Question

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

Answer 1

For this, you're going to have to complete the square.

`x^2 - 9x - 7`

`7 = x^2 - 9x`

`28/4 + 81/4 = x^2 - 9x + 81/4`

`109/4 = (x - 9/2)^2`

Now bring it back into the problem.

`int 1/(sqrt((x - 9/2)^2 - 109/4))dx`

Let:
`u = x - 9/2`
`du = dx`

`= int 1/(sqrt(u^2 - 109/4))du`

Now it looks more like a `sqrt(x^2 - a^2)` form. If we let `a = sqrt(109)/2`, then:

`u = sqrt(109)/2sectheta`
`du = sqrt(109)/2secthetatanthetad theta`
`sqrt(u^2 - 109/4) = sqrt(109/4sec^2theta - 109/4) = sqrt(109)/2tantheta`

What we now have is:

`= int 1/(cancel(sqrt(109)/2)cancel(tantheta))*cancel(sqrt(109)/2)secthetacancel(tantheta)d theta`

`= int secthetad theta`

Remember the trick to do this? See the following:

`color(green)(intsecxdx)`

`= int (secx(secx+tanx))/(secx+tanx)dx`

`= int (sec^2x+secxtanx)/(secx+tanx)dx`

Let:
`u = secx+tanx`
`du = secxtanx+sec^2xdx`

`=> int 1/udu = ln|u| + C = color(blue)(ln|secx + tanx| + C)`

Therefore, what we have is `ln|sectheta + tantheta|` for the result. Now what we can do is substitute in the proper variables.

`tantheta = (2sqrt(u^2 - 109/4))/sqrt(109) = (2sqrt(x^2 - 9x - 7))/sqrt(109)`
`sectheta = (x - 9/2)*2/sqrt(109) = (2x - 9)/sqrt(109)`

So the answer is:

`= color(green)(ln|(2x - 9)/sqrt(109) + (2sqrt(x^2 - 9x - 7))/sqrt(109)| + C)`

or, embedding the `sqrt(109)` into the `C`:

`= ln|2x - 9 + 2sqrt(x^2 - 9x - 7)| - ln sqrt(109) + C`

`= color(blue)(ln|2x - 9 + 2sqrt(x^2 - 9x - 7)| + C)`