Results 1 to 5 of 5

Math Help - Constructing Antiderivatives!!

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    112

    Constructing Antiderivatives!!

    Find the general antiderivative of the function below. Use theta for θ.





    P(θ) = + C

    need help with another one please, just dont know where to begin, any help would be fantastic, thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Some hints:
    \frac{d}{dx} \sec \theta = \sec \theta \tan \theta
    \frac{d}{dx} \tan \theta = \sec^{2} \theta

    Can you see how these apply to your integral? Basically you're going backwards from derivatives you learned in differentiation.
    Last edited by o_O; April 24th 2008 at 04:54 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    Quote Originally Posted by o_O View Post
    Some hints:
    \frac{d}{dx} \sec \theta = \sec \theta \tan \theta
    \frac{d}{dx} \cot \theta = -\csc^{2} \theta

    Can you see how these apply to your integral? Basically you're going backwards from derivatives you learned in differentiation.

    Hey i have a problem similar to this and I dont see how that applies to the question.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by bluejewballs View Post
    Hey i have a problem similar to this and I dont see how that applies to the question.
    the first part would be because the derivative of the integral or vice versa is the original function
    Follow Math Help Forum on Facebook and Google+

  5. #5
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Let me colour code this:

    p(\theta) = {\color{red} \sec \theta \tan \theta } + {\color{blue}\frac{6}{\cos^{2} \theta}}

    \frac{d}{dx} \sec \theta = {\color{red} \sec \theta \tan \theta }
    \frac{d}{dx} \tan \theta = {\color{blue}sec^{2} \theta}**

    ** Should've been this one in the first post
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. antiderivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 28th 2009, 03:14 PM
  2. Constructing Antiderivatives Analytically!!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 23rd 2008, 11:22 AM
  3. antiderivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2008, 10:55 PM
  4. antiderivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 23rd 2008, 03:09 PM
  5. Constructing Antiderivatives and areas
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2007, 01:40 PM

Search Tags


/mathhelpforum @mathhelpforum