Question

What's the integral of `int (tan(x))^2 * sec(x) dx`?

Answer 1

`(secxtanx-ln(abs(secx+tanx)))/2+C`

Explanation:

We have:

`I=inttan^2xsecxdx`

Write `tan^2x` as `sec^2x-1`.

`I=int(sec^2x-1)secxdx`

`I=intsec^3xdx-intsecxdx`

The second is a commonly known integral:

`I=intsec^3xdx-ln(abs(secx+tanx))`

Now, for the remaining integral, we will try to use integration by parts, which takes the form:

`intudv=uv-intvdu`

So, let:

`{(u=secx" "=>" "du=secxtanxdx),(dv=sec^2xdx" "=>" "v=tanx):}`

Thus:

`I=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))`

Now, notice that we have `intsecxtan^2xdx` in the problem here, which is what we started with. So, we also know that:

`intsecxtan^2xdx=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))`

Add `intsecxtan^2xdx` to both sides:

`2intsecxtan^2xdx=secxtanx-ln(abs(secx+tanx))`

Divide both sides by `2`:

`intsecxtan^2xdx=(secxtanx-ln(abs(secx+tanx)))/2+C`