Results 1 to 2 of 2

Math Help - Integration problem

  1. #1
    Senior Member
    Joined
    Nov 2012
    From
    Hawaii
    Posts
    285
    Thanks
    2

    Integration problem

    i gave it a shot and this is what i got

    = (1/cosx)(1/cosx)(sinx/cosx) (I REWROTE IT)
    = (secx)(secxtanx) (TIME TO INTEGRATE)
    = (secxtanx)(secx) + C

    did i do it right?
    Attached Thumbnails Attached Thumbnails Integration problem-screen-shot-2012-12-13-10.05.22-pm.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Integration problem

    No, I think I would approach it with the substitution:

    u=\cos(x)\,\therefore\,du=-\sin(x)\,dx and we have:

    -\int u^{-3}\,du=\frac{1}{2u^2}+C=\frac{1}{2\cos^2(x)}+C

    Using your approach, we could let:

    u=\sec(x)\,\therefore\,du=\sec(x)\tan(x)\,dx and we have the easier to integrate:

    \int u\,du=\frac{u^2}{2}+C=\frac{\sec^2(x)}{2}+C

    The two results are equivalent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration problem
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 20th 2010, 08:10 AM
  2. Problem with integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 19th 2010, 12:30 AM
  3. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  4. integration problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 8th 2009, 02:16 AM
  5. Yet another integration problem.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 26th 2008, 01:11 PM

Search Tags


/mathhelpforum @mathhelpforum