Question

How do you integrate `int 1/sqrt(9x^2-18x+13)` using trigonometric substitution?

Answer 1

`sinh^-1(3/2(x-1))+C`

Explanation:

I can give you a solution but it uses a hyperbolic substitution as oppose to a standard trig substitution but it is still very similar.

We begin by re writing the quadratic under the square root in completed square / vertex form.

`9x^2-18x+13=9(x^2-2x)+13`
`=9(x-1)^2+4`

We can re write the integral as:

`intdx/sqrt(9x^2-18x+13) = intdx/sqrt(9(x-1)^2+4)`

At this point we may like to consider the substitution:

`3(x-1) = 2sinh(u) -> 9(x-1)^2=4sinh^2(u)`

It also follows from this substitution that:

`3dx = 2cosh(u)du`

Notice it is not a standard trig function but a hyperbolic function

Now putting this into the integral we get:

`2/3intcosh(u)/sqrt(4sinh^2(u)+4)du`

We can factor `4` out of the square root on the bottom (leaving us with `2`) which we will put on the front of the integral:

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

From the hyperbolic identity: `cosh^2(u)-sinh^2(u) =1`
we can replace the expression under the square root with:

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

Which simplifies to:

`1/3intcosh(u)/cosh(u)du=1/3intdu=1/3u+C`

Now reverse the substitution for `u` and we get:

`1/3sinh^-1(3/2(x-1))+C`

You can leave it here or if you want to go a bit further, re-write this in terms of the definition of the `sinh` inverse function which would give:

`1/3ln{3/2(x-1)+sqrt(9/4(x-1)^2+1) }+C`