Question

Calcule a integral indefinida por substituição trigonométrica.

HELP.

Answer 1

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$\sf observe~o~tri\hat{a}ngulo~anexo$

$\sf vamos~usar~uma~substituic_{\!\!,}\tilde{a}o~da~forma~x=a~tg(\theta)\\\sf para~simplificar~o~integrando:\\\sf x=2tg(\theta)\implies dx=2sec^2(\theta)~d\theta\\\sf\sqrt{4+x^2}=\sqrt{4+4tg^2(\theta)}\\\sf=\sqrt{4\cdot(1+tg^2(\theta))}=\sqrt{4sec^2(\theta)}=2sec(\theta)\\\displaystyle\sf\int\dfrac{dx}{\sqrt{4+x^2}}=\int\dfrac{\diagup\!\!\!2~~\diagdown\!\!\!\!\!sec^2(\theta)~d\theta}{\diagup\!\!\!2~~\diagdown\!\!\!\!sec(\theta)}=\int sec(\theta)~d\theta\\\sf=\huge\ell n|sec(\theta)+tg(\theta)|+k$

$\sf pelo~tri\hat{a}ngulo~auxiliar\\\sf sec(\theta)=\dfrac{2}{\sqrt{4+x^2}}\\\sf tg(\theta)=\dfrac{x}{\sqrt{4+x^2}}\\\sf portanto\\\boxed{\displaystyle\sf\int\dfrac{dx}{\sqrt{4+x^2}}=\huge\ell n\left|\dfrac{2+x}{\sqrt{4+x^2}}\right|+k}$