Results 1 to 9 of 9

Thread: Trig Antiderivative

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    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?
    Last edited by topsquark; Feb 27th 2008 at 04:16 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    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?.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by galactus View Post
    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 $\displaystyle sec^3(x)$???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by larson View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Jhevon View Post
    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...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2008
    Posts
    102
    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...

    $\displaystyle \frac {sec(x)tan(x) - ln|sec(x)+tan(x)|}{2}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by larson View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Jhevon View Post
    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 ^^^.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    I suggest lookin' for $\displaystyle \int\sec^nx\,dx$ for $\displaystyle n\ge2.$

    It makes more sense than solvin' these particular cases.
    Last edited by Krizalid; Feb 28th 2008 at 05:37 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The antiderivative of dx is x?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 14th 2009, 11:25 AM
  2. Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 28th 2008, 06:37 AM
  3. Trig Antiderivative Once Again...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 27th 2008, 08:14 PM
  4. Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 27th 2008, 08:03 PM
  5. Another Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Feb 27th 2008, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum