Question

How do you integrate `int sqrt(-x^2-6x-18)/xdx` using trigonometric substitution?

Answer 1

Complete the square under the `sqrt`.

`int sqrt(-(x^2 + 6x + 9 - 9) - 18)/xdx`

`int sqrt(-(x^2 + 6x + 9) + 9 - 18)/xdx`

`int sqrt(-(x + 3)^2 - 9)/xdx`

Let `u = x + 3`. Then `du = dx`.

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

But the expression under the `sqrt` clearly has no real values of `x` that satisfy. So techniqually this problem has no solution.

Hopefully this helps!