How do you integrate #int 1/sqrt(9x^2+6x-8)# by trigonometric substitution?

1 Answer
Feb 15, 2017

The answer is #1/3ln|x +1/3 + sqrt((x + 1/3)^2 - 1)| + C#. See below for details.

Explanation:

Complete the square under the square root.

#int 1/sqrt(9(x^2 + 2/3x + 1/9 - 1/9)- 8)dx#

#int 1/sqrt(9(x + 1/3)^2 - 1 - 8)dx#

#int 1/sqrt(9(x + 1/3)^2 - 9)dx#

Now let #u = x + 1/3#. Then #du = dx#.

#int 1/sqrt(9u^2 - 9)du#

#int 1/sqrt(9(u^2 - 1))du#

We are of the form #sqrt(x^2- a^2)#. This means using the substitution #x = asectheta#. Now, let #u = sectheta#. Then #du = secthetatantheta d theta#.

#int 1/sqrt(9(sec^2theta - 1)) * secthetatantheta d theta#

#int 1/sqrt(9tan^2theta) * sec thetatantheta d theta#

#int 1/(3tantheta) * secthetatantheta d theta#

#1/3intsectheta d theta#

This is a relatively well known integral which can be derived here

#1/3ln|sectheta + tantheta| + C#

Recall the original trig substitution was #u/1 = sectheta#. This means that the side opposite #theta# measures #sqrt(u^2 - 1)#. Therefore, #tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)#.

#1/3ln|u + sqrt(u^2 - 1)| + C#

Now reverse the other substitution.

#1/3ln|x +1/3 + sqrt((x + 1/3)^2 - 1)| + C#

Hopefully this helps!