% (find-LATEX "2020-1-C2-int-pots-sen-cos.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2020-1-C2-int-pots-sen-cos.tex" :end))
% (defun D () (interactive) (find-pdf-page      "~/LATEX/2020-1-C2-int-pots-sen-cos.pdf"))
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% (defun v () (interactive) (find-2a '(e) '(d)) (g))
% (find-pdf-page   "~/LATEX/2020-1-C2-int-pots-sen-cos.pdf")
% (find-sh0 "cp -v  ~/LATEX/2020-1-C2-int-pots-sen-cos.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2020-1-C2-int-pots-sen-cos.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2020-1-C2-int-pots-sen-cos.pdf
%               file:///tmp/2020-1-C2-int-pots-sen-cos.pdf
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% http://angg.twu.net/LATEX/2020-1-C2-int-pots-sen-cos.pdf
% (find-LATEX "2019.mk")
% (find-C2-aula-links "2020-1-C2-int-pots-sen-cos" "18" "ipsc")

% «.exemplo-1»		(to "exemplo-1")
% «.exemplo-1-verif»	(to "exemplo-1-verif")
% «.exemplo-2»		(to "exemplo-2")
% «.exercicio-1»	(to "exercicio-1")
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\usepackage{amsfonts}
\usepackage{amssymb}
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\begin{document}

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% (find-angg ".emacs" "c2q192")


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% «title»  (to ".title")

\thispagestyle{empty}

\begin{center}

\vspace*{1.2cm}

{\bf \Large Cálculo 2 - 2020.1}

\bsk

Aula 18: Integrais de potências de senos e cossenos

\bsk

Eduardo Ochs - RCN/PURO/UFF

\url{http://angg.twu.net/2020.1-C2.html}

\end{center}

\newpage

% «exemplo-1»  (to ".exemplo-1")
% (c2m201ipscp 2 "exemplo-1")
% (c2m201ipsca   "exemplo-1")
% (c2q192 32 "20190906 gde aula 7: integração por partes; potências de sen e cos")
% (c2q192 34 "20190906 peq aula 7: integração por partes; potências de sen e cos")
% (c2q192 35 "20190906 peq aula 7: integração por partes; potências de sen e cos")

\def\S{\sen x}
\def\C{\cos x}
\def\D{\displaystyle}
\def\und#1#2{\underbrace{#1}_{#2}}

{\bf Exemplo 1}
%
$$\begin{array}[t]{l}
  \D \intx{(\S)^5 (\C)^3} \\
  \D = \;\; \intx{(\S)^5 (\C)^2 (\C)} \\
  \D = \;\; \intx{(\und{\S}{s})^5 \und{(\C)^2}{1-s^2} \und{(\C)}{\frac{ds}{dx}}} \\
  \D = \;\; \ints{s^5 (1-s^2)} \\
  \D = \;\; \ints{s^5 - s^7} \\
  \D = \;\; \frac{s^6}{6} - \frac{s^8}{8} \\
  \D = \;\; \frac{(\S)^6}{6} - \frac{(\S)^8}{8} \\
  \end{array}
  \qquad
  \begin{array}[t]{c}
  \\ \\
    \bmat{s = \sen x \\
          \frac{ds}{dx} = \cos x \\
          \sen x = s \\
          (\cos x)^2 = 1 - s^2 \\
          \cos x \, dx = ds
    }
  \end{array}
$$

\newpage

% «exemplo-1-verif»  (to ".exemplo-1-verif")
% (c2m201ipscp 3 "exemplo-1-verif")
% (c2m201ipsc    "exemplo-1-verif")

{\bf Exemplo 1: verificação}

$$\begin{array}[t]{l}
  \D \ddx \left(\frac{(\S)^6}{6} - \frac{(\S)^7}{7}\right) \\
  \D = \frac{6(\S)^5\C}{6} - \frac{8(\S)^7\C}{8} \\
  \D = ((\S)^5 - (\S)^7) \C \\
  \D = ((\S)^5(1-(\S)^2)) \C \\
  \D = (\S)^5(\C)^2 \C \\
  \D = (\S)^5(\C)^3 \\
  \end{array}
$$

\newpage

% «exemplo-2»  (to ".exemplo-2")
% (c2m201ipscp 4 "exemplo-2")
% (c2m201ipsc    "exemplo-2")

{\bf Exemplo 2}
%
$$\begin{array}[t]{l}
  \D \intx{(\S)^5 (\C)^2} \\
  \D = \;\; \intx{(\S)^4 (\C)^2 \S} \\
  \D = \;\; \intx{(\und{(\S)^2}{1-c^2})^2 \und{(\C)^2}{c^2} \und{(\S)}{-\frac{dc}{dx}}} \\
  \D = \;\; \intc{(1-c^2)^2 c^2} \\
  \D = \;\; \intc{(c^4 - 2c^2 + 1) c^2 (-1)} \\
  \D = \;\; \intc{-c^6 + 2c^4 - c^2} \\
  \D = \;\; \ldots \\
  \end{array}
  \qquad
  \begin{array}[t]{c}
  \\ \\
    \bmat{c = \cos x \\
          \frac{dc}{dx} = - \sen x \\
          \cos x = c \\
          (\sen x)^2 = 1 - c^2 \\
          \sen x \, dx = (-1)\,dc
    }
  \end{array}
$$

\newpage

% «exercicio-1»  (to ".exercicio-1")
% (c2m201ipscp 5 "exercicio-1")
% (c2m201ipsc    "exercicio-1")

{\bf Exercício 1.}

\ssk

a) Calcule a integral do exemplo 1 -- $\D \intx{(\S)^5 (\C)^3}$ --

usando a substituição $c=\cos x$ ao invés de $s=\sen x$.

\msk

b) Teste o seu resultado.



\newpage

% «pares-e-impares»  (to ".pares-e-impares")
% (c2m201ipscp 6 "pares-e-impares")
% (c2m201ipsc    "pares-e-impares")

{\bf Dica importante}

\ssk

Pra integrar algo como:
%
$$ \intx {(\S)^α (\C)^β}
$$

Se tanto $α$ quanto $β$ são ímpares as duas subtituições,

$s=\S$ e $c=\C$, funcionam.

\msk

Se só um dos dois é ímpar só uma delas funciona

(não vou dizer qual).

\msk

Se tanto $α$ quanto $β$ são \ColorRed{pares} aí \ColorRed{nenhuma

das duas substituições funciona}, e a gente vai precisar

de técnicas mais avançadas que vamos ver depois.





%\printbibliography

\end{document}

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% «video»  (to ".video")
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% (code-eevvideo      "c2m201ipcs" "2020_int_pots_sen_cos")
% (code-eevlinksvideo "c2m201ipcs" "2020_int_pots_sen_cos")
% (find-c2m201ipcsvideo "0:00")

 (eepitch-shell)
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cp -v  ~/LATEX/2020-1-C2-int-pots-sen-cos.pdf /tmp/
cd /tmp/
xournalpp 2020-1-C2-int-pots-sen-cos.pdf


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