Question

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

Answer 1

`=1/3sinh^-1(x-1)+C`

Explanation:

First of all, we complete the square/ re write in vertex form the quadratic under the square root:

`9x^2-18x+18`
`=9(x^2-2)+18`
`=9(x-1)^2+9`

So the integral can be re written as:

`int1/sqrt(9(x-1)^2+9)dx=1/3int1/sqrt((x-1)^2+1)dx`

(Note the 9 under the square root has been factored out to the front)

Now consider the substitution `sinh(u)=x-1`
It will follow that `cosh(u)du=dx`

Now substitute this into the integral to get:

`1/3intcosh(u)/sqrt(sinh^2(u)+1)dx`

Use the identity: `cosh^2(u)-sinh^2(u)=1` to re write the bottom as:

`1/3intcosh(u)/sqrt(cosh^2(u))dx=1/3intcosh(u)/cosh(u)du = 1/3intdu`

Now evaluating the integral and reversing the substitution:

`1/3u +C`
`=1/3sinh^-1(x-1)+C`