Question

Boa tarde , me ajudem a resolver esta integral por partes ?só falto esta para finalizar o trabalho.
$\int\limits{x^{3} . \sqrt{1-x^{2} } } \, dx$

Answer 1

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


Método de integração por partes:

$\displaystyle\begin{array}{lcl} u=-\dfrac{1}{2}\,x^2&~\Rightarrow~&du=-x\,dx\\\\ dv=\sqrt{1-x^2}\cdot (-2x)\,dx&~\Leftarrow~&v=\dfrac{2}{3}\,\big(1-x^2\big)^{3/2} \end{array}$


Dessa forma,

$\displaystyle\int u\,dv=uv-\int v\,du\\\\\\ \int \left(-\frac{1}{2}\,x^2 \right )\cdot \sqrt{1-x^2}\,(-2x)\,dx=-\dfrac{1}{2}\,x^2\cdot \dfrac{2}{3}\,\big(1-x^2\big)^{3/2}-\int \dfrac{2}{3}\,\big(1-x^2\big)^{3/2}\cdot (-x)\,dx\\\\\\ \int x^3\sqrt{1-x^2}\,dx=-\dfrac{1}{3}\,x^2\,\big(1-x^2\big)^{3/2}-\frac{1}{3}\int \big(1-x^2\big)^{3/2}\cdot (-2x)\,dx\\\\\\ \int x^3\sqrt{1-x^2}\,dx=-\dfrac{1}{3}\,x^2\big(1-x^2\big)^{3/2}-\frac{1}{3}\cdot \dfrac{\big(1-x^2\big)^{(3/2)+1}}{\frac{3}{2}+1}+C\\\\\\ \int x^3\sqrt{1-x^2}\,dx=-\dfrac{1}{3}\,x^2(1-x^2)^{3/2}-\frac{1}{3}\cdot \dfrac{\big(1-x^2\big)^{5/2}}{\frac{5}{2}}+C\\\\\\ \int x^3\sqrt{1-x^2}\,dx=-\dfrac{1}{3}\,x^2\big(1-x^2\big)^{3/2}-\frac{1}{3}\cdot \dfrac{2}{5}\,\big(1-x^2\big)^{5/2}+C\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int x^3\sqrt{1-x^2}\,dx=-\dfrac{1}{3}\,x^2\big(1-x^2\big)^{3/2}-\frac{2}{15}\,\big(1-x^2\big)^{5/2}+C \end{array}}$


Bons estudos! :-)