Question

How do you integrate `int 1/sqrt(9x^2-18x+25)` using trigonometric substitution?

Answer 1

`1/3ln|(3(x - 1) + sqrt(9(x - 1)^2 + 16))/4| +C`

Explanation:

Your first step will be to complete the square in the denominator.

`int 1/sqrt(9(x^2 - 2x) + 25)dx`

`int 1/sqrt(9(x^2 - 2x + 1 - 1) + 25)dx`

`int 1/sqrt(9(x^2 - 2x + 1) - 9 + 25)dx`

`int 1/sqrt(9(x - 1)^2 + 16)dx`

We want to convert the radical to the form `sqrt(a^2 + x^2)` in order to make the trig substitution. We can do this by making the u-substitution `u = x - 1`. Then `du = dx`.

`int 1/sqrt(9u^2 + 16)du`

Let `u = 4/3tantheta`. Then `du = 4/3sec^2theta d theta`.

`int 1/sqrt(9(4/3tantheta)^2 + 16) 4/3sec^2theta d theta`

`int 1/sqrt(16tan^2theta + 16) * 4/3sec^2theta d theta`

`int 1/sqrt(16(tan^2theta + 1)) * 4/3sec^2theta d theta`

The whole point of making the trig substitution was to get rid of the radical. Here is our chance: apply the pythagorean identity `1 + tan^2beta = sec^2beta`.

`int 1/sqrt(16sec^2theta) * 4/3sec^2theta d theta`

`int 1/(4sectheta) * 4/3sec^2theta d theta`

`int 1/3sectheta d theta`

This is a known integral, whose derivation can be found here

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

Now comes the last part of the solution--reversing the substitutions. From our initial substitution, we know that `(3u)/4 = tantheta`, so the hypotenuse of the imaginary triangle would be `sqrt(9u^2 + 16)`. Therefore, `sectheta = sqrt(9u^2 + 16)/4`.

`1/3ln|(3u)/4 + sqrt(9u^2 + 16)/4| + C`

`1/3ln|(3(x - 1))/4 + sqrt(9(x - 1)^2 + 16)/4| + C`

`1/3ln|(3(x - 1) + sqrt(9(x - 1)^2 + 16))/4| +C`

Hopefully this helps!