How do you integrate $int 1/sqrt(9x^2-18x+8)$ using trigonometric substitution?

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Answer 1 · 2016-03-09T12:54:38

$1/3 cosh^-1 (3x-3) + C$

Since: $9x^2 - 18x + 8 = (3x-3)^2 - 1$

We use the following substitution:

Hence:
$sqrt(9x^2 - 18x + 8) = sqrt((3x-3)^2 - 1) = sqrt(cosh^2 u - 1) = sqrt(sinh^2 u) = sinh u$

So we have:

$int 1/sqrt(9x^2 - 18x + 8) dx = int 1/sinhu * 1/3 sinhu * du = 1/3u + C = 1/3 cosh^-1 (3x-3) + C$

How do you integrate int 1/sqrt(9x^2-18x+8) int 1/sqrt(9x^2-18x+8) using trigonometric substitution?