Question

resolva integral ∫³₀ (3/raiz 9- x^2) dx

Answer 1

Calcular a integral:

    $\displaystyle\int_0^3\frac{3}{\sqrt{9-x^2}}\,dx$

Faça a seguinte substituição trigonométrica:

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

com $0\le x\le \frac{\pi}{2}.$

Além disso, temos que

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

Novos limites de integração:

     $\begin{array}{lcl} \mathsf{Quando~~}x=0&\quad\Longrightarrow\quad &\theta=\mathrm{arcsen}\Big(\dfrac{0}{3}\Big)\\\\ &&\theta=\mathrm{arcsen}(0)\\\\ &&\theta=0\\\\\\ \mathsf{Quando~~}x=3&\quad\Longrightarrow\quad &\theta=\mathrm{arcsen}\Big(\dfrac{3}{3}\Big)\\\\ &&\theta=\mathrm{arcsen}(1)\\\\ &&\theta=\dfrac{\pi}{2} \end{array}$

Substituindo, a integral fica

    $\displaystyle=\int_0^{\pi/2}\frac{3}{3\cos\theta}\cdot 3\cos\theta\,d\theta\\\\\\ =\int _0^{\pi/2} 3\,d\theta\\\\\\ =3\theta\Big|_0^{\pi/2}\\\\\\ =3\cdot \frac{\pi}{2}-3\cdot 0\\\\\\ =\frac{3\pi}{2}\quad\longleftarrow\quad \mathsf{resposta.}$

Bons estudos! :-)

Answer 2

Resposta:

3pi/2

Explicação passo a passo:

AVA