Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - Another integration

  1. #1
    Junior Member
    Joined
    Oct 2013
    From
    Australia
    Posts
    45

    Another integration

    Int[(cos(3x))^2]

    Can someone help me with this too? I'll return the thumbsup tomorrow. I'm really exhausted.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,879
    Thanks
    742

    Re: Another integration

    By the double angle formula for cosine: \cos(2\theta) = 2\cos^2 \theta-1. Add one to both sides, then divide both sides by two. So, \cos^2 \theta = \dfrac{1+\cos(2\theta)}{2}. Plug in 3x for \theta. Now, you have:

    \int \dfrac{1+\cos(6x)}{2}dx = \int\left( \dfrac{1}{2} + \dfrac{\cos(6x)}{2}\right)dx = \int \dfrac{1}{2}x^0 dx + \int \dfrac{\cos(6x)}{2}dx

    Now, you can apply the power rule on the first integral and use the substitution u=6x for the second.
    Thanks from Darrylcwc
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum