How do you integrate #int 1/(x^4sqrt(16+x^2))# by trigonometric substitution?

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Answer 1 · 2018-03-23T14:21:23

The answer is #=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C#

Let #x=4tantheta#, #=>#, #dx=4sec^2thetad theta#

#sqrt(16+x^2)=sqrt(16+16tan^2theta)=4sectheta#

Therefore, the integral is

#I=int(dx)/(x^4sqrt(16+x^2))=int(4sec^2thetad theta)/(256tan^4theta*4sectheta)#

#=1/256int(secthetad theta)/(tan^4theta)#

#tantheta=sintheta/costheta#

#sectheta=1/costheta#

#I=1/256int(d theta)/(costheta*sin^4theta/cos^4 theta)#

#=1/256int(cos^3thetad theta)/(sin^4theta)#

#=1/256int(costheta(1-sin^2theta)d theta)/sin^4theta#

Let #v=sintheta#, #=>#, #dv=costheta d theta#

#I=1/256int((1-v^2)dv)/(v^4)#

#=1/256int(1/v^4-1/v^2)dv#

#=-1/256*1/(3v^3)+1/256*(1/v)#

#=1/256*1/sintheta-1/768*1/sin^3theta#

#=1/256*1/(x/sqrt(16+x^2))-1/768*1/(x^3/(x^2+16)^(3/2))+C#

#=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C#