How do you integrate $\int x {(1 - x^{2})}^{\frac{1}{4}} d x$ $int x(1-x^2)^(1/4)dx$ ?

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How do you integrate $\int x {(1 - x^{2})}^{\frac{1}{4}} d x$ $int x(1-x^2)^(1/4)dx$ ?

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Answer 1 · 2018-04-06T05:20:26

$\int x {(1 - x^{2})}^{\frac{1}{4}} d x = - \frac{2}{5} {(1 - x^{2})}^{\frac{5}{4}} + C$ $intx(1-x^2)^(1/4)dx=-2/5(1-x^2)^(5/4)+C$

Explanation:

A substitution will do.

$u = 1 - x^{2}$ $u=1-x^2$

$\frac{d u}{d x} = - 2 x$ $(du)/dx=-2x$

$d u = - 2 x d x$ $du=-2xdx$

We see $x d x$ $xdx$ shows up in the integral. So, solving the above for $x d x$ $xdx$ yields

$- \frac{1}{2} d u = x d x$ $-1/2du=xdx$

We then have

$- \frac{1}{2} \int u^{\frac{1}{4}} d u = - \frac{1}{2} (\frac{u^{\frac{5}{4}}}{\frac{5}{4}}) + C = (- \frac{1}{2}) (\frac{4}{5}) u^{\frac{5}{4}} + C = - \frac{2}{5} u^{\frac{5}{4}} + C$ $-1/2intu^(1/4)du=-1/2(u^(5/4)/(5/4))+C=(-1/2)(4/5)u^(5/4)+C=-2/5u^(5/4)+C$ , as $\int x^{a} d x = \frac{x^{a + 1}}{a + 1} + C$ $intx^adx=x^(a+1)/(a+1)+C$

Rewriting in terms of $x$ $x$ yields

$\int x {(1 - x^{2})}^{\frac{1}{4}} d x = - \frac{2}{5} {(1 - x^{2})}^{\frac{5}{4}} + C$ $intx(1-x^2)^(1/4)dx=-2/5(1-x^2)^(5/4)+C$

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How do you integrate $\int x {(1 - x^{2})}^{\frac{1}{4}} d x$ $int x(1-x^2)^(1/4)dx$ ?

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Explanation:

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How do you integrate int x(1-x^2)^(1/4)dx∫x(1−x2)14dxint x(1-x^2)^(1/4)dx?

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How do you integrate $\int x {(1 - x^{2})}^{\frac{1}{4}} d x$ $int x(1-x^2)^(1/4)dx$ ?