Question

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

Answer 1

`int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|`

Explanation:

Substitute:

`x=3sect`
`dx = 3sect tant dt`

and you have for `t in (0,pi/2)`:

`int dx/sqrt(x^2-9) = int (3sect tant dt)/sqrt(9sec^2t-9)= int (sect tant dt)/sqrt(1/cos^2t-1) = int (sect tant dt)/sqrt((1-cos^2t)/cos^2t) = int (sect tant dt)/sqrt(sin^2t/cos^2t) =int (sect tant dt)/sqrt(tan^2t)`

In this interval for `tant > 0`, so:

`int dx/sqrt(x^2-9) = int (sect tant dt)/tant =intsectdt= log |sect+ tant|`

To go back to x we note that:

`sect = x/3`

and `tant= sint*sect=sectsqrt(1-1/sec^t2t)= x/3sqrt(1-9/x^2)=1/3sqrt(x^9-9)`.

Finally:

`int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|`

Answer 2

Given: `int 1/sqrt(x^2 - 9)dx`

Let `x = 3sec(u)", then "dx = 3tan(u)sec(u)du`

Substitute this into the integral:

`int (3tan(u)sec(u))/sqrt((3sec(u))^2 - 9)du =`

`int (3tan(u)sec(u))/sqrt(9sec^2(u) - 9)du =`

`int (3tan(u)sec(u))/(3sqrt(sec^2(u) - 1))du =`

`int (tan(u)sec(u))/(sqrt(sec^2(u) - 1))du =`

`int (tan(u)sec(u))/(sqrt(tan^2(u)))du =`

`int (tan(u)sec(u))/tan(u)du =`

This integral is in any good list of integrals:

`int sec(u)du = ln(tan(u) + sec(u)) + C`

`int 1/sqrt(x^2 - 9)dx = ln(sqrt(sec^2(u) - 1) + sec(u)) + C`

Substitute `x/3` for `sec(u)`:

`int 1/sqrt(x^2 - 9)dx = ln(sqrt(x^2/9 - 1) + x/3) + C`

Answer 3

`cosh^{-1)|x/3|+C`

Explanation:

Substitute `x=3cosh u`, `dx=3sinh u du` and use `cosh^2u -sinh^2u = 1` to get the integral as being `u`, which is `cosh^{-1}(x/3)`
which can be written as `ln|x+sqrt(x^2-9)|+C` which is equivalent to
`ln |(x/3+(1/3)sqrt(x^2-)9)|` given above, as the two answers differ by a constant of `ln 3` which is subsumed into the constant of integration.