Results 1 to 7 of 7

Thread: Very basic antidifferentiation problem

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    101

    Very basic antidifferentiation problem

    I still don't know the math symbols, so please bear with me!

    I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

    I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

    Thanks, I hope I'm making SOME sense.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Remember that

    $\displaystyle \int \sin nx ~dx = -\frac{1}{n}\cos nx +C$

    Therefore

    $\displaystyle \int \sin 2x ~dx = -\frac{1}{2}\cos 2x +C$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    Quote Originally Posted by Wolvenmoon View Post
    I still don't know the math symbols, so please bear with me!

    I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

    I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

    Thanks, I hope I'm making SOME sense.
    formally ...

    $\displaystyle \int \sin(2x) \, dx
    $

    $\displaystyle u = 2x $

    $\displaystyle du = 2 \, dx$

    $\displaystyle \frac{1}{2} \int \sin(2x) \cdot 2 \, dx$

    substitute ...

    $\displaystyle \frac{1}{2} \int \sin(u) \, du
    $

    integrate ...

    $\displaystyle -\frac{1}{2} \cos(u) + C
    $

    back-substitute ...

    $\displaystyle -\frac{1}{2} \cos(2x) + C$


    informally ...

    $\displaystyle \int \sin(kx) \, dx $

    the antiderivative of $\displaystyle \sin(kx)$ has to be a form of $\displaystyle -\cos(kx)$, but taking the derivative of $\displaystyle -\cos(kx)$ yields $\displaystyle \sin(kx) \cdot k$ because of the chain rule ... now, what do you think will get rid of the constant factor $\displaystyle k$ ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2009
    Posts
    101
    Okay Skeeter, I went back over this and there are two steps in particular that have me. When the 1/2 shows up in front of everything, how did it get there and why?

    My instructor has taken a completely different approach to this than my book did, hers is an easier way but I think I've managed to drop something somewhere.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by Wolvenmoon View Post
    When the 1/2 shows up in front of everything, how did it get there and why?
    Find $\displaystyle \frac{d}{dx}\left(-\frac{1}{n}\cos nx +C\right)$

    That should make things clearer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    When you are substituting into $\displaystyle \int sin(2x) dx$, you have to substitute for every "x", even the one is "dx".

    As skeeter said, he let u= 2x. Differentiating both sides of that, du= 2 dx.

    There are two ways of thinking about that but they give the same result:

    1) Multiply and divide the integral by 2-
    $\displaystyle \int sin(2x)\left(\frac{1}{2}\right)(2x) dx= \frac{1}{2}\int sin(2x) 2dx$ so your integral becomes $\displaystyle \frac{1}{2}\int sin(u)du$.

    2) Divide both sides of du= 2 dx by 2 to get $\displaystyle \frac{1}{2} du= dx$ and replace dx by that-
    $\displaystyle \int sin(u) \frac{1}{2} du= \frac{1}{2}\int sin(u) du$.

    Notice that it is the fact that "$\displaystyle \frac{1}{2}$" is a constant that allows us to move it in or out of the integral at will.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2009
    Posts
    101
    So essentially that's 2/2 or a form of 1 that we're adding into the picture to force du = 2dx, breaking up into 2/1 and 1/2, and moving the 1/2 over while we work?


    Okay. Think I've cleared the brick wall here. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Antidifferentiation problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 16th 2010, 01:38 AM
  2. antidifferentiation problem. is this correct?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 8th 2010, 01:11 PM
  3. Help with antidifferentiation
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Aug 17th 2009, 08:24 AM
  4. Antidifferentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 8th 2009, 08:57 PM
  5. Antidifferentiation
    Posted in the Calculus Forum
    Replies: 11
    Last Post: Apr 16th 2008, 06:29 PM

Search Tags


/mathhelpforum @mathhelpforum