Question

Ajuda nessa integral..

$\int\limits \, \sqrt[]{ \frac{9}{1 - x^{2} } } dx$

Agradecida

Answer 1

$I=\displaystyle\int{\sqrt{\dfrac{9}{1-x^{2}}}\,dx}\\ \\ \\ =\int{\dfrac{\sqrt{9}}{\sqrt{1-x^{2}}}\,dx}\\ \\ \\ =\int{\dfrac{3}{\sqrt{1-x^{2}}}\,dx}$


Substituição trigonométrica:

$x=\mathrm{sen\,}t~\Rightarrow~\left\{ \begin{array}{l} dx=\cos t\,dt\\ \\ t=\mathrm{arcsen\,}x~\Rightarrow~-\frac{\pi}{2}<t<\frac{\pi}{2} \end{array} \right.$


Sendo assim, temos que

$\sqrt{1-x^{2}}\\ \\ =\sqrt{1-\mathrm{sen^{2}\,}t}\\ \\ =\sqrt{\cos^{2}t}\\ \\ =\left|\cos t\right|=\cos t\,,\text{ pois }-\frac{\pi}{2}<t<\frac{\pi}{2}.$


Substituindo na integral, ficamos com

$I=\displaystyle\int{\dfrac{3}{\cos t}\cdot \cos t\,dt}\\ \\ \\ =\int{3\,dt}\\ \\ \\ =3t+C\\ \\ =3\,\mathrm{arcsen\,}x+C$