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\begin{document}

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Cálculo II

PURO-UFF

Notas sobre duas técnicas de integração:

substituição trigonométrica e frações parciais

Prof: Eduardo Ochs

8/maio/2009

\bsk


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% (find-kopkadaly4text            "\nStraight lines\n")

\def\sen{\operatorname{sen}}
\def\sec{\operatorname{sec}}
\def\bhbox{}

Abreviações: quando $$ é uma variável,

$s = \sen $

$c = \cos $

$t = \tan  = \frac{\sen }{\cos } = \frac{s}{c}$

$z = \sec  = \frac{1}{\cos } = \frac{1}{c}$

\msk

Identidades:

$t^2 = \frac{s^2}{c^2} = \frac{1 - c^2}{c^2}$

$t^2c^2 = 1 - c^2$

$t^2c^2 + c^2 = 1$

$(1 + t^2)c^2 = 1$

$1 + t^2 = \frac{1}{c^2} = z^2$

$z^2 = 1 + t^2$

$z = \sqrt{1 + t^2}$

$t^2 = z^2 - 1$

$t = \sqrt{z^2 - 1}$

\msk

Derivadas e diferenciais:

$\frac{ds}{d} = \frac{d\sen}{d} = \cos  = c$

$\frac{dt}{d} = \frac{d}{d} \frac{s}{c}
               = \frac{s'c - sc'}{c^2}
               = \frac{c^2 + s^2}{c^2}
               = \frac{1}{c^2}
               = z^2
               = 1 + t^2
$

$\frac{dz}{d} = \frac{d}{d}c^{-1} = -c^{-2}c' = -c^{-2}(-s)
               = \frac{1}{c} \frac{s}{c} = zt$

$ds = c \, d = \sqrt{1 - s^2}d$

$dt = z^2 d = (1 + t^2) d$

$dz = zt\, d$

\newpage

Caso 1:

\bhbox{%
\setlength{\unitlength}{1cm}%
\begin{picture}(5,3)(0,-1)
  \thicklines
  \put(0.8,0.1){$\theta$}
  \put(0,0){\line(2,1){4}}
  \put(0,0){\line(1,0){4}}
  \put(4,0){\line(0,1){2}}
  \put(1.9,1.4){1}
  \put(4,0.9){$\begin{array}{c}
    \sen  \\ = s
    \end{array}$}
  \put(1.4,-.6){$\begin{array}{c}
    \cos  = c = \\ \sqrt{1-s^2}
    \end{array}$}
\end{picture}%
}

Caso 2:

\bhbox{%
\setlength{\unitlength}{1cm}%
\begin{picture}(5.7,3.1)(0,-0.6)
  \thicklines
  \put(0.8,0.1){$\theta$}
  \put(0,0){\line(2,1){4}}
  \put(0,0){\line(1,0){4}}
  \put(4,0){\line(0,1){2}}
  \put(1.1,1.7){$\begin{array}{l}
    \frac{1}{\cos } = \\ \sec  = z = \\ \sqrt{1 + t^2}
    \end{array}$}
  \put(4,0.9){$\begin{array}{l}
    \frac{\sen }{\cos } = \\ \, \tan  = t
    \end{array}$}
  \put(1.4,-.4){$\begin{array}{c}
    \frac{\cos }{\cos } = 1
    \end{array}$}
\end{picture}%
}

\msk

Caso 3:

\bhbox{%
\setlength{\unitlength}{1cm}%
\begin{picture}(5.9,2.8)(0,-0.6)
  \thicklines
  \put(0.8,0.1){$\theta$}
  \put(0,0){\line(2,1){4}}
  \put(0,0){\line(1,0){4}}
  \put(4,0){\line(0,1){2}}
  \put(0.9,1.6){$\begin{array}{r}
    \frac{1}{\cos } = \\ \sec  = z
    \end{array}$}
  \put(4,0.9){$\begin{array}{l}
    \frac{\sen }{\cos } = \\ \, \tan  = t \\ \, = \sqrt{z^2 - 1}
    \end{array}$}
  \put(1.4,-.4){$\begin{array}{c}
    \frac{\cos }{\cos } = 1
    \end{array}$}
\end{picture}%
}

\newpage

Exemplos (adaptados do Munem, pp.492--493):

\begin{eqnarray*}
 \int \frac{s^2}{(1-s^2)^{3/2}} \, ds
   & = & \int \frac{s^2}{(1-s^2)^{3/2}} (1-s^2)^{1/2} \, d \\
   & = & \int \frac{s^2}{1-s^2} \, d \\
   & = & \int \frac{(\sen )^2}{(\cos )^2} \, d \\
   & = & \int (\tan )^2 \, d \\
\end{eqnarray*}

\begin{eqnarray*}
 \int \frac{1}{t^2 \sqrt{1 + t^2}} \, dt
   & = & \int \frac{1}{t^2 \sqrt{1 + t^2}} (1 + t^2) \, d \\
   & = & \int \frac{\sqrt{1 + t^2}}{t^2} \, d \\
   & = & \int \frac{(1/c)}{(s^2/c^2)} \, d
     =   \int \frac{1}{c} \frac{c^2}{s^2} \, d
     =   \int \frac{c}{s^2} \, d \\
\end{eqnarray*}

\begin{eqnarray*}
 \int \frac{1}{z^3 \sqrt{z^2 - 1}} \, dz
   & = & \int \frac{1}{z^3 \sqrt{z^2 - 1}} zt \, d \\
   & = & \int \frac{zt}{z^3 t} \, d \\
   & = & \int \frac{1}{z2} \, d \\
   & = & \int c^2 \, d \\
\end{eqnarray*}


\newpage

\def\qv#1{\left[\begin{matrix}#1\end{matrix}\right]}
\def\sqv#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}
\def\displayfrac{\displaystyle\frac}


Um exercício de frações parciais:

usando esta notação,

$\sqv{a_2 \\ a_1 \\ a_0} = [a_2,a_1,a_0] = a_2x^2 + a_1x + a_0$,

$\sqv{a_3 \\ a_2 \\ a_1 \\ a_0} = [a_3,a_2,a_1,a_0] = a_3x^3 + a_2x^2 + a_1x + a_0$, etc,

descubra que valores de $k_1, k_0, c_1, c_2$ e $c_3$ fazem a conta abaixo fazer sentido:

$$\begin{array}{rcl}
  \multicolumn{3}{l}{
    \displaystyle
    k_1x + k_0 + \frac{c_1}{x+2} + \frac{c_2}{x+1} + \frac{c_3}{(x+1)^2} =
  } \\ \\
  \qquad\qquad\qquad
  & = & \displaystyle\frac{
           k_1 \qv{1 \\ 5 \\ 7 \\ 3 \\ 0}
         + k_0 \qv{0 \\ 1 \\ 5 \\ 7 \\ 3}
         + c_1 \qv{0 \\ 0 \\ 1 \\ 2 \\ 1}
         + c_2 \qv{0 \\ 0 \\ 1 \\ 3 \\ 2}
         + c_3 \qv{0 \\ 0 \\ 0 \\ 1 \\ 2}
        }{ [1, 5, 7, 3]
        } \\ \\
   & = & \displaystyle\frac{
            \begin{array}{crl}
                & 10000 & x^4 \\
              + & 51000 & x^3 \\
              + & 75110 & x^2 \\
              + & 37231 & x   \\
              + &  3122 &     \\
            \end{array}
         }{ x^3 + 5x^2 + 7x + 3 }
  \end{array}
$$

E agora encontre uma primitiva para:
%
$$\int \frac{10000 x^4 +
             51000 x^3 +
             75110 x^2 +
             37231 x   +
              3122} 
            { x^3 + 5x^2 + 7x + 3 } \, dx
$$





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