Question

Integrate ?

`(secxtanx)/(secx-1)dx`

Answer 1

`ln|sec(x)-1|+C`

Explanation:

Let
`u=sec(x)`

Differentiating, we get:

`(du)/dx=sec(x)tan(x)dx`

`du=sec(x)tan(x)dx`

`sec(x)tan(x)dx` appears in the numerator, so we can make this substitution.

Rewrite the integral:

`int(du)/(u-1)`

This is a common integral ( `intdx/(x+-a)=ln|x+-a|+C`) :

`int(du)/(u-1)=ln|u-1|+C`

Revert back to `x:`
`ln|sec(x)-1|+C`