Results 1 to 9 of 9

Math Help - Trig Antiderivative

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    Trig Antiderivative

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

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

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

     sec^2(x) = 1 + tan^2(x)

    which means...

     tans^2(x) = sec^2(x) - 1

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

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

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

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

    Now:
     \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  \frac {du}{sec(x)tan(x)} part.
    Any help?
    Last edited by topsquark; February 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
    3,001
    Thanks
    1
    You are right down too 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.

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

    But

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

    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 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.

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

    But

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

    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...
     \int sec^3(x) - sec(x)

    which equals:

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

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

    So lets put that aside... now

     \int sec^3(x)dx

    You said that we can let  u = sec(x) which means  \frac {du}{dx} = sec(x)tan(x) so  du = \frac {sec(x)tan(x)}{dx} and then you said  dv = sec^2(x)dx so  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  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...
     \int sec^3(x) - sec(x)

    which equals:

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

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

    So lets put that aside... now

     \int sec^3(x)dx

    You said that we can let  u = sec(x) which means  \frac {du}{dx} = sec(x)tan(x) so  du = \frac {sec(x)tan(x)}{dx} and then you said  dv = sec^2(x)dx so  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  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...
    \int sec^3(x)dx

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

    which also equals
    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

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

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

     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...

     \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...
    \int sec^3(x)dx

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

    which also equals
    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 - \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 - \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
    3,654
    Thanks
    13
    I suggest lookin' for \int\sec^nx\,dx for n\ge2.

    It makes more sense than solvin' these particular cases.
    Last edited by Krizalid; February 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: June 14th 2009, 11:25 AM
  2. Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 28th 2008, 06:37 AM
  3. Trig Antiderivative Once Again...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 27th 2008, 08:14 PM
  4. Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 27th 2008, 08:03 PM
  5. Another Trig Antiderivative
    Posted in the Calculus Forum
    Replies: 9
    Last Post: February 27th 2008, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum