Question

Resolva a integral ∫ 3 -3 4/√(9-x²) Escolha uma:
a. 2π
b. π
c. 6π
d. 4π
e. 0

Answer 1

Calcular a integral definida:

     $\displaystyle\int_{-3}^3 \frac{4}{\sqrt{9-x^2}}\,dx$


Faça a seguinte substituição trigonométrica:

     $x=3\,\mathrm{sen\,}\theta\quad\Longrightarrow\quad \left\{ \begin{array}{l} dx=3\cos\theta\,d\theta\\\\ \theta=\mathrm{arcsen}\!\left(\dfrac{x}{3}\right) \end{array}\right.$

com  $-\,\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}.$


Dessa forma, temos que

     $\sqrt{9-x^2}=\sqrt{9-(3\,\mathrm{sen\,}\theta)^2}\\\\ \sqrt{9-x^2}=\sqrt{9-9\,\mathrm{sen^2\,}\theta}\\\\ \sqrt{9-x^2}=\sqrt{9\cdot (1-\mathrm{sen^2\,}\theta)}\\\\ \sqrt{9-x^2}=\sqrt{9\cos^2\theta}\\\\ \sqrt{9-x^2}=3\left|\cos\theta\right|=3\cos\theta$

pois no intervalo em que θ se encontra, o cosseno nunca é negativo. Logo, o módulo do cosseno é o próprio cosseno.


Novos limites de integração em θ:

     $\begin{array}{lcl} \mathsf{Quando~~}x=-3&\quad\Longrightarrow\quad&\theta=\mathrm{arcsen}\!\left(\!-\,\dfrac{3}{3}\right)\\\\ &&\theta=\mathrm{arcsen}(-1)\\\\ &&\theta=-\,\dfrac{\pi}{2} \end{array}$


     $\begin{array}{lcl} \mathsf{Quando~~}x=3&\quad\Longrightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{3}{3}\right)\\\\ &&\theta=\mathrm{arcsen}(1)\\\\ &&\theta=\dfrac{\pi}{2} \end{array}$


Substituindo, a integral fica

     $\displaystyle\int_{-\pi/2}^{\pi/2}\;\frac{4}{3\cos \theta}\cdot 3\cos\theta\,d\theta\\\\\\ =\int_{-\pi/2}^{\pi/2}\;4\,d\theta\\\\\\ =4\theta\Big|_{-\pi/2}^{\pi/2}\\\\\\ =4\cdot \left[\frac{\pi}{2}-\Big(\!-\frac{\pi}{2}\Big) \right]$

     $=4\cdot \left[\dfrac{\pi}{2}+\dfrac{\pi}{2}\right]\\\\\\ =4\cdot \dfrac{\diagup\!\!\!\! 2\pi}{\diagup\!\!\!\! 2}$

     $=4\pi\quad\longleftarrow\quad\mathsf{esta~\acute{e}~a~resposta.}$


Resposta:  alternativa  d.  4π.


Bons estudos! :-)