Question

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

Answer 1

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!