# How do you integrate int 1/sqrt(9x^2-6x+5)  using trigonometric substitution?

May 13, 2018

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \ln \left\mid 3 x - 1 + \sqrt{9 {x}^{2} - 6 x + 5} \right\mid + C$

#### Explanation:

Complete the square at the denominator:

$9 {x}^{2} - 6 x + 5 = 9 {x}^{2} - 6 x + 1 + 4 = {\left(3 x - 1\right)}^{2} + 4$

then write the integrand as:

$\frac{1}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{\sqrt{{\left(3 x - 1\right)}^{2} + 4}} = \frac{1}{2} \frac{1}{\sqrt{{\left(\frac{3 x - 1}{2}\right)}^{2} + 1}}$

Substitute now:

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

with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

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

so that:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \int \frac{{\sec}^{2} t \mathrm{dt}}{\sqrt{{\tan}^{2} t + 1}}$

Use now the trigonometric identity:

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

and as for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the secant is positive:

$\sqrt{{\tan}^{2} t + 1} = \sec t$

So:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \int \frac{{\sec}^{2} t \mathrm{dt}}{\sec} t$

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \int \sec t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \ln \left\mid \sec t + \tan t \right\mid + C$

Undo the substitution, considering that:

$\sec t = \sqrt{{\tan}^{2} t + 1} = \sqrt{{\left(\frac{3 x - 1}{2}\right)}^{2} + 1} = \frac{1}{2} \sqrt{9 {x}^{2} - 6 x + 5}$

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \ln \left\mid \frac{1}{2} \sqrt{9 {x}^{2} - 6 x + 5} + \frac{3 x - 1}{2} \right\mid + C$

and as multiplying the argument of the logarithm by a constant is equivalent to adding a constant to the expression:

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} - 6 x + 5}} = \frac{1}{3} \ln \left\mid 3 x - 1 + \sqrt{9 {x}^{2} - 6 x + 5} \right\mid + C$