Question

integral de dx/(raiz(7+5x-2x^2))

Answer 1

Primeiramente temos que expressar o polinômio quadrático como uma diferença de quadrados. Para isso, vamos usar a técnica de completamento de quadrados:

     $7+5x-2x^2=\dfrac{1}{8}\cdot 8(7+5x-2x^2)\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot (56+40x-16x^2)\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot \big[56+2\cdot 5\cdot 4x-(4x)^2\big]$


Some e subtraia 25 = 5² dentro dos colchetes:

     $7+5x-2x^2=\dfrac{1}{8}\cdot \big[56+25-25+2\cdot 5\cdot 4x-(4x)^2\big]\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot \big[81-25+2\cdot 5\cdot 4x-(4x)^2\big]\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot \big[81-(25-2\cdot 5\cdot 4x+(4x)^2)\big]\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot \big[81-(5-4x)^2\big]\\\\\\ 7+5x-2x^2=\dfrac{1}{8}\cdot \big[9^2-(5-4x)^2\big]$


Portanto, a integral fica

     $\displaystyle\int\dfrac{dx}{\sqrt{7+5x-2x^2}}\\\\\\ =\int\dfrac{dx}{\sqrt{\frac{1}{8}\cdot \big[9^2-(5-4x)^2\big]}}\\\\\\ =\frac{1}{\sqrt{\frac{1}{8}}}\int\dfrac{dx}{\sqrt{9^2-(5-4x)^2}}\\\\\\ =\sqrt{8}\int\dfrac{dx}{\sqrt{9^2-(5-4x)^2}}\\\\\\ =2\sqrt{2}\int\dfrac{dx}{\sqrt{9^2-(5-4x)^2}}\qquad\quad\mathsf{(i)}$


Faça a seguinte substituição trigonométrica:

     $5-4x=9\,\mathrm{sen\,}\theta\\\\\\ \Longrightarrow\quad \left\{\begin{array}{l} \theta=\mathrm{arcsen}\!\left(\dfrac{5-4x}{9}\right)\\\\ -\,4\,dx=9\cos\theta\,d\theta\quad\Longrightarrow\quad dx=-\,\dfrac{9}{4}\cos\theta\,d\theta \end{array}\right.$

com −π/2 < θ < π/2.


Além disso, temos que

     $\sqrt{9^2-(5-4x)^2}=\sqrt{9^2-(9\,\mathrm{sen\,}\theta)^2}\\\\ \sqrt{9^2-(5-4x)^2}=\sqrt{9^2-9^2\,\mathrm{sen^2\,}\theta}\\\\ \sqrt{9^2-(5-4x)^2}=\sqrt{9^2\cdot (1-\mathrm{sen^2\,}\theta)}\\\\ \sqrt{9^2-(5-4x)^2}=\sqrt{9^2\cos^2\theta}\\\\ \sqrt{9^2-(5-4x)^2}=9\cos\theta$


Substituindo, a integral fica

     $\displaystyle =2\sqrt{2}\int\dfrac{1}{\diagup\!\!\!\! 9\cos\theta}\cdot \Big(\!-\frac{\diagup\!\!\!\! 9}{4}\Big)\cos\theta\,d\theta\\\\\\ =2\sqrt{2}\cdot \Big(-\frac{1}{4}\Big)\int\dfrac{1}{\cos\theta}\cdot \cos\theta\,d\theta\\\\\\ =-\,\frac{\sqrt{2}}{2}\int 1\,d\theta\\\\\\ =-\,\frac{\sqrt{2}}{2}\,\theta+C$


Substitua de volta para a variável x, e finamente você obtém o resultado procurado:

     $=-\,\dfrac{\sqrt{2}}{2}\,\mathrm{arcsen}\!\left(\dfrac{5-4x}{9}\right)+C\quad\longleftarrow\quad \mathsf{resposta.}$


Dúvidas? Comente.


Bons estudos! :-)