INTEGRAL:Calcule a seguinte integral:.... Pergunta de ideia deRonny06

Ronny06 @Ronny06

January 2020 1 114 Report

INTEGRAL:

Calcule a seguinte integral:

$\int\frac{1+\frac{y}{x}}{\sqrt{1+(\frac{y}{x}})^{2}} \, dx$

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Resolver a integral indefinida:

$\mathsf{\displaystyle\int\!\frac{1+\frac{y}{x}}{\sqrt{1+\left(\frac{y}{x}\right)^2}}\,dx\qquad\quad(y~constante\ne 0)\qquad(i)}$

Substituição:

$\mathsf{\dfrac{y}{x}=u}\quad\Rightarrow\quad\mathsf{x=\dfrac{y}{u}\quad\Rightarrow\quad dx=-\,\dfrac{y}{u^2}\,du}$

Substituindo, a integral $\mathsf{(i)}$ fica

$\mathsf{\displaystyle\int\!\frac{1+u}{\sqrt{1+u^2}}\cdot \left(-\,\frac{y}{u^2}\right)du}\\\\\\
=\mathsf{\displaystyle -y\int\!\frac{1+u}{u^2\sqrt{1+u^2}}\,du\qquad\quad(ii)}$

Substituição trigonométrica:

$\mathsf{u=tg\,t\quad\Rightarrow\quad du=sec^2\,t\,dt\qquad\left(-\,\frac{\pi}{2}<t<\frac{\pi}{2} \right )}\\\\\\ \mathsf{\sqrt{1+u^2}}\\\\ \mathsf{=\sqrt{1+tg^2\,t}}\\\\ \mathsf{=\sqrt{sec^2\,t}}\\\\ =\mathsf{sec\,t}\qquad\mathsf{(pois~sec\,t>0)}$

Substituindo a integral $\mathsf{(ii)}$ fica

$=\mathsf{\displaystyle -y\int\!\frac{1+tg\,t}{(tg\,t)^2\cdot sec\,t}\cdot sec^2\,t\,dt}\\\\\\ =\mathsf{\displaystyle -y\int\!\frac{1+tg\,t}{tg^2\,t}\cdot sec\,t\,dt}\\\\\\ =\mathsf{\displaystyle -y\int\!\frac{1}{tg^2\,t}\cdot sec\,t\,dt-y\int\!\frac{tg\,t}{tg^2\,t}\cdot sec\,t\,dt}\\\\\\ =\mathsf{\displaystyle -y\int\!\frac{cos^2\,t}{sen^2\,t}\cdot \frac{1}{cos\,t}\,dt-y\int\!\frac{cos\,t}{sen\,t}\cdot \frac{1}{cos\,t}\,dt}$

$=\mathsf{\displaystyle -y\int\!\frac{1}{sen^2\,t}\cdot cos\,t\,dt-y\int\!cossec\,t\,dt}\\\\\\ =\mathsf{\displaystyle -y\int\!(sen\,t)^{-2}\cdot cos\,t\,dt-y\int\!cossec\,t\,dt}\\\\\\ =\mathsf{\displaystyle -y\cdot \frac{(sen\,t)^{-2+1}}{-2+1}-y\cdot \ell n|cossec\,t-cotg\,t|+C}\\\\\\ =\mathsf{\displaystyle -y\cdot \frac{(sen\,t)^{-1}}{-1}-y\cdot \ell n|cossec\,t-cotg\,t|+C}$

$=\mathsf{\displaystyle \frac{y}{sen\,t}-y\cdot \ell n|cossec\,t-cotg\,t|+C}\\\\\\ =\mathsf{\displaystyle y\cdot cossec\,t-y\cdot \ell n|cossec\,t-cotg\,t|+C\qquad\quad(iii)}$

Para voltar à variável u, devemos expressar as razões trigonométricas envolvidas acima em termos de $\mathsf{tg\,t}$ e $\mathsf{sec\,t.}$

Note que

$\mathsf{cossec\,t=\dfrac{1}{sen\,t}}\\\\\\ \mathsf{cossec\,t=\dfrac{~~\frac{1}{cos\,t}~~}{\frac{sen\,t}{cos\,t}}}\\\\\\ \mathsf{cossec\,t=\dfrac{sec\,t}{tg\,t}}\\\\\\ \mathsf{cossec\,t=\dfrac{\sqrt{1+u^2}}{u}\qquad\quad\checkmark}$

$\mathsf{cotg\,t=\dfrac{1}{tg\,t}}\\\\\\ \mathsf{cotg\,t=\dfrac{1}{u}\qquad\quad\checkmark}$

Então, a integral $\mathsf{(iii)}$ fica

$=\mathsf{y\cdot \dfrac{\sqrt{1+u^2}}{u}-y\cdot \ell n\left|\dfrac{\sqrt{1+u^2}}{u}-\dfrac{1}{u}\right|+C}\\\\\\ =\mathsf{y\cdot \dfrac{\sqrt{1+u^2}}{u}-y\cdot \ell n\left|\dfrac{\sqrt{1+u^2}-1}{u}\right|+C}\\\\\\ =\mathsf{y\cdot \dfrac{\sqrt{1+u^2}}{u}-y\cdot \left(\ell n\big|\sqrt{1+u^2}-1\big|-\ell n\big|u\big| \right )+C}\\\\\\ =\mathsf{y\cdot \dfrac{\sqrt{1+u^2}}{u}-y\cdot \ell n\big|\sqrt{1+u^2}-1\big|+y\cdot \ell n\big|u\big|+C}$

Voltando à variável x,

$=\mathsf{y\cdot \dfrac{\sqrt{1+\left(\frac{y}{x} \right )^2}}{\frac{y}{x}}-y\cdot \ell n\left|\sqrt{1+\left(\dfrac{y}{x} \right )^2}-1\right|+y\cdot \ell n\left|\dfrac{y}{x}\right|+C}\\\\\\ =\mathsf{x\cdot \sqrt{1+\left(\dfrac{y}{x}\right )^2}-y\cdot \ell n\left|\sqrt{1+\left(\dfrac{y}{x} \right )^2}-1\right|+y\cdot \ell n\left|\dfrac{y}{x}\right|+C\qquad\checkmark}$

Bons estudos! :-)

Tags: integral indefinida substituição trigonométrica cálculo integral

Ronny06 Matematico !!

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