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% (find-LATEX "2019-2-C2-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2019-2-C2-P1.tex" :end))
% (defun d () (interactive) (find-pdf-page "~/LATEX/2019-2-C2-P1.pdf"))
% (defun d () (interactive) (find-pdftools-page "~/LATEX/2019-2-C2-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2019-2-C2-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2019-2-C2-P1"))
% (setq revert-without-query '("pdf$"))
% (find-pdf-page "~/LATEX/2019-2-C2-P1.pdf")
% (find-sh0 "cp -v ~/LATEX/2019-2-C2-P1.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2019-2-C2-P1.pdf /tmp/pen/")
% file:///home/edrx/LATEX/2019-2-C2-P1.pdf
% file:///tmp/2019-2-C2-P1.pdf
% file:///tmp/pen/2019-2-C2-P1.pdf
% http://angg.twu.net/LATEX/2019-2-C2-P1.pdf
% (find-LATEX "2019.mk")
% (find-CN-aula-links "2019-2-C2-P1" "2" "c2m192p1" "c2p1o")
% «.gabarito» (to "gabarito")
% «.gabarito-maxima» (to "gabarito-maxima")
\documentclass[oneside]{book}
\usepackage[colorlinks,urlcolor=DarkRed]{hyperref} % (find-es "tex" "hyperref")
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{pict2e}
\usepackage[x11names,svgnames]{xcolor} % (find-es "tex" "xcolor")
%\usepackage{colorweb} % (find-es "tex" "colorweb")
%\usepackage{tikz}
%
% (find-dn6 "preamble6.lua" "preamble0")
%\usepackage{proof} % For derivation trees ("%:" lines)
%\input diagxy % For 2D diagrams ("%D" lines)
%\xyoption{curve} % For the ".curve=" feature in 2D diagrams
%
\usepackage{edrx15} % (find-LATEX "edrx15.sty")
\input edrxaccents.tex % (find-LATEX "edrxaccents.tex")
\input edrxchars.tex % (find-LATEX "edrxchars.tex")
\input edrxheadfoot.tex % (find-LATEX "edrxheadfoot.tex")
\input edrxgac2.tex % (find-LATEX "edrxgac2.tex")
%
% (find-es "tex" "geometry")
\begin{document}
\catcode`\^^J=10
\directlua{dofile "dednat6load.lua"} % (find-LATEX "dednat6load.lua")
% %L dofile "edrxtikz.lua" -- (find-LATEX "edrxtikz.lua")
% %L dofile "edrxpict.lua" -- (find-LATEX "edrxpict.lua")
% \pu
\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)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar
% _____
% |_ _| _ _ __ _ __ ___ __ _ _ __ ___ __ _
% | || | | | '__| '_ ` _ \ / _` | | '_ \ / _ \/ _` |
% | || |_| | | | | | | | | (_| | | |_) | __/ (_| |
% |_| \__,_|_| |_| |_| |_|\__,_| | .__/ \___|\__, |
% |_| |_|
{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2019.2
\par P1 - turma pequena - 30/outubro/2019 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.
}
\bsk
\bsk
% \bsk
% \bsk
% (find-TH "2019.1-C2" "provas-antigas")
% (find-es "sympy" "trig-subst-questions")
% (c2q)
1) \T(Total: 2.5 pts) Calcule $$\intx {(2x+3)\sqrt{4x+5}}.$$
\bsk
2) \T(Total: 2.5 pts) Calcule $$\intx {x^3 \sqrt{1-x^2}}.$$
\bsk
3) \T(Total: 2.5 pts) Calcule $$\intx {(\sen 5x)^2(\cos 6x)^2}.$$
\bsk
4) \T(Total: 2.5 pts) Calcule $$\intx {\frac{x^3}{x^2 + 9x + 20}}.$$
\bsk
\bsk
Algumas definições, fórmulas e substituições:
$\begin{array}[t]{l}
c = \cos θ \\
s = \sen θ \\
t = \tan θ \\
z = \sec θ \\
E = e^{iθ} \\
\end{array}
%
\begin{array}[t]{l}
c^2+s^2=1 \\
z^2=t^2+1 \\
\sqrt{1-s^2} = c \\
\sqrt{t^2+1} = z \\
\sqrt{z^2-1} = t \\
\end{array}
%
\begin{array}[t]{l}
\frac{ds}{dθ} = c \\
\frac{dc}{dθ} = -s \\
\frac{dt}{dθ} = z^2 \\
\frac{dz}{dθ} = zt \\
\end{array}
%
\begin{array}[t]{l}
E = c+is \\
c = \frac{E+E¹}{2} \\
s = \frac{E-E¹}{2i} \\
e^{ikθ} + e^{-ikθ} = 2 \cos kθ \\
e^{ikθ} - e^{-ikθ} = 2i \sen kθ \\
\end{array}
$
\newpage
% _____ _
% |_ _| _ _ __ _ __ ___ __ _ __ _ __| | ___
% | || | | | '__| '_ ` _ \ / _` | / _` |/ _` |/ _ \
% | || |_| | | | | | | | | (_| | | (_| | (_| | __/
% |_| \__,_|_| |_| |_| |_|\__,_| \__, |\__,_|\___|
% |___/
{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2019.1
\par P1 - turma grande - 31/outubro/2019 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.
}
\bsk
\bsk
1) \T(Total: 2.5 pts) Calcule $$\intx {(2x+3)\sqrt{4x+5}}.$$
\bsk
2) \T(Total: 2.5 pts) Calcule $$\intx {x^3 \sqrt{1-x^2}}.$$
\bsk
3) \T(Total: 2.5 pts) Calcule $$\intx {(\sen 5x)^2(\cos 6x)^2}.$$
\bsk
4) \T(Total: 2.5 pts) Calcule $$\intx {\frac{x^3}{x^2 + 9x + 20}}.$$
\bsk
\bsk
Algumas definições, fórmulas e substituições:
$\begin{array}[t]{l}
c = \cos θ \\
s = \sen θ \\
t = \tan θ \\
z = \sec θ \\
E = e^{iθ} \\
\end{array}
%
\begin{array}[t]{l}
c^2+s^2=1 \\
z^2=t^2+1 \\
\sqrt{1-s^2} = c \\
\sqrt{t^2+1} = z \\
\sqrt{z^2-1} = t \\
\end{array}
%
\begin{array}[t]{l}
\frac{ds}{dθ} = c \\
\frac{dc}{dθ} = -s \\
\frac{dt}{dθ} = z^2 \\
\frac{dz}{dθ} = zt \\
\end{array}
%
\begin{array}[t]{l}
E = c+is \\
c = \frac{E+E¹}{2} \\
s = \frac{E-E¹}{2i} \\
e^{ikθ} + e^{-ikθ} = 2 \cos kθ \\
e^{ikθ} - e^{-ikθ} = 2i \sen kθ \\
\end{array}
$
\newpage
% _____ _
% |_ _| _ _ __ _ __ ___ __ _ __ _ __| | ___
% | || | | | '__| '_ ` _ \ / _` | / _` |/ _` |/ _ \
% | || |_| | | | | | | | | (_| | | (_| | (_| | __/
% |_| \__,_|_| |_| |_| |_|\__,_| \__, |\__,_|\___|
% |___/
{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2019.1
\par P1 - versão pra Thais Knupp - 6/novembro/2019 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.
}
\bsk
\bsk
1) \T(Total: 2.5 pts) Calcule $$\intx {(2+3x)\sqrt{4+5x}}.$$
\bsk
2) \T(Total: 2.5 pts) Calcule $$\intx {x^3 \sqrt{1-x^2}}.$$
\bsk
3) \T(Total: 2.5 pts) Calcule $$\intx {(\sen 4x)^2(\cos 5x)^2}.$$
\bsk
4) \T(Total: 2.5 pts) Calcule $$\intx {\frac{x^3}{x^2 + 7x + 12}}.$$
\bsk
\bsk
Algumas definições, fórmulas e substituições:
$\begin{array}[t]{l}
c = \cos θ \\
s = \sen θ \\
t = \tan θ \\
z = \sec θ \\
E = e^{iθ} \\
\end{array}
%
\begin{array}[t]{l}
c^2+s^2=1 \\
z^2=t^2+1 \\
\sqrt{1-s^2} = c \\
\sqrt{t^2+1} = z \\
\sqrt{z^2-1} = t \\
\end{array}
%
\begin{array}[t]{l}
\frac{ds}{dθ} = c \\
\frac{dc}{dθ} = -s \\
\frac{dt}{dθ} = z^2 \\
\frac{dz}{dθ} = zt \\
\end{array}
%
\begin{array}[t]{l}
E = c+is \\
c = \frac{E+E¹}{2} \\
s = \frac{E-E¹}{2i} \\
e^{ikθ} + e^{-ikθ} = 2 \cos kθ \\
e^{ikθ} - e^{-ikθ} = 2i \sen kθ \\
\end{array}
$
\newpage
% «gabarito» (to ".gabarito")
% (c2m192p1p 4 "gabarito")
% (c2m192p1a "gabarito")
Gabarito (incompleto e não revisado):
\def\d{\displaystyle}
\def\t{\textstyle}
\bsk
% Versão da prova da Thais:
%
% 1)
% $\begin{array}[t]{rcll}
% \d\intx {(2+3x)\sqrt{4+5x}}
% &=& \d\intu {\t (2+3(\frac u5 - \frac 45)) \sqrt{u} (\frac15)}
% & \subst{u = 4+5x \\ x=\frac{u-4}5 \\ dx=\frac15 du} \\
% &=& \d\intu {\t (\frac15) (\frac35u + \frac{10}5 - \frac{12}5) \sqrt{u}} \\
% &=& \d\intu {\t (\frac3{25} u - \frac2{25}) \sqrt{u}} \\
% &=& \d\intu {\t \frac3{25} u^{3/2} - \frac2{25} u^{1/2}} \\
% &=& \frac3{25} \frac52 u^{5/2} - \frac2{25} \frac23 u^{3/2} \\
% &=& \frac3{10} u^{5/2} - \frac4{75} u^{3/2} \\
% &=& \frac3{10} (4+5x)^{5/2} - \frac4{75} (4+5x)^{3/2} \\
% \end{array}
% $
% (find-es "ipython" "2019.2-C2-P1")
% (find-es "ipython" "2019.2-C2-P1" "Questao 1")
1)
$\begin{array}[t]{rcll}
\d\intx {(2x+3)\sqrt{4x+5}}
&=& \d\intu {\t (2(\frac u4 - \frac 54)+3) \sqrt{u} (\frac14)}
& \subst{u = 4x+5 \\ x=\frac{u-5}4 \\ dx=\frac14 du} \\
&=& \d\intu {\t (\frac14) (\frac{2u}4 - \frac{10}4 - \frac{12}4) \sqrt{u}} \\
&=& \d\intu {\t (\frac{u}8 - \frac{11}8) \sqrt{u}} \\
&=& \d\intu {\t \frac1{8} u^{3/2} - \frac{11}8 u^{1/2}} \\
&=& \frac18 \frac52 u^{5/2} - \frac{11}8 \frac23 u^{3/2} \\
&=& \frac5{16} u^{5/2} - \frac{11}{12} u^{3/2} \\
&=& \frac5{16} (4x+5)^{5/2} - \frac{11}{12} (4x+5)^{3/2} \\
\end{array}
$
% (xz "~/tmp/2019nov18_gab_c2_p1_1.jpg")
% (xz "~/tmp/2019nov18_gab_c2_p1_2.jpg")
\bsk
2) $\begin{array}[t]{lll}
\intx {x^3 \sqrt{1-x^2}} & \subst{s=x} \\
= \;\; \ints {s^3 \sqrt{1-s^2}} & \subst{s=\senθ \\ ds = \cosθdθ \\ \sqrt{1-s^2} = \cosθ} \\
= \;\; \ints {(\senθ)^3 \cosθ\,\cosθ} \\
= \;\; \ints {(\senθ)^2 (\cosθ)^2 \senθ}
& \subst{\cosθ=c \\ (\senθ)^2=1-c^2 \\ \senθdθ=(-1)dc} \\
= \;\; \intc {(1-c^2)c^2(-1)} \\
= \;\; \intc {c^4-c^2} \\
= \;\; \frac15 c^5 - \frac13 c^3 \\
= \;\; \frac15 {\sqrt{1-s^2}}^5 - \frac13 {\sqrt{1-s^2}}^3 \\
= \;\; \frac15 {\sqrt{1-x^2}}^5 - \frac13 {\sqrt{1-x^2}}^3 \\
\end{array}
$
\bsk
3) $\def\E#1 {E^{#1}}
\begin{array}[t]{lll}
(\sen 5θ)^2 (\cos 6θ)^2 \\
= \;\; \left(\frac{\E5 - \E-5 }{2i}\right)^2
\left(\frac{\E6 + \E-6 }{2}\right)^2 \\
= \;\; -\frac1{16}(\E10 - 2 +\E-10 )(\E12 + 2 +\E-12 ) \\
= \;\; -\frac1{16}\pmat{
(\E10 - 2 +\E-10 ) \E12 + \\
(\E10 - 2 +\E-10 ) · 2 + \\
(\E10 - 2 +\E-10 ) \E-12 \\
} \\
= \;\; -\frac1{16}\pmat{
\E22 - 2\E12 +\E2 + \\
2\E10 - 4 +2\E-10 + \\
\E-2 - 2\E-12 +\E-22 \\
} \\
= \;\; -\frac1{16}( (\E22 + \E-22 )
-2 (\E12 + \E-12 )
+2 (\E10 + \E-10 )
+ (\E2 + \E-2 )
- 4
) \\
= \;\; -\frac1{16}( 2\cos22θ
-4\cos12θ
+4\cos10θ
+2\cos2θ
- 4
) \\
= \;\; -\frac18 \cos22θ
+\frac14 \cos12θ
-\frac14 \cos10θ
-\frac18 \cos2θ
\frac14 \\
\end{array}
$
$$\begin{array}[t]{rcl}
\intx {(\sen 5x)^2 (\cos 6x)^2}
&=& \intx {-\frac18 \cos22x
+\frac14 \cos12x
-\frac14 \cos10x
-\frac18 \cos2x
+\frac14} \\
&=& -\frac1{8·22} \sen22x
+\frac1{4·12} \sen12x
-\frac1{4·10} \sen10x
-\frac1{8·2} \sen2x
+\frac14x \\
\end{array}
$$
\newpage
% (find-es "ipython" "2019.2-C2-P1")
4) $\begin{array}[t]{rcl}
\d\intx {\frac{x^3}{x^2 + 7x + 12}}
&=& \d\intx {x - 7 + \frac{37x + 84}{x^2 + 7x + 12}} \\
&=& \d\intx {x - 7 + \frac{37x + 84}{(x+3)(x+4)}} \\
&=& \d\intx {x - 7 - \frac{27}{x+3} + \frac{64}{x+4}} \\
&=& \d\frac{x^2}{2} - 7x + 27\ln|x+3| + 64\ln|x+4| \\
\end{array}
$
% 3
% x
% -------------
% 2
% x + 7*x + 12
%
% 37*x + 84
% x - 7 + ---------------
% (x + 3)*(x + 4)
%
% 64 27
% x - 7 + ----- - -----
% x + 4 x + 3
% «gabarito-maxima» (to ".gabarito-maxima")
% (c2m192p1p 4 "gabarito-maxima")
% (c2m192p1a "gabarito-maxima")
% (find-es "maxima" "trig-ids")
% (setq eepitch-preprocess-regexp "^")
% (setq eepitch-preprocess-regexp "^% ")
%
% * (eepitch-maxima)
% * (eepitch-kill)
% * (eepitch-maxima)
% load("/usr/share/emacs/site-lisp/maxima/emaxima.lisp")$
% display2d:'emaxima$
% **
% ** Questao 1:
% **
% f : (2*x + 3) * sqrt(4*x + 5);
% F : integrate (f, x);
% r : (4*x + 5)^(3/2);
% F / r;
% expand(F / r);
% ratsimp(expand(F / r));
% F2 : ratsimp(expand(F / r)) * r;
% ratsimp(F2 - F);
% **
% ** Questao 2:
% **
% f : x^3 * sqrt(1 - x^2);
% F : integrate (f, x);
% r : (1 - x^2)^(3/2);
% F2 : ratsimp(expand(F / r)) * r;
% ratsimp(F2 - F);
% **
% ** Questao 3:
% **
% f : sin(5*x)^2 * cos(6*x)^2;
% g : expand(demoivre(expand(exponentialize(f))));
% G : integrate(g, x);
% **
% ** Questao 4:
% **
% f : x^3 / (x^2 + 9*x + 20);
% g : partfrac(f, x);
% G : integrate(g, x);
\end{document}
% Local Variables:
% coding: utf-8-unix
% ee-tla: "c2p1"
% End: