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

Feb 17, 2016

${\sinh}^{-} 1 \left(\frac{3}{2} \left(x - 1\right)\right) + C$

#### 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.

$9 {x}^{2} - 18 x + 13 = 9 \left({x}^{2} - 2 x\right) + 13$
$= 9 {\left(x - 1\right)}^{2} + 4$

We can re write the integral as:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 18 x + 13}} = \int \frac{\mathrm{dx}}{\sqrt{9 {\left(x - 1\right)}^{2} + 4}}$

At this point we may like to consider the substitution:

$3 \left(x - 1\right) = 2 \sinh \left(u\right) \to 9 {\left(x - 1\right)}^{2} = 4 {\sinh}^{2} \left(u\right)$

It also follows from this substitution that:

$3 \mathrm{dx} = 2 \cosh \left(u\right) \mathrm{du}$

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

Now putting this into the integral we get:

$\frac{2}{3} \int \cosh \frac{u}{\sqrt{4 {\sinh}^{2} \left(u\right) + 4}} \mathrm{du}$

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:

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

From the hyperbolic identity: ${\cosh}^{2} \left(u\right) - {\sinh}^{2} \left(u\right) = 1$
we can replace the expression under the square root with:

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

Which simplifies to:

$\frac{1}{3} \int \cosh \frac{u}{\cosh} \left(u\right) \mathrm{du} = \frac{1}{3} \int \mathrm{du} = \frac{1}{3} u + C$

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

$\frac{1}{3} {\sinh}^{-} 1 \left(\frac{3}{2} \left(x - 1\right)\right) + C$

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:

$\frac{1}{3} \ln \left\{\frac{3}{2} \left(x - 1\right) + \sqrt{\frac{9}{4} {\left(x - 1\right)}^{2} + 1}\right\} + C$