Question

5) Resolva a integral e diga o método que está usando:

$\int\limits { \sqrt{1-4x^2} } \, dx$

Answer 1

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!\sqrt{1-4x^2}\,dx}$


Faça a seguinte substituição trigonométrica  (ver figura em anexo):

     $\mathsf{2x=sen\,\theta\qquad \quad com~~-\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}\\\\\\ \Rightarrow \quad \left\{ \begin{array}{l} \mathsf{2\,dx=cos\,\theta\,d\theta\quad \Rightarrow \quad dx=\dfrac{1}{2}\,cos\,\theta\,d\theta}\\\\ \mathsf{\theta=arcsen(2x)} \end{array} \right.$


e temos também que

     $\mathsf{\sqrt{1-4x^2}}\\\\ =\mathsf{\sqrt{1-(2x)^2}}\\\\ =\mathsf{\sqrt{1-sen^2\,\theta}}\\\\ =\mathsf{\sqrt{cos^2\,\theta}}\\\\ =\mathsf{|cos\,\theta|}\\\\ =\mathsf{cos\,\theta\qquad \quad pois~-\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}\,.}$

Nesse intervalo para  θ,  o cosseno nunca é negativo.


Substituindo, a integral fica

     $=\mathsf{\displaystyle\int\!cos\,\theta\cdot \frac{1}{2}\,cos\,\theta\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!cos^2\,\theta\,d\theta}$


Aplique a identidade do cosseno do arco duplo

     cos² θ = (1/2) · (1 + cos 2θ)


e a integral fica

     $=\mathsf{\displaystyle\frac{1}{2}\int\!\frac{1}{2}\cdot (1+cos\,2\theta)\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{4}\int\!(1+cos\,2\theta)\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{4}\int\!(1+cos\,2\theta)\cdot \frac{1}{2}\cdot 2\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{8}\int\!(1+cos\,2\theta)\cdot 2\,d\theta}$


Faça uma substituição comum  (mudança de variável):

     $\mathsf{2\theta=u\quad \Rightarrow \quad 2\,d\theta=du}$


e a integral fica

     $=\mathsf{\displaystyle\frac{1}{8}\int\!(1+cos\,u)\,du}\\\\\\ =\mathsf{\displaystyle\frac{1}{8}\int\!1\,du+\frac{1}{8}\int\!cos\,u\,du}\\\\\\ =\mathsf{\displaystyle\frac{1}{8}\,u+\frac{1}{8}\,sen\,u+C}$


Substituindo de volta de  u  para  θ,

     $=\mathsf{\displaystyle\frac{1}{8}\cdot (2\theta)+\frac{1}{8}\,sen\,2\theta+C}$


Aplique a identidade do seno do arco duplo:

     sen 2θ = 2 sen θ cos θ


e a integral fica

     $=\mathsf{\displaystyle\frac{1}{8}\cdot (2\theta)+\frac{1}{8}\cdot (2\,sen\,\theta\,cos\,\theta)+C}\\\\\\ =\mathsf{\displaystyle\frac{1}{4}\,\theta+\frac{1}{4}\,sen\,\theta\,cos\,\theta+C}$


Substituindo de volta de  θ  para  x,  obtemos

     $=\mathsf{\displaystyle\frac{1}{4}\,arcsen(2x)+\frac{1}{4}\cdot (2x)\sqrt{1-4x^2}+C}$

     $=\mathsf{\displaystyle\frac{1}{4}\, arcsen(2x)+\frac{1}{2}\, x\sqrt{1-4x^2}+C}$   <————   esta é a resposta.


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \sqrt{1-4x^2}dx\\\\\\\int \frac{1}{2}cos ^2\left(u\right)du\\\\\\\frac{1}{2}\cdot \int cos ^2\left(u\right)du\\\\\\=\frac{1}{2}\cdot \int \frac{1+cos \left(2u\right)}{2}du\\\\\\=\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:1+cos \left(2u\right)du\\\\\\=\frac{1}{2}\cdot \frac{1}{2}\left(\int \:1du+\int cos \left(2u\right)du\right)\\\\\\=\frac{1}{2}\cdot \frac{1}{2}\left(u+\frac{1}{2}sin \left(2u\right)\right)$

$\sf \displaystyle =\frac{1}{2}\cdot \frac{1}{2}\left(arcsin \left(2x\right)+\frac{1}{2}sin \left(2arcsin \left(2x\right)\right)\right)\\\\\\=\frac{1}{4}\left(arcsin \left(2x\right)+\frac{1}{2}sin \left(2arcsin \left(2x\right)\right)\right)\\\\\\\to \boxed{\sf =\frac{1}{4}\left(arcsin \left(2x\right)+\frac{1}{2}sin \left(2arcsin \left(2x\right)\right)\right)+C}$