Question

Ensino superior
Calcule o valor da integral usando o método da substituição trigonométrica

Answer 1

$\large\boxed{\begin{array}{l}\rm Quando~integrando~\acute e~da~forma~\sqrt{b^2+x^2}\\\rm use~a~substituic_{\!\!,}\tilde ao~x=b\,tg(\theta)\\\rm de~modo~que~dx=b\,sec^2(\theta)~d\theta\\\rm e~o~radical~\sqrt{b^2+x^2}~se~torna~ b\,sec(\theta)\end{array }}$

$\boxed{\begin{array}{l}\rm fac_{\!\!,}a~x=\dfrac{3}{5} tg(\theta)\\\rm de~modo~que~dx=\dfrac{3}{5}sec^2(\theta)~d\theta\\\rm e~\sqrt{x^2+\dfrac{9}{25}}=\dfrac{3}{5}sec(\theta)\\\underline{\rm substituindo~temos:}\\\displaystyle\sf\int\dfrac{dx}{\sqrt{x^2+\dfrac{9}{25}}}=\int\dfrac{\diagup\!\!\!\frac{3}{5}~\backslash\!\!\!\!sec^2(\theta)}{\diagup\!\!\!\frac{3}{5}~\backslash\!\!\!sec(\theta)}~d\theta\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int sec(\theta)~d\theta=\ell n|sec(\theta)+tg(\theta)|+C\\\underline{\rm usando~o~tri\hat angulo~auxiliar~temos:}\\\sf tg(\theta)=\dfrac{cat~op}{cat~adj}=\dfrac{x}{\frac{3}{5}}=\dfrac{5x}{3}\\\sf sec(\theta)=\dfrac{hip}{cat~adj}=\dfrac{\sqrt{x^2+\dfrac{9}{25}}}{\frac{3}{5}}=\dfrac{5\sqrt{x^2+\dfrac{9}{25}}}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm substituindo~temos:}\\\displaystyle\sf\int\dfrac{dx}{\sqrt{x^2+\dfrac{9}{25}}}=\ell n\bigg|\dfrac{5\bigg[\sqrt{x^2+\dfrac{9}{25}}+x\bigg]}{3}\bigg|+C\end{array}}$