Question

How do you integral `int [secx(x)tan(x)]/(sec(x)-1)dx`?

Answer 1

`int \ (secx \ tanx )/ (secx - 1) \ dx = ln|secx+1| + C`

Explanation:

We seek:

`I =int \ (secx \ tanx )/ (secx - 1) \ dx`

If we perform the substitution:

`u =secx-1 => (du)/dx = secx \ tanx`

Applying this substitution to the integral we get:

`I =int \ 1/u \ du`

Which is a trivial standard integral, so if we integrate we get:

`I = ln|u| + C`

And restoration of the earlier substitution gives:

`I = ln|secx+1| + C`