How do you integrate #int 1/sqrt(9x^2-6x+2) # using trigonometric substitution?

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Answer 1 · 2018-03-13T08:55:37

The answer is #=1/3ln(|(3x-1)+sqrt((3x-1)^2+1)|)+C#

Let complete the square in the denominator

#9x^2-6x+2=9x^2-6x+1+2-1=(3x-1)^2+1#

Perform the substitution

#u=3x-1#, #=>#, #du=3dx#

Therefore,

#int(dx)/(sqrt(9x^2-6x-2))=1/3int(du)/sqrt(u^2+1)#

Let #u=tantheta#, #=>#, #du=sec^2thetad theta#

Therefore,

#int(dx)/(sqrt(9x^2-6x-2))=1/3int(sec^2thetad theta)/(sectheta)#

#=1/3intsecthetad theta#

#=1/3int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)#

Let #v=sectheta+tantheta#

#dv=(sec^2theta +secthetatantheta)d theta#

So,

#int(dx)/(sqrt(9x^2-6x-2))=1/3int(dv)/v#

#=1/3lnv#

#=1/3ln(sectheta+tantheta)#

#=1/3ln(u+sqrt(u^2+1))#

#=1/3ln(|(3x-1)+sqrt((3x-1)^2+1)|)+C#