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

Mar 13, 2018

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

#### Explanation:

Let complete the square in the denominator

$9 {x}^{2} - 6 x + 2 = 9 {x}^{2} - 6 x + 1 + 2 - 1 = {\left(3 x - 1\right)}^{2} + 1$

Perform the substitution

$u = 3 x - 1$, $\implies$, $\mathrm{du} = 3 \mathrm{dx}$

Therefore,

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{\mathrm{du}}{\sqrt{{u}^{2} + 1}}$

Let $u = \tan \theta$, $\implies$, $\mathrm{du} = {\sec}^{2} \theta d \theta$

Therefore,

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{{\sec}^{2} \theta d \theta}{\sec \theta}$

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

$= \frac{1}{3} \int \frac{\sec \theta \left(\sec \theta + \tan \theta\right) d \theta}{\sec \theta + \tan \theta}$

Let $v = \sec \theta + \tan \theta$

$\mathrm{dv} = \left({\sec}^{2} \theta + \sec \theta \tan \theta\right) d \theta$

So,

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{\mathrm{dv}}{v}$

$= \frac{1}{3} \ln v$

$= \frac{1}{3} \ln \left(\sec \theta + \tan \theta\right)$

$= \frac{1}{3} \ln \left(u + \sqrt{{u}^{2} + 1}\right)$

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