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05/09/2017

Integração por substituição: seno hiperbólico(II)
Integração por substituição: tangente(I)

Exercício Calcular o integral \[ \int\limits_1^{\sqrt 3 } {\frac{1}{x^2 \sqrt {1 + x^2 } }dx} \] Recorrendo à substituição: $x=\tg t$ ou $x=\sh t$
(Nota: $\sh$ é uma notação para seno hiperbólico ... também se representa por $\sinh$ )

\[ \frac{3\sqrt 2 - 2\sqrt 3 }{3} \]

Fazendo a substituição $x=\sh t$ temos, \begin{eqnarray*} {t}&{=}&{\arg\sh x}\\ {x}&{=}&{1\Rightarrow t=\argsh(1)}\\ {x}&{=}&{\sqrt 3\Rightarrow t=\argsh(\sqrt 3)} \end{eqnarray*} e \[ \frac{{dx}}{{dt}} = \ch t \] Portanto \begin{eqnarray*} {\int\limits_1^{\sqrt 3 } {\frac{1}{x^2 \sqrt {1 + x^2 } }dx} }&{=}&{\int\limits_{\argsh 1}^{\argsh \sqrt 3 } {\frac{1}{\shq t \sqrt {1 + \shq t } }\ch t dt} }\\ {}&{=}&{\int\limits_{\argsh 1}^{\argsh \sqrt 3 } {\frac{1}{\shq t}dt}}\\ {}&{=}&{\int\limits_{\argsh 1}^{\argsh \sqrt 3 } {\cosechq tdt}}\\ {}&{=}&{\left[-\cotgh t\right]_{\argsh 1}^{\argsh \sqrt 3 }}\\ {}&{=}&{\left[-\frac{\ch t}{\sh t}\right]_{\argsh 1}^{\argsh \sqrt 3 }}\\ {}&{=}&{\left[-\frac{\sqrt {1 + \shq t }}{\sh t}\right]_{\argsh 1}^{\argsh \sqrt 3 }}\\ {}&{=}&{-\frac{2}{\sqrt{3}}+\sqrt{2}}\\ {}&{=}&{\frac{3\sqrt 2 - 2\sqrt 3 }{3}} \end{eqnarray*}


Fazendo a substituição $x=\tg t$, com $t\in \left]-\frac{\pi}{2},\frac{\pi}{2}\right[$ temos, \begin{eqnarray*} {t}&{=}&{\arctg x}\\ {x}&{=}&{1\Rightarrow t=\frac{\pi}{4}}\\ {x}&{=}&{\sqrt 3\Rightarrow t=\frac{\pi}{3}} \end{eqnarray*} e \[ \frac{{dx}}{{dt}} = \sec^2 t \] Portanto \begin{eqnarray*} {\int\limits_1^{\sqrt 3 } {\frac{1}{x^2 \sqrt {1 + x^2 } }dx} }&{=}&{\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3} } {\frac{1}{\tgq t \sqrt {1 + \tgq t } }\sec^2 t dt} }\\ {}&{=}&{\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3} } {\frac{\sec^2 t}{\senq t \sec^2 t\sec t}dt}}\\ {}&{=}&{\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3} } \frac{1}{\senq t \sec t }dt}\\ {}&{=}&{\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3} } \frac{\cos t}{\senq t }dt}\\ {}&{=}&{\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3} }{\cotg t}{\cosec t }dt}\\ {}&{=}&{\left[-\cosec t\right]_{\frac{\pi}{4}}^{\frac{\pi}{3} }}\\ {}&{=}&{-\cosec \frac{\pi}{3} + \cosec \frac{\pi}{4}}\\ {}&{=}&{-\frac{2}{\sqrt{3}}+\sqrt{2}}\\ {}&{=}&{\frac{3\sqrt 2 - 2\sqrt 3 }{3}} \end{eqnarray*}

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