# How do you integrate int 1/sqrt(x^2-16x+37)  using trigonometric substitution?

Dec 30, 2015

$\ln | x - 8 + \sqrt{{\left(x - 8\right)}^{2} - 27} | + C$

You may also see the answer written as follows

$\ln | x - 8 + \sqrt{{x}^{2} - 16 x + 37} | + C$

#### Explanation:

We begin by completing the square of the denominator.

${x}^{2} - 16 x + 64 + 37 - 64$

${x}^{2} - 16 x + 64 - 27$

${\left(x - 8\right)}^{2} - 27$

The integral becomes

$\int \frac{1}{\sqrt{{\left(x - 8\right)}^{2} - 27}} \mathrm{dx}$

Now we choose a substitution

Let $\sec \setminus \theta = \frac{x - 8}{\sqrt{27}}$

Solve for $x$

$x - 8 = \sqrt{27} \sec \setminus \theta$

$x = \sqrt{27} \sec \setminus \theta + 8$

Differentiate both sides

$\mathrm{dx} = \sqrt{27} \sec \setminus \theta \tan \setminus \theta d \setminus \theta$

Now make the substitution into the integral

$\int \frac{\sqrt{27} \sec \setminus \theta \tan \setminus \theta}{\sqrt{{\left(\sqrt{27} \sec \setminus \theta + 8 - 8\right)}^{2} - 27}} d \setminus \theta$

$\int \frac{\sqrt{27} \sec \setminus \theta \tan \setminus \theta}{\sqrt{27 {\sec}^{2} \setminus \theta - 27}} d \setminus \theta$

$\int \frac{\sqrt{27} \sec \setminus \theta \tan \setminus \theta}{\sqrt{27 \left({\sec}^{2} \setminus \theta - 1\right)}} d \setminus \theta$

$\frac{\sqrt{27}}{\sqrt{27}} \int \frac{\sec \setminus \theta \tan \setminus \theta}{\sqrt{{\tan}^{2} \setminus \theta}} d \setminus \theta$

$\int \frac{\sec \setminus \theta \tan \setminus \theta}{\tan \setminus \theta} d \setminus \theta$

$\int \sec \setminus \theta d \setminus \theta$

Now we can integrate

$\ln | \sec \setminus \theta + \tan \setminus \theta |$

We have to get things back in terms of $x$

$\sec \setminus \theta = \frac{x - 8}{\sqrt{27}}$

$\tan \setminus \theta = \frac{\sqrt{{\left(x - 8\right)}^{2} - 27}}{\sqrt{27}}$

Back substituting we have

$\ln | \frac{x - 8 + \sqrt{{\left(x - 8\right)}^{2} - 27}}{\sqrt{27}} | + C$

Using properties of logarithms we can rewrite

$\ln | x - 8 + \sqrt{{\left(x - 8\right)}^{2} - 27} | - \ln | \sqrt{27} | + C$

$\ln | \sqrt{27} |$ is just a constant so we can write

$\ln | x - 8 + \sqrt{{\left(x - 8\right)}^{2} - 27} | + C$

You may also see the answer written as follows

$\ln | x - 8 + \sqrt{{x}^{2} - 16 x + 37} | + C$