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Evaluate the integral (help) sin2x dx b=pi/8 a=0

  1. May 12, 2012 #1
    Evaluate the integral (help)!! sin2x dx b=pi/8 a=0

    ∫sin2x dx b=pi/8 a=0

    I'm having a little trouble finishing out the problem. This is what i've got so far.

    u=2x
    du=2dx
    du= 1/2dx

    1/2∫sin(u) du
    (-1/2)cos u
    (-1/2) cos 2x

    [(-1/2)cos2(pi/8)]-[(-1/2)cos(2(0))]
    ????? + (1/2)


    So... The homework question gave me the answer already, but I do not understand how the answer is .... (2 - sqrt(2) / 4)

    Can someone please "spell it out for me"...
    Thanks for your help!!
     
  2. jcsd
  3. May 12, 2012 #2
    Re: Evaluate the integral (help)!! sin2x dx b=pi/8 a=0

    You are asking the wrong question.

    Ok, so what is the equivalent to sin(2x), in terms of sin(x), cos(x), equally,

    WHY

    u=2x
    HOW DID YOU GO FROM

    du=2dx

    to
    du= 1/2dx?

    and why? It seems this is the problem, if you take a look.

    Cheers,
    yus310
     
  4. May 12, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Evaluate the integral (help)!! sin2x dx b=pi/8 a=0

    First, this is a question about Calculus, not "differential equations" so I will move it.

    I assume this was a typo since you don't continue it. If du= 2x, then dx= (1/2)du, not "du= (1/2)dx" as you have.

    But here you have replaced dx with (1/2) du so you have done it correctly.

    [itex]2(\pi/8)= \pi/4[/itex]. Surely you know that [itex]cos(\pi/4)= \sqrt{2}/2[/itex]?

     
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