\( \newcommand{\combin}[2]{{}^{#1}C_{#2} } \newcommand{\cmod}[3]{#1 \equiv #2\left(\bmod {}{#3}\right)} \newcommand{\mdc}[2]{\left( {#1},{#2}\right)} \newcommand{\mmc}[2]{\left[ {#1},{#2}\right]} \newcommand{\cis}{\mathop{\rm cis}} \newcommand{\sen}{\mathop{\rm sen}} \newcommand{\tg}{\mathop{\rm tg}} \newcommand{\cotg}{\mathop{\rm cotg}} \newcommand{\cosec}{\mathop{\rm cosec}} \newcommand{\cotgh}{\mathop{\rm cotgh}} \newcommand{\cosech}{\mathop{\rm cosech}} \newcommand{\sech}{\mathop{\rm sech}} \newcommand{\sh}{\mathop{\rm sh}} \newcommand{\ch}{\mathop{\rm ch}} \newcommand{\th}{\mathop{\rm th}} \newcommand{\senEL}[1]{\mathop{\rm sen}^{#1}} \newcommand{\tgEL}[1]{\mathop{\rm tg}^{#1}} \newcommand{\cotgEL}[1]{\mathop{\rm cotg}^{#1}} \newcommand{\cosecEL}{\mathop{\rm cosec}^{#1}} \newcommand{\shEL}[1]{\mathop{\rm sh^{#1}}} \newcommand{\chEL}[1]{\mathop{\rm ch^{#1}}} \newcommand{\thEL}[1]{\mathop{\rm th^{#1}}} \newcommand{\cotghEL}[1]{\mathop{\rm cotgh^{#1}}} \newcommand{\cosechEL}[1]{\mathop{\rm cosech^{#1}}} \newcommand{\sechEL}[1]{\mathop{\rm sech^{#1}}} \newcommand{\senq}{\senEL{2}} \newcommand{\tgq}{\tgEL{2}} \newcommand{\cotgq}{\cotgEL{2}} \newcommand{\cosecq}{\cosecEL{2}} \newcommand{\cotghq}{\cotghEL{2}} \newcommand{\cosechq}{\cosechEL{2}} \newcommand{\sechq}{\sechEL{2}} \newcommand{\shq}{\shEL{2}} \newcommand{\chq}{\chEL{2}} \newcommand{\arctg}{\mathop{\rm arctg}} \newcommand{\arcsen}{\mathop{\rm arcsen}} \newcommand{\argsh}{\mathop{\rm argsh}} \newcommand{\argch}{\mathop{\rm argch}} \newcommand{\vect}[1]{\overrightarrow{#1}} \newcommand{\tr}[1]{ \textnormal{Tr}\left({#1}\right)} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\H}{\mathbb{H}} \newcommand{\vect}[1]{\overrightarrow{#1}} \newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)} \)

03/09/2017

Integração por substituição: seno hiperbólico (I)

Exercício
Determinar o valor exacto do integral: \[ \int\limits_{ - 1}^1 {\sqrt {1 + x^2 } dx} \] Exprimir o valor na forma $\sqrt{m}+\arg\sinh{(n)}$, com $m,n\in \N$

\[\sqrt {2}+\arg \sinh \left( 1 \right)\]

Começamos por fazer a substituição $x=\sinh u$ então, \begin{eqnarray*} {u}&{=}&{\arg\sinh x}\\ {x}&{=}&{-1\Rightarrow u=\arg\sinh(-1)=-\arg\sinh(1)}\\ {x}&{=}&{1\Rightarrow u=\arg\sinh(1)} \end{eqnarray*} e \[ \frac{{dx}}{{du}} = \cosh u \] Na restante resolução é também útil recordar a fórmula fundamental das funções hiperbólicas \[\cosh^2 u - \sinh^2 u = 1 \] que $\cosh u>0$    $\forall u \in \R$
e as fórmulas \[\cosh^2 u=\frac{1+\cosh(2u)}{2}\] \[\sinh (2 u) =2\sinh u \cosh u\] Assim sendo, temos que: \begin{eqnarray} {\int\limits_{ - 1}^1 {\sqrt {1 + x^2 } dx} }&{ =}&{ \int\limits_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)} {\sqrt {1 + \left( {\sinh u} \right)^2 } \cosh udu}}\\ {}&{=}&{ \int\limits_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)} {\cosh ^2 udu} }\\ {}&{=}&{ \int\limits_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)} {\frac{{1 + \cosh \left( {2u} \right)}}{2}du} } \\ {}&{=}&{ \frac{1}{2}\int\limits_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)} {1du} + \frac{1}{4}\int\limits_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)} {2\cosh \left( {2u} \right)du}} \\ {}&{=}&{ \arg \sinh \left( 1 \right) + \frac{1}{4}\left[ {\sinh \left( {2u} \right)} \right]_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)}} \\ {}&{=}&{ \arg \sinh \left( 1 \right) + \frac{1}{4}\left[ {2\sinh \left( u \right)\cosh \left( u \right)} \right]_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)}} \\ {}&{=}&{ \arg \sinh \left( 1 \right) + \frac{1}{4}\left[ {2\sinh \left( u \right)\sqrt {1 + \left( {\sinh u} \right)^2 } } \right]_{ - \arg \sinh \left( 1 \right)}^{\arg \sinh \left( 1 \right)}} \\ {}&{=}&{\arg \sinh \left( 1 \right) + \frac{1}{4}\left( {2\sqrt {1 + 1} + 2\sqrt {1 + 1} } \right) }\\ {}&{=}&{ \sqrt {2}+\arg \sinh \left( 1 \right) } \end{eqnarray}

Sem comentários:

Enviar um comentário