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Trigonometric Applications - complex numbers

  1. Jan 7, 2012 #1
    any help with me understanding this problem would be very much appreciated.

    1. The problem statement, all variables and given/known data

    show,
    [itex]^{π/2}_{0}[/itex][itex]\int[/itex] cos[itex]^{5}[/itex]xdx = 8/15
    hence show
    [itex]^{π/2}_{0}[/itex][itex]\int[/itex] sin[itex]^{5}[/itex]xdx = [itex]^{π/2}_{0}[/itex][itex]\int[/itex] cos[itex]^{5}[/itex]xdx

    where,
    cos[itex]^{5}[/itex]θ = [itex]\frac{cos5θ + 5cos3θ + 10cosθ}{16}[/itex]
    sin[itex]^{5}[/itex]θ = [itex]\frac{sin5θ - 5sin3θ + 10sinθ}{16}[/itex]


    2. Relevant equations

    [itex]^{x}_{0}[/itex][itex]\int[/itex] cos(t)dt = [sin(t)][itex]^{x}_{0}[/itex]
    = sin(x) - sin(0)
    = sin(x)


    3. The attempt at a solution

    [itex]^{π/2}_{0}[/itex][itex]\int[/itex] cos[itex]^{5}[/itex]xdx = [itex]\frac{1}{16}[/itex] [itex]^{π/2}_{0}[/itex][itex]\int[/itex] (cos5θ + 5cos3θ + 10cosθ)dθ
    = [itex]\frac{1}{16}[/itex] [sin5θ + 5sin3θ + 10sinθ][itex]^{π/2}_{0}[/itex]
    = [itex]\frac{1}{16}[/itex] (1 + 5 + 10)


    the answers say from the 2nd line of my attempt it should be...
    = [itex]\frac{1}{16}[/itex] [[itex]\frac{sin5θ}{5}[/itex] + [itex]\frac{5sin3θ}{3}[/itex] + 10sinθ][itex]^{π/2}_{0}[/itex]
    = [itex]\frac{1}{16}[/itex] ([itex]\frac{1}{5}[/itex] - [itex]\frac{5}{3}[/itex] + 10)

    but i don't understand why the first term was divided by 5 and the second by 3,
    or why the sign changed from plus to minus.
     
  2. jcsd
  3. Jan 7, 2012 #2

    jedishrfu

    Staff: Mentor

    isnt that because you have cos(3 . theta) and so you have to divide by 1/3

    try differentiating the 5sin(3 theta) with respect to theta 5 . cos(3 theta) . 3 and you see you get the 3 factor from 3 theta

    from the chain rule.
     
  4. Jan 7, 2012 #3
    I don't think you have to concern yourself with complex numbers for this problem. Although the relation between trig functions and complex numbers is fun to study.

    Also I would recommend just looking up the reduction formulas for integrating the trig functions to some power. It's much easier to just memorize these and then just use them to obtain the correct answer without thinking much at all.
     
  5. Jan 7, 2012 #4
    thanks for your help!

    but i still don't get it.

    can anyone explain step by step?
     
  6. Jan 7, 2012 #5
    for integral[0.pi/2] cos^5(x) dx
    use the reduction formula you can't go wrong and will get the right answer
     
  7. Jan 7, 2012 #6
    thanks figured it out both ways.

    i did it with the complex numbers as this is what we are focusing on at the moment.

    i was just forgetting basic integration rules.

    integral: cos.5x.dx

    let,
    u=5x
    du=5.dx
    dx=1/5.du

    so,
    integral: cos.5x.dx = integral: cos.u.1/5.du
    =1/5.integral: cos.u.du
    =1/5.sin.u + c
    =1/5.sin.5x + c

    integral (with limits 0, pi/2): cos.5x.dx = 1/5.sin.5.pi/2 - 1/5.sin.5.0
    =1/5.1 - 1/5.0
    =1/5

    with,
    integral (limits 0, pi/2): 5.cos.3x
    = 5/3.sin.3(pi/2)
    = -5/3
    integral (limits 0, pi/2): 10.cos.x
    = 10

    also thanks for introducing me to the reduction formula.

    managed to work it out with the same result.
     
    Last edited: Jan 7, 2012
  8. Jan 7, 2012 #7
    I'm not seeing how this has to do with complex numbers at all =(
     
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