Question

find integral of {sec(x)tan(x)/sec(x)-1

Answer 1

could you write it out using the equation button?

Answer 2

\[\int\limits \frac{\sec(x) \tan(x)}{\sec(x) -1} dx\]?

Answer 3

so du = sec(x)tan(x) ^_^ and you'll get the following:

\[\int\limits \frac{1}{u} du = \ln|u| + c = \ln|\sec(x)-1| + c\]let u = sec(x) -1

Correct me if I'm wrong ^_^

Answer 4

oh dear, the steps are messed up , I apologize

Answer 5

thank you im trying to figure it out ;-)

Answer 6

oh, you're welcome, let me know if you don't get it :)