Question

How do you integrate `int 1/sqrt(x^2-25)` by trigonometric substitution?

Answer 1

`cosh^-1(x/5)+C`

Explanation:

The integrand is defined only for `x>=5`.
Substitute `x=5cosh u`, `dx=5 sinh u du`.
The integral becomes `int1/sqrt(25 cosh^2u-25)5sinh u du`.
Since `cosh^2u-sinh^2=1`, this becomes:
`=int 1 du`
`=u+C`
`=cosh^-1(x/5)+C`
which has an alternative representation in the form of a logarithm.