# How do you integrate int 1/sqrt(9x^2-6x-3)  using trigonometric substitution?

May 17, 2018

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \ln \left\mid 3 x - 1 + \sqrt{9 {x}^{2} - 6 x - 3} \right\mid + C$

#### Explanation:

Complete the square at the denominator:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \int \frac{\mathrm{dx}}{\sqrt{{\left(3 x - 1\right)}^{2} - 4}} = \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{3 x - 1}{2}\right)}^{2} - 1}}$

Note that the integrand is defined only for:

$\left\mid 3 x - 1 \right\mid > 2$

that is for $x > 1$ or $x < - \frac{1}{3}$.

Let $x > 1$ and substitute:

$\frac{3 x - 1}{2} = \sec t$

$\mathrm{dx} = \frac{2}{3} \sec t \tan t \mathrm{dt}$

with $t \in \left(0 , \frac{\pi}{2}\right)$. Then

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Using the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

and considering that for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\tan}^{2} t}}$

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t$

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

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \ln \left\mid \sec t + \tan t \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \ln \left\mid \frac{3 x - 1}{2} + \sqrt{{\left(\frac{3 x - 1}{2}\right)}^{2} - 1} \right\mid + C$

Multiplying the argument of the logarithm by $2$ which is equivalent to adding a constant:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x - 3}} = \frac{1}{3} \ln \left\mid 3 x - 1 + \sqrt{9 {x}^{2} - 6 x - 3} \right\mid + C$

and by direct differentiation we can see that this is valid also for $x < - \frac{1}{3}$