A maior rede de estudos do Brasil

# como integrar x^2/raiz(2-x^2)dx

## 1 resposta(s) - Contém resposta de Especialista

RD Resoluções

Há mais de um mês

Devemos encontrar a integral dada e para isso realizaremos os cálculos abaixo:

\begin{align} & \int_{{}}^{{}}{f(x)}=\int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}} \\ & x=\sqrt{2}\sin u \\ & dx=\sqrt{2}\cos udu \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=\int_{{}}^{{}}{\frac{{{(\sqrt{2}\sin u)}^{2}}}{\sqrt{2-{{(\sqrt{2}\sin u)}^{2}}}}\sqrt{2}\cos u}du \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=\int_{{}}^{{}}{\frac{{{\left( \sqrt{2}\sin u \right)}^{2}}\cos u}{\sqrt{2-{{\left( \sqrt{2}\sin u \right)}^{2}}}}du} \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}-\int_{{}}^{{}}{-\sqrt{2-{{x}^{2}}}}dx \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}+2\left( \frac{{{\sin }^{-1}}\left( \frac{x}{\sqrt{2}} \right)}{2} \right)+\frac{x\sqrt{1-\frac{{{x}^{2}}}{2}}}{2\sqrt{2}} \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}+2\left( \frac{{{\sin }^{-1}}\left( \frac{x}{\sqrt{2}} \right)}{2} \right)+\frac{x\sqrt{1-\frac{{{x}^{2}}}{2}}}{2\sqrt{2}}+C \\ \end{align}

Portanto, o valor da integral será $$\boxed{\int_{}^{} {\frac{{{x^2}}}{{\sqrt {2 - {x^2}} }}} = - x\sqrt {2 - {x^2}} + 2\left( {\frac{{{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right)}}{2}} \right) + \frac{{x\sqrt {1 - \frac{{{x^2}}}{2}} }}{{2\sqrt 2 }} + C}$$.

Devemos encontrar a integral dada e para isso realizaremos os cálculos abaixo:

\begin{align} & \int_{{}}^{{}}{f(x)}=\int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}} \\ & x=\sqrt{2}\sin u \\ & dx=\sqrt{2}\cos udu \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=\int_{{}}^{{}}{\frac{{{(\sqrt{2}\sin u)}^{2}}}{\sqrt{2-{{(\sqrt{2}\sin u)}^{2}}}}\sqrt{2}\cos u}du \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=\int_{{}}^{{}}{\frac{{{\left( \sqrt{2}\sin u \right)}^{2}}\cos u}{\sqrt{2-{{\left( \sqrt{2}\sin u \right)}^{2}}}}du} \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}-\int_{{}}^{{}}{-\sqrt{2-{{x}^{2}}}}dx \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}+2\left( \frac{{{\sin }^{-1}}\left( \frac{x}{\sqrt{2}} \right)}{2} \right)+\frac{x\sqrt{1-\frac{{{x}^{2}}}{2}}}{2\sqrt{2}} \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}}{\sqrt{2-{{x}^{2}}}}}=-x\sqrt{2-{{x}^{2}}}+2\left( \frac{{{\sin }^{-1}}\left( \frac{x}{\sqrt{2}} \right)}{2} \right)+\frac{x\sqrt{1-\frac{{{x}^{2}}}{2}}}{2\sqrt{2}}+C \\ \end{align}

Portanto, o valor da integral será $$\boxed{\int_{}^{} {\frac{{{x^2}}}{{\sqrt {2 - {x^2}} }}} = - x\sqrt {2 - {x^2}} + 2\left( {\frac{{{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right)}}{2}} \right) + \frac{{x\sqrt {1 - \frac{{{x^2}}}{2}} }}{{2\sqrt 2 }} + C}$$.

Gabriela de Oliveira

Há mais de um mês

Podemos integrar usando ∫u.dv = u.v - ∫v.du #

chamando u= 1/(2-x²)½  e dv= x².dx

Derivando u e integrando dv, temos:  du= x.dx e v=x³/3

Substituindo em #: ∫x²/(2-x²)½ .dx = x³/3(2-x²)½ - ∫x^4/3dx

= x³/3(2-x²)½ - x^5/15