How do you integrate $int x(1-x^2)^(1/4)dx$ ?

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Answer 1 · 2018-04-06T05:20:26

$intx(1-x^2)^(1/4)dx=-2/5(1-x^2)^(5/4)+C$

A substitution will do.

$u=1-x^2$

$(du)/dx=-2x$

$du=-2xdx$

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

$-1/2du=xdx$

We then have

$-1/2intu^(1/4)du=-1/2(u^(5/4)/(5/4))+C=(-1/2)(4/5)u^(5/4)+C=-2/5u^(5/4)+C$ , as $intx^adx=x^(a+1)/(a+1)+C$

Rewriting in terms of $x$ yields

$intx(1-x^2)^(1/4)dx=-2/5(1-x^2)^(5/4)+C$

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