Question

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

Answer 1

The answer is `=ln(|tanx+secx|)-sinx+C`

Explanation:

`"Reminder"`

`intsecxdx=ln(tanx+secx)+C`

Therefore, the integral is

`int(tan^2xdx)/(secx)= intcosxtan^2xdx`

`=intcosx(sec^2x-1)dx`

`=int(secx-cosx)dx`

`=ln(|tanx+secx|)-sinx+C`

Answer 2

`int tan^2(x)/(sec(x))dx=lnabs(sec(x)+tan(x))-sin(x)+C`

Explanation:

We have:

`int tan^2(x)/(sec(x))dx`

We can rewrite this as:

`int tan^(2)(x)*(sec(x))^-1dx` Remember that:

`tan^2(x)=sec^2(x)-1`

`=>int (sec^2(x)-1)*(sec(x))^-1dx`

`=>int sec(x)-(sec(x))^-1dx`

`=>int sec(x)dx-int(sec(x))^-1dx`

Here, remember that `intsec(x)dx=lnabs(sec(x)+tan(x))+C`

`=>lnabs(sec(x)+tan(x))-int(sec(x))^-1dx` (You can ignore the `C`.)

Now...

What does `sec^-1(x)` equal to?

First, `(cos(x))^-1=sec(x)`

Therefore, `sec^-1(x)=((cos(x))^-1)^-1=>cos(x)`

Therefore, we now have:

`=>lnabs(sec(x)+tan(x))-intcos(x)dx`

Another thing to remember is that `intcos(x)dx=sin(x)`

Therefore, we now have:

`=>lnabs(sec(x)+tan(x))-sin(x)` Do you `C` why this is incomplete?

`int tan^2(x)/(sec(x))dx=lnabs(sec(x)+tan(x))-sin(x)+C`