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

Mar 1, 2016

$= \frac{1}{3} {\sinh}^{-} 1 \left(x - 1\right) + C$

#### Explanation:

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

$9 {x}^{2} - 18 x + 18$
$= 9 \left({x}^{2} - 2\right) + 18$
$= 9 {\left(x - 1\right)}^{2} + 9$

So the integral can be re written as:

$\int \frac{1}{\sqrt{9 {\left(x - 1\right)}^{2} + 9}} \mathrm{dx} = \frac{1}{3} \int \frac{1}{\sqrt{{\left(x - 1\right)}^{2} + 1}} \mathrm{dx}$

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

Now consider the substitution $\sinh \left(u\right) = x - 1$
It will follow that $\cosh \left(u\right) \mathrm{du} = \mathrm{dx}$

Now substitute this into the integral to get:

$\frac{1}{3} \int \cosh \frac{u}{\sqrt{{\sinh}^{2} \left(u\right) + 1}} \mathrm{dx}$

Use the identity: ${\cosh}^{2} \left(u\right) - {\sinh}^{2} \left(u\right) = 1$ to re write the bottom as:

$\frac{1}{3} \int \cosh \frac{u}{\sqrt{{\cosh}^{2} \left(u\right)}} \mathrm{dx} = \frac{1}{3} \int \cosh \frac{u}{\cosh} \left(u\right) \mathrm{du} = \frac{1}{3} \int \mathrm{du}$

Now evaluating the integral and reversing the substitution:

$\frac{1}{3} u + C$
$= \frac{1}{3} {\sinh}^{-} 1 \left(x - 1\right) + C$