Question

How do you integrate `int 1/sqrt(t^2-6t+13)` by trigonometric substitution?

Answer 1

`sinh^-1((t-3)/2)+C`, by completing the square and then either using a hyperbolic substitution, or by citing a standard integral.

Explanation:

`=int1/sqrt((t-3)^2+4)dt`
which upon substituting `t-3=x` is the standard integral `int1/sqrt(x^2+a^2)dx = sinh^-1(x/a)` with `a=2`.

If the standard integral is not accessible, substitute `x=2 sinh u`, `dx=2 cosh u du` and use `cosh^2u - sinh^2u=1`:

`=int (2 cosh u)/sqrt(4 sinh^2u+4)du`
`=int 1du`
`=u+C`
`=sinh^-1(x/2)+C`
`=sinh^-1((t-3)/2)+C`.

If the `+4` had been `-4` you would have substituted `x= 2cosh u`.

The final answer can also be expressed using logarithms instead of inverse hyperbolic functions, because `sinh^-1x=ln(x+sqrt(x^2+1))`, not forgetting also that `ln (x/2+sqrt(x^2/4+1))` differs from `ln(x+sqrt(x^2+4))` by a constant which can be merged with the arbitrary constant of integration.

You use hyperbolic trig if the quadratic under the square root has a positive coefficient of `x^2` and circular trig if it is negative.