How do you integrate #int x^2 /sqrt( 16+x^4 )dx# using trigonometric substitution?

1 Answer
Sep 10, 2016

This cannot be integrated using elementary functions.

Explanation:

Use the substitution #x^2=4tantheta#. This implies that #2xdx=4sec^2thetad theta#. Also keep in mind that #x=2sqrttantheta#.

We have:

#intx^2/sqrt(16+x^4)dx=1/2int(x(2xdx))/sqrt(16+x^4)#

#=1/2int(2sqrttantheta(4sec^2thetad theta))/sqrt(16+16tan^2theta)=int(sqrttantheta(sec^2theta)d theta)/sqrt(1+tan^2theta)#

Note that #1+tan^2theta=sec^2theta#, so #sectheta=sqrt(1+tan^2theta)#:

#=int(sqrttantheta(sec^2theta)d theta)/sectheta=intsqrttanthetasecthetad theta#

The more we continue, we see that this cannot be integrated using elementary functions.