Question

What is the Integral of `tan^2(x)sec(x)`?

Answer 1

`1/2sec(x)tan(x)-1/2ln(abs(sec(x)+tan(x)))+C`

Explanation:

`I=inttan^2(x)sec(x)dx`

Let `tan^2(x)=sec^2(x)-1` which comes from the Pythagorean identity:

`I=int(sec^2(x)-1)sec(x)dx=intsec^3(x)dx-intsec(x)dx`

The integral of `sec(x)` is well known:

`I=intsec^3(x)dx-ln(abs(sec(x)+tan(x)))`

The integral of `sec^3(x)` can be found through integration by parts with `u=sec(x)` and `dv=sec^2(x)dx` at this link. You can also see how to integrate `sec(x)` there.

`I=(1/2sec(x)tan(x)+1/2ln(abs(sec(x)+tan(x))))-ln(abs(sec(x)+tan(x)))`

`I=1/2sec(x)tan(x)-1/2ln(abs(sec(x)+tan(x)))+C`