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{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2017.1
\par P1 - 17/jul/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.

}

\bsk
\bsk

\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


\bsk
\bsk

% (c2q)

1) \T(Total: 1.0 pts) Demonstre que $\ddx \arctan x = \frac {1}
{x^2+1}$. Você vai precisar do truque pra derivar funções inversas ---
se $f(g(x))=x$ então $\ddx(f(g(x))) = \ddx x = 1$ --- e de um outro
truque extra... dica: na demonstração de que $\ddx \arcsen x = \frac
{1} {\sqrt{1-x^2}}$ o truque extra é $\cos^2 x + \sen^2 x = 1$.

\bsk

2) \T(Total: 1.0 pts) Sejam $f(x) = x^{-4}$ e $F(x) =
\frac{x^{-3}}{-3}$.

a) \B(0.1 pts) Mostre que a área $\Intx{-1}{1}{f(x)}$ é positiva.

b) \B(0.2 pts) Calcule $\Difx{-1}{1}{F(x)}$.

c) \B(0.3 pts) Calcule $G(ε) = \Intx{-1}{-ε}{f(x)} + \Intx{ε}{1}{f(x)}$. Obs: $ε>0$.

d) \B(0.4 pts) Calcule $\lim_{ε→0^+} G(ε)$.

\bsk



3) \T(Total: 2.0 pts) Calcule $$\Intx{-1}{2}{\frac{1}{3+|x|}}.$$

4) \T(Total: 2.0 pts) Calcule $$\intx {\frac{x^2}{x^2 + 4x - 5}}.$$

5) \T(Total: 2.0 pts) Calcule $$\intx {\frac{x}{1+x^2}}.$$

% 6) \T(Total: 1.0 pts) Calcule $$\intx {(2x+3) \sen (4x+5)}.$$
6) \T(Total: 2.0 pts) Calcule $$\intx {(\cos x)^4}.$$



\bsk
\bsk

Dicas:
$\bsm{s=\senθ \\ \sqrt{1-s^2}=\cosθ   \\ ds=\cosθ\,dθ \\ θ=\arcsen s}$,
$\bsm{t=\tanθ \\ \sqrt{1+t^2}=\secθ=z \\ dt=  z^2\,dθ \\ θ=\arctan t}$,
$\bsm{z=\secθ \\ \sqrt{z^2-1}=\tanθ=t \\ dz=   zt\,dθ \\ θ=\arcsec z}$,

$E = e^{iθ} = \cosθ+i\senθ = c + is$,

$c=(E+E^{-1})/2$,

$s=(E-E^{-1})/2i$.



\newpage

{\bf Gabarito parcial:}

% (find-es "ipython" "2017.1-C2-P1")


1) $\begin{array}[t]{l}
    \ddx \arctan(\tan x) = \ddx x = 1 \\
    \ddx \arctan(\tan x) = \arctan'(\tan x) \tan' x \\
    \arctan'(\tan x) = \frac 1 {\tan' x} \\
    \ddth \tanθ = \ddth \frac sc = \frac{s'c-sc'}{c^2} = \frac{c^2+s^2}{c^2} = \frac{c^2}{c^2} + \frac{s^2}{c^2} = 1 + t^2 \\
    \ddx \tan x = (\tan x)^2 + 1 \\
    \arctan'(\tan x) = \frac 1 {(\tan x)^2 + 1} \\
    \arctan'(t) = \frac 1 {t^2 + 1} \\
    \arctan'(x) = \frac 1 {x^2 + 1} \\
    \end{array}
   $

\bsk

2a) (um desenho)

2b) $\Difx{-1}{1}{F(x)} = F(1)-F(-1)
     = \frac{1^{-3}}{-3} - \frac{(-1)^{-3}}{-3}
     = \frac{1}{-3} - \frac{-1}{-3}
     = \frac{2}{-3}
     = - \frac{2}{3}
    $

2c) $\begin{array}[t]{rcl}
     G(ε) &=& \Difx{-1}{-ε}{F(x)} + \Difx{ε}{1}{F(x)} \\
          &=& F(-ε)-F(-1) + F(1)-F(ε) \\
          &=& \frac{(-ε)^{-3}}{-3} - \frac{(-1)^{-3}}{-3} + \frac{1^{-3}}{-3} - \frac{ε^{-3}}{-3} \\
          &=& \frac{(-ε)^{-3} + 1 + 1 - ε^{-3}}{-3} \\
          &=& \frac{2 - 2(ε^{-3})}{-3} \\
          &=& \frac{2(ε^{-3}) - 2}{3} \\
          &=& \frac{2}{3}(ε^{-3} - 1) \\
     \end{array}
    $

2d) $\lim_{ε→0^+} G(ε) = \frac{2}{3}(\lim_{ε→0^+} ε^{-3} - \lim_{ε→0^+} 1) = \frac23(+∞-1) = +∞$ 

\bsk

\def\mycases#1#2{\begin{cases}
    #1 & \text{quando $x<0$} \\
    #2 & \text{quando $x≥0$} \\
 \end{cases}
}

3) Seja $f(x) = \frac{1}{3+|x|}$. Então $f$ é contínua e:
%
$$f(x) = \mycases{\frac{1}{3+|x|}}{\frac{1}{3+|x|}}
       = \mycases{\frac{1}{3-x}}{\frac{1}{3+x}}
       = \mycases{-\frac{1}{x-3}}{\frac{1}{x+3}}
$$
$$\begin{array}{rcl}
    \Intx{-1}{2}{f(x)} &=& \Intx{-1}{0}{-\frac{1}{x-3}} + \Intx{0}{2}{\frac{1}{x+3}} \\ 
                       &=& \Difx{-1}{0}{(-\ln|x-3|)} + \Difx{0}{2}{\ln|x+3|} \\
                       &=& (-\ln|-1-3|) - (-\ln|0-3|) + (-\ln|2+3|) - (-\ln|0+3|) \\
                       &=& (-\ln 4) - (-\ln 3) + (- \ln 5) - (-\ln 3) \\
                       &=& 2\ln 3 - \ln 4 - \ln 5 \\
  \end{array}
$$

\msk

4) $\intx {\frac{x^2}{x^2 + 4x - 5}} =
    \intx {1 + \frac{1}{6}\frac{1}{x-1} - \frac{25}{6}\frac{1}{x+5} } = 
    x + \frac{1}{6}\ln|x-1| - \frac{25}{6}\ln|x+5|
   $






\end{document}

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