# How do you integrate int 1/sqrt(9x^2-18x-7)  using trigonometric substitution?

May 4, 2018

$- \frac{1}{12} \ln | \csc \theta + \cot \theta | + C$

#### Explanation:

First, you can complete the square in the bottom

$\int \frac{1}{\sqrt{{\left(3 x - 3\right)}^{2} - 16}} \mathrm{dx}$

Then, use the substitution $3 x - 3 = 4 \sec \theta$
Differentiating this gives $3 \mathrm{dx} = 4 \sec \theta \tan \theta \cdot d \left(\theta\right)$
Substituting this in gives:
$\int \frac{1}{\sqrt{16 {\sec}^{2} \theta - 16}} \cdot \frac{4}{3} \sec \theta \tan \theta \cdot d \left(\theta\right)$
Since $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$
$= \frac{1}{3} \int \sec \theta \cdot d \left(\theta\right)$
$= \frac{1}{3} \ln | \csc \theta + \cot \theta | + C$
Then, you can draw a right triangle based on your intial substitution to turn $\theta$ back into $x$.