How do you integrate #int 1/sqrt(9x^2-18x) # using trigonometric substitution?

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Answer 1 · 2018-05-26T05:57:23

Use the substitution #x-1=sectheta#.

Let

#I=int1/sqrt(9x^2-18x)dx#

Complete the square in the square root:

#I=1/3int1/sqrt((x-1)^2-1)dx#

Apply the substitution #x-1=sectheta#:

#I=1/3intsecthetad theta#

Integrate directly:

#I=1/3ln|sectheta+tantheta|+C#

Reverse the substitution:

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