Question

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

Answer 1

Calcular a integral indefinida

    $\mathsf{\displaystyle\int \sqrt{1-4x^2}\,dx}\\\\\\ \mathsf{\displaystyle=\frac{1}{2}\int \sqrt{1-4x^2}\cdot 2\,dx\qquad(i)}$

Faça uma substituição trigonométrica:

    $\begin{array}{rcl} \mathsf{2x=sen\,\theta}&\quad\Longrightarrow\quad &\mathsf{2\,dx=cos\,\theta\,d\theta}\\\\ &&\mathsf{\theta=arcsen(2x)} \end{array}$

com $\mathsf{-\frac{\pi}{2}\le \theta\le \frac{\pi}{2}}.$

Além disso, temos que

    $\mathsf{\sqrt{1-4x^2}=\sqrt{1-(2x)^2}}\\\\ \mathsf{\sqrt{1-4x^2}=\sqrt{1-sen^2\,\theta}}\\\\ \mathsf{\sqrt{1-4x^2}=\sqrt{cos^2\,\theta}}\\\\ \mathsf{\sqrt{1-4x^2}=|cos\,\theta|}$

Como o cosseno nunca é negativo para $\mathsf{-\frac{\pi}{2}\le \theta\le \frac{\pi}{2},}$ podemos dispensar o módulo e ficamos com

     $\mathsf{\sqrt{1-4x^2}=cos\,\theta}$

Substituindo em (i), a integral fica

    $\mathsf{\displaystyle =\frac{1}{2}\int cos\,\theta\cdot cos\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle =\frac{1}{2}\int cos^2\,\theta\,d\theta\qquad(ii)}$

Vamos aplicar uma das identidades para o cosseno do arco duplo:

    $\mathsf{cos^2\,\theta=\dfrac{1}{2}\big[1+cos(2\theta)\big]}$

e a integral (ii) fica

    $\mathsf{\displaystyle=\frac{1}{2}\int \frac{1}{2}\big[1+cos(2\theta)\big]\,d\theta}\\\\\\ \mathsf{\displaystyle=\frac{1}{4}\int\big[1+cos(2\theta)\big]\,d\theta}\\\\\\ \mathsf{\displaystyle=\frac{1}{4}\int d\theta+\frac{1}{4}\int cos(2\theta)\,d\theta}\\\\\\ \mathsf{\displaystyle=\frac{1}{4}\int d\theta+\frac{1}{4}\cdot \frac{1}{2}\int cos(2\theta)\cdot 2\,d\theta}\\\\\\ \mathsf{\displaystyle=\frac{1}{4}\int d\theta+\frac{1}{8}\int cos(2\theta)\cdot 2\,d\theta\qquad (iii)}$

Faça uma substituição simples na integral que aparece na segunda parcela:

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

e a integral fica

    $\mathsf{\displaystyle=\frac{1}{4}\int d\theta+\frac{1}{8}\int cos\,u\,du}\\\\\\ \mathsf{=\dfrac{1}{4}\,\theta+\dfrac{1}{8}\,sen\,u+C}\\\\\\ \mathsf{=\dfrac{1}{4}\,\theta+\dfrac{1}{8}\,sen(2\theta)+C\qquad(iv)}$

Aplique a identidade do seno do arco duplo:

    $\mathsf{sen(2\theta)=2\,sen\,\theta\,cos\,\theta}$

e a integral (iv) fica

    $\mathsf{=\dfrac{1}{4}\,\theta+\dfrac{1}{8}\cdot 2\,sen\,\theta\,cos\,\theta+C}\\\\\\ \mathsf{=\dfrac{1}{4}\,\theta+\dfrac{1}{4}\,sen\,\theta\,cos\,\theta+C}$

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

    $\left\{\begin{array}{l}\mathsf{\theta=arcsen(2x)}\\\\ \mathsf{sen\,\theta=2x}\\\\ \mathsf{cos\,\theta=\sqrt{1-4x^2}} \end{array}\right.$

e a integral fica

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

    $\mathsf{=\dfrac{1}{4}\,arcsen(2x)+\dfrac{1}{2}\,x\sqrt{1-4x^2}+C\quad \longleftarrow\quad resposta.}$

Bons estudos! :-)