Question

How do you use substitution to integrate `(1+2tanx)`?

Answer 1

I found: `int(1+2tan(x))dx=x-2ln|cos(x)|+c`

Explanation:

I would try this:
`int(1+2tan(x))dx=int(1+2sin(x)/cos(x))dx=`
`=intdx+2intsin(x)/cos(x)dx=`
but: `d[cos(x)]=-sin(x)dx` so substituting you get:
`=intdx-2int(d[cos(x)])/cos(x)=`

`=x-2ln|cos(x)|+c`