Question

How do you integrate `int 1/sqrt(9x^2-18x-7)` using trigonometric substitution?

Answer 1

`-1/12ln|csctheta+cottheta|+C`

Explanation:

First, you can complete the square in the bottom

`int1/sqrt((3x-3)^2-16)dx`

Then, use the substitution `3x-3=4sectheta`
Differentiating this gives `3dx=4secthetatantheta*d(theta)`
Substituting this in gives:
`int1/sqrt(16sec^2theta-16)*4/3secthetatantheta*d(theta)`
Since `1+tan^2theta=sec^2theta`
`=1/3intsectheta*d(theta)`
`=1/3ln|csctheta+cottheta|+C`
Then, you can draw a right triangle based on your intial substitution to turn `theta` back into `x`.