Question

Calcular integral por substituição trigonométrica.

Answer 1

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$\displaystyle\sf\int\dfrac{dx}{x\cdot\sqrt{x^2+4}}\\\sf fac_{\!\!,}a~x=2tg(\theta)\implies dx=2sec^2(\theta)d\theta\\\sf \sqrt{x^2+4}=\sqrt{4tg^2(\theta)+4}=\sqrt{4\cdot(1+tg^2(\theta)}\\\sf \sqrt{x^2+4}=\sqrt{4sec^2}(\theta)=2sec(\theta)\\\displaystyle\sf\int\dfrac{dx}{x\cdot\sqrt{x^2+4}}=\int\dfrac{\diagup\!\!\!2\diagup\!\!\!\!\!sec^2(\theta)~d\theta}{\diagup\!\!\!2tg(\theta)2\diagup\!\!\!\!sec(\theta)}=\dfrac{1}{2}\int\dfrac{sec(\theta)~d(\theta)}{tg(\theta)}$

$\displaystyle\sf\dfrac{1}{2}\int\dfrac{1}{\diagup\!\!\!\!\!cos(\theta)}\cdot\dfrac{\diagup\!\!\!\!cos(\theta)}{sent(\theta)}~d\theta=\dfrac{1}{2}\int cosec(\theta)~d\theta=\ell n |cosec(\theta)-cotg(\theta)|+k\\\sf usando~o~tri\hat angulo~auxiliar~temos:\\\sf cosec(\theta)=\dfrac{2}{x}~~cotg(\theta)=\dfrac{\sqrt{x^2+4}}{x}\\\sf logo\\\Large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\dfrac{dx}{x\cdot\sqrt{x^2+4}}=\ell n\left|\dfrac{2-\sqrt{x^2+4}}{x}\right|+k}}}}$