How do you integrate #int 1/sqrt(4x^2-12x+8) # using trigonometric substitution?

We start by transforming the integral

The integral is a standard integral and evaluates to

To integrate ( \int \frac{1}{\sqrt{4x^2 - 12x + 8}} ) using trigonometric substitution, you can follow these steps:

1. Complete the square under the square root. ( 4x^2 - 12x + 8 = 4(x^2 - 3x + 2) = 4((x - \frac{3}{2})^2 - \frac{1}{4}) )

2. Perform the substitution: ( x - \frac{3}{2} = \frac{1}{2}\sec(\theta) ) (This choice is made because the derivative of secant is a multiple of secant and tangent, which will help simplify the expression).

3. Calculate ( dx ) using the chosen substitution: ( dx = \frac{1}{2}\sec(\theta)\tan(\theta) d\theta )

4. Rewrite the integral in terms of ( \theta ): ( \int \frac{1}{\sqrt{4((\frac{1}{2}\sec(\theta))^2 - 3(\frac{1}{2}\sec(\theta)) + 2)}} \cdot \frac{1}{2}\sec(\theta)\tan(\theta) d\theta ) Simplify and rearrange to get: ( \int \frac{1}{\sqrt{\frac{1}{4}(\sec^2(\theta) - 3\sec(\theta) + 2)}} \cdot \frac{1}{2}\sec(\theta)\tan(\theta) d\theta ) ( = \int \frac{1}{\sqrt{\sec^2(\theta) - 3\sec(\theta) + 2}} \cdot \sec(\theta)\tan(\theta) d\theta )

5. Simplify the expression under the square root: ( \sec^2(\theta) - 3\sec(\theta) + 2 = \tan^2(\theta) - 3\tan(\theta) + 2 = (\tan(\theta) - 1)(\tan(\theta) - 2) )

6. Rewrite the integral in terms of ( \tan(\theta) ): ( \int \frac{1}{\sqrt{(\tan(\theta) - 1)(\tan(\theta) - 2)}} \cdot \sec(\theta)\tan(\theta) d\theta )

7. Now, make another substitution, such as ( u = \tan(\theta) - 1 ) or ( u = \tan(\theta) - 2 ), and proceed with the integration in terms of ( u ).

8. After integrating with respect to ( u ), convert back to the variable ( x ) using the original substitution.