Question

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

Answer 1

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

Explanation:

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

We use the following substitution:

• `3x -3 = cosh u`
• `dx = 1/3 (sinh u ) du`

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`