Results 1 to 4 of 4

Thread: Definite integral over a period

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Definite integral over a period

    Hi,

    how do I integrate,

    $\displaystyle \int^{\pi}_0 (8-8cos(2t))^{1/2}dt $?

    I've tried by defining

    $\displaystyle u=8-8cos(2t)$ and then

    $\displaystyle \frac{du}{dt}=16sin(2t) \Rightarrow dt=\frac{1}{16sin(2t)}du,$

    but that doesn't get me very far.

    I think I remember that there are some trick when dealing with definite integrals over periods and when the function is trigonometric. I don't remember how it works though..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    $\displaystyle 8-8\cos{2t} = 8(1-\cos{2t}) = 16\sin^2{t}. $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Remember that $\displaystyle \cos{2t} = 1 - 2\sin^2{t}$, so

    $\displaystyle \int{\sqrt{8 - 8\cos{2t}}\,dt} = \int{\sqrt{8 - 8(1 - 2\sin^2{t})}\,dt}$

    $\displaystyle = \int{\sqrt{8 - 8 + 16\sin^2{t}}\,dt}$

    $\displaystyle = \int{\sqrt{16\sin^2{t}}\,dt}$

    $\displaystyle = \int{4\sin{t}\,dt}$

    $\displaystyle = -4\cos{t} + C$.


    Therefore

    $\displaystyle \int_0^\pi{\sqrt{8 - 8\cos{2t}}\,dt} = \left[-4\cos{t}\right]_0^\pi$

    $\displaystyle = \left[-4\cos{\pi}\right] - \left[-4\cos{0}\right]$

    $\displaystyle = -4(-1) + 4(1)$

    $\displaystyle = 4 + 4$

    $\displaystyle = 8$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Ah, all those trig identities!!

    Thank you both very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 5th 2011, 05:21 PM
  2. Replies: 4
    Last Post: Apr 13th 2011, 02:08 AM
  3. Replies: 2
    Last Post: Aug 17th 2010, 02:37 AM
  4. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 22nd 2010, 04:00 AM
  5. another definite integral
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Apr 14th 2009, 07:22 AM

Search Tags


/mathhelpforum @mathhelpforum