Question

Calcule a integral definida $\displaystyle \int^{a}_{0} x^{2} \sqrt{a^{2} - x^{2}} \, dx$

Answer 1

Calcular a integral definida:

     $\displaystyle\int_0^a x^2\sqrt{a^2-x^2}\,dx$


Caso a constante  a  seja nula,  caímos em um caso trivial, no qual o valor da integral definida dá zero.

     $\displaystyle\int_0^0 x^2\sqrt{0^2-x^2}\,dx=0$

pois o limite inferior é igual ao limite superior da integral.


Consideremos que  a  é uma constante não-nula. Aqui, podemos fazer uma substituição trigonométrica:

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

com  − π/2 ≤ θ ≤ π/2,  ou seja,  θ  é um arco do  4º  ou do  1º  quadrante.


Além disso,

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

Como  θ  é um arco do  4º  ou do  1º  quadrante,  o cosseno de  θ  nunca será negativo. Por isso, o módulo do cosseno será o próprio cosseno.


Novos limites de integração em  θ:

     $\begin{array}{lcl} \textsf{Quando~~}x=0&\quad\Rightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{0}{a}\right)\\\\ &&\theta=\mathrm{arcsen}(0)\\\\ &&\theta=0 \end{array}$


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


Substituindo tudo, a integral fica

     $\displaystyle=\int_0^{\pi/2}(a\,\mathrm{sen\,}\theta)^2\cdot |a|\cos \theta\cdot a\cos \theta\,d\theta\\\\\\ =\int_0^{\pi/2}a^2\,\mathrm{sen^2\,}\theta\cdot |a|\cos \theta\cdot a\cos \theta\,d\theta\\\\\\ =a^3|a|\int_0^{\pi/2}\mathrm{sen^2\,}\theta\cdot \cos^2 \theta\,d\theta\\\\\\ =a^3|a|\int_0^{\pi/2}(\mathrm{sen\,}\theta\cdot \cos \theta)^2\,d\theta$


Mas  sen θ · cos θ = (1/2) sen 2θ.  E a integral fica

     $\displaystyle=a^3|a|\int_0^{\pi/2}\left(\frac{1}{2}\,\mathrm{sen\,}2\theta\right )^{\!2}d\theta\\\\\\ =a^3|a|\int_0^{\pi/2}\frac{1}{4}\,\mathrm{sen^2\,}2\theta\,d\theta\\\\\\=\frac{a^3|a|}{4}\int_0^{\pi/2}\mathrm{sen^2\,}2\theta\,d\theta$


Para a integral acima, use uma das identidades trigonométricas do cosseno do arco duplo:

     $\mathrm{sen^2\,}\alpha=\dfrac{1}{2}-\dfrac{1}{2}\cos 2\alpha$


Para  α = 2θ,  a integral fica

     $\displaystyle=\frac{a^3|a|}{4}\int_0^{\pi/2}\left(\frac{1}{2}-\frac{1}{2}\cos(2\cdot 2\theta) \right )\!d\theta\\\\\\ =\frac{a^3|a|}{4}\int_0^{\pi/2}\left(\frac{1}{2}-\frac{1}{2}\cos 4\theta\right)\!d\theta\\\\\\ =\frac{a^3|a|}{4}\cdot \left[\frac{1}{2}\,\theta-\frac{1}{2}\cdot \left(\frac{1}{4}\,\mathrm{sen\,}4\theta \right )\right]_0^{\pi/2}\\\\\\ =\frac{a^3|a|}{4}\cdot \left[\frac{1}{2}\,\theta-\frac{1}{8}\,\mathrm{sen\,}4\theta\right]_0^{\pi/2}$

     $=\dfrac{a^3|a|}{4}\cdot \left[\dfrac{1}{2}\cdot \dfrac{\pi}{2}-\dfrac{1}{8}\,\mathrm{sen\!}\left(4\cdot \dfrac{\pi}{2}\right)\right]-\dfrac{a^3|a|}{4}\cdot \left[\dfrac{1}{2}\cdot 0-\dfrac{1}{8}\,\mathrm{sen}(4\cdot 0)\right]\\\\\\ =\dfrac{a^3|a|}{4}\cdot \left[\dfrac{\pi}{4}-\dfrac{1}{8}\,\mathrm{sen}(2\pi)\right]-0\\\\\\ =\dfrac{a^3|a|}{4}\cdot \left[\dfrac{\pi}{4}-\dfrac{1}{8}\cdot 0\right]\\\\\\ =\dfrac{a^3|a|}{4}\cdot \dfrac{\pi}{4}$

     $=\dfrac{\pi a^3|a|}{16}$          ✔

—————

Portanto,

     $\displaystyle\int_0^a x^2\sqrt{a^2-x^2}=\left\{ \begin{array}{rl} \dfrac{\pi a^4}{16}\,,&\textsf{se~~}a\ge 0\\\\ -\,\dfrac{\pi a^4}{16}\,,&\textsf{se~~}a<0 \end{array} \right.\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Bons estudos! :-)