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#