Question

resolva a integral (x^2+4/raiz9-x^2)dx limite 3 e -3

Answer 1

Calcular a integral definida

     $\displaystyle\int_{-3}^3\left(x^2+\frac{4}{\sqrt{9-x^2}}\right)dx\\\\\\ =\int_{-3}^3\left(x^2+\frac{4}{\sqrt{3^2-x^2}}\right)dx\\\\\\ =\int_{-3}^3 x^2\,dx+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx$


Computando primeiro apenas a integral que aparece na 1ª parcela:

     $\displaystyle=\frac{x^{2+1}}{2+1}\bigg|_{-3}^3+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\\\\\\ =\frac{x^3}{3}\bigg|_{-3}^3+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx$

     $\displaystyle=\left(\frac{3^3}{3}-\frac{(-3)^3}{3}\right)+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\\\\\\ =\left(\frac{27}{3}-\frac{(-27)}{3}\right)+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\\\\\\ =(9-(-9))+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\\\\\\ =(9+9)+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\\\\\\ =18+\int_{-3}^3\frac{4}{\sqrt{3^2-x^2}}\,dx\qquad\quad\mathbf{(i)}$


Para a integral da 2ª parcela, fazemos uma substituição trigonométrica:

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

com  − π/2 ≤ θ ≤ π/2.


Além disso,

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

pois neste intervalo para  θ  o cosseno nunca é negativo, de modo que o módulo do cosseno é o próprio cosseno.


Novos limites de integração em  θ:

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


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


Substituindo em  (i),  a expressão fica

     $\displaystyle=18+\int_{-\pi/2}^{\pi/2}\frac{4}{3\cos\theta}\cdot 3\cos\theta\,d\theta\\\\\\ =18+\int_{-\pi/2}^{\pi/2}4\,d\theta\\\\\\ =18+4\theta\Big|_{-\pi/2}^{\pi/2}\\\\\\ =18+\left(4\cdot \frac{\pi}{2}-4\cdot \big(\!-\frac{\pi}{2}\big)\right)\\\\\\ =18+(2\pi-(-2\pi))\\\\ =18+(2\pi+2\pi)$

     $=18+4\pi$    <————    esta é a resposta.


Bons estudos! :-)