1. ## Trig Antiderivative

Ok, I'm having a bit of trouble with this one:

$\displaystyle \int sec(x)tan^2(x)dx$

I'm guessing that I would want to use this trig identity:

$\displaystyle sec^2(x) = 1 + tan^2(x)$

which means...

$\displaystyle tans^2(x) = sec^2(x) - 1$

So, the integral with the identity would now look like this:

$\displaystyle \int sec(x)(sec^2(x) - 1)dx$
Then it would be:

$\displaystyle \int sec^3(x) - sec(x)dx$
Now, with u sub, I would let $\displaystyle u = sec(x)$ so this means $\displaystyle du = sec(x)tan(x)dx$

So $\displaystyle dx = \frac {du}{sec(x)tan(x)}$

Now:
$\displaystyle \int u^3 - u (\frac {du}{sec(x)tan(x)})$

Now I'm kind of stuck here, I know how to take the antiderivative of the first part, but not the $\displaystyle \frac {du}{sec(x)tan(x)}$ part.
Any help?

2. You are right down too $\displaystyle sec^{3}(x)-sec(x)$, but then I don't know what you're doing.
Some big no-nos. Do not generally mix up x and u. There is no place to sub in the sec(x)tan(x)
Parts may be better.

$\displaystyle \int{sec(x)}dx=ln|sec(x)+tan(x)|$

But

$\displaystyle \int{sec^{3}(x)}dx$

$\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ du=sec(x)tan(x)dx, \;\ v=tan(x)$

Now, can you put it together?.

3. Originally Posted by galactus
You are right down too $\displaystyle sec^{3}(x)-sec(x)$, but then I don't know what you're doing.
Some big no-nos. Do not generally mix up x and u. There is no place to sub in the sec(x)tan(x)
Parts may be better.

$\displaystyle \int{sec(x)}dx=ln|sec(x)+tan(x)|$

But

$\displaystyle \int{sec^{3}(x)}dx$

$\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ du=sec(x)tan(x)dx, \;\ v=tan(x)$

Now, can you put it together?.
Honestly, I'm still really confused, gah.

Ok so I'm right up until this...
$\displaystyle \int sec^3(x) - sec(x)$

which equals:

$\displaystyle \int sec^3(x)dx - \int sec(x)dx$

You said that this:
$\displaystyle \int sec(x)dx$ equals $\displaystyle ln|sec(x)+tan(x)|$

So lets put that aside... now

$\displaystyle \int sec^3(x)dx$

You said that we can let $\displaystyle u = sec(x)$ which means $\displaystyle \frac {du}{dx} = sec(x)tan(x)$ so $\displaystyle du = \frac {sec(x)tan(x)}{dx}$ and then you said $\displaystyle dv = sec^2(x)dx$ so $\displaystyle v = tan(x)$

Why would you need to use u, v, and du for this question when theres only one thing to worry about... the $\displaystyle sec^3(x)$???

4. Originally Posted by larson
Honestly, I'm still really confused, gah.

Ok so I'm right up until this...
$\displaystyle \int sec^3(x) - sec(x)$

which equals:

$\displaystyle \int sec^3(x)dx - \int sec(x)dx$

You said that this:
$\displaystyle \int sec(x)dx$ equals $\displaystyle ln|sec(x)+tan(x)|$

So lets put that aside... now

$\displaystyle \int sec^3(x)dx$

You said that we can let $\displaystyle u = sec(x)$ which means $\displaystyle \frac {du}{dx} = sec(x)tan(x)$ so $\displaystyle du = \frac {sec(x)tan(x)}{dx}$ and then you said $\displaystyle dv = sec^2(x)dx$ so $\displaystyle v = tan(x)$

Why would you need to use u, v, and du for this question when theres only one thing to worry about... the $\displaystyle sec^3(x)$???
galactus is doing integration by parts. are you familiar with that method?

5. Originally Posted by Jhevon
galactus is doing integration by parts. are you familiar with that method?
Yeah, I had a lot of questions about integration by parts a couple days ago... but I didn't think you would have to do this method with only 1 thing to deal with... but I'll give it a shot anyways...

6. So to find this...
$\displaystyle \int sec^3(x)dx$

would it be like this with the integration by parts method...
$\displaystyle sec(x)tan(x) - \int tan(x)sec(x)tan(x)dx$

which also equals
$\displaystyle sec(x)tan(x) - \int tan^2(x)sec(x)dx$

I'm going to try and do the rest by hand...
WAIT WAIT I THINK I GOT IT... Ok, haha...

So OVERALL

$\displaystyle \int tan^2(x)sec(x) = sec(x)tan(x) - ln|sec(x)+tan(x)| - \int tan^2(x)sec(x)$

move this over $\displaystyle - \int tan^2(x)sec(x)$ to get...

$\displaystyle 2 \int tan^2(x)sec(x) = sec(x)tan(x) - ln|sec(x)+tan(x)|$

Then divide both sides by 2 to get your final answer of...

$\displaystyle \frac {sec(x)tan(x) - ln|sec(x)+tan(x)|}{2}$

7. Originally Posted by larson
So to find this...
$\displaystyle \int sec^3(x)dx$

would it be like this with the integration by parts method...
$\displaystyle sec(x)tan(x) - \int tan(x)sec(x)tan(x)dx$

which also equals
$\displaystyle sec(x)tan(x) - \int tan^2(x)sec(x)dx$
you're ok so far.

I'm going to try and do the rest by hand...
that would be a good idea. (don't forget to carry the $\displaystyle - \int \sec x ~dx$ along)

8. Originally Posted by Jhevon
you're ok so far.

that would be a good idea. (don't forget to carry the $\displaystyle - \int \sec x ~dx$ along)
Yep... I almost did forget about that actually haha... but I did find the right answer finally which is just up... I edited my last post so its back up there ^^^.

9. I suggest lookin' for $\displaystyle \int\sec^nx\,dx$ for $\displaystyle n\ge2.$

It makes more sense than solvin' these particular cases.