Give the correct answer and solve the given equation Evaluate displaystyleint{x}^{3}{left({sqrt[{3}]{{{1}-{x}^{2}}}}right)}{left.{d}{x}right.}

Give the correct answer and solve the given equation Evaluate $\int {x}^{3}\left(\sqrt[3]{1-{x}^{2}}\right)dx$
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l1koV
We have to find $\int {x}^{3}\left(\sqrt[3]{1-{x}^{2}}\right)dx$
Aplly substitution $u=1-{x}^{2}$ gives us $du=-2xdx\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=\left(1-u\right)$
$\int {x}^{2}\sqrt[3]{1-{x}^{2}}dx=\int {x}^{2}\sqrt[3]{1-{x}^{2}}\left(xdx\right)$
$=\int -\frac{\left(1-u\right)\sqrt[2]{u}}{2}du$

$=\frac{1}{2}\cdot \int \sqrt[3]{u}-{u}^{\frac{4}{3}}du$
$=-\frac{1}{2}\left(\int \sqrt[3]{u}du-\int {u}^{\frac{4}{3}}du\right)\left[\text{Apply the sum rule of integration}\right]$
$=-\frac{1}{2}\left(\frac{{u}^{\frac{1}{3}+1}}{\frac{1}{3}+1}-\frac{{u}^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right)\left[\therefore {x}^{n}dx=\frac{{x}^{n+1}}{n+1},n\ne -1\right]$
$=-\frac{1}{2}\left(\frac{3}{4}{u}^{\frac{4}{3}}-\frac{3}{7}{u}^{\frac{7}{3}}\right)$
$=-\frac{1}{2}\left(\frac{3}{4}\frac{{\left(1-{x}^{2}\right)}^{4}}{3}-\frac{3}{7}\frac{{\left(1-{x}^{2}\right)}^{4}}{3}\right)\left[\therefore u=1-{x}^{2}\right]$
$=\frac{3\frac{{\left(1-{x}^{2}\right)}^{7}}{3}}{14}-\frac{3\frac{{\left(1-{x}^{2}\right)}^{4}}{3}}{8}$
$=C-{\left(1-{x}^{2}\right)}^{\frac{4}{3}}\frac{12{x}^{2}+9}{56}$
where C is an integrating constant.