Question

$\int\ { \sqrt{5-4x^2} } \, dx$ passo a passo por favor e explica como foi encontrado

Answer 1

$\displaystyle\mathtt{I=\int\sqrt{5-4x^2}\,dx}\\\\\\ =\mathtt{\int\frac{1}{2}\cdot 2\sqrt{5-4x^2}\,dx}\\\\\\ =\mathtt{\frac{1}{2}\int\sqrt{5-4x^2}\cdot 2\,dx\quad\quad(i)}$


Substituição trigonométrica:

$\mathtt{2x=\sqrt{5}\,sen\,t}\quad\Rightarrow\quad \left\{ \begin{array}{l} \mathtt{2\,dx=\sqrt{5}\,cos t\,dt}\\\\ \mathtt{t=arcsen\!\left(\dfrac{2x}{\sqrt{5}} \right )} \end{array} \right.$

$\mathtt{5-4x^2}\\\\ =\mathtt{5-(2x)^2}\\\\ =\mathtt{5-(\sqrt{5}\,sen\,t)^2}\\\\ =\mathtt{5-5\,sen^2\,t}\\\\ =\mathtt{5(1-sen^2\,t)}\\\\ =\mathtt{5\,cos^2\,t}$


Portanto,

$\mathtt{\sqrt{5-4x^2}}\\\\ =\mathtt{\sqrt{5\,cos^2\,t}}\\\\ =\mathtt{\sqrt{5}\,|cos\,t|}\\\\ =\mathtt{\sqrt{5}\,cos\,t}\quad\quad\texttt{pois }\mathtt{-\pi/2\le t\le \pi/2}$

_________

Substituindo, a integral $\mathtt{(i)}$ fica

$\displaystyle\mathtt{\frac{1}{2}\int\sqrt{5}\,cos\,t\cdot \sqrt{5}\,cos t\,dt}\\\\\\ =\mathtt{\frac{1}{2}\int 5\,cos^2 t\,dt}\\\\\\ =\mathtt{\frac{5}{2}\int cos^2 t\,dt}\\\\\\ =\mathtt{\frac{5}{2}\int \left(\dfrac{1}{2}+\dfrac{1}{2}\,cos\,2t \right )dt}$

$=\displaystyle\mathtt{\frac{5}{2}\cdot \left(\dfrac{1}{2}\,t+\dfrac{1}{2}\cdot \dfrac{1}{2}\,sen\,2t\right)+C}\\\\\\ =\mathtt{\frac{5}{4}\,t+\dfrac{5}{8}\,sen\,2t+C}\\\\\\ =\mathtt{\frac{5}{4}\,t+\dfrac{5}{4\cdot 2}\cdot 2\,sen\,t\,cos\,t+C}\\\\\\ =\mathtt{\frac{5}{4}\,t+\dfrac{5}{4}\,sen\,t\,cos\,t+C\quad\quad(ii)}$


Substituindo de volta para a variável $\mathtt{x}:$

(Veja triângulo retângulo em anexo para auxiliar na compreensão)

$=\mathtt{\dfrac{5}{4}\,arcsen\!\left(\dfrac{2x}{\sqrt{5}} \right )+\dfrac{5}{4}\cdot \dfrac{2x}{\sqrt{5}}\cdot \dfrac{\sqrt{5-4x^2}}{\sqrt{5}}+C}\\\\\\ =\mathtt{\dfrac{5}{4}\,arcsen\!\left(\dfrac{2x}{\sqrt{5}} \right )+\dfrac{\diagup\!\!\!\! 5}{2\cdot \diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2x}{\diagup\!\!\!\! 5}\cdot \sqrt{5-4x^2}+C}\\\\\\ =\boxed{\begin{array}{c}\mathtt{\dfrac{5}{4}\,arcsen\!\left(\dfrac{2x}{\sqrt{5}} \right )+\dfrac{1}{2}\,x\sqrt{5-4x^2}+C} \end{array}}$