# How do you integrate int 1/sqrt(x^2-6x+4)  using trigonometric substitution?

Mar 30, 2018

Hint:
$I = \int \frac{1}{\sqrt{{\left(x - 3\right)}^{2} - \left({\sqrt{5}}^{2}\right)}} \mathrm{dx}$
Use:
color(red)(int1/sqrt(X^2-A^2)dX=ln|X+sqrt(X^2-A^2)|+c
$I = \ln | \left(x - 3\right) + \sqrt{{\left(x - 3\right)}^{2} - {\left(\sqrt{5}\right)}^{2}} | + c$
$I = \ln | \left(x - 3\right) + \sqrt{{x}^{2} - 6 x + 4} | + c$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{{x}^{2} - 6 x + 4}} \mathrm{dx}$

$I = \int \frac{1}{\sqrt{{x}^{2} - 6 x + 9 - 5}} \mathrm{dx}$

$I = \int \frac{1}{\sqrt{{\left(x - 3\right)}^{2} - \left({\sqrt{5}}^{2}\right)}} \mathrm{dx}$

Taking ,$x - 3 = \sqrt{5} \sec u \implies x = 3 + \sqrt{5} \sec u$

$\therefore \mathrm{dx} = \sqrt{5} \sec u \tan u \mathrm{du}$

So,

$I = \int \frac{\sqrt{5} \sec u \tan u}{\sqrt{{\left(\sqrt{5} \sec u\right)}^{2} - {\left(\sqrt{5}\right)}^{2}}} \mathrm{du}$

$I = \int \frac{\sqrt{5} \sec u \tan u}{\sqrt{5} \sqrt{{\sec}^{2} u - 1}} \mathrm{du}$

$= \int \left(\frac{\sec u \tan u}{\tan} u\right) \mathrm{du}$

$= \int \sec u \mathrm{du}$

$= \ln | \sec u + \tan u | + c$

$= \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | + c$

$= \ln | \frac{x - 3}{\sqrt{5}} + \sqrt{{\left(\frac{x - 3}{\sqrt{5}}\right)}^{2} - 1} | + c$

$= \ln | \frac{x - 3}{\sqrt{5}} + \frac{\sqrt{{\left(x - 3\right)}^{2} - 5}}{\sqrt{5}} | + c$

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

$= \ln | \left(x - 3\right) + \sqrt{{x}^{2} - 6 x + 4} | - \ln \sqrt{5} + c$

$= \ln | \left(x - 3\right) + \sqrt{{x}^{2} - 6 x + 4} | + C , w h e r e , C = c - \ln \sqrt{5}$