Question

How do you integrate `sec2x tan2x dx`?

Answer 1

I would write the integral as:

`intsec(2x)tan(2x)dx=int1/cos(2x)sin(2x)/cos(2x)dx=`

we can see that:

`d[cos(2x)]=-2sin(2x)dx`
and so:
`sin(2x)dx=(d[cos(2x)])/-2`

with this in mind I write the integral as:

`int1/cos^2(2x)*1/-2d[cos(2x)]=`
`=1/-2intcos^(-2)(2x)d[cos(2x)]=` (treating `cos(2x)` as if it were `x` in a normal integration where you get `x^(n+1)/(n+1)`)
`=1/2*1/cos(2x)+c`