Question

integral por substituição trigonometrica:

integral x^3 raiz de (9-x^2) dx

Answer 1

Calcular a integral indefinida

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


Façamos uma substituição trigonométrica:

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

com  − π/2 ≤ θ ≤ π/2.


Além disso, temos que

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

pois como  − π/2 ≤ θ ≤ π/2,  neste intervalo o cosseno nunca é negativo, de modo que o módulo do cosseno é igual ao próprio cosseno.


Substituindo tudo, a integral fica

     $\displaystyle\int x^3 \sqrt{9-x^2}\,dx\\\\\\ =\int (3\,\mathrm{sen\,}\theta)^3\cdot 3\cos \theta\cdot 3\cos \theta\,d\theta\\\\\\ =\int 27\,\mathrm{sen^3\,}\theta\cdot 9\cos^2 \theta\,d\theta\\\\\\ =243\int \mathrm{sen^3\,}\theta\cdot \cos^2 \theta\,d\theta\\\\\\ =\,243\int \mathrm{sen^2\,}\theta\cdot \cos^2 \theta\cdot \mathrm{sen\,\theta}\,d\theta\\\\\\ =-\,243\int \mathrm{sen^2\,}\theta\cdot \cos^2 \theta\cdot (-\,\mathrm{sen\,\theta})\,d\theta$


Mas  sen² θ = 1 − cos² θ:

     $\displaystyle=-\,243\int(1-\cos^2\theta)\cdot \cos^2\theta\cdot (-\,\mathrm{sen\,}\theta)\,d\theta$


Agora, faça outra substituição comum:

     $\cos \theta=u\quad\Rightarrow\quad -\,\mathrm{sen\,}\theta\,d\theta=du$


e a integral fica

     $\displaystyle=-\,243\int (1-u^2)\cdot u^2\,du\\\\\\ =-\,243\int (u^2-u^4)\,du\\\\\\ =-\,243\cdot \left(\frac{u^{2+1}}{2+1}-\frac{u^{4+1}}{4+1}\right)+C\\\\\\ =-\,243\cdot \left(\frac{u^3}{3}-\frac{u^5}{5}\right)+C\\\\\\ =-\,\frac{243}{3}u^3+\frac{243}{5}\,u^5+C\\\\\\ =-\,81u^3+\frac{243}{5}\,u^5+C$

     $=-\,81\cos^3\theta+\dfrac{243}{5}\cos^5\theta+C$


Substitua de volta para a variável  x,  usando as relações obtidas na substituição trigonométrica:

      •   $\sqrt{9-x^2}=3\cos\theta\quad\Rightarrow\quad \cos\theta=\dfrac{\sqrt{9-x^2}}{3}$


e a integral fica

     $=-\,81\cdot \bigg(\dfrac{\sqrt{9-x^2}}{3}\bigg)^{\!3}+\dfrac{243}{5}\cdot \bigg(\dfrac{\sqrt{9-x^2}}{3}\bigg)^{\!5}+C\\\\\\ =-\,81\cdot \dfrac{(\sqrt{9-x^2})^3}{3^3}+\dfrac{243}{5}\cdot \dfrac{(\sqrt{9-x^2})^5}{3^5}+C\\\\\\ =-\,81\cdot \dfrac{[(9-x^2)^{1/2}]^3}{27}+\dfrac{243}{5}\cdot \dfrac{[(9-x^2)^{1/2}]^5}{243}+C\\\\\\ =-\,\dfrac{81}{27}\,(9-x^2)^{3/2}+\dfrac{243}{5\cdot 243}\,(9-x^2)^{5/2}+C$

     $=-\,3\cdot (9-x^2)^{3/2}+\dfrac{1}{5}\,(9-x^2)^{5/2}+C\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Bons estudos! :-)