How do you integrate #int x /sqrt( 81 - x^4 )dx# using trigonometric substitution?

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Answer 1 · 2016-09-19T20:00:07

#1/2arcsin(x^2/9)+C#

We should apply the substitution #x^2=9sintheta#. Note that this implies that #2xdx=9costhetad theta#.

#intx/sqrt(81-x^4)dx=1/2int(2xdx)/sqrt(81-(x^2)^2)=1/2int(9costhetad theta)/sqrt(81-81sin^2theta)#

Note that #9/sqrt81=1#:

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

Recall that #1-sin^2theta=cos^2theta#:

#=1/2intd theta=1/2theta+C#

From #x^2=9sintheta# we see that #theta=arcsin(x^2/9)#:

#=1/2arcsin(x^2/9)+C#