How do you integrate #int x/(sqrt(1-x^4))# using substitution?

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Answer 1 · 2016-09-12T19:04:38

#1/2arcsin(x^2)+C#

#intx/sqrt(1-x^4)dx#

Apply the substitution #x^2=sintheta#. This implies that #2xdx=costhetad theta#. Rearranging then substituting:

#=1/2int(2xdx)/sqrt(1-(x^2)^2)=1/2int(costhetad theta)/sqrt(1-sin^2theta)#

Note that #sqrt(1-sin^2theta)=costheta#:

#=1/2intd theta#

#=1/2theta+C#

From #x^2=sintheta# we see that #theta=arcsin(x^2)#:

#=1/2arcsin(x^2)+C#