâ1} \frac {-x+2} {x+2} = B = \frac 1 3$.

Conferindo:

$\begin{array}[t]{rcl}
 \frac A {x+2} + \frac B {x-1} &=& \frac {-4/3} {x+2} + \frac {1/3} {x-1} \\
                               &=& \frac {- \frac 4 3 (x-1) + \frac 1 3 (x+2)} {(x+2)(x-1)} \\
                               &=& \frac {-x+2} {(x+2)(x-1)} \\
 \end{array}
$

DaÃ:

$\begin{array}[t]{rcl}
 \intx {\frac {x^2} {x^2 + x - 2}} &=& \intx {1 + {-4/3} {x+2} + \frac {1/3} {x-1}} \\
                                   &=& x - \frac 4 3 \ln |x+2| + \frac 1 3 \ln |x-1| \\
 \end{array}
$


\newpage

%  ____  
% | ___| 
% |___ \ 
%  ___) |
% |____/ 
%        

\def\uf #1{\underbrace {#1}_{f}}
\def\uff#1{\underbrace {#1}_{f'}}
\def\ug #1{\underbrace {#1}_{g}}
\def\ugg#1{\underbrace {#1}_{g'}}

5) Sejam $c = \cos x$, $s = \sen x$. EntÃo

$\intx {\uf{e^x} \ugg{c}} = \uf{e^x} \ug{s} - \intx {\uff{e^x} \ug{s}}$

$\intx {\uf{e^x} \ugg{s}} = \uf{e^x} \ug{(-c)} - \intx {\uff{e^x} \ug{(-c)}} = -e^x c + \intx {e^x c}$

\msk

$\intx {e^x c} = e^x s - \intx {e^x s} = e^x s - (-e^x c + \intx {e^x c})
               = e^x s + e^x c - \intx {e^x c}
$

$2 \intx {e^x c} = e^x s + e^x c$

$\intx {e^x c} = (e^x s + e^x c)/2$

\msk

$\intx {e^x s} = -e^x c + \intx {e^x c} = -e^x c + (e^x s + e^x c)/2 = (e^x s - e^x c)/2$

\msk

$\begin{array}{rcl}
 \intx {\uf{x} \ugg{e^x c}}
    &=& \uf{x} \ug{(e^x s + e^x c)/2} - \intx {\uff{1} \ug{(e^x s + e^x c)/2}} \\
    &=& \frac12 (x e^x s + x e^x c) - \frac12 \intx {e^x s} - \frac12 \intx{e^x c} \\
    &=& \frac12 (x e^x s + x e^x c) - \frac14 (e^x s - e^x c) - \frac14 (e^x s + e^x c) \\
    &=& \frac12 (x e^x s + x e^x c) - \frac12 e^x s \\
    &=& \frac12 (x e^x s + x e^x c - e^x s) \\
 \end{array}
$


\bsk

%   __   
%  / /_  
% | '_ \ 
% | (_) |
%  \___/ 
%        

6) $\begin{array}[t]{rclcl}
    \sqrt{4 - x^2} &=& \sqrt{4 - 4(x/2)^2} \\
                   &=& \sqrt{4(1 - (x/2)^2)} \\
                   &=& 2 \sqrt{1 - (x/2)^2} \\
    \intx {\sqrt{4 - x^2}} &=& \intx {2 \sqrt{1 - (x/2)^2}} && \bsm{s = x/2 \\ x = 2s \\ dx = 2ds} \\
                           &=& \ints {2 \sqrt{1 - s^2} Â 2} \\
                           &=& 4 \ints {\sqrt{1 - s^2}} &&
                                 \bsm{s = \senθ \\ θ = \arcsen s \\ ds = \cosθ  dθ} \\
                           &=& 4 \intth {\sqrt{1 - (\senθ)^2} \cosθ} \\
                           &=& 4 \intth {(\cosθ)^2} \\
                           &=& 4 \intth {\frac {1 + \cos 2θ} 2 } \\
                           &=& 4 ({\frac θ2 + \frac {\sen 2θ} 4 }) \\
                           &=& 2θ + \sen 2θ \\
    \end{array}
   $


\bsk

%  _____ 
% |___  |
%    / / 
%   / /  
%  /_/   
%        

7) $\begin{array}[t]{rclcl}
    \Intx{-1}{2} {|e^x-1|} &=& \Intx{-1}{0} {|e^x-1|} + \Intx{0}{2} {|e^x-1|} \\
                           &=& \Intx{-1}{0} {1-e^x} + \Intx{0}{2} {e^x-1} \\
                           &=& \difx{-1}{0} {(x-e^x)} + \difx{0}{2} {(e^x-x)} \\
                           &=& (0-e^0) - (-1-e^{-1}) + (e^2-2) - (e^0-0) \\
                           &=& -1 + 1 + e^{-1} + e^2 - 2 - 1 \\
                           &=& -3 + e^{-1} + e^2 \\
    \end{array}
   $




\end{document}

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