Integration Help

Discussion in 'Community Discussion' started by Shaun.P, Feb 26, 2006.

  1. Shaun.P macrumors 68000

    Shaun.P

    Joined:
    Jul 14, 2003
    Location:
    Omicron Persei 8
    #1
    Hi all,

    I have a maths test on Tuesday at University, and I have a practice test. One of the questions I cannot find the solution to in my notes, it is:

    Integrate (Cos[2x])^3, that is, Cos2x all cubed.

    I don't see how to go about this problem and it is really starting to bug me! My notes only have examples for stuff like 'Sinx^3' integrate, but not with the '2x' just x.

    Any help is greatly appreciated. Apparently the answer should be:

    1/24 (9Sin2x + Sin6x).
     
  2. mlw1235 macrumors 6502

    Joined:
    Jul 16, 2004
    Location:
    Milwaukee, WI
    #2
    You have to use a power reducing formula (see this page

    Applying the cos^3 rule you get

    .25 Int(3cos(2x) + cos(6x))

    split up that integral setting u = 2x and u=6x

    You the get

    (3/8)sin2x - (1/24)sin6x which is equivalent to your answer.
     
  3. CanadaRAM macrumors G5

    CanadaRAM

    Joined:
    Oct 11, 2004
    Location:
    On the Left Coast - Victoria BC Canada
    #3
    My son keeps coming to me with math problems in his homework (in college doing University prep) and I can't remember these ^#**ers that I supposedly learned 30 years ago. Thanks for the link, it helps Dads too.. ;)
     
  4. iGav macrumors G3

    Joined:
    Mar 9, 2002
    #4
    what the hell does it mean???? :eek: (keep in mind you're talking to someone that doesn't know all their times tables. :eek: :p )
     
  5. Macaddicttt macrumors 6502a

    Macaddicttt

    Joined:
    Apr 22, 2004
    Location:
    San Diego, CA
    #5
    I wouldn't recommend that method since on a test you aren't going to be able to remember every power reduction formula. I suggest using the double angle formulas and 1 - (sinx)^2 = (cosx)^2 and 1 - (cosx)^2 = (sinx)^2 and just playing around with it until you come to something you can integrate.
     
  6. Shaun.P thread starter macrumors 68000

    Shaun.P

    Joined:
    Jul 14, 2003
    Location:
    Omicron Persei 8
    #6
    No there should be an easier way to do it.

    Is there a substitution method?

    I have tried writing (Cos2x)^2 x Cos2x, and tried using integrating by parts, but I don't think this is correct.

    The paper is out of 30, and this question is out of 4.

    I am also stuck on two other questions:

    (ii) Prove the following statement in which m is an integer.

    m odd => 4|(m^2 + 3) (3 marks)

    I have written: If m is odd, then has forumula m = 2k + 1. m^2 + 3 = 4k^2 + 4k + 4. But I don't know how to finish this proof - Do I take out the 4 as a common factor, hence 4 (K^2 + K + 1) but how do I actually prove 4 is divisible into this?

    Also,

    Consider the following statement in which x is a postive real number. (3 marks)

    x^(1/2) (i.e. root x) => x rational

    Write down the converse of this and the contrapositive. Give a counter example to show that the converse of above is false.

    Any help with any of these questions is greatly appreciated. I have all my notes - I haven't missed lectures but the examples in class are nothing like the questions asked for in the test.
     
  7. Shaun.P thread starter macrumors 68000

    Shaun.P

    Joined:
    Jul 14, 2003
    Location:
    Omicron Persei 8
    #7

    I don't actually know what an integral means, other than it is an anti-derivative. If you integrate with limits, you can find the maximum area under a curve. Differentiation on the other hand is rate of change.
     
  8. mlw1235 macrumors 6502

    Joined:
    Jul 16, 2004
    Location:
    Milwaukee, WI
    #8
    This is quick, but I have to take a test quick here but.

    m is odd.
    Let m = 2a + 1 (def of odd number)
    Let there be a c such that c = a^2 + a + 1 (you'll see why...)
    4c = 4a^2 + 4a + 4
    4c = 4a^2 + 4a + 1 + 3
    4c = (2a+1)^2 + 3
    4c = m^2 + 3
    4 | m^2 + 3


    fyi....start at the top of the proof, and go as far as you can, then go to the bottom and work your way up.
     

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