Question

3) Calcule a integral

$\int\ {} \frac{dx}{(6-x^2)^\frac{3}{2} } \,$

Answer 1

$I=\displaystyle\int\!\frac{dx}{(6-x^2)^{3/2}}~~~~~~\mathbf{(i)}$


Faça a seguinte substituição (trigonométrica)

$x=\sqrt{6}\,\mathrm{sen\,}t~~~~(-\pi/2<t<\pi/2)~~\Rightarrow~~\left\{\!\begin{array}{l} dx=\sqrt{6}\cos t\,dt\\\\ t=\mathrm{arcsen}\left(\dfrac{x}{\sqrt{6}} \right ) \end{array}\right.\\\\\\ 6-x^2=6-\big(\sqrt{6}\,\mathrm{sen\,}t\big)^2\\\\ 6-x^2=6-6\,\mathrm{sen^2\,}t\\\\ 6-x^2=6\,(1-\mathrm{sen^2\,}t)\\\\ 6-x^2=6\cos^2 t\\\\ (6-x^2)^{3/2}=(6\cos^2 t)^{3/2}\\\\ (6-x^2)^{3/2}=6\sqrt{6}\cos^3 t~~~~~~(\text{pois }-\pi/2<t<\pi/2)$


Substituindo em $\mathbf{(i)}$ a integral fica

$=\displaystyle\int\!\frac{\sqrt{6}\cos t\,dt}{6\sqrt{6}\cos^3 t}\\\\\\ =\frac{1}{6}\int\!\frac{dt}{\cos^2 t}\\\\\\ =\frac{1}{6}\int\!\sec^2 t\,dt\\\\\\ =\frac{1}{6}\,\mathrm{tg\,}t+C~~~~~~\mathbf{(ii)}$


Voltando à variável $x$

Observando o triângulo retângulo em anexo, vemos que

$\mathrm{tg\,}t=\dfrac{x}{\sqrt{6-x^2}}$


Substituindo em $\mathbf{(ii)},$ finalmente chegamos a

$=\dfrac{1}{6}\cdot \dfrac{x}{\sqrt{6-x^2}}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{dx}{(6-x^2)^{3/2}}=\frac{x}{6\sqrt{6-x^2}}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{dx}{\left(6-x^2\right)^{\frac{3}{2}}}\\\\\\=\int \frac{1}{6\cos ^2\left(u\right)}du\\\\\\=\frac{1}{6}\cdot \int \frac{1}{\cos ^2\left(u\right)}du\\\\\\=\frac{1}{6}\tan \left(u\right)\\\\\\\frac{1}{6}\tan \left(\arcsin \left(\frac{1}{6^{\frac{1}{2}}}x\right)\right)\\\\\\=\frac{x}{6\left(6-x^2\right)^{\frac{1}{2}}}\\\\\\\to \boxed{\sf =\frac{x}{6\left(6-x^2\right)^{\frac{1}{2}}}+C}$