How do you integrate int 1/sqrt(9x^2+6x-8) by trigonometric substitution?

Feb 15, 2017

The answer is $\frac{1}{3} \ln | x + \frac{1}{3} + \sqrt{{\left(x + \frac{1}{3}\right)}^{2} - 1} | + C$. See below for details.

Explanation:

Complete the square under the square root.

$\int \frac{1}{\sqrt{9 \left({x}^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9}\right) - 8}} \mathrm{dx}$

$\int \frac{1}{\sqrt{9 {\left(x + \frac{1}{3}\right)}^{2} - 1 - 8}} \mathrm{dx}$

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

Now let $u = x + \frac{1}{3}$. Then $\mathrm{du} = \mathrm{dx}$.

$\int \frac{1}{\sqrt{9 {u}^{2} - 9}} \mathrm{du}$

$\int \frac{1}{\sqrt{9 \left({u}^{2} - 1\right)}} \mathrm{du}$

We are of the form $\sqrt{{x}^{2} - {a}^{2}}$. This means using the substitution $x = a \sec \theta$. Now, let $u = \sec \theta$. Then $\mathrm{du} = \sec \theta \tan \theta d \theta$.

$\int \frac{1}{\sqrt{9 \left({\sec}^{2} \theta - 1\right)}} \cdot \sec \theta \tan \theta d \theta$

$\int \frac{1}{\sqrt{9 {\tan}^{2} \theta}} \cdot \sec \theta \tan \theta d \theta$

$\int \frac{1}{3 \tan \theta} \cdot \sec \theta \tan \theta d \theta$

$\frac{1}{3} \int \sec \theta d \theta$

This is a relatively well known integral which can be derived here

$\frac{1}{3} \ln | \sec \theta + \tan \theta | + C$

Recall the original trig substitution was $\frac{u}{1} = \sec \theta$. This means that the side opposite $\theta$ measures $\sqrt{{u}^{2} - 1}$. Therefore, $\tan \theta = \frac{\sqrt{{u}^{2} - 1}}{1} = \sqrt{{u}^{2} - 1}$.

$\frac{1}{3} \ln | u + \sqrt{{u}^{2} - 1} | + C$

Now reverse the other substitution.

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

Hopefully this helps!