Question

Calcule as seguintes integrais utilizando o método da substituição de variável
b) 5x$\sqrt{4-3 x^{2} } dx$

Answer 1

$\displaystyle\int \!5x\sqrt{4-3x^2}\,dx\\\\\\ =\int \!\frac{1}{-6}\cdot (-6)\cdot 5x\sqrt{4-3x^2}\,dx\\\\\\ =-\frac{5}{6}\int \!\sqrt{4-3x^2}\,\cdot (-6x)\,dx~~~~~~\mathbf{(i)}$


Substituição:

$4-3x^2=u~~\Rightarrow~~-6x\,dx=du$


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

$\displaystyle=-\,\frac{5}{6}\int \!\sqrt{u}\,du\\\\\\ =-\,\frac{5}{6}\int u^{1/2}\,du\\\\\\ =-\,\frac{5}{6}\cdot \dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =-\,\frac{5}{6}\cdot \dfrac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =-\,\frac{5}{6}\cdot \dfrac{2}{3}\,u^{3/2}+C\\\\\\ =-\,\frac{10}{18}\,u^{3/2}+C\\\\\\ =-\,\frac{5}{9}\,u^{3/2}+C$

$=-\,\dfrac{5}{9}\cdot (4-3x^2)^{3/2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int \!5x\sqrt{4-3x^2}\,dx=-\,\dfrac{5}{9}\cdot (4-3x^2)^{3/2}+C \end{array}}$


Bons estudos! :-)