Question

como resolver a integral de x³ √1-x² dx

Answer 1

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


Faça a seguinte substituição:

$1-x^2=u~~\Rightarrow~~\left\{ \!\begin{array}{l} -2x\,dx=du\\\\ x^2=1-u \end{array} \right.$


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

$=\displaystyle-\frac{1}{2}\int\!(1-u)\sqrt{u}\,du\\\\\\ =-\frac{1}{2}\int\!\big(\sqrt{u}-u\sqrt{u}\big)\,du\\\\\\ =-\frac{1}{2}\int\!\big(u^{1/2}-u^{3/2}\big)\,du\\\\\\ =-\frac{1}{2}\cdot \left(\frac{u^{(1/2)+1}}{\frac{1}{2}+1}-\frac{u^{(3/2)+1}}{\frac{3}{2}+1} \right )+C$

$=-\dfrac{1}{2}\cdot \left(\dfrac{u^{3/2}}{\frac{3}{2}}-\dfrac{u^{5/2}}{\frac{5}{2}} \right )+C\\\\\\ =-\dfrac{1}{2}\cdot \left(\dfrac{2}{3}\,u^{3/2}-\dfrac{2}{5}\,u^{5/2} \right )+C\\\\\\ =-\dfrac{1}{3}\,u^{3/2}+\dfrac{1}{5}\,u^{5/2}+C\\\\\\ =-\dfrac{1}{3}\,(1-x^2)^{3/2}+\dfrac{1}{5}\,(1-x^2)^{5/2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int\!x^3\sqrt{1-x^2}\,dx=-\frac{1}{3}\,(1-x^2)^{3/2}+\frac{1}{5}\,(1-x^2)^{5/2}+C \end{array}}$


Bons estudos! :-)