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:

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

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